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+2 CHEMISTRY , PART-1_490-896 Flipbook PDF
+2 CHEMISTRY , PART-1_490-896
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476
+2 CHEMISTRY (VOL. - I)
11.8.2 Manufacture of Hydrogen Peroxide : 1.
By electrolysis of 50% H2SO4 solution : Hydrogen peroxide (80 – 90%) is manufactured by electrolysing a cold 50% solution of sulphuric acid in an electrolytic cell using platinum as anode and graphite as cathode. The following reactions occur. 2H2SO4 ® 2H+ + 2HSO–4 At cathode, 2H+ + 2e– ® H2 1 At anode, 2HSO–4 ® H2S2O8 + 2e– Peroxydisulphuric acid.
Peroxydisulphuric acid thus formed around anode is then distilled with water under reduced pressure. The low boiling hydrogen peroxide distils over leaving behind high boiling sulphuric acid which is recovered and recycled. H2S2O8 + 2H2O u H2O2 + 2H2SO4 . It has also been observed recently that equimolar mixture of sulphuric acid and ammonium sulphate on electrolysis gives a more concentrated solution of hydrogen peroxide as per the following reaction. (NH4)2 SO4 + H2SO4
®
2NH4HSO4 At cathode, 2H + + 2e– At anode, 2NH4SO–4
® ® ®
2NH4HSO4 Ammonium hydrogen sulphate. 2H+ + 2NH4SO–4 H2(NH4)2 S2O8 + 2e– Ammonium persulphate.
Ammonium persulphate thus formed is taken out and distilled with water to get hydrogen peroxide. (NH4)2 S2O8 + 2H2O u 2NH4HSO4 + H2O2. 2. By Auto oxidation of Anthraquinol : This is the most recent method widely used in America. 2 - Ethyl anthraquinone is dissolved in benzene and hydrogen gas is passed through the solution in presence of palladium catalyst. On passing air through the resulting solution (anthraquinol then formed by reduction), 20% solution of hydrogen peroxide is formed alongwith reproduction of anthraquinone derivative. 11.8.3 Concentration of Hydrogen Peroxide solution : Hydrogen peroxide obtained by the above process is in the form of dilute solution. Concentration of dilute solution of hydrogen peroxide is undertaken by the following steps. i.
Slow evaporation in water bath : Dilute solution of hydrogen peroxide is taken in an evaporating dish and evaporated carefully on a water bath until the solution contains 30% solution of hydrogen peroxide.
ii.
Evaporation in vacuum desiccator : The resulting solution of hydrogen peroxide obtained after the first step, is kept in vacuum desiccator over concentrated sulphuric acid. The water vapour gets absorbed by conc. H2SO4. The residual solution thus obtained contains about 90% solution of hydrogen peroxide.
HYDROGEN
477
iii.
Distillation under reduced pressure : The above solution is then distilled under reduced pressure (10 – 15mm). Water distils over at 303 – 313 K leaving behind 99% of H2O2 solution.
iv.
Removal of last traces of water : This is carried out by cooling the solution in a freezing mixture consisting of dry ice (solid carbon dioxide) and ether when crystals of hydrogen peroxide separate out. These crystals are removed, dried and melted to obtain pure hydrogen peroxide.
Precautions taken for storing hydrogen peroxide : Decomposition of hydrogen peroxide (2H2O2 ® 2H2O + O2) is catalysed by the presence of heavy metal ion, dust, strong bases, rough surfaces and light. Hence, the following precautions should be taken while storing hydrogen peroxide solution. i.
It is to be stored in coloured paraffin wax coated plastic or teflon bottle. Glass bottles are not used since the rough surface of glass, alkali oxides present in it and exposure to light catalyse the decomposition of hydrogen peroxide.
ii.
Some stabilisers such as glycerine, acetanilide, phosphoric acid etc. must be used to check further decomposition of hydrogen peroxide.
Strength of Hydrogen peroxide solution : The strength of hydrogen peroxide is expressed in terms of weight or volume. i.
As weight percentage. The weight percentage of hydrogen peroxide shows the weight of H2O2 in 100 gms of solution For example, a 15% solution by weight of hydrogen peroxide means 15 grams of hydrogen peroxide are present in 100 gm. of solution.
ii.
As volume. Hydrogen peroxide is usually available as solution labelled with a volume strength. (Example, 40 volumes H2O2 ) 40 volumes of H2O2 means that one litre of this solution will give 40 litres oxygen at N.T.P.
It is possible to calculate the strength of hydrogen peroxide from its volume strength. A 40 volume solution of hydrogen peroxide means one litre of the solution gives 40 litres of oxygen at N.T.P. Hydrogen peroxide decomposes as : 2H2O2 2 x 34g.
®
2H2O
+
O2 (22.4 litres at N.T.P)
22.4 litres of oxygen is obtained from 68g of H2O2 40 litres of oxygen is obtained from 6822´.440 g of H2O2 So, 1 litre of H2O2 solution contains 6822´.440 g H2O2 = 121.4g of H2O2. Hence, 1 litre of 40 volume hydrogen peroxide contains 121.4 g of H 2O2.
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Problem : Calculate the normality of 10 volume of hydrogen peroxide solution. Solution : Step 1. To calculate the strength in g/litre of 10 volume H 2O2 solution. 10 volume of H2O2 means that one litre of the solution liberates 10 litres of O 2 at N.T.P. H2O2 decomposes as : 2H2O2 ® 2H2O + O2 2 x 34 = 68g 22.4 litres at N.T.P. 22.4 litre of O2 is liberated from 68g of H2O2 at N.T.P. 10 litres of O2 is liberated from 6822´.410 = 30.357g of H2O2 at N.T.P Hence, the strength of 10 volume of H2O2 solution = 30.357 g/litre Step 2. To calculate the equivalent mass of H2O2. 32 parts by weight of O2 are obtained from 68 parts by weight of H2O2. 8 parts by weight of O2 are obtained from 6832´ 8 = 17 parts by weight of H2O2. Equivalent mass of H2O2 = 17 Step 3. To calculate the normality of 10 volume H2O2 solution Normality
Gms/litre.
= Equivalent mass =
30.357 = 1.79. 17
Hence, the strength of 10 volume H2O2 solution = 1.79 N. Problem : Calculate volume 1.5 N H2O2 solution. Solution : mass = 34 = 17 Eq. mass of H2O2 = Molecular 2 2
Normality
= Gms/litre. Eq.mass
\ Gms/litre
= Normality x Eq. mass. = 1.5 x 17 = 25.5 2H2O2 ® 2H2O + O2 2moles 1mole. 68 gms of H2O2 liberate 22.4 litres of O2 at N.T.P. 25.5g of H2O2 will liberate 22.4 ´ 25.5 = 8.4 litre of O2 at N.T.P.. 68
Hence, the strength of hydrogen peroxide solution is 8.4 volume.
HYDROGEN
479
11.8.4 Properties of Hydrogen Peroxide : Physical properties : 1.
Pure hydrogen peroxide is a colourless, syrupy liquid.
2.
Its aqueous solution has a bitter taste.
3.
It is soluble in water, alcohol and ether in all proportions.
4.
Its density is 1.448g/cm3 at 293 K. The high density is attributed to association of its molecules by intermolecular hydrogen bonds.
5.
It produces blister in contact with skin.
6.
Its melting point is 272.54 K .
Chemical properties : 1.
Decomposition : Pure hydrogen peroxide is not very stable and decomposes to oxygen and water on standing or on heating. 2H2O2 ® 2H2O + O2.
The decomposition is accelerated in the presence of heavy metal ions, dust, strong bases, light or even in contact with rough surface. 2.
Acidic nature : Its aqueous solution behaves as weak acid. It forms two types of salts, hydroperoxide (NaO2H) and normal peroxide (Na2O2) with sodium hydroxide. H2O2 + H2O ® H3O+ + HO–2 HO2– + H2O ® H3O+ + O22– NaOH + H2O2 ® NaHO2 + H2O 2NaOH + H2O2
3.
® Na2O2 + 2H2O
Oxidising property : Hydrogen peroxide acts as an oxidising agent (electron acceptor) in neutral, acidic and alkaline medium.
(a) In neutral medium it oxidises iodides to iodine, sulphites to sulphates, sulphides to sulphates, nitrites to nitrates. i.
ii.
Potassium iodide is oxidised to iodine. 2I– + H2O2
® I2 + 2OH–
2KI + H2O2
® 2KOH + I2 .
Sodium sulphite is oxidised to sodium sulphate. SO32– + H2O2 ® SO42– + H2O Na2SO3 + H2O2
® Na2SO4 + H2O
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+2 CHEMISTRY (VOL. - I)
iii.
Lead sulphide is oxidised to lead sulphate S2– +4H2O2
®
SO42– + 4H2O
PbS + 4H2O2 ® PbSO4 + 4H2O (Black) (White) Hydrogen peroxide restores the colour of old oil paintings. In these paintings white basic lead carbonate is used which is turned black by hydrogen sulphide present in air due to the formation of Pbs. By treatment with H2O2 is changed again to white PbSO4 . iv.
Potassium nitrite is oxidised to potassium nitrate NO2– + H2O2 ® NO3– + H2O KNO2 + H2O2 ® KNO3 + H2O
(b)
In acidic medium hydrogen peroxide oxidises ferrous salts to ferric salts, ferrocyanide to ferricyanide, and dichromate to chromium pentoxide. i.
Ferrous sulphate in acidic medium is oxidised to ferric sulphate. 2Fe2+ + H2O + 2H+ ® 2Fe3+ + 2H2O H2O2 ® H2O + O 2FeSO4 + H2SO4 + O® Fe2(SO4)3 + 2H2O . ——————————————————— 2FeSO4 + H2SO4 + H2O2 ® Fe2(SO4)3 + 2H2O.
ii.
Potassium ferrocyanide in acidic medium is oxidised to K 3 [Fe(CN)6] 2[Fe(CN)6]4–+ 2H+ + H2O2
® 2[Fe(CN)6]3– + 2H2O
H2O2 ® H2O + O 2K4[Fe(CN)6] + H2SO4 + [O] ® 2K3[Fe(CN)6] + K2SO4 + 2H2O ———————————————————————————— 2K4[Fe(CN)6] + H2SO4 + H2O2 ® 2K3[Fe(CN)6] + K2SO4 + 2H2O iii.
Acidified potassium dichromate is oxidised to blue chromium pentoxide. K2Cr2O7 + H2SO4 + 4H2O2 ® K2SO4 + 2CrO5 + 5H2O. Chromium pentoxide.
(c)
In alkaline medium hydrogen peroxide oxidises chromium salts and manganese salts to chromates and manganese dioxide respectively. 3Cr3+ + 4H2O2 + 10 OH– ® 2CrO42– + 8H2O Cr2(SO4)3 + 3H2O2 + 10NaOH ® 2Na2CrO4 + 3Na2SO4 + 8H2O. Mn2+ + H2O2 + 2OH– ® MnO2 + 2H2O MnSO4 + H2O2 + 2NaOH ® Na2SO4 + MnO2 + 2H2O .
4.
Reducing property. Hydrogen peroxide also acts as a reducing agent ( electron donor) in neutral, acidic and alkaline medium.
HYDROGEN
481
(a) In neutral medium it reduces halogens to halogen acids, silver oxide to metallic silver, lead dioxide to lead monoxide and ozone to oxygen. Cl2 + H2O2 ® 2HCl + O2 Ag2O + H2O2 ® 2Ag + H2O + O2 PbO2 + H2O2 ® PbO + H2O + O2 . (b) In acidic medium hydrogen peroxide reduces potassium permanganate solution, potassium dichromate solution and manganese dioxide. (i)
It reduces pink coloured acidified KMnO4 solution to colourless manganous sulphate. 2MnO4– + 5H2O2 + 6H+
® 2Mn2+ + 8H2O + 5O2
2KMnO4 + 3H2SO4
® K2SO4 + 2MnSO4 + 3H2O + 5O
[H2O2 + O
® H2O + O2 ] x 5
———————————————————————————— 2KMnO4 + 3H2SO4 + 5H2O2 ® K2SO4 + 2MnSO4 + 8H2O + 5O2 (ii)
Acidified K2Cr2O7 solution (orange colour) is reduced to green chromic sulphate. Cr2O72– + 3H2O2 + 8H+
® 2Cr3+ + 7H2O + 3O2
K2Cr2O7 + 4H2SO4
® K2SO4 + Cr2(SO4)3 + 4H2O + 3O
[H2O2 + O
® H2O + O2] x 3
K2Cr2O7 + 4H2SO4 + 3H2O2 ® K2SO4 + Cr2(SO4)3 + 7H2O + 3O2 (iii) In presence of dil. H2SO4, hydrogen peroxide reduces manganese dioxide to manganous sulphate. MnO2 + H2O2 + H2SO4 ®MnSO4 + 2H2O + O2 . c. In alkaline medium hydrogen peroxide reduces potassium ferricyanide to potassium ferrocyanide and ferric salts to ferrous salts. 2[Fe(CN6)]3– + 2OH– + H2O2 ® 2[Fe(CN)6]4– + 2H2O + O2 2K3[Fe(CN)6] + 2KOH + H2O2 ® 2K4[Fe(CN)6] + 2H2O + O2 2Fe3+ + H2O2 + 2OH– ® 2Fe2+ + 2H2O + O2 Fe2(SO4)3 + H2O2 + 2KOH ® 2FeSO4 + K2SO4 + 2H2O + O2 5.
Bleaching action. It acts as a bleaching agent. It bleaches delicate materials such as silk, feather, ivory etc by oxidation. H2O2 ® H2O + O Colouring matter + O ® Colourless matter.
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Summary : H2O + O2 H 2O
H3O+ + HO2–
NaOH
NaHO2 + H2O
KI
KOH + I2
Na2SO3
Na2SO4
PbS
PbSO4
KNO2
KNO3 + H2O
FeSO4 / H + K4 (Fe (CN)6)
H2 O2
K2 Cr2O7 /H+ 3+
Fe2 (SO4)3 K3[Fe(CN)6] CrO5
Cr /OH CrO42– Mn2+/OH–
MnO2
Fe3+/OH–
Fe2+
Cl2
HCl
Ag2O
Ag
MnO4 –/H+ Cr2O72–/H+ [Fe(CN)6) OH–
Mn2+ Cr3+
3–/
(Fe(CN)64–
Uses of hydrogen peroxide : It is used i.
As an antiseptic for cleaning wounds, teeth and ears.
ii.
For bleaching hair, silk, wool, ivory etc.
iii.
For restoring the colour of lead paintings.
iv.
In the laboratory for detection of chromium, vanadium and titanium.
v.
As a fuel in submarines and rockets.
HYDROGEN
483
Structure of Hydrogen peroxide. Hydrogen peroxide molecule has a non-planar and non-linear structure often described as open book structure. The molecule contains a peroxy bond (O–O) and the two hydrogen atoms are attached to two different oxygen atoms. The O–O–H remains in one plane and the other hydrogen atom remains in another plane at an angle of 94 0 as shown below. The H–O– O bond angle is 970. The O–O bond distance is around 1.48A0 while that of each O–H bad is 0.95A0.
970
1.48A0
O
H
A0
970
0.95
O
0
940
11.9
A 0.95
H
HYDROGEN AS A FUEL
Dihydrogen on combustion produces large quantities of heat (Table 11.5). Also the pollutants released in this case are only the oxides of nitrogen (due to the presence of dinitrogen as impurity with dihydrogen) which can be minimised by injecting a small amount of water into the cylinder. However, cylinder of compressed dihydrogen is about 30 times as heavy as a cylinder containing petrol which is a great disadvantage. Liquid dihydrogen requires expensive insulated tanks made up of alloys like NaNi5, Ti–TiH2, Mg – MgH2 etc. to store it at very low temp like 20K. These limitations have prompted researchers to search for alternative techniques to use dihydrogen efficiently.
Table 11.5 Energy released by combustion of various fuels in moles, mass and volume. Energy released on combustion in kJ
Dihydrogen (in gaseous state)
Dihydrogen (in liquid)
LPG
CH4 gas
Per mole
286
285
2220
880
Per gram
143
142
50
53
Per litre
12
9968
25590
35
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+2 CHEMISTRY (VOL. - I)
CHAPTER (11) AT A GLANCE 1. 2. 3. 4. 5. 6.
7.
8.
Hydrogen, due to its electronic configuration, resembles both alkali metals of Gr 1 and halogens of Gr 17. There are three isotopes of hydrogen. These are protium ( 1H1), deuterium or heavy hydrogen (1H2) and tritium (1H3). Heavy water - It is deuterium oxide (D2O). It is used as a moderator in some nuclear reactors. Isotopic effect - The difference in properties due to the mass difference in isotopes of an element is known as Isotopic effect. Nascent hydrogen - Hydrogen which is just liberated as a result of chemical reaction. It is represented by H. Ortho and para hydrogen -These are two isomers of hydrogen. In ortho hydrogen the spins of the protons are in the same direction (parallel) whereas in para hydrogen the spins of the protons are in the opposite direction (antiparallel). Orthoform is more stable than paraform. Hydrogen peroxide is a colourless, syrupy liquid, soluble in water, alcohol, ether. It is not very stable, readily decomposes to water and oxygen upon standing or heating. It acts as an oxidising and reducing agent in acidic, alkaline, neutral medium. and also a bleaching agent. The bleaching action is due to oxidation. Hydrogen peroxide is prepared in the laboratory by the reaction of sodium or barium peroxide with 20% sulphuric acid. Na2O2 + H2SO4 ® Na2SO4 + H2O2
9.
10.
11.
BaO2.8H2O + H2SO4 ® BaSO4 + H2O2 + 8H2O. Strength of hydrogen peroxide is expressed in terms of weight or volume. (a) 15% solution of H2O2 means 15 gms of H2O2 are present in 100 gms of solution. (b) 40 volumes of H2O2 means that one litre of this solution will give 40 litres of oxygen at NTP. Hydrogen peroxide molecule has a non-planar and nonlinear structure. O – O bond length is 1.48 A0 and O – H bond length is 0.95 A0. H – O – O bond angle is 970 and the interplanar angle is 940. Hydrogen on combustion produces large amount of heat and hence can be used as a fuel. QUESTIONS
(A) Short answer type (2 marks each) 1.
Old paintings of lead are generally washed with dilute solution of hydrogen peroxide in order to regain its colour. Explain.
2.
What is the trade name of hydrogen peroxide used as an antiseptic ?
HYDROGEN
485
3.
Name one compound each in which hydrogen exists in (i) positive oxidation state and (ii) negative oxidation state.
4.
Which isotope of hydrogen (i) does not contain neutron (ii) is radio active ?
5.
Explain why H2O2 cannot be stored for a prolonged period ?
6.
Show how can hydrogen peroxide function both as an oxidising and reducing agent.
7.
What is the structure of hydrogen peroxide ?
8.
Give four uses of hydrogen peroxide.
9.
Why is anhydrous barium peroxide not used in the preparation of H 2O2 ?
10.
Distinguish between ortho and para hydrogen.
11.
Name the isomers of hydrogen.
12.
Name two isotopes of hydrogen.
13.
Give the products of electrolysis of H2O are D2O
14.
Write the formula of heavy water. What is the mass number of heavy hydrogen?
15.
What is the oxidation number of oxygen in hydrogen peroxide ?
16.
Give examples showing electropositive and electronegative character of hydrogen.
17.
How hydrogen peroxide restores the colour of old oil paintings ?
18.
Name the isotopes of hydrogen with mass numbers.
19.
Name one reaction in which water acts (i)
as an oxidising agent and (ii) as a reducing agent.
(B)
Long answer type (7 marks each)
1.
Discuss the position of hydrogen in the periodic table.
2.
Mention one method of preparation of hydrogen peroxide. Show with equations what happens when it reacts with (a) Lead sulphide (b) Ozone (c) Potassium iodide solution.
3.
Describe the preparation, properties and uses of heavy water.
4.
What are ortho and para hydrogen ? How is para hydrogen isolated ? Explain the term - Isotopic effect.
5.
Explain how is dilute silution of hydrogen peroxide concentrated ? Write chemical equation for reactions of hydrogen peroxide with the following. (i) Chlorine (ii) Acidified ferrous sulphate solution
6.
Describe one method of preparation of hydrogen peroxide. What happens when it reacts with (a) Acidified K2Cr2O7 solution (b) FeSO4?Give two of its important uses.
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+2 CHEMISTRY (VOL. - I)
7.
Write note on- Heavy water (5 marks)
8.
Describe one method of preparation of hydrogen peroxide. Explain with example how it acts both as oxidising and reducing agent. What is meant by '10 volume' of hydrogen peroxide ? MULTIPLE CHOICE TYPE QUESTIONS
1. 2. 3.
4. 5.
6. 7. 8. 9. 10. 11. 12. 13.
In which of the following compounds does hydrogen exhibit a negative oxidation state? (a) LiH (b) H2O (c) HCl (d) none of these. The number of neutrons in deuterium is (a) 2 (b) 3 (c) 1 (d) 0. Which of the following statements is correct ? (a) Deuterium has two electrons. (b) Deuterium has two protons. (c) Deuterium has the same mass as that of hydrogen. (d) Deuterium has the same atomic number as that of hydrogen. Heavy water is (a) Water obtained by electrolysis. (b) H 2O (c) D2O (d) T2O. The composition of nucleus of deuterium is (a) One electron and one proton (b) One proton and one neutron. (c) One neutron and one electron. (d) Two protons and one electron. Hydrogen peroxide does not act as : (a) A reducing agent (b) an oxidising agent (c) a dehydrating agent (d) a bleaching agent. The oxide that gives hydrogen peroxide on treatment with a dilute acid is (a) PbO2 (b) NaO2 (c) MnO2 (d) TiO2. Ortho hydrogen differs from para hydrogen in (a) atomic mass (b) atomic number (c) nuclear spins. Electronic configuration of deuterium atom is (a) Is1, (b) 2s2 (c) 2s1 (d) Is2 Decomposition of hydrogen peroxide is prevented by (a) NaOH (b) MnO2 (c) Glycerol (d) Oxalic acid Which of the following is based as a moderator in neclear reactors ? (a) Heavy hydrogen (b) Ozone (c) Heavy water (d) Hydrogen peroxide. The strength of 20 volume of H2O2 is per litre or 100 ml ? (a) 13.6 g / litre (b) 60 g / litre (c) 160 g / litre (d) 20 g / litre. When hydrogen peroxide is oxidised the product formed is (a) O2– (b) OH– (c) HO2– (d) O2.
HYDROGEN
14.
15. 16.
17. 18. 19.
20.
487
Heavy water is manufactured by (a) Fractional distillation of water (b) exhaustive electrolysis of water (c) Fractional diffusion of steam (d) fractional crystallisation of ice. Hydrogen peroxide involves (a) polar bond. (b) Non-polar bond. (c) Semi polar bond (d) hydrogen bond. Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RbH, CsH, the correct order of increasing ionic character is (a) LiH > NaH > CsH > KH > RbH (b) LiH < NaH < KH < RbH < CsH (c) RbH > CsH > NaH > KH > LiH (d) NaH > CsH > RbH > LiH > KH The radioactive isotope of hydrogen is : (a) Protium (b) Deuterium (c) Tritium (d) Hydronium Which of the following ions will cause hardness of water sample ? (a) Na+ (b) K+ (c) Ca2+ (d) Li+ Which of the following statements are correct ? (a) Hydrides of group 13 act as Lewis acids (b) Hydrides of group 14 act as Lewis acids (c) Hydrides of group 15 act as Lewis bases. (d) Hydrides of group 14 act as Lewis bases. Permanent hardness of water is due to the presence of (a) Chlorides of Ca and Mg in water. (b) Sulphates of Ca and Mg in water. (c) Hydrogen carbonates of Ca and Mg in water. (d) Cabonates of alkali metals in water. ANSWERS TO MULTIPLE CHOICE TYPE QUESTIONS 1. (a)
6. (c)
11. (c)
16. (b)
2. (c)
7. (a)
12. (b)
17. (c)
3. (d)
8. (c)
13. (d)
18. (c)
4. (c)
9. (a)
14. (b)
19. (a, c)
5. (b)
10. (c)
15. (c)
20. (a, b)
qqq
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UNIT - X
S-BLOCK ELEMENTS : ALKALI AND ALKALINE-EARTH METALS CHAPTER - 12
ALKALI METALS (GROUP – 1) 12.1
INTRODUCTION :
The elements belonging to Group 1 of the periodic table are called Alkali metals. The six elements are- Lithium(Li), Sodium (Na), Potassium (K), Rubidium (Rb), Caesium (Cs) and Francium (Fr). These are called alkali metals because they readily dissolve in water to form hydroxides which are strongly alkaline in nature. They also form alkaline oxides. The element Francium (Fr) is radioactive. 2 He
1 H 4 82
3 L: 11 Na 19 K
22 Ti 40 Zr 72 Hf
23 V 41 Nb 73 Ta
24 Cr 42 Mo 74 W
25 Mn 43 Tc 75 Re
26 Fe 44 Ru 76 Os
27 Co 45 Rh 77 Ir
28 Ni 46 Pd 78 Pt
29 Cu 47 Ag 79 Au
30 Zn 48 Cd 80 Hg
5 B 13 A1 31 Ga 49 In 81 Tl
6 C 14 Si 32 Ge 50 Sn 82 Pb
7 N 15 P 33 As 51 Sb 83 Bi
37 Rb 55 Cs 87 Fr Fig 12.1 - Position of the alkali metals in the periodic table.
8 O 16 S 34 Se 52 Te 84 Po
9 F 17 Cl 35 Br 53 I 85 At
10 Ne 18 Ar 36 Kr 54 Xe 86 Rn
THE ALKALI METALS
489
12.2 ELECTRONIC CONFIGURATIONS All the alkali metals have one electron in their outermost s- orbital preceded by the noble gas (inert gas) configuration. Thus the general configuration of alkali metals may be written as (Noble gas) ns1 where n represents the valence shell. The electronic configurations of alkali metals are given below. Element
Table – 12.1 Symbol At. No Electronic Configuration
Configuration of the outermost shell
Lithium
Li
3
1s2,2s1 or [He] 2s1
2s1
Sodium
Na
11
1s2,2s2,2p6, 3s1 or [Ne] 3s1
3s1
Potassium
K
19
1s2,2s2,2p6, 3s2,3p6, 4s1 or [Ar] 4s1
4s1
Rubidium
Rb
37
1s2,2s2,2p6, 3s2,3p6, 3d10,4s2,4p6,5s1 or [Kr] 5s1 5s1
Caesium
Cs
55
1s2,2s2,2p6, 3s2,3p6, 3d10,4s2,4p6, 4d10,5s2 ,5p6,6s1 or [Xe] 6s1
Francium
Fr
87
6s1
1s2,2s2,2p6, 3s2,3p6, 3d10,4s2,4p6,4d10, 4f14,5s2 ,5p6,5d10,6s1, 6p6, 7s1 or [Rn] 7s1
7s1
Anomalous behaviour of lithium: Though lithium belongs to the alkali metal group (Group 1) and has most of the characteristic properties of the metals of group 1, it differs from the remaining members of the group in many respects. This is mainly due to the following reasons: (i)
The size of the lithium atom and its ion is very small.
(ii)
It has high ionisation energy and least electropositive character.
(iii)
The polarising power of Li+ ion is quite high due to its small size, which result in the covalent character of its compounds.
(iv)
It has no vacant d-orbital in its valence shell.
(v)
It is acting as strongest reducing agent even if it is expected to be the least reducing agent due to its high consisation energy
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+2 CHEMISTRY (VOL. - I)
Reason:- This anomaly can be explained if its hydration energy is to be taken into account. It is known that ionisation is the property of isolated atoms in the gaseous state. M+(g)+e- (ionisation energy)
M(g)
However, oxidation potential is the property when the metal goes into the solution as M (aq) ions. +
M+(aq) + e- (oxidation potential)
M(s)
It involves several steps. (i)
M(s)
M(g) DH Sublimation
(ii)
M(g)
M+(g) + e- Ionisation energy
(iii)
M+(g) + H2O
M+(aq)
Hydration energy
The over-all tendency for the charge depends upon the net effect of three steps Since Li+ has the smallest size , it gets hydrated to a maximum extent and a large amount of hydration energy is released in the third step which compensates the higher energy needed to remove electron in the second step. The net effect is that it has greater tendency to lose electrons in solution than other alkalimetals. Thus, the reaction can be represented as Li+(aq)+ e- and
Li(s)
Therefore, Lithium is the strongest reducing agent due to greater hydration energy. Diagonal relationship It has been observed that some elements of second period show similarities with the elements of the third period present diagonally to each others, though belonging to different groups. This is called diagonal relationship. Table 12.2 Group- 1
Group- 2
Group – 13
Group- 14
Second period
Li
Be
B
C
Third period
Na
Mg
Al
Si
For example, Lithium (Gr. 1) resembles Magnesium (Gr. 2). Similarly, Be with Al and B with Si have resemblances in respect of their physical and chemical properties.
THE ALKALI METALS
491
Cause of Diagonal relationship The cause of diagonal relationship is the similarity in properties such as electronegativity, ionisation energy, size etc. between the diagonal elements, For example, on moving from left to right across a period, the electronegativity increases, while on moving down a group, electronegativity decreases. Therefore, on moving diagonally, the two opposing tendencies almost cancel out and the electronegativity values remain almost same, thus the diagonal pairs have many similar properties. Electronegativity increases
re la te da to m s
Electronegativity decreases
PERIOD GROUP
D ia go na lly
Similar Electronegativity
12.3 GENERAL CHARACTERSTICS OF ALKALI METALS Physical Characteristics : The important physical properties of alkali metals are given below. Table - 12.3 Physical Characteristics of Alkali metals Property
Li
Atomic Radius (Ao) 1.23 o Ionic radius (A ) 0.68 Ionization Potential(kJ/mol) 520 Electronegativity 1.00 Density(g/cm3) 0.53 Melting Point(K) 453 Boiling Point(K) 1609 Abundance in Earth’s Crust(ppm) 65
Na 1.57 0.92 496 0.9 0.97 371 1156 28300
K 2.03 1.3 418 0.8 0.86 336.5 1032 25904
Rb 2.16 1.48 403 0.8 1.53 312 973 300
Cs 2.35 1.69 374 0.7 1.09 301.5 943 7
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The properties of francium have not been listed in the above table, as it is a recently discovered element. A very little is known about this radioactive element. 1.
Atomic and Ionic Radii
Alkali metals have the largest atomic and ionic radii in their respective periods. On moving down the group the atomic and ionic radii increase. Explanation : The alkali metals are the first elements of each period. As we move in a period, the atomic radius and ionic radius tend to decrease due to increase in the effective nuclear charge. Therefore, the alkali metals have the largest atomic and ionic radii in their respective periods. On moving down the group there is increase in the number of shells and therefore atomic and ionic radii increase. 2.
Ionisation Energies (i)
Alkali metals have the lowest ionization energy in each period.Within the group, the ionization energies of alkali metals decrease down the group.
Explanation : The atoms of alkali metals are largest in their respective periods and therefore the valence electrons, are loosely held by the nucleus.By losing the valence electrons, theyacquire stable noble gas configurations. This accounts for their ease to lose electrons and hence they have low ionization energies. On moving down the group, the atomic size increases and the magnitude of screening effect (number of inner shells) also increases and consequently, the ionization energy decreases down the group. (ii) The second ionization energies of alkali metals are very high. Explanation : When an electron is removed from the atoms of alkali metals, they form monovalent cations which have very stable electronic configuration (same as that of noble gases). Therefore, it becomes very difficult to remove the second electron from the stable noble gas configurations and hence their second ionisation energy values (IE2) are very high. 3.
Melting and Boiling Points All these metals are soft and have low melting and boiling points. Explanation : The alkali metals have only one valence electron per metal atom and therefore the energy binding the atoms in the crystal lattice of the metal is low. Thus, the metallic bonds in these metals are very strong and consequently their melting and boiling points decrease on moving down from Li to Cs.
4.
Density
The densities of alkali metals are quite low as compared to other metals. Li, Na and K are even lighter than water. The densities increase on moving down the group. However, K is lighter than Na.
THE ALKALI METALS
493
Explanation : The densities of metallic elements depend upon the type of packing of atoms in metallic state and also on their size. The alkali metals have close packing of metal atoms in their lattice .Because of the large size of their atoms, they have low densities. As we move down the group from Li to Cs, there is increase in atomic size as well as atomic mass. But the increase in atomic mass is more to compensate the increase in atomic size. As a result, the densities (mass/volume) of alkali metals gradually increase from Li to Cs. However, potassium is lighter than sodium probably due to increase in atomic size of potassium. 5.
Electropositive or Metallic character All the alkali metals are strongly electropositive or metallic in character. Explanation: The electropositive character of an element is expressed in terms of the tendency of its atom to release electrons: M
®
M+ + e
¾
As alkali metals have low inonisation energies, their atoms readily lose their valence electron. These elements are, therefore, said to have strong electropositive or metallic character. Since the ionisation energies decrease down the family, the electron releasing tendency or electropositive character is expected to increase down the family. 6.
Oxidation states All the alkali metals exhibit an oxidation state of +1 in their compounds. Explanation: The alkali metals have only one electron in their valence shell and therefore they can lose the single valence electron readily to acquire the stable configuration of a noble gas. Thus, they form monovalent ions, M+ (e.g., Li+, Na+, K+, Rb+, Cs+). Since the second ioniza tion energies are very high, they cannot form divalent ions. Thus alkali metals are univalent and form ionic compounds.
7.
Characteristic Flame colouration All the alkali metals and their salts impart characteristic flame colouration Explanation: The alkali metals have very low ionization energies. The energy from the flame of bunsen burner is sufficient to excite the electrons of alkali metals to higher energy levels. The excited state is quite unstable and therefore when these excited electrons come back to their original energy levels, they emit extra energy, which fall in the visible region of the electromagnetic spectrum and thus appear coloured.
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The following colours are given by alkali metals: Li Crimson red
Na
K
Rb
Cs
Yellow
Pale violet
Violet
Bluish
The different colours of the alkali metals can be explained on the basis of amount of energy absorbed for excitation of the valence electron. 8.
Photoelectric effect The alkali metals emit electrons when radiation strikes on their surfaces. This phenomenon of emission of electrons when electromagnetic radiation strikes against them is called photoelectric effect. Therefore, alkali metals exhibit photoelectric effect. Explanation: Alkali metals have low ionization energies and therefore the electrons are easily ejected when exposed to light. Among alkali metals, caesium has lowest ionization energy and hence it can show photoelectric effect to the maximum extent.
9.
Nature of the compounds The compounds of the alkali metals are ionic in nature. Explanation: Due to the low ionization energies and large atomic sizes the atoms of alkali metals form cations readily by losing the valence electrons. Consequently, they form ionic bonds with the non-metals of the p-block.
10.
Lattice energies The alkali metal salts consist of cations and anions which are held together by strong electrostatic force of attraction. Therefore, the lattice energies of alkali metal salts are very high. The lattice energy is defined as the amount of energy required to break one mole of a crystal into its free ions MX(s)
Lattice energy
M+(g) + X (g) ¾
Therefore lattice energy gives a measure of the forces of attraction between the ions. Larger the forces of attraction, the greater will be lattice energy. The lattice energy also depends upon the size of the ion and its charge. For the cation of same valency the lattice energy of ionic solids having the same anion decreases with increase in size of the cation due to decrease in force of attraction between them as shown below.
THE ALKALI METALS
495
Table - 12.4 Salt
Lattice energy
Lattice energy
12.4
1.
DECREASES
LiCl NaCl KCl RbCl CsCl
(kJ mol ) 802.6 758.7 681.4 660.6 618.7
NaF NaCl NaBr NaI
(kJ mol-1) 894.5 758.7 714.8 668.8
DECREASES
Salt
-1
CHEMICAL CHARACTERISTICS : The alkali metals exhibit high chemical reactivity. This is due to (i) their low ionisation energies. (ii) low heat of atomisation As the value of ionisation energy decreases down the group from Li to Cs, the reactivity of alkali metals increases from Li to Cs. The alkali metals are highly reactive towards the more electronegative elements such as oxygen and halogens. Some of their characteristic chemical properties are given below. Action with air. All the alkali metals get tarnished on exposure to air due to the formation of oxides on the surface. They burn vigorously in air or oxygen forming oxides. Lithium forms monoxide (Li2O), sodium forms peroxide(Na2O2), the other elements form superoxides(MO 2:M = K, Rb, Cs). 4Li + O2
2Li2O(Monoxide)
2Na + O2
Na2O2(Peroxide)
M + O2 MO2(Superoxides) (M = K, Rb, Cs). Explanation: The formation and stability of these oxides can be explained on the basis of the fact that a small cation can stabilise a small anion and a large cation can stabilise a large anion. The size of Li+ ion is very small and it has a strong positive field around it. It can, therefore, combine with only small anion, O 2- ion. This results in the formation of monoxide (Li2O). On the other hand, Na+ ion is larger cation and has a weak positive field around it. Therefore, Na+ ion can stabilize a bigger peroxide ion, O22- or [-O-O-]2- which has also
weak negative field around it. Similarly, the other ions K +, Rb+, Cs+ are still larger and
have very weak positive field and can therefore, stabilise a bigger superoxide, O2- anion and form superoxides.
496
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+2 CHEMISTRY (VOL. - I)
Action with hydrogen The alkali metals react with hydrogen to form hydrides which are ionic in nature (M + H). 2M + H2
2M+ H- (M = Li, Na, K, Rb or Cs) Metal hydride
3.
i)
The reactivity of alkali metals with hydrogen increases from Li to Cs.
ii)
The ionic character of the hydrides increases from Li to Cs. This is because of decrease in ionization energy down the group so that electrons become easily available to hydrogen for forming H- ion.
iii)
The stability of hydrides decreases from Li to Cs. This is due to the fact that as the size of alkali metal increases from Li to Cs, the M – H bond becomes weak. Therefore, the stability of hydrides decreases.
iv)
The hydrides behave as strong reducing agents and their reducing nature increases down the group.
Action with water
2Na + 2H2O
2NaOH + H2
2K + 2H2O
2KOH + H2
Lithium reacts somewhat slowly. But sodium and the other members of the family react so rapidly with water that the hydrogen gas evolved immediately catches fire. Thus alkali metals cannot be kept either in air or in water. They are normally kept in kerosene oil.
LiOH NaOH KOH RbOH CsOH
Basic character increases
Alkali metals readily react with water to form hydroxides and hydrogen gas.
The hydroxides of alkali metals are strongly basic and the strength of the base increases down the group. Explanation: The M—OH bond in the hydroxides of alkali metals is very weak and it can easily ionize to M+ and OH ions. This accounts for their basic character. Since the ionization energy decreases down the group, the bond between metal and oxygen becomes weak. Therefore, the base strength of the hydroxides increases accordingly. Thus, NaOH is a stronger base than LiOH and so on.
THE ALKALI METALS
4.
497
Action with Halogens -
The alkali metals readily combine with halogens to form ionic halides M +X . e.g. 2M + X2
2MX (X = halogen)
2Na + Cl2
2NaCl
The reactivity of halogen towards alkali metals increases on moving down the group. This is due to the decrease in ionisation energy or increase in the electropositive character as we move down the group. All the metal halides are ionic crystals. However, LiI is slightly covalent because of the polarisation (Li being the smallest cation has maximum polarizing power and iodide ion being largest anion can be polarised to the maximum extent). All the alkali metal halides except LiF are soluble in water. However LiF is insoluble in water due to its high lattice energy because of small cation and small anion. 5.
Solution in liquid Ammonia. Alkali metals dissolve in liquid ammonia to give blue solutions which are conducting in nature. Explanation: In solution the alkali metal atom readily loses the valence electron. Both the cation and the electron combine with ammonia o form ammoniated cation and ammoniated electron. M + (x+y) NH3
[M(NH3)x]+ + [e(NH3)y]Ammoniated Ammoniated cation electron
The ammoniated electron is responsible for the blue colour of the solution. Both ammoniated cation and electron make the solution conducting in nature. On standing, the solution slowly liberates hydrogen as: 2M + 2NH3
2MNH2
+ H2
Metal amine 6.
Hydration of Ions The alkali metal ions are highly hydrated. The smaller the size of the ion the greater is the degree of hydration. Thus, Li+ ion gets much more hydrated than Na+ ion which is more hydrated than K+ ion and so on. Therefore, the extent of hydration decreases from Li + to Cs+. As a result of larger hydration of Li+ ions than Na+ ion, the effective size of Li+ ions is more than that of Na+ ion and their ionic radii in water (called hydrated ionic radii) decrease in the order: Li+ > Na+ > K+ > Rb+ > Cs+
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Table - 12.5 Ion
Li+
Na+
K+
Rb+
Cs+
Ionic radius (PM)
76
102
138
152
167
Hydrated radius(PM)
340
276
232
228
226
Ionic mobility
33.5
43.5
64.5
67.5
68.0
(ohm -1 cm2 mol-1) +
As a result, the hydrated Li ion being largest ion in size has the lowest mobility in water. On the other hand the hydrated Cs+ ion being smallest ion in size has the highest mobility in water. 7.
Reducing nature Alkali metals are strong reducing agents. This is due to their greater ease to lose electrons. This is also indicated by the larger negative values of their reduction potentials . All of them are better reducing agents than hydrogen (E ° = Zero). Therefore, these metals react with compounds containing acidic hydrogen atoms such as alcohols and acetylene liberating hydrogen. 2Li + 2C2H5OH
2C2H5O Li
+ H2
2Na + HC º CH
NaC ºC.Na
+ H2
Acetylene
Sodium acetylide
12.5
PREPARATION AND PROPERTIES OF SOME IMPORTANT COMPOUNDS :
1.
SODIUM CARBONATE (Na2CO3. 1OH2O) Sodium carbonate finds a lot of commercial use and is manufactured by Solvay process.
Preparation by Solvay ammonia soda process : In this process, raw materials like common salt, ammonia and lime stone are used. It involves the reaction of sodium chloride and ammonium hydrogen carbonate to form sparingly soluble sodium hydrogen carbonate, where the latter decomposes on heating to yield sodium carbonate. The equations for Solvay process are NH3 + H2O + CO2
NH4 HCO3
NaCl + NH4HCO3
NaHCO3 + NH4Cl
°C 2NaHCO3 ¾250 ¾¾ ¾® Na2CO3 + CO2 + H2O
THE ALKALI METALS
499
Carbon dioxide used in the carbonating tower is produced by heating lime stone. CaCO3
CaO + CO2
This process involves recovery of ammonia, when the filtrate from the vacuum filter containing ammonium chloride and ammonium bicarbonate are treated as follows: D NH + H O + CO NH4HCO3 ¾¾® 3 2 2
2NH4Cl + Ca(OH)2
2NH3 + CaCl2 + H2O
Ammonia, thus recovered, is reused in the process. Proprties : Physical :
(i)
Sodium carbonate is a white crystalline solid, which mostly exists as decahydrate, Na2CO3.IOH2O (washing soda)
(ii)
The decahydrate loses its water of crystallisation on heating to form monohydrate, Na2CO3.H2O, which on heating changes to anhydrous form, a white powder called soda ash (M.P.852°C)
(iii) Na2CO3 dissolves fairly in water, forming an alkaline solution due to hydrolysis. Na2CO3 + 2H2O l 2NaOH + H2CO3 Chemical :
(i)
Action of acids : Acids react with evolution of CO2 Na2CO3 + HCl NaHCO3 + HCl
(ii)
NaHCO3 + NaCl NaCl + H2O + CO2
Action of CO2 : By passing CO2 to a concentrated solution of Na2CO3, Sodium hydrogen carbonate is precipitated. Na2CO3 + H2O + CO2
2NaHCO3
(iii) Action of Silica : On heating sodium carbonate with silica strongly sodium silicate, also called as water glass or soluble glass is produced. Na2CO3 + SiO2
Na2SiO3 + CO2
(iv) Formation of metal carbonates : Metal salts, when treated with Na2CO3, form insoluble carbonates. BaCl2 + Na2CO3 CuSO4 + Na2CO3
BaCO3 + 2NaCl CuCO3 + Na2SO4
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Uses : (i) It is used in softening of water and cleaning, as washing soda. (ii) It is used in the manufacture of glass, borax, paper and soap. (iii)
It is used as an important laboratory reagent both in qualitative and quantitative analysis.
2.
SODIUM CHLORIDE (NaCl) (Common salt)
Sodium chloride is the most common of the salts of sodium. It is widely distributed in nature. Sea water contains about 2.95% of NaCl. In the Dead sea it is found up to 9.2 percent. As rock salt it is found in England, Australia, Germany, Canada, Pakistan and in Himachal Pradesh in India. Preparation. 1)
From seawater: In India and also in other tropical countries where sea water runs deep into the coast line is evaporated in tanks. The solid crust so left after evaporation is collected. The heaps of crystals of sodium chloride are given a fine spray of water which dissolves magnesium chloride. The remaining product contains about 98% of NaCl and the rest is MgCl2, CaCl2 etc. In very cold countries the brine is concentrated by freezing out water. The concen trated solution is then evaporated in big iron pans. Purification: It is purified by passing hydrogen chloride through a saturated solution of the commercial common salt. Sodium chloride precipitates due to common ion effect. Only KCl is present as an impurity which is removed by repeated crystallisation.
Properties: 1. Sodium chloride is a white crystalline substance which is slightly hygroscopic 2. It dissolves in water with the absorption of heat. The solubility changes slightly with the change in temperature. o
3.
It melts at 800 C.
4.
Action of sulphuric acid- It gives hydrogen chloride with hot concentrated H 2SO4. 2NaCl + H2SO4
5.
Action of silver nitrate – It gives a white precipitate with silver nitrate. NaCl + AgNO3
6.
Na2SO4 + 2HCl NaNO3 + AgCl¯
Action on sand (SiO2) – When fused with sand in presence of moisture it gives sodium silicate. This reaction is used in salt glazing 2NaCl + SiO2 + H2O
Na2SiO3 + 2HCl sodium silicate
Uses; it is used 1) As a preservative for food articles like fish, meat etc. 2) For the manufacture of sodium metal, caustic soda, washing soda etc.
THE ALKALI METALS
3) 4) 5) 6) 3.
501
For the salting out of soap. In freezing mixtures. For glazing pottery. It is an essential constituent of food.
SODIUM HYDROXIDE (CAUSTIC SODA) (NaOH):
Sodium hydroxide is manufactured by the electrolysis of sodium chloride solution (brine) in a specially designed cell called Castner –Kellner cell (Fig. 12.6). Castner – Kellner or Mercury Cathode Process. The apparatus consists of a large rectangular iron vessel divided into three compartments by state partitions dipping into a layer of mercury which runs along the floor and is brought into circulation with the help of an eccentric wheel. The two outer compartments are filled with brine and are fitted with stout graphite electrodes which form the anodes. The middle compartment is filled with water and a little NaOH and is fitted with a bunch of iron rods serving as the cathode. By induction, mercury acts as cathode in the outer compartments and as anode in the middle compartment. On passing electric current chlorine is liberated at the anodes in the outer compartments and escapes through the outlets. The sodium liberated dissolves in the flowing mercury forming sodium amalgam, reacts with water, forming a solution of NaOH and hydrogen escapes through an outlet above and the solution of caustic soda, the concentration of which is not allowed to exceed 20%, is taken out from time to time.
H2 Cl2
DILUTE NaOH SOLUTION
IRON CATHODE
Cl2
NaCl SOLUTION
NaCl SOLUTION
+
+
GRAPHITE ANODE
GRAPHITE ANODE
MERCURY
ECCENTRIC GROOVE WHEEL
Fig. 12.6 Castner-Kellner process for the manufacture of caustic soda.
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Reactions: Outer compartments: In these compartments mercury layer acts as cathode by induction. NaCl
Na+ + Cl-
At the mercury cathode:
Na + + e
Na
Na + Hg Na - amalgam Sodium dissolves in mercury to form sodium amalgam, which due to rocking motion of the cell travels to the central compartment. ¾ At the carbon anode : Cl Cl + e Cl2-
Cl + Cl
Central compartment: In this compartment sodium amalgam acts as anode by induction. Na + + OH
¾
NaOH At the iron cathode: Na + + e
¾
Na + H2O
Na NaOH + ½ H2-
So at cathode hydrogen gas is liberated and sodium hydroxide is formed. At the mercury anode: At sodium amalgam anode, sodium undergoes oxidation and passes into solution as Na + ions. Na -- amalgam
Na + + e -
Na+ + OH -
NaOH
+ Hg
With the progress of electrolysis, concentration of NaOH in the central compartment continues to increase. When it reaches 20%, the solution is withdrawn, evaporated to dryness so that fused sodium hydroxide, is obtained. This is then cast into sticks or pellets. Properties of NaOH : Physical properties : It is a white crystaline solid having bitter taste. It is soluble in water and is corrosive in nature. Its density is 2.13 g/ml and melting points is 318°. Chemical properties : 1. It reacts with atmospheric carbon dioxide to form sodium carbonate 2NaOH + CO2 Na2CO3 + H2O 2. As a strong alkali, it reacts with acids to form salts NaOH + HCl NaCl + H2O 2NaOH + H2SO4 Na2SO4 + 2H2O 3. Action on metals : It reacts with metals like Zn, Al etc. to liberate hydrogen. Zn + 2NaOH Na2ZnO2 + H2 2Al + 2NaOH + 2H2O 2NaAlO2 + 3H2
THE ALKALI METALS
503
4.
Action on non-metals : With cold and dil.NaOH, halides and hypohalites are formed, when treated with halogens. NaCl + NaClO + H2O Cl2 + 2NaOH With hot and conc. NaOH solution, halogens form halates and halides 3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O When NaOH is heated with sulphur, sodium thiosulphate is formed. 4S + 6 NaOH Na2S2O3 + 2Na2S + 3H2O With boron, hydrogen is evolved. 2B + 6 NaOH 2Na3BO3 + 3H2 Action on Salts : When treated with metallic salts, NaOH forms the corresponding hydroxides. FeSO4 + 2NaOH Fe(OH)2 + Na2SO4 CrCl3 + 3 NaOH Cr(OH)3 + 3NaCl Green ppt Zn(OH)2 + 2NaOH Na2ZnO2 + 2H2O Sodium Zincate Uses: Causic soda is extensively used 1. In the manufacture of sodium in many industries such as: hard soap, paper and viscose rayon (artificial silk) 2. In the manufacture of many dye – stuffs. 3. In the petroleum industry for the bleaching and refining of oils. 4. In the manufacture of hypochlorites, chlorates and nitrates. 4.
SODIUM HYDROGEN CARBONATE (BAKING SODA) (NaHCO3) : It is called as baking soda for its use in the preparation of cakes and pastries, because of its property to evolve CO2 on heating. Preparation : It is prepared by passing CO2 through the saturated solution of sodium carbonate. Na2CO3 + H2O + CO2
2NaHCO3
Properties : Physical
: It is a white crystalline solid, sparingly soluble in water.
Chemical
: (i)
On hydrolysis, it forms an alkaline solution. NaHCO3 + H2O l NaOH + H2CO3
(ii)
On heating, it forms sodium carbonate with evolution of CO 2 2NaHCO3
(iii)
Na2CO3 + H2O + CO2
It reacts with metal salts to form normal carbonates ZnSO4 + 2NaHCO3
ZnCO3 + Na2SO4 + H2O + CO2
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Uses : Sodium hydrogen carbonate is used : (i)
in fire extinguishers
(ii)
for making Baking powder
(iii)
as a medicine, as mild antiseptic for skin infections and to neutralise acidity in the stomach.
(iv) 12.6
ANALYTICAL TESTS FOR SODIUM (a)
(b)
12.7
in making effervescent drinks.
Flame Test: A thin paste of sodium salt with conc. Hydrochloric acid imparts persistent golden yellow colour to the flame. Golden yellow colour will not be visible when seen through double blue glass. (i) Action with Magnesium uranyl acetate solution: To a little sodium salt solution, 2 ml of magnesium uranyl acetate solution is added; a yellow crystalline ppt. of sodium magnesium uranyl acetate results. (ii) Action with Potassium pyroantimonate solution: To a little sodium salt solution, 2ml of potassium pyroantimonate solution is added; a white ppt of sodium pyroantimonate results.
BIOLOGICAL IMPORTANCE OF SODIUM AND POTASSIUM : An individual, averagely healthy and with normal dietary habit, seldom suffers from deficiency of sodium or potassium. Sodium and potassium occur in plants and animals as the salts (chlorides, phosphates and carbonates) of inorganic acids and salts of proteins and organic acids. Sodium ions are found outside the cells, as extracellular cations, while potassium ions are abudantly present in the cell fluids as main intracellular cations. Generally, sodium and potassium are almost totally absorbed from the gastrointestinal tracts. Functions : The metabolism of sodium and potassium are related to some fundamental physiological mechanism, as described below : 1. Maintaining osmotic pressure : Sodium and potassium function to maintain the normal osmotic pressure of the different body fluids and thus protect the body against excessive loss of fluids. 2. Maintaining proper viscosity of blood : Sodium and potassium chlorides present in the blood plasma help in keeping the globulins in physical solution and in regulating the degree of hydration of the plasma proteins, which is very vital in maintaining proper viscosity of blood. 3. Secretion of digestive fluids : Gastric HCl is derived from NaCl present in the blood, while the base in pancreatic juice and bile is derived from blood sodium and potassium salts. Although quite similar in their chemical properties, sodium and potassium differ quantitatively in their ability to activate enzymes, in their transport mechanism and in penetrating cell membranes.
THE ALKALI METALS
505
CHAPTER (12) AT AGLANCE 1. 2. 3.
Baking powder Borax Carnallite
4. 5. 6. 7. 8. 9.
Chile salt petre Caustic soda Caustic potash Nitre Sylvinite Common salt
- The chemical is sodium bicarbonate (NaHCO3) - The chemical composition is Na2B4O7.10H2O - It is an ore of potassium and magnesium and its composition is KCl.MgCl2.6H2O - The chemical is Na NO3 (sodium nitrate) - The chemical is sodium hydroxide (NaOH) - The chemical is potassium hydroxide (KOH) - The chemical is potassium nitrate (KNO 3) - Its chemical composition is NaCl. KCl - Its chemical composition is sodium chloride (NaCl.)
10.
Potash alum
- Its chemical composition is K2SO4.Al2 (SO4)3.24H2O QUESTIONS
Very short answer type. (1 mark each) 1.
Give the electronic configuration of potassium atom.
2.
Give one use of potassium chlorate.
3.
(a) Write the formula of Borax. (b) What is the composition of carnallite?
4.
What happens when CO gas is passed through aqueous solution of sodium carbonate?
5.
Between Lithium and Sodium which is more electropositive ?
6.
What is Borax? Give one use of borax.
7.
Oxide of an alkali metal is – (amphoteric, basic, acidic) Hint. basic.
8.
The element in which the outer most electron is held most loosely is ..(Li, Na, K, Rb)
9.
Soda lime is a mixture of —— and ——.
10.
Why alkali metals do not from M2+ion?
11.
Write the electronic configuration of Sodium.
12.
Sodium metal is stored under …………
13.
Why are alkali metals soft ?
14.
Sodium nitrate on heating gives ———— gas.
15.
Why sodium is less electropositive than potassium ?
2
506
16.
+2 CHEMISTRY (VOL. - I)
In sodium chloride — Sodium ion is smaller than the sodium atom Sodium ion is larger than sodium atom Sodium ion is larger than the chloride ion Sodium ion and chloride ion have the same size.
17.
Using s,p,d,f notations, write the electronic configuration of K.
18.
Which one of the following is an ore of potassium ? (Cryolite, Carnallite, Bauxite, Dolomite)
19.
———— is manufactured by Solvay process
20.
Why alkali metals are used as reducing agents ?
21.
What product is obtained at the anode during electrolysis of fused sodium hydride ?
22.
Give one use of potassium chlorate.
23.
Which is the most electropositive element among the Group IA elements ?
24.
Aqueous solution of sodium bicarbonate is——— (a)
acidic (b)
strongly acidic (c)
alkaline
(d) neutral.
25.
Name the elements in the group I-A of the periodic table.
26.
What is the composition of carnallite?
27.
Between sodium and potassium which is more electropositive?
28.
Why alkali metals have low densities?
29.
What is the chemical formula of washing soda?
30.
Write the chemical formula of Glauber’s salt.
31.
Write the electronic configuration of sodium atom.
Short answer type. (2 mark each) 1.
Why is the size of sodium ion less than that of potassium ion?
2.
Give the principle of extraction of sodium by electrolytic method.
3.
Give the principle of preparation of sodium hydroxide by Solvay’s process.
4.
How is Glauber’s salt prepared?
5.
Why is solid NaCl a bad conductor of electricity?
6.
What are the products of electrolysis when an aqueous solution of sodium chloride is electrolysed?
THE ALKALI METALS
7.
507
What happens when NaOH is added to Al2(SO4)3 solution drop by drop and finally in excess. Give balanced equations only. Hints: Al2(SO4)3 + 6NaOH Al(OH)3 + NaOH
2Al(OH)3 + 3Na SO4 2
NaAlO2 + 2H2O
8.
In the extraction of sodium by electrolytic method, fused sodium chloride is used instead of aqueous solution of sodium chloride. What is the reason?
9.
What happens when sodium bromide is heated with hot conc. H 2SO4?
10. 11. 12.
Give two uses of potassium chlorate. What happens to the pH of the sodium chloride solution on electrolysis? (a) Give the formula of borax (b) Give one use of potassium chlorate (c) Give electronic configuration of potassium. What happens when sodium metal is heated with ammonia gas? 2NaNH2 + H2 Hint : - 2Na + 2NH3 What happens when the lime water is added to sodium carbonate solution ? Hint : - Ca(OH)2 + Na2CO3 2NaOH + CaCO3 The alkali metals are good reducing agents. How do you explain it? Account for the fact that sodium metal cannot be obtained by the electrolysis of aqueous sodium chloride solution. How will you obtain the following compounds from sodium chloride ? (a) Sodium metal (b) NaOH Give the uses of ach of the following. How do the following properties of alkali metals vary with rise in atomic number ? (a) Ionisation energy (b) Electropositive character.
13. 14. 15. 16. 17. 18. 19.
Short answer type. (3 mark each) 1. How does metallic sodium react with (i) Ammonia and (ii) Methyl alcohol 2. In the extraction of sodium by electrolytic method, fused sodium chloride is used instead of aqueous solution of sodium chloride. What is the reason? 3. Give the principle of preparation of sodium hydroxide by Solvay’s process. 4.
What happens when sodium is heated with ammonia ?
5.
Give the principle of manufacture of sodium hydroxide.
6.
Give the principle of extraction of sodium by electrolytic method.
7.
What happens when sodium metal is exposed the moist air?
8.
What happens when sodium reacts with water?
9.
Name 4 elements of Group 1.
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10.
How Glauber salt is prepared?
11.
Write one method of preparation and uses of potassium nitrate.
12. 13. 14.
Write with equation what happens when sodium is heated with carbon dioxide gas. Give the principle of manufacture of sodium carbonate. “On exposure to air solid NaOH becomes a liquid and after sometime it changes to a white powder”.. Explain why it happens. Give two uses of potassium cyanide. Give the principle of manufacture of sodium hydroxide. What happens when sodium is heated with ammonia? Give equation. Give the principle of manufacture of sodium hydroxide & give one of its uses. Write the chemical formula for the Chile salt petre & brown ring produced during test for Nitrate. What happens when sodium reacts with water. Give equation.
15. 16. 17. 18. 19. 20.
Long answer type (7 marks each) 1. Name the element of Group 1 in the periodic table. Write one ore of each of Na, K, Li and also write their electronic configuration. 2. Write the principles by which sodium is extracted by electrolytic method. Mention any three properties and two uses of sodium metal. 3. Give an account of properties and uses of alkali metal. 4. What are alkali metals? Suggest a method for extraction of sodium from one of its compounds? 5. Give the principle of manufacturing sodium carbonate and sodium chlorate. Give the uses of potassium chlorate. 6. Explain why the alkali metals have low densities, are electrically conducting, good reducing agents, form monovalent ions, soluble salts and soluble hydroxides, which become stronger alkalis on descending the group. 7. Why are potassium and caesium, rather than lithium, used in photoelectric cells? 8. List some important ores of sodium ? 9. Write the preparation and uses of potassium chlorate. 10. Describe two methods for the preparation of potassium cyanide. What are its uses ? 11. Explain the following : (i) Sodium metal cannot be obtained by the electrolysis of aqueous sodium chloride solution. (ii) The ionization energy decreases from lithium to caesium. (iii) Alkali metals are difficult to reduce. (iv) The elements of Group IA are called alkali metals.
THE ALKALI METALS
12.
13.
14.
15.
16. 17. 18.
19.
20.
21.
509
How is sodium isolated by Down’s process ? Sketch a diagram for the Down’s cell and lable the electrodes, electrolyte and the direction of flow of electrons and ions. Write the equations for the reactions involved in the process. (a) Discuss in detail the manufacture of sodium carbonate by Solvay process. What is the other important product formed in this process ? State the principle involved in the process. (b) Why can potassium carbonate not be prepared by Solvay’s process ? (a) Explain any two of the following : (i) Which is the most reactive metal of alkali metals and why ? (ii) Sodium cannot be extracted by the electrolysis of aqueous solution of sodium chloride. (iii) Alkali metals have low densities. (iv) Alkali metals are strong reducing agents. (b) Describe the Solvay process for the manufacture of washing soda. Give its two important uses. (i) Write equations for the reactions involved in making sodium bicarbonate from sodium chloride. (ii) For any two of the following state one large scale use of : (a) Sodium metal (b) NaOH. Describe the manufacture of sodium carbonate by the solvay process. Discuss Solvay process. (a) Sodium metal cannot be obtained by the electrolysis of an aqueous solution of sodium chloride. Why ? (b) How is sodium hydroxide manufactured from Castner Kellner’s process ? Write the equations for the reactions which occur at each electrode ? Comment on the following : (i) Caesium is the most reactive among the alkali metals. (ii) Alkali metals impart characteristic colours to the flame. (a) Giving chemical equaltion state what happens when ammonia reacts on a mixture of K CO and carbon at a high temperature. 2 3 (b) How is sodium isolated by Down’s process ? Write equations for the reactions involved. Why can sodium metal not be isolated by electrolysis of an aqueous solution of sodium chloride ? Explain the following : (a) Alkali metals are obtained by the electrolysis of their molten salts but not by the electrolysis of their aqueous solution.
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(b)
22. 23. 24.
The softness of Group IA metals decreases down the Group with increasing atomic number. (c) Aqueous solution of sodium chloride is a good conductor of electricity while solid sodium chloride is a non-conductor. Give the general trends in the properties of alkali metals. Write the principle of manufacture of potassium chlorate. Give the comparative account of properties of alkali metals. Write any two uses of potassium chlorate. Name the elements of group 1 of the periodic table. Give the general trends in the properties of these elements. How is Glauber’s salt prepared? qqq
CHAPTER - 13
ALKALINE EARTH METALS (GROUP - 2) 13.1
INTRODUCTION : The elements Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba) and Radium (Ra) constitute Group 2 of the periodic table. The oxides of these elements were known much earlier than the metals themselves and they were named alkaline earths since (a) they were alkaline in nature like alkali metal oxides and (b) they were found in earth’s surface.
2 He
1 H 3 Li 11 Na 19 K 37 Rb 55 Cs 87 Fr
4 Be 12 Mg 20 Ca
23 V 41 Nb 73 Ta
24 Cr 42 Mo 74 W
25 Mn 43 Tc 75 Re
26 Fe 44 Ru 76 Os
27 Co 45 Rh 77 Ir
28 Ni 46 Pd 78 Pt
29 Cu 47 Ag 79 Au
30 Zn 48 Cd 80 Hg
5 B 13 A1 31 Ga 49 In 81 Tl
6 C 14 Si 32 Ge 50 Sn 82 Pb
7 N 15 P 33 As 51 Sb 83 Bi
8 O 16 S 34 Se 52 Te 84 Po
38 Sr 56 Ba 88 Ra
Fig 13.1 - Position of the alkaline earth metals in the periodic table.
9 F 17 Cl 35 Br 53 I 85 At
10 Ne 18 Ar 36 Kr 54 Xe 86 Rn
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ELECTRONIC CONFIGURATIONS : The atoms of all the alkaline earth metals have two electrons in their valence shell preceded by the noble gas configuration. Their general configuration may be written as [Noble gas] ns2 where n represents the valence shell (Table-13.1).
Table - 13.1 Element
Symbol
At.No.
Electronic configuration
Configuration of outermost shell
4
1s2, 2s2 or [He] 2s2
2s2
Magnesium Mg
12
1s2, 2s2, 2p6, 3s2 or [Ne] 3s2
3s2
Calcium
Ca
20
1s2, 2s2, 2p6, 3s2, 3p6, 4s2 or [Ar] 4s2
4s2
Strontium
Sr
38
1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6 5s2
Beryllium
Be
or [Kr] 5s2 Barium
Ba
5s2
1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10,
56
5s2, 5p6, 6s2, or [Xe] 6s2 Radium
Ra
6s2
1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10,
88
4f14, 5s2, 5p6, 6s2, 6p6, 7s2 or [Rn] 7s2 13.3
7s2
GENERAL CHARACTERISTICS OF ALKALINE EARTH METALS (GROUP-2) : Physical characteristics: The important physical constants of alkaline earth metals are given below. The other physical properties are discussed thereafter. Table - 13.2 Physical constants of Alkaline Earth Metals
Property
Be
Mg
Ca
Sr
Ba
Ra
Atomic radius (A0) Ionic radius (M++) (A0) Ionisation energy E1 (kJ mole-1)E2
1.05 1.31 900 1757
1.36 0.65 737 1450
1.74 0.99 590 146
1.91 1.13 549 1060
1.98 1.35 502 965
1.50 510 975
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ALKALINE EARTH METALS
Density (g/cm3)
1.86
1.75
1.55
2.60
3.59
5.00
Melting point (K)
1551
924
1116
1026
998
973
6.00
2.1´104
3.6´104 3.0´102 2.5´102
Abudance in Earth’s crust (ppm)
1.
1.3´10-6
Atomic and ionic radii : The atomic and ionic radii of the members of the family are smaller than the corresponding members of the alkali metals. Explanation : The alkaline earth metals have a higher nuclear charge and therefore, the electrons are attracted more towards the nucleus. As a result, their atomic and ionic radii are smaller than those of alkali metals. On moving down the group, the radii increase due to gradual increase in the number of the shells and the screening effect.
2.
Melting and boiling points: Alkaline earth metals have low melting and boiling points which are higher than the corrresponding alkali metals in the same period due to comparatively smaller size. However, melting and boiling points do not show regular trends. Explanation : The atoms of alkaline earth metals have smaller size as compared to alkali metals. Therefore, they are more closely packed in their crystal lattices and hence they have higher melting and boiling points.
3.
Ionisation energy : The alkaline earth metals have low ionisation energies due to fairly large size of the atoms. Since the atomic size increases down the group, the ionisation energy decreases (Table 13.2). A comparison of the ionisation energies of the members of Groups 1 and 2 shows that the members present in the second group have higher values as compared to those of Group 1, because they have smaller size and electrons are more attracted towards the nucleus of the atoms. It may be noted that although I E 1 values of alkaline earth metals are higher than those of alkali metals, the IE 2 values of alkaline earth metals are much smaller than
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those of alkali metals. The ionisation energy values of sodium (alkali metal) and magnesium (alkaline earth metal) are given below.
Element
IE1
IE2
Na
496 kJ mol-1
4562 kJ mol-1
Mg
737 kJ mol-1
1450 kJ mol-1
Explanation: In case of alkali metals (i.e. Na) the second electron is to be removed from a cation which has already acquired a noble gas configuration. On the other hand, in the alkaline earth metals (i.e. Mg), the second electron is to be removed from a monovalent cation, Mg+ : (1s2 2s2 3s1) which still has one electron in the outermost shell. Thus, the second electron in Mg can be removed easily. Na(g) (1s2, 2s2, 2p6, 3s2)
IE1 -e
Na+(g)
—
(1s2, 2s2, 2p6)
IE2 -e
—
Na2+(g)
Stable configuarion Mg(g) (1s2, 2s2, 2p6, 3s2)
IE1 -e
Mg+(g)
—
(1s2, 2s2, 2p6, 3s1) Not very stable
IE2 -e
—
Mg2+(g) (1s2, 2s2, 2p6) Stable
The IE3 of Mg will be very high because the electron is to be removed from stable noble gas configuration. 4.
Electropositive and metallic character : Because of the low ionisation energies of alkaline earth metals, they are strongly electropositive in nature. However, these are not as strongly electropositive as the alkali metals of Group 1 because of comparatively higher ionisation energies. The electropositive character increases down the group i.e. from Be to Ba.
5.
Characteristic flame colouration : Except Be and Mg, the alkaline earth metal salts impart characteristic colours to the flame.
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ALKALINE EARTH METALS
Metal
Be
Mg
Ca
Sr
Ba
Ra
Colouration
—
—
Brick red
Crimson red
Grassy green
Crimson red
Beryllium and magnesium do not give any characteristic flame colour action. Explanation : The alkaline earth metals give characteristic flame colouration because their ionisation energies are low. Therefore the valence electrons in these atoms can be easily excited to higher energy states by the energy of flame of bunsen burner. When these excited electrons (excited state is unstable state) come back to ground state they emit radiations which fall in the visible region. Therefore, they give colours to the flame. On the otherhand, beryllium and magnesium atoms are comparatives smaller and their ionisation energies are very high. Hence the energy of the flame is not sufficient to excite their electrons to higher energy levels. These elements, therefore, do not give any colour to bunsen flame. 6.
Tendency to form bivalent ions: The alkaline earth metals exhibit a valency of 2 as they can lose two electrons and form bivalent ions. Inspection of Table 13.2 reveals that the second ionisation energy (IE2) of alkaline earth metals is greater than the first ionisation energy (IE 1). It, thus appears that if ionisation energy is the only factor involved in the formation of divalent ions, we would expect that alkaline earth metals would prefer to form + 1 ions (M +) rather than +2 ions (M 2+). Actually, they predominantly show +2 oxidation state e.g. Mg2+, Ca2+, Ba2+ etc. Explanation: This can be explained as : (i)
Divalent ions have the stable noble gas configuration.
(ii)
In solution, the +2 ions of alkaline earth metals are extensively hydrated and the high hydration energies of M 2+ ions make them more stable than M + ions. It is observed that the amount of enegy released when M 2+ ion is dissolved in water is much more than that for M + ion. This large amount of extra energy released in the hydration of +2 ions is more to compensate the second ionisation energy required for the formation of such ions.
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(iii)
In the solid state, the divalent cation form stronger lattices than monovalent cations and therefore, a lot of energy called lattice energy is released. It is the greater lattice energy of M 2+ ion which compensates the high second ionisation energy and is responsible for its greater stability as compared to M+ ion.
Chemical properties : Because of their low ionisation energies and high electropositive character the alkaline earth metals have strong tendency to lose valence electrons. Therefore, they are very reactive. The reactivity of these elements increases on going down the group. However, alkaline earth metals are, in general less reactive than alkali metals. The chemistry is mainly dominated by the dipositive oxidation state (M 2+). Some of the general trends are discussed below. 1.
Action with air or oxygen - Formation of oxides : The alkaline earth metals being less electropositive than alkali metals react with air or oxygen slowly upon heating to form oxides, MO. However, Ba and Ra form peroxides.
2 M + O2
Heat
2MO (M = Be, Mg or Ca) Monoxide
M + O2
Heat
MO2 (M = Ba or Ra) Peroxide
The reactivity with oxygen increases as we move down the group due to increasing electropositive character of the elements. The monoxides can also be prepared by decomposition of their carbonates. MCO3 Carbonate
MO + CO2 Monoxide
(M = Be, Mg, Ca, Sr or Ba)
Among the oxides BeO is amphoteric while oxides of other elements are basic in nature. BeO Amphoteric
MgO Weakly basic
(CaO, SrO, BaO) Basic
The amphoteric character of BeO is supported by the fact that it reacts with acids as well as alkalies. BeO + HCl
BeCl2 + H2O (Basic nature)
BeO + NaOH
Na2BeO2 + H2O (Acidic nature) Sod. beryllate
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ALKALINE EARTH METALS
The oxides of Be and Mg (BeO, MgO) are almost insoluble in water while the oxides of rest of the metals dissolve in water to form hydroxides. CaO + H2O
Ca(OH)2 + Heat
The insolubility of BeO and MgO in water is due to their large lattice energies. 2.
Combination with hydrogen — Formation of hydrides : All the elements except beryllium combine with hydrogen upon heating to form ionic hydrides, MH2.
Ca + H2
Heat
CaH2 Calcium hydride
BeH2, however, can be prepared by the reaction of BeCl 2 with LiAlH4. 2 BeCl2 + LiAlH4
2 BeH2 + LiCl + AlCl3
MgH2 is covalent while all other hydrides are ionic. Calcium hydride is known as hydrolith. Thus, the hydrides are highly reactive with water and form hydroxides and liberate hydrogen. CaH2 + 2H2O 3.
Ca(OH)2 + 2 H2
Action with water — Formation of hydroxides : The alkaline earth metals have lesser tendency to react with water as compared to alkali metals. They combine with water slowly to form hydroxides. Ca + 2H2O
Ca(OH)2 +H2
Like alkali metals, the hydroxides are basis in nature and basic strength increases down the family. The basic character of hydroxides is due to the low ionisation energies of these metals. Because of low ionisation energies the M-O bond in MOH is weak and it can cleave to give OH— ions in solution. Since the ionisation energy decreases down the family, the basic strength of the hydroxides increases. For example, Be(OH) 2 is amphoteric, Mg(OH)2 is mildly basic while others are strong bases.
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Hydroxides of Alkali & Alkaline Earth Metals LiOH
Be(OH)2 Amphoteric
NaOH
Mg(OH)2 Mild base
KOH
Ca(OH)2 Strong base
RbOH
Sr(OH)2 Strong base
CsOH
BASIC CHARACTER INCREASES
518
Ba(OH)2 Very strong base
BASIC CHARACTER DECREASES
All alkali metal hydroxides are strong bases. The hydroxides of alkaline earth metals are less basic than those of alkali metals of the corresponding periods. The lesser basic strength of hydroxides of alkaline earth metals is due to their (i)
High ionisation energies
(ii)
Small ionic sizes and
(iii)
Dipositive charge on the ions.
As a result, the M-O bond in these hydroxides is relatively stronger than that of corresponding alkali metals and therefore, does not break. Therefore, they are less basic than corresponding alkali metals. The hydroxides of alkaline earth metals are less soluble than alkali metal hydroxides. However, the solubility of hydroxides in water increases with increase in size of the metal. Be(OH)2 and Mg (OH)2 are almost insoluble. Ca(OH) 2 is sparingly soluble while Sr(OH)2 is fairly soluble. The trend is due to increase is size of the cation on moving down the group so that their latttice energies decrease. Due to decrease in lattice energies, the solublility of hydroxide in water increases. 4.
Formation of Carbonates : The carbonates of alkaline earth metals can be prepared by passing carbon dioxide in limited supply through the solution of their hydroxides. Ca(OH)2 + CO2
CaCO3 + H2O
519
ALKALINE EARTH METALS
All carbonates decompose on heating to give carbon dioxide and metal oxide. Heat
CaCO3
CaO + CO2
However, the stability of these carbonates increases down the group. For exmaple, BeCO3 is least stable while BaCO 3 is most stable. The solubilities of the metal carbonates in water decrease down the family. For example, BeCO 3 is slightly soluble in water while BaCO3 is completely insoluble. 5.
Formation of Sulphates : These can be formed by the action of dil. H 2SO4 on metals, metal oxides, metal hydroxides and metal carbonates. Ca + H2SO4 CaO + H2SO4
CaSO4 + H2 CaSO4 + H2O
Ca(OH)2 + H2SO4
CaSO4 + 2H2O
CaCO3 + H2SO4
CaSO4 + CO2 + H2O
The sulphates of alkaline earth metals are less soluble than the corresponding salts of alkali metals. Their solubilities decrease on going down the group. This is due to higher lattice energies of alkaline earth metal sulphates than alkali metal sulphates (as already explained). On moving down the group, the hydration energy decreases with increase in size of the metal ion and consequently their solubilities decrease. 6.
Formation of Halides : All the alkaline earth metals combine with halogens at higher temperatures forming their halides. M + X2
MX2 (X = Cl, Br)
The metal halides can also be obtained by the action of halogen acids on metals, their oxides, carbonates and hydroxides. MO + 2HX M(OH)2 + 2HX
MX2 + H2O MX2 + 2H2O
i)
Beryllium halides are covalent while all other halides are ionic and are readily soluble in water.
ii)
BeCl2 is relatively low melting solid and volatile while others have high melting points. BaCl 2 has very high melting point.
iii)
Being covalent, BeCl2 is soluble in organic solvents.
iv)
The anhydrous halides of alkaline earth metals are hygroscopic i.e. they absorb
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water molecules and form hydrates such as BeCl 2. 4H2O, CaCl2. 6H2O. BeCl2 gets hydrolysed by moisture and therefore gives fumes of hydrochloric acid. Be(OH)2 +2HCl
BeCl2 + 2H2O v)
7.
Except BeCl 2 and MgCl 2, the chlorides of the other members impart characteristic flame colours. CaCl2
SrCl2
BaCl2
Brick red
Crimson
Grassy green (Pea green)
Tendency to form complexes: The Group 2 elements have tendency to form some stable complexes. Among these, beryllium and magnesium have maximum tendency to form complexes. This is due to their small size and higher charge density. For example, Beryllium forms complexes of the type (BeF4]2
—
[BeF4]2
BeF2 + 2F
—
Chlorophyll is an important complex of magnesium. Difference between Beryllium and Magnesium Beryllium differs from other alkaline earth metals due to the following reasons. (i)
Beryllium has extremely small size.
(ii)
It has comparatively high electronegativity and ionisation potential than magnesium and other members of the family.
(iii)
It has no vacant d-orbital in the valence shell. A comparative account of the difference between beryllium and magnesium is given below.
1. 2. 3.
Property Hardness M.P & B.P Compounds
Beryllium Hard High Covalent
Magnesium Soft Low Mainly ionic
4.
Action of water
No action
Forms MgO & H2with boiling
Amphoteric Forms methane
water. Basic Forms acetylene
5. 6.
Oxides & Hydroxides Nature of carbides
(Be2C+4H2O ® 2Be(OH) 2+CH4)
(MgC 2+2H2O ® Mg(OH) 2+C2H2)
7.
Action of acid
No action
Forms H2
8.
Complex formation
Forms complexes
Does not form complexes.
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ALKALINE EARTH METALS
13.4
SOME IMPORTANT COMPOUNDS OF CALCIUM :
(i)
Calcium oxide, CaO (Quick lime) Lime is a very important commodity of chemical industry. It is also known as quick lime. Manufacture : Quick lime is obtained when lime stone is heated in a lime kiln. The modern lime kiln is a long cylindrical tower (Fig.13.2) fitted system. The lumps of lime stone and coal are fed from the top. Producer gas is introduced from the inlet near the base. The temperature near the middle of the tower is about 8000C. (a)
Lime stone decomposes to give quick lime. CaCO3
CaO + CO2 — 42 cals.
CO2 LIME STONE
Carbon dioxide gas so formed is at once taken out otherwise back reaction may also start. The gas is taken out of the kiln by an upward drought. The product quick lime is drawn out from the bottom through the discharge door. Properties :
FIRE BOX
1.
Pure calcium oxide is a white amorphous powder.
2.
It melts at 2600 C and is extremely stable.
3.
AIR
0
LIME CARRIER
QUICK LIME
Fig. 13.2 A lime kiln
It reacts vigorously with water forming calcium hydroxide and releasing tremendous amount of heat. CaO + H2 O
Ca(OH)2 + 15,000 cals.
The process is called slaking and the product is also known as slaked lime. 4.
It reacts with acids forming corresponding salts. CaO + 2HCl
CaCl2 + H2O
CaO + H2SO4
CaSO4 +H2 O
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5.
Being a basis oxide it reacts with acidic oxides to form salts. CaO + SO2
CaSO3 (Calcium sulphite)
CaO + SiO2
CaSiO3 (Calcium silicate)
3CaO + P2O5
Ca3(PO4)2 (Calcium phosphate)
Uses : It is used
(ii)
1.
an the laboratory as a drying agent.
2.
for the preparation of slaked lime.
3.
for the preparation of bleaching powder.
Calcium carbonate, CaCO3 Calcium carbonate occurs in large quantities in nature as chalk, marble and lime stone. Preparation : It is obtained in the laboratory by the action of a soluble carbonate on a calcium salt or by passing CO2 through lime water. CaCl2 + Na2CO3 Ca(OH)2 + CO2
CaCO3 + 2NaCl CaCO3 +H2O
Properties : 1.
Calcium carbonate is a white fluffy powder.
2.
It is almost insoluble in water.
3.
It is soluble in water containing carbon dioxide forming calcium bicarbonate CaCO3 + H2O + CO2
Ca(HCO3)2
Uses : It is used
13.5
1.
For the manufacture of lime.
2.
As a flux in the smelting of ores.
3.
In the preparation of tooth pastes and face powder.
BIOLOGICAL IMPORTANCE OF MAGNESIUM AND CALCIUM :
Mg+2 and Ca+2 ions are found to be essential for human body, which contains about 25g of Mg and 1200g of Ca.
523
ALKALINE EARTH METALS
Functions : (i)
Chlorophyll, the green pigments of the plant, contains magnesium, which makes photosynthesis possible.
(ii)
All enzymes, undergoing phosphate transfer by using ATP, need magnesium as the cofactor.
(iii)
While Mg+2 ions are found to be concentrated in animal cells, Ca +2 ions are concentrated in the body fluid outside the cell.
(iv)
While most of the body calcium is pesent in bones and teeth, Ca +2 ions are required to maintain the regular beating the heart.
(v)
Calcium plays important role in clotting of blood and stabilisation of protein structure.
(vi)
Calcium finds its use in the functioning of neuro-muscular and interneuronal systems.
CHAPTER (13) AT A GLANCE 1.
Quick line
-
Its chemicals composition is CaO
2.
Slaked line
-
Its chemical composition is Ca(OH)2
3.
The green pigment of the plant, chlorophyll contains magnesium.
4.
Calcium and magnesium are essential for human body.
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QUESTIONS A.
Very short answer type : (1 mark each)
1.
Name the following compounds. CaO, Ca(HCO3)2
2.
Why magnesium is less electropositive than calcium ?
3.
What oxidation state is shown by alkaline earth elements is their compounds ?
4.
Which element has the least atomic radius among the alkaline earth elements ?
5.
Write the electronic configuration of Group 2 element.
6.
Write the electronic configuration of calcium.
7.
Why is ionisation energy of Mg greater than Al?
8.
Which is bigger in size between Na+ and Mg2+?
B.
Short answer type : (2 marks each)
1.
An element has atomic number 12. Find out its electronic structure. Name the element and quantum number of its valence shell electrons.
2.
Magnesium metal burns in air to give a white ash. When this ash is treated with water the odour of ammonia can be detected. Suggest the explanation for this observation. Δ Mg N ; Mg N +6H O 3Mg(OH) + 2NH ] Hint : [3Mg + N ¾¾® 2
3 2
3 2
2
2
3
3.
State with equation what happens when carbon dioxide gas is passed in excess through clear lime water.
4.
Name the alkaline earth metals and write their electronic configurations.
5.
Name some of the important properties of the alkaline earth metals which make them similar.
6.
Account for the fact that alkaline earth metals exhibit only +2 oxidation state.
7.
Discuss the gradation in properties of group 2 elements with reference to i) Atomic radii ii) Action as reducing agent iii) Ionisation energies iv) Electropositive character v) Metallic character.
8.
Mention the properties in which Beryllium differs from other members of the family.
9.
Compare the properties of Beryllium with that of aluminium.
C.
Short answer type : (3 marks each)
1.
Mention some important uses of magnesium and calcium.
2.
How will you detect magnesium and calcium in separate salts by qualitative analysis ?
3.
Why calcium can not be extracted from calcium oxide by carbon reduction process ?
4.
Give two biological importance of Ca2+ and Mg2+
5.
State with equation what happens when excess carbon dioxide gas is passed through lime water.
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ALKALINE EARTH METALS
6.
Why do beryllium halides fume in air ? (Hint : Due to formation of halogen acids by hydrolysis as BeCl2 + 2H2O
Be (OH)2 +2HCl)
D.
Long answer type questions. (7 marks each)
1.
Account for the following :
2.
(i)
Energy required to remove the first electron in alkaline earth metals is higher than that in alkali metals.
(ii)
Alkaline earth metals are less electropositive than alkali metals.
(iii)
Alkaline earth metals are denser, harder and have higher melting points than alkali metals.
Compare the chemical properties of alkaline earth metals with those of alkali metals with reference to i) Action with water ii) Action with oxygen iii) Action with dilute H 2SO4 (iv) Strength of hydroxides v) Solubility of carbonates vi) Solubility of sulphates.
3.
Give one method for the preparation and one important application of each of the following: a) Quick lime b) Lime stone c) Slaked lime
4.
Give formula of two sets of magnesium and calcium compounds and their properties which resemble each other and that of one set of compounds whose properties differ.
5.
(a)
The barium ion is poisonous. Why then can barium sulphate is used safely in making X-ray photographs of the digestive tract ?
(b)
Explain the following phenomenon by means of balanced equations. (i)
When exhaling is made through a tube passing into a solution of lime water, solution becomes turbid.
(ii)
The turbidity of the solution (i) eventually disappears when continued exhaling is made through it.
(iii) When solution obtained in ii) is heated, turbidity reappears. (iv) Why are alkaline earth metals good reducing agents ? (v)
Why is anhydrous CaSO4 used as a drying agent ? Why not Plaster of Paris ?
6.
Discuss the chemistry of alkaline earth metals. Why sodium metal is soft, though metals are hard generally ?
7.
Write the general trends in properties of Group 2 elements (alkaline earth metals), What is Epsom salt ? Mention one of its uses.
8.
Discuss the chemistry of alkaline earth metals. Why sodium metal is soft, though metals are hard generally?
9.
Discuss about two characteristic properties of alkaline earth metals. What is quick lime? Give one method of its preparation and one of its uses. qqq
526
+2 CHEMISTRY (VOL. - I)
UNIT - XI
THE P-BLOCK ELEMENTS CHAPTER - 14
(GROUP 13 ELEMENTS) INTRODUCTION : Elements in which the last electron enters the outermost p-orbital are called p-block elements. As there are three number of p-orbitals six electrons can be accomodated in a set of porbitals. Consequently there are six groups of p-block elements in the periodic table numbering from 13 to 18. Boron, Carbon, Nitrogen, Oxygen, Fluorine and Neon are the first member of these groups respectively. 14.1
GROUP 13 ELEMENTS : THE BORN FAMILY :
Group 13, collectively called Boron family or Boron group elements consists of five elements namely Boron (B), Aluminium(Al), Gallium(Ga), Indium(In) and Thallium(Tl). Boron is a nonmetal whereas other members of the group are silvery-white metals. 14.2
GENERAL CHARACTERISTICS OF P-BLOCK ELEMENTS :
(a) Electronic congfiguration : The p-Block elements have general valence shell electronic configuration ns2np1–6 (except for He). But they may differ in their inner core electronic configuration, which in turn influences their physical properties. The presence of vacant d-orbitals in valence shell of heavier elements (starting from the 3rd period onwards) also influence the chemical properties. (b) Oxidation state : The p-Block elements show a common oxidation state, also known as group oxidation state which is in general equal to number of s– and p– electrons in the valence shell. This oxidation state is the maximum value shown by elements of a particular group. In addition to this value p-Block elements may show other oxidation states which normally but not necessarily differ from the total number of valence electrons by unit of two. This trend becomes prominent as we move down the group which is commonly attributed to the ‘Inert Pair Effect’. Table 14.1 shows a list of the important oxidation states exhibited by p-Block elements.
527
BORON FAMILY
Table 14.1 General electronic configuration and oxidation states of p-Block elements Group
13
14
15
16
17
18
ns2np1
ns2np2
ns2np3
ns2np4
ns2np5
ns2np6
First members of the group
B
C
N
O
F
He
Group oxidation state
+3
+4
+5
+6
+7
+8
Other oxidation states
+1
+2, –4
+3 –3
+4 +2 –2
+5 +3 +1 –1
+6 +4 +2
General electronic configuration
(c) Metallic character : The p-Block of the periodic table accomodates non-metals, metalloids as well as the metals. As we go down a particular group the non-metallic character decreases and metallic character inceases. The trend down the group is non-metal-metalloidmetal. Thus the heaviest elements of each group is metallic in nature. The change from nonmetallic to metallic character is also illustrated by the change in their oxides from acidic behaviour to basic behaviour down a group. (d) Covalency : The non-metals present at the top of each group have higher electronegativities and higher ionisation enthalpies suitable for the formation of covalent bond. The second period elements of p-Groups starting from Boron are restricted to a maxium covalence of four (one 2s and three 2p orbitals). But the presence of d-orbitals in the valence shell of heavier elements enable them to expand their covalence above four. For example, Boron forms only [BF4]¯ whereas Aluminium gives [AlF6]3– ion. The first member of a group differs from heavier members in its ability to form pp–pp multiple bonds (e.g., C = C, C º C , N º N, C = O, C º N, N = O). The heavier elements form dp – pp or dp – dp bonds.
14.3
OCCURRENCE : Boron occurs in two isotopic forms,
10 5
B (19%) and
crust is only 0.0001% by mass. It mainly occurs as (1)
Borax (Na2B4O7.10H2O)
(2)
Kernite (Na2B4O7.4H2O)
(3)
Colemanite (Ca2B6O11.5H2O)
11 5
B (81%). Its abundance in Earth’ss
528
+2 CHEMISTRY (VOL. - I)
Aluminium is the third most abundant element (8.3%) found in the earth’s crust after oxygen (45.5%) and silicon (27.7%). The important ores of aluminium are : 1.
Oxide : Bauxite (Al 2O 32H 2O, Hydrated aluminium oxide), Diaspore (Al2O3H2O) Corundum (Al2O3), Ruby (Al2O3), Sapphire (Al2O3).
2.
Fluoride : Cryolite (Na3AlF6 or AlF3, 3NaF)
3.
Silicates: Porcelain and China clay or Kaoline (Al 2O3, 2SiO2. 2H2O), Felspar KAlSi3O8
4.
Aluminates : Chrysoberyl (BeO.Al2O3), Spinel (MgO, Al2O3)
5.
Sulphate: Alunite K2SO4, Al2(SO4)3. 4Al(OH)3
Gallium, indium and thallium are less abudant than aluminium. Gallium (0.1 – 1%) is found in the rare mineral Germanite. Traces of indium are found in the sulphide ores of zinc while that of thallium are found in the sulphide ore of lead.
14.4
GENERAL CHARACTERISTICS OF THE ELEMENTS
The atomic numbers and some of the important physical properties of the elements are given below. Table 14.2 Physical properties of the elements of Group 13 Element
B
AI
Ga
In
TI
Atomic number
5
13
31
49
81
Atomic Mass
10.81
26.98
69.72
114.8
204.4
Electronic configuration [He]2s22p1 [Ne]3s23p1 [Ar]3d104s24p1 [Kr]4d105s25p1 [Ar]4f145d106s26p1 I.E.(E1), kJ mol—1
780
578
578
557
589
I.E.(E2), kJ mol—1
2428
1817
1980
1821
1972
I.E.(E3), kJ mol—1
3659
2747
2964
2704
2876
Electronegativity
2.0
1.5
1.6
1.7
1.8
Atomic radius, (A0)
0.80
1.25
1.24
1.50
1.55
Ionic radius(M3+), (A0) —
0.45
0.60
0.81
0.95
Density(g/cm3)
2.30
2.70
5.90
7.31
11.85
Melting point(K)
2300
932
312
429
577
Boiling point(K)
4000
2700
2500
2300
1740
BORON FAMILY
1.
529
Electronic Configuration The electronic configuration of these elements (Table 14.2) shows the presence of three electrons (ns2 np1) in their outer-most shell, two in ns and one in np orbitals. On the basis of the electronic configuration of the penultimate shell (second last), these elements are categorised into two groups. B and Al are of one type and Ga, In and Tl, of other type. The first type, B and Al have inert gas type configuration (s2 or s2 p6) in the penultimate shell. The second type Ga, In and Tl have s2p1d10 type configuration in the penultimate shell. This difference in electronic configuration is responsible for the difference in properties of the elements B and Al with the rest of the elements (Ga, In, Tl).
2.
Oxidation states All these elements form compounds in their +3 oxidation state due to the presence of 3 electrons in their valence shells. Oxidation states shown by the elements of this group are B=+3, Al=+3, Ga=+3, +1, In=+3, +1, and Tl=+3, +1. B and Al show +3 oxidation states. Heavier elements of this group Ga, In and Tl show +1 oxidation state as well as +3 oxidation state. +1 Oxidation state becomes more and more stable as we move down the group (Ga+ < In+ < Tl+). This is due to inert pair effect. Therefore Tl, the last element of the group prefers to exist in lower oxidation state that is +1 than to exist in higher oxidation state of +3, Thallium forms stable monovalent ion Tl +. Boron, due to its small size and high ionisation energy cannot lose three valence electrons to form B3+ ion and therefore forms covalent compounds. Al also has tendency to form covalent compounds. Example : AlCl3.
3.
Atomic and ionic radii As usual the atomic radii of these elemetns are expected to increase down the group due the successive addition of new orbits. But moving from Al to Ga the atomic radius, instead of increasing slightly decreases (Al = 1.25 A 0, Ga=1.24A0). The outer-shell configurations of Al and Ga are 3s2, 3p1 and 3d10, 4s2, 4p1 respectively. Since the 3d orbitals present in Ga are larger in size and poorly screen the nucleus of Ga, the effective nuclear charge in Ga becomes slightly more than that in Al. Therefore, the atomic radius of Ga becomes slightly less than that of Al. In and Tl show the expected increasing trend in atomic radii. The ionic radii of M 3+ and M+ cations (M=Al, Ga, In Tl) increase down the group as expected.
530
+2 CHEMISTRY (VOL. - I)
Atomic radii of the elements of boron family are comparatively smaller than atomic radii of corresponding members of Group 1 and Group 2. This is because, the boron family elements have higher value of nuclear charge. 4.
Ionisation energies The ionisation energies of the elements show a fluctuation with the increase in the atomic number. The I.E. of Al is less than B as expected. This is due to an appreciable increase in atomic radius and effective shielding of the nuclear charge in case of Al. The trend reverses while going from Al to Ga. I.E. of Al is less than Ga. In gallium the d-subshell is being filled. These intervening d-electrons do not screen the nuclear charge effectively and the effective nuclear charge becomes greater than screening effect in Ga than Al. Similarly, the fluctuating tendencies of I.E. can be explained in case of Ga to Tl on the basis of effective nuclear charge and screening effect. The first ionisation energies of these elements are less in comparison to the corresponding values of Group 2 elements. In case of these elements an electron is removed from porbital and less energy is required to remove a p-electron than removing from s-orbital. The second and third ionisation energies of these elements are considerably higher since the successive electrons are removed from the s-orbital.
5.
Electropositive or Metallic Character Boron is less metallic than aluminium since the ionisation energy of B is more than that of Al. B is regarded as a semi-metal or metalloid. It has more properties of a nonmetal. It has high ionisation energy. It is crystalline, hard and also a poor conductor of electricity. Al and the other remaining elements are metallic in nature. Their metallic character is more or less the same, since their ionisation energy values are slightly different from each other. The elements of this group have less electropositive or metallic character than the elements of Groups 1 and 2. This is due to the smaller size and higher ionisation energies of these elements as compared to the elements of Groups 1 and 2.
6.
Electronegativity On moving down the group from B to Al, the electronegativity values first decreases and then increases marginally (Table 14.2). This is because of the discrepancies in the atomic size of the elements.
531
BORON FAMILY
7.
Density Densities of the elements of this group increase regularly on moving down the group from B to Tl. Boron and aluminium have relatively very low densities whereas other members of the group have high densities.
8.
Melting point and Boiling point. The melting points decrease from B to Ga and then increase from Ga to Tl. This irregular trend in melting points may be attributed to structural changes. Boron atoms are firmly bound to one another and thus it has a very high melting point. The low value of melting point of Ga is due to the fact that its structure consists of Ga 2 molecules. The boiling points decrease regularly down the group from B to Al. Gallium has unusual low melting point (303K). It exists in liquid state in summer. Its high B.P. (2676K) makes it a useful material for measuring high temperature.
Chemical properties : All the elements of the group form M 3+ compounds. With the increase of the size of to M3+ to Tl3+, the tendency of these ions to form covalent compounds decreases. The nature of compounds of M 3+ ions is decided by Fajan’s rule (the smaller the cation, the greater is its tendency to form covalent compounds). Therefore the compounds of B3+ are predominantly covalent while Al 3+, Ga3+ and Tl3+ give ionic compounds. Boron does not form B + ion. Ga+, In+ and Tl+ ions exit. Tl+ ion is quite stable. General trend in chemical properties 1.
Hydrides : The elements of group IIIA do not react directly with hydrogen. Their hydrides are prepared by indirect way. Boron forms a large number of stable covalent hydrides which are called boranes. Boranes are of two types. a)
Boranes of BnHn+4 type
:
B2H6, B5H9 B6H10 etc.
b)
Boranes of BnHn+6 type
:
B4H10, B5H11, B6H12 etc.
B2H6 is the most important borane. It is used for preparing high energy fuels and for welding torches. It is used to prepare organic boron compounds. It is also used as a reducing agent and as a catalyst in polymerisation reactions. Boranes are electron deficient compounds. Al, Ga and In give high molecular mass polymeric hydrides, such as (AlH3)2, (GaH3)2, (InH3)2 respectively. Tl does not form a hydride. B, Al and Ga form complex hydrides which contain [AH 4]— ion (A=B, Al, Ga). Example : NaBH4, sodium borohydride, LiAlH 4, lithium aluminium hydride etc.
532
+2 CHEMISTRY (VOL. - I)
The hydrides of the elements of this group are electron-deficient and hence act as lewis acids. 2.
Oxides and Hydroxides All the elements of this group form sesquoxides of general formula M 2O3. These elements form trihydroxides of M(OH) 3 type. On moving down the group, there is gradual change from acidic to amphoteric and then to basic character of the oxides and hydroxides. B2O3 (acidic), Al 2O3 (amphoteric) Ga 2O3 (amphoteric), In 2O3 (basic) and Tl 2O3 (strongly basic). D 2M O (s) 4M(s) + 3O2(g) ¾¾® 2 3
3.
Halides All the elements of this group except thallium react with halogens to form trihalides MX3.BF3, BCl3 and BBr3 are covalent compounds. These compounds are called Lewis acids since they accept lone pair of electrons from groups containing donor atoms like S, O, N etc. F | F—B | F
+
Boron trifluoride Planar
F | F—B | F
H | :N—H | H Ammonia
H | N—H | H
Boron trifluoride - ammonia adduct tetrahedral
Aluminium halides except AlF 3 are covalent in nature. However, aluminium chloride hydrolyses in water to give Al 3+ (aq) ion. In the trivalent state, most of the compounds being covalent are hydrolysed by water to form either tetrahedral species, [M(OH) 4)¯ in which the central elements is sp3– hybridized or octahedral species, [M(H 2O)6]3+ in which the central element is sp3d2– hybridized. For example, BCl3 on hydrolysis forms tetrahedral [B(OH) 4]¯ species, as Boron can not expand its covalency beyond four (due to absence of d-orbitals) whereas, Al2Cl6 on hydrolysis gives octahedral [Al(H 2O)6]3+ species, where the Al 3+ is sp3d2 hybridized.
533
BORON FAMILY
Cl
Cl
Cl Al
Al Cl
Cl
Cl
Tetrahedral Boron trihalides exist as monomers whereas trihalides of Al, Ga and In exist as dimers, e.g. Al2 Cl6 in the vapour state. Boron being small in size cannot coordinate with four large halide ions whereas aluminium atom due to its size is capable of coordinating with four halide ions. Al, Ga and In form unstable monohalides. Thallium monohalides, Tl +X-- are stable ionic compounds. The bonding characteristics of halides change from covalent character to ionic character as we go down the group from boron to thallium. 4.
Nitrides and Carbides All the elements of this group combine with nonmetals like nitrogen and carbon forming nitrides and carbides respectively. 2B + N2
2 BN
4Al + 3C
Boron nitride 5.
Al4C3 Aluminium carbide
Reaction with water All the members of Group III A except boron decompose water when heated at high temperature. 2Al + 3H2O
6.
Al2O3 + 3H2
Reaction with acids With HCl : Boron does not react with hydrochloric acid. Aluminium readily dissolves in dilute as well as concentrated hydrochloric acid liberating hydrogen. 2Al + 6HCl
2AlCl3 + 3H2
With H2SO4: Boron reacts with hot concentrated sulphuric acid and is oxidised to boric acid. 2B + 3H2 SO4
2H3BO3 +3SO2 Boric acid
Aluminium does not react with dilute sulphuric acid at ordinary temperature. Hot dilute H2SO4 reacts slowly liberating hydrogen. Sulphur dioxide is evolved when conc. H2SO4 acts on aluminium metal.
534
+2 CHEMISTRY (VOL. - I)
2Al + 3H2SO4 (dil)
Al2(SO4)3 +3H2
2Al + 6H2SO4(conc.)
Al2(SO4)3 +3SO2 +6H2O
With HNO3: Hot concentrated nitric acid oxidises boron to boric acid B + 3HNO3
heat
H3BO3+3NO2
Aluminium becomes passive when reacts with concentrated HNO 3. This is because a thin layer of its oxide gets coated on the metal which prevents further action of nitric acid. Gallium has similar behaviour towards acids as that of aluminium. Indium reacts with all acids. Thallium does not react with HCl. However, the metal reacts with sulphuric acid and nitric acid. 7.
Reaction with alkalies : Boron does not react with alkalies even at moderate temparature. Aluminium being amphoteric in nature reacts with aqueous alkali liberating hydrogen gas. 2 Al (s) + 2 NaOH (aq) + 6 H2O(l) ® 2 Na+ [Al(OH)4]¯(aq) +3H2O Tetrahydroxoaluminate (III)
Behaviour of Boron Anomalous properties : Boron differs from others of its group members in many properties. because of its small atomic size and absence of orbitas. Table 14.3: Differences between boron and other members of the Group 13 Property 1. Nature of compounds
B Boron has small atomic radius, small size, high charge and hence high charge density for which all its
Al, Ga, In and TI Form covalent as well as ionic compounds
compounds are covalent. B 3+ ion does not exist. 2. Maximum covalency
Boron shows a maxium covalency of four, due to absence of d– orbitals.
Show covalency of six or more, as the d-orbitals are available.
3. Electron accepting power of compounds
Boron has less than four valence electrons for which it has a great tendency to accept electrons. Therefore, its compounds behave as strong Lewis acids and form large number of complex compounds.
Less tendency to accept electrons.
4. Inert pair effect
Does not exhibit inert pair effect
Exhibit inert pair effect.
535
BORON FAMILY
Difference between Boron and Aluminium The difference in the properties between boron and aluminium is due to the large difference in their size and ionisation potential. Table 14.4: Differences in properties between Boron and Aluminium Property
Boron
Aluminium
1. Nature
Non-metal
Metal
2. Hardness
Its crystalline form is very hard
It is sufficiently soft
3. Melting point
2300K
932K
4. Electrical conductivity Bad conductor 5. Nature of compounds
Forms covalent compounds only
Good conductor Forms both covalent and electrovalent compounds
hydrides
Does not form stable hydrides.
7. Nature of hydroxide
B(OH)3 or H3BO3 is weakly acidic.
Al(OH)3 is amphoteric
8. Action on steam
Does not decompose water or steam.
Decomposes steam to liberate hydrogen.
9. Action with acids
Does not react with dilute acids
2Al+6H2O 2Al(OH)3+3H2 Reacts with dilute acids to form hydrogen.
6. Nature of hydrides
Forms number of stable covalent
10.Combination with Combines with metals to form metals. borides 3Mg + 2B Mg3B3 (Magnesium boride)
2Al+H2SO4 Al2(SO4)3+3H2 Combines with metals to form alloys.
Similarities in properties of Boron and Aluminium In Group 13 of the periodic table aluminium is lying just below boron and exhibit many properties similar to those of boron. Following are the properties in which boron resembles aluminium: 1.
Both have similar electronic configurations having three electrons in their outer most orbits. B(5) = 1s 2, 2s2, 2p1; Al (13) = 1s2, 2s2, 2p6, 3s2, 3p1
2.
Both elements are trivalent. Since the penultimate shell of both elements are complete and stable, they do not show variable valencies. The common oxidation state of these elements is +3. Boron however shows -3 oxidation state in the metal borides.
536
3.
+2 CHEMISTRY (VOL. - I)
Both form amphoteric oxides when heated with oxygen at high temperature. 975K
4B + 3O2 4.
B 2O 3
2BCl3
2Al + 3Cl2
2BN
2AlCl3
2Al + N2
2AlN
2B + 2NH3
2BN + 3H2
2Al + 2NH3
2AlN + 3H2
BN + 3H2O
H3BO3 + NH3
AlN + 3H2O
Al(OH)3 + NH3
Both react with conc. H2SO4 to form SO2. 2B + 3H2SO4
2H3BO3 +3SO2
2Al + (6H2SO4) 7.
2Al2O3
Both, when heated with nitrogen or ammonia form nitrides. These nitrides are decomposed by steam to form ammonia. 2B + N 2
6
1075K
Both form covalent trichlorides when heated with chlorine. 2B + 3Cl2
5.
4Al + 3O2
Al2(SO4)3 + 3SO2 + 6H2O
Both react with alkalis to form H 2 2B + 6NaOH
2Na3 BO3 + 3H2
2Al + 2NaOH + 6H2O
2Na [Al(OH)4] + 3H2O
Position of Boron in the periodic table We know that boron is present as 1st element in Group 13 and Silicon as 2nd element in Group 14 of the periodic table. 13
14
The element silicon is diagonally related to boron in its properties.
B
C
Both the elements have almost the same electronegativity values.
Al
Si
(B=2.0. Si = 1.8) and they have identical values of ionic potentials
Ga
Ge
charge (= ————), B+3 = 0.073 and Si+4 = 0.074. radius
According to diagonal relationship, the elements of the 2nd period of the periodic table show similarities in properties with the elements which are lying diagonally opposite right in 3rd period. Thus, B-Si pair is called diagonal pair and these two elements show many similar properties.
537
BORON FAMILY
Similarities in properties of Boron and Silicon: The following similarities in properties justify the diagonal relationship between B and Si. 1.
Both of the elements are non-metals and hence bad conductors of heat and electricity.
2.
Both the elements exist in two allotropic forms viz. amorphous and crystalline. Their crystalline forms are hard.
3.
Both elements have almost the same electronegativity, ionisation potential and density. Their boiling points are very high and close to each other. Both have high melting points.
4.
Most of the compounds of these elements are colvalent in nature.
5.
Both with hydrogen form gaseous hydrides; B 2H6 and SiH4
6.
Both elements form stable oxides B2O3 and SiO2 respectively. These oxides are weakly acidic in nature, since in water they give weak acids. B2O3 + 3H2O
H3BO3
SiO2 + H2O
Boric Acid 7.
H2SiO3 Metasilicic acid
The halides of both the elements are covalent in nature and have many similar properties. BCl3 and SiCl4 both are liquids and get hydrolysed by water to form H3BO3 and H2SiO3. BCl3 + 3H2O
8.
H3 BO3 + 3HCl
SiCl4 + 3H2O
H2SiO3 +4HCl
Both the elements can be prepared by reducing their oxides with magnesium metal. B2O3 +3Mg
2B + 3MgO
SiO2 +2Mg
Si + 2MgO
Some important common chemical characteristics of Group 13 elements 2M + 3X2
2MX3
4M + 3O2
2M2O3
2M + 3Y
M 2Y 3
2M + N2 2M + 6H+
(X = F, Cl, Br, I)
(Y = S, Se, Te)
2MN 2M3+ + 3H2
(With Al, Ga, In)
Gallium, Indium and Thallium Gallium is one of the rarest elements. It is silvery - white, hard and low melting metal. Next to mercury it has the lowest melting point of all the metals. This property makes it useful as a thermometric liquid for high temperature thermometers.
538
+2 CHEMISTRY (VOL. - I)
The chemistry of gallium is similar to that of aluminium. Gallium is a good reducing agent like aluminium. Ga 3+ ion exists exclusively in solution. Aqueous solutions of its salts undergo hydrolysis like solutions of aluminium salts. Uses: 1.
Just as mercury, gallium is used in arc lamps employed for spectroscopic analysis.
2.
Liquid gallium, like mercury does not wet the glass or quartz and it is used in quartz thermostats for measuring high temperatures.
Indium Indium is a rare metal. In aqueous solutions In 3+ is hydrolysed to a lesser extent than Al3+ and Ga3+ ions. In addition to +3 oxidation state indium exists in +1 oxidation state. Its compounds like InCl, InBr, InI and In 2O are known. Thallium Thallium is a soft grey metal and differs from other metals in Group III A, that the +1state is quite stable in addition to +3 state. The behaviour of Tl + salts are similar to those of alkali metal salts. However, TlCl is insoluble in water. Uses Though thallium as a metal does not have any worthwhile application, its salts are very useful. 1.
Thallium salts are used as germicide, fungicide and for rat poison.
2.
Thallium salts have high refractive index for which these are used in the making of optical glasses like Crooke’s glasses.
3.
It salts are used in ointments for the treatment of ringworms.
14.5 BORON Boron is an important non-metal. Its electrical conductivity is very low and increases with the increase in temperature like a typical semiconductor. Boron is used in making boron- steels which are very hard and are used as control rods in atomic reactors. Boron compounds have numerous industrial applications. Boron and boric acid are industrially important. Metal borides are used in nuclear industry as protective shields and control rods since boron-10(B 10) isotope has high ability to absorb neutrons.
539
BORON FAMILY
Some important compounds of boron: Boranes: Boron forms covalent hydrides like carbon. These are called boranes. Two series of boranes are known having molecular formulae BnHn+4 (B2H6, B5H9, B6H10 etc) and BnHn+6 (B4H10, B5H11, B6H12, etc) All boranes are electron deficient compounds like diborane (B2H6) and therefore, have hydrogen−bridge structures. Boranes are inflammable and liberate considerable amount of energy during combustion. B2H6 + 3O2
B2O3 + 3H2O :
H = –2018kJ
There is a considerable interest in boranes today as possible rocket fuels. Structure of Diborance : Diborane, B2H6: Diborane is the simplest of boranes and analogous in molecular formula to ethane (C2H6). Each B atom is sp3 hybridized. Boron atoms and the four of the six hydrogen atoms lie in one plane with the remaining two hydrogen atoms lie outside it and occupy ‘bridge’ positions between the boron atoms. The four terminal B–H bonds are regular two centre-two electron bonds while the two bridge (B–H–B) bonds are three centre-two electron bonds referred as banana bonds. The electron-dot structure of diborance can be written as:
.. H
H
B
H.
H
H .. . B.
or ..
..
B
. .H
H
H
H
. .
H.
H. .
or
B
.. H
H
H B
H
B H
H
H
Thus, we find there are one electron bonds between boron and the bridging hydrogen atoms. This arises as a result of resonance between the structures given below. H B H
H
H
H and
B H
H
H B
H
H B
H
H
In 1976, William Discount was awarded Nobel Prize for elucidating the structrures of boranes.
540
+2 CHEMISTRY (VOL. - I)
Preparation of Diborance : 1.
It is prepared by treating boron trifluoride with LiAlH 4 in diethyl ether. 4BF3 + 3LiAlH4
2.
2B2H6 + 3LiF + 3AlF3
It is prepared in the laboratory by the oxidation of sodium borohydride with iodine. 2NaBH4 + I2
3.
B2H6 + 2NaI + H2
It is produced in the industry by the reaction of BF 3 with sodium hydride. 2BF3 + 6NaH ¾450K ¾¾ ¾® B2H6 + 6NaF..
Properties : 1.
Physical :
Diborane is a colourless, highly toxic gas with a B.P. of 180K.
Chemical :
(1) Reaction with Oxygen :
On exposure to air Diborane catches fire spontaneously releasing an enormous amount of energy. B2H6 + 3O2
Hydrolysis : Diborane gets readily hydrolysed by water to give boric acid. 2B(OH)3 (aq) + 6H2(g)
B2H6(g) + 6H2O (l)
Reaction with Lewis bases : With Lewis bases they give borane adducts. B2H6 + 2NMe3
Reaction with Ammonia : Diborane combines with ammonia to give an addition product B2H6. 2NH3 which when heated at 473K decomposes to give a volatile compound called ‘borazine’. 2B2H6 + 6NH3 ® 3[BH2 (NH3)2] + [BH4]¯ 437K ¯ Heat 2B3N3H6 + 12 H2 Borazine Borazine is isoelectronic with benzene. Its ring structure is like that of benzene with alternate ‘BH’ and ‘NH’ groups, therefore known as ‘inorganic benzene’. H
..
B
N B H
..
N
..
N H
Borazine
B
B
H
H H
H
H
N
N
B
B N H
Borazine
H
4.
2BH3.NMe3
H
3.
H
2.
B2O3 + 3H2O + 1976 kJmol–1
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BORON FAMILY
5.
Formation of complex borohydrides : Diborane reacts with metal hydrides in diethylether to form tetrahydridoborates, commonly known as borohydrides. B2H6 + 2NaH
Diethyl ether
2Na+ [BH4]¯
Sodium borohydride B2H6 + 2LiH
Diethyl ether
2Li+ [BH4]¯
Lithium borohydride Sodium borohydride and lithium borohydride are used as reducing agents in organic synthesis. They are also starting materials for many other borohydrides. Boric oxide (B2O3) It is nonvolatile and an extremely stable compound, acidic in nature and dissolves in alkalis to give borates. It is so stable that it cannot be even reduced by carbon at high temperature. It is prepared by heating boric acid to red heat. 2H3BO3
B2O3 + 3H2O
It reacts with metal oxides to form transparent metaborates. NiO + B2O3
Ni(BO2)2 (brown)
These reactions are used to detect the presence of metals in qualitative analysis.
14.6 BORAX (NA2B4O7.10H2O, SODIUM TETRABORATE DECAHYDRATE) Borax (Na2B4O7.10H2O) is the most important compound amongst borates. It is sodium salt of tetraboric acid (H 2B4O7) Occurrence and preparation : In India it is obtained in the name Tincal from dried up lakes in eastern Kashmir valley. Borax is also found in Tibet and California (U.S.A). Tincal is the crude form of borax and contains many impurities. The finely powdered Tincal mineral is digested in hot water which dissolves borax leaving the impurities undissolved. The borax solution is collected as filtrate, concentrated and cooled to obtain the crystals of borax. Borax can also be prepared from the mineral colemanite. The mineral is boiled with sodium carbonate when the following reaction takes place.
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2CaCO3 ¯ + Na2B4O7+ 2NaBO2 (Borax) (Sodium metaborate)
Ca2B6O11 + 2 Na2CO3 (Colemanite)
Calcium carbonate precipitates. The mother liquor contains borax and sodium metaborate. Sodium metaborate is also converted to borax by passing carbon dioxide through the mother liquor. 4NaBO2 + CO2
Na2B4O7 + Na2CO3
Properties: Borax is a white crystalline solid, sparingly soluble in cold water and fairly soluble in hot water. It contains the tetranuclear units [B 4O5 (OH)4]2– and correct formula therefore is Na 2[B4O5(OH)4], 8H2O. In water it gets hydrolysed and results NaOH and H3BO3 (orthoboric acid). Na2B4O7 + 7H2O
2NaOH + 4H3BO3 Orthoboric acid
Borax Bead Test : Borax when heated finally gives a colourless glassy bead, (NaBO 2 + B2O3). This bead further acquires a characteristic colour when heated along with a coloured salt. Borax bead test is used for the detection of cations of coloured salts. In this test, borax is heated on a platinum loop when a glassy bead is obtained. Na2B4O7.10H2O
heat
Na2B4O7
Red Heat
–10H2O
2NaBO2 + B2O3 Boric anhydride (Glassy bead)
This colourless glassy bead when brought in contact with coloured salt (to be identified) and heated again in the oxidising flame, the bead acquires a characteristic colour due to the formation of metaborate. CuSO4 + B2O3
SO3 + Cu(BO2)2 (bluish green)
CoSO4 + B2O3
SO3 + Co(BO2)2 (dark blue)
Similarly, for Cr 3+ and Ni2+ ions, Cr(BO2)3 (green) and Ni(BO2)2 (reddish brown) beads are respectively obtained. When the coloured bead is heated in reducing flame the colour of the bead changes.
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BORON FAMILY
2 Cu (BO2)2 Cupric metaborate (Bluish green) 2CuBO2 + C
Reducing flame Containing C
2 Cu
2CuBO2
+ B2O3 + CO
Cuprous metaborate (Colourless)
+ B2O3 + CO
Metal(red) Uses : Borax is used
14.7
1.
As a flux in soldering, welding and certain metallurgical operations.
2.
In enamelling and in making glazes for pottery and tiles.
3.
In leather industry for cleaning hides and skins and in leather-dyeing.
4.
In making optical glasses and borosilicate glass (pyrex glass) which is resistant to heat and shock.
5.
As a good cleansing agent, since it hydrolyses to form mildly alkaline solution. As a water-softening agent.
6.
In borax bead test to detect cations of coloured salts.
7.
For impregnating match sticks to prevent after-glow.
8.
As a mild antiseptic.
BORIC ACID (H3BO3, ORTHOBORIC ACID) Orthoboric acid, commonly known as boric acid is an important oxy-acid of boron. It is found in the volcanic regions of Italy. Preparation: 1.
Orthoboric acid is prepared by acidifying an aqueous solution of borax with dil.HCl or dil.H2SO4. Na2B4O7 + 2HCl + 5H2O
4B(OH)3 + 2NaCl
Orthoboric acid crystallises out by cooling the resulting solution. 2.
Orthoboric acid is also obtained by the hydrolysis (reacting with water or dilute acids) of boron compounds such as hydrides, halides etc. B2H6 + 6H2O
2H3BO3 + 6H2 -
BCl3 + 3H2O
H3BO3 + 3HCl -
Properties: It has a layer structure in which planar BO 3 units are joined by hydrogen bonds as shown in Fig. 14.1.
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+2 CHEMISTRY (VOL. - I)
OH B O
H
H
O
O B
O
H
H
O
H
H
O
O
HO
O
H
B
H
B
O
H
O H
Fig. 14.1 Structure of Orthoboric Acid 1. 2.
It forms soft pearly-white, needle like crystals with a greasy feel. It is moderately soluble in water and completely soluble in hot water.
3.
H3BO3 is a very weak acid and ionises as a monobasic acid. Although it is not a proton donor, it behaves a a Lewis acid by accepting an electron pair. [B(OH)4]-- + H3O+
H3BO3 or B(OH)3 + H2O
4.
Here H3BO3 accepts an electron pair from OH — (that is obtained from the ionisation of H 2O). It reacts with sodium hydroxide forming sodium metaborate, Na+ [B(OH)4]-- or NaBO2. 2H2O sodium metaborate
H3BO3 or B(OH)3 + NaOH 5.
Orthoboric acid on heating slowly loses water molecules and finally forms boron trioxide B2O3. 375K 4H3BO3
–4H2O
Orthoboric acid
Red heat
435K 4HBO2 Metaboric acid
–H2
H 2B 4O 7
–H2O
Tetraboric acid
2B2O3 Boron trioxide or Boric anhydride
Uses: Boric acid is used 1.
As an antiseptic and as an eye-wash (eye lotion) in medicine. A solution of boric acid in water is commonly used as a mild antiseptic.
2.
As a food preservative, a preservative for milk and other food.
3.
In making enamels and borate glazes in pottery.
4.
In candle industry to stiffen wicks and in tanning industry.
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14.8
ALUMINIUM Aluminium is a very important and strategic metal. It is one of the most common metals in use. Reaction of Aluminium with Acids and and Alkalis : (1)
With dil.HCl: Aluminium dissolves in dil. HCl and liberates dihydrogen. 2Al(s) + 6HCl (aq) ® 2Al3+(aq) + 2Cl¯(aq) +3H 2(g)
(2)
With dil.H2SO4: Alumunium with dil. H 2SO4 also liberates hydrogen gas. 2Al(s) + 3H2SO4 ® Al2(SO4)3 + 3H2
(3)
With conc. H2SO4 : Aluminium with conc. H 2SO4 liberates SO2 gas. 2Al + 6H2SO4 (conc) ® Al2 (SO4)3 + 3SO2 + 6H2O
(4)
With conc. HNO3 : With conc HNO3 acid, the metal gets coated with oxide and becomes passive.
(5)
With NaOH : Aluminium reacts with alkalis liberating hydrogen gas. 2Al(s) + 2NaOH(aq) + 6H2O)(l) ®
2Na [Al(OH)4] (aq)
+ 3H2(g)
Sodium tetrahydroxo aluminate (III) Uses : 1.
Being light and cheap, aluminium and its alloys are extensively used for the construction of aircraft, railway coach, bus bodies, household utensils etc.
2.
With the increasing cost of copper in India, aluminium wires are progressively used in transmission cables.
3.
Aluminium foils are used as wrappers for food materials, tobacco etc. Its powder is used to give shining paints to metallic structures.
4.
The reducing property of the metal is used in thermite welding (Fig.14.2) and for the extraction of metals like chromium, manganese etc. This is also called alumino-thermic process. The following reactions are involved in the process.
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+2 CHEMISTRY (VOL. - I)
D
Cr2O3+2Al 3Mn3O4+8Al Fe2O3 + 2Al 5.
6.
D D
Al2O3+ 2Cr 4Al2O3+ 9Mn Al2O3 + 2Fe
It is used in making house-hold utensils, for winding moving coils for dynamos and motors. Many salts of aluminium like aluminium chloride, alums etc. are useful in various ways. A mixture of aluminium powder and aluminium nitrate is used in bomb making.
BaO2+Mg-powder
Magnesium ribbon Magnesium lining Funnel-shaped Crucible
Molten iron
Fe2O3 + Al (powder) (Thermite mixture)
Fireclay mould
Iron pieces to be welded
Tapping plug (Removable)
Crack in an iron piece
Fig.14.2. Thermite welding process
Alloys of aluminium
Aluminium alloys are useful because they, in general, possess high tensile strength and resistance towards corrosion. Table 14.5. Some important alloys of aluminium Name
Composition
Uses
Magnalium
Al 98%, Mg 2%
Being tough can be used in lathes, used for construction of balance beams etc.
Duralumin
Al 95%, Cu 4%
Used to make bodies of airships.
Mn 0.5% Mg 0.5% Aluminium bronze
Al 90%, Cu10%
Used for making coins, picture frames, trays etc.
Y-alloy
Al 92.5%, Cu 4%,
Used for making some aeroplane parts.
Mg 1.5%, Mn 2%
14.9
ALUM Alums are double salts having the general formula, M2ISO4.M2III(SO4)3.24H2O
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BORON FAMILY
(where M is a monovalent ion like Na +, K+, Rb+, Cs+, NH4+, Tl+ etc, and MIII is a trivalent ion like Al 3+, Fe3+, Cr3+, Ga3+, Co3+ etc.). An alum is prepared by dissolving equimolar quantities of M 2ISO4 and M2III (SO4)3 salts in water and crystallising. Potash, ferric, chrome and sodium alums are the most commonly used alums. Potash alum : K2SO4.Al2(SO4)3.24H2O is prepared by dissolving alunite ore in dil. H2SO4 or bauxite in dil. H 2SO4 and adding K2SO4 to the solution. It is a colourless crystal, soluble in water and acidic in nature. Sodium alum
:
Na2SO4.Al2(SO4)3.24H2O
Ferric alum
:
(NH4)2SO4.Fe2(SO4)3.24H2O
Chrome alum
:
K2SO4.Cr2(SO4)3.24H2O
Uses: i)
Alums are used as mordants in dyeing and calicoprinting
ii)
For purification of drinking water.
iii)
In tanning of leather, sizing paper and making water proof clothes.
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CHAPTER (14)AT A GLANCE 1.
Group 13 or Boron family or Boron group elements consist of boron (B), aluminium (Al), gallium (Ga), indium (In) and thallium (Tl).
2.
Boron family elements have three electrons in their outermost shell (ns2 np1)
3.
The oxidation states shown by the elements of boron group are B=+3, Al=+3, Ga = +3, + 1, In = +3, + 1 and Tl = +3, +1.
4.
Boron is a semimetal or metalloid. It has more properties of non-metal. Boron forms covalent compounds due to its small size and high ionisation energy. Al and other elements of Group IIIA are metallic in nature.
5.
The atomic radii and ionic radii of elemetns of Boron group increase down the group particularly in case of Al, Ga, In and Tl, from Al to Ga atomic radius slightly decreases.
6.
The ionisation energies of these elements show fluctuation with the increase in the atomic number down the group.
7.
On moving down the group from B to Tl, the electronegativity values donot show the expected decrease. Group 13 or Boron group elements are more electronegative as compared to those of Group 12 elements.
8.
Densities of the elements of Boron group increase regularly on moving down the group.
9.
The melting points decrease from B to Al and then increase from Ga to Tl. The boiling points decrease regularly down the group from B to Al.
10.
All the elements of the Boron group form M3+ compounds. The tendency of M3+ ions to form covalent compounds decreases. Boron does not form B + ion.
11.
Boron forms a large number of stable covalent hydrides called boranes. Boranes of BnHn+4 type are B2H6, B5H9, B6H10 etc. Boranes of BnHn+6 type are B4H10, B5H11, B6H12 etc.
12.
Trihalides of boron are electron deficient compounds and are Lewis acids, since they accept lone pair of electrons from donor atoms.
13.
All these elements form oxides of the type M 2O3 and hydroxides of the type M(OH)3. The basic character of oxides and hydroxides increases from B to Tl.
14.
Boron differs from other members of the Group 13 elements. All compounds of boron are covalent. B+ ion does not exist. It has greater tendency to accept electrons in comparison to other elements of the group. Boron does not exhibit inert pair effect.
15.
Boron, the first element of Group 13 has many similar properties with those of silicon, the second element of group 14. This is due to the diagonal relationship between B and Si.
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BORON FAMILY
16.
Some important common chemical properties of Group 13 elements are as follows. (M = element of Group 13) (a)
2M +3X2
2MX3 (X=F, Cl, Br, I)
(b)
4M + 3O2
2M2O3
(c)
2M + 3Y
M2Y3 (Y = S, Se, Te)
(d)
2M + N2
2MN
(e)
2M + 6H+
2M3+ + 3H2 (With Al, Ga, In)
17.
Gallium is one of the rarest elements and used as a thermometric liquid for high temperature thermometers.
18.
Since thallium salts have high refractive index these are used in the making of optical glasses like Crookes glass.
19.
Boron is an important semimetal and its electrical conductivity is very low and increases with increase in temperature. Boron steels are used as control rods in atomic reactors. Metal borides are also used as control rods since they absorb neutrons.
20.
Boron mainly occurs as borax (Na2B4O7.10H2O)
21.
Boranes are covalent hydrides of boron. Since boranes are inflammable and liberate considerable amount of energy during combustion, these are considered as possible rocket fuels.
22.
Borax, Na2B4O7.10H2O is the most important compound of boron. It is sodium tetraborate decahydrate. It is a white crystalline solid, gets hydrolysed to give boric acid H3BO3. Borax has numerous uses in metallurgical operations, making optical glasses and pyrex glass, leather industry, medicine etc.. Borax is also used in borax bead test to detect basic part (cations) of coloured salts.
23.
Boric acid or Orthoboric acid, H3BO3 is an important compound of boron. It is soft pearly-white needle like crystals prepared from borax. It is used in medicine as an antiseptic and eye-wash. It is used in food preservation, pottery and pigment.
24.
Some important ores of aluminium are Bauxite, Al2O3.2H2O, Corundum, Al2O3, Cryolite, Na3AlF6 and Alunite, K2SO4.Al2(SO4)3.4Al(OH)3.
25.
Aluminium and its alloys are extensively used in construction of aircraft, railway coach, bus bodies, utensils etc. Also aluminium wires are used in transmisssion cables.
26.
Aluminium metal is used to prepare metals like Cr, Mn etc. by reducing their oxides in the process called thermite welding (Alumino-thermic process).
27.
Potash alum is K2SO4.Al2(SO4)3.24H2O.
32.
Alums are used as mordant in dyeing and calico printing. It is used for purifying drinking water and tanning leather.
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+2 CHEMISTRY (VOL. - I)
QUESTIONS A. 1. 2. 3. 4. 5.
Very short answer type. (1 mark each) Write the formula of orthoboric acid. What is borax ? Give one use of borax. Name one important ore of aluminium and its composition. Name any two elements of group 13. Write the formula of cryolite.
6.
Complete the equation: Al2O3 + NaOH
7. 8. 9.
11.
Between boron and aluminium, which metal has a higher value of atomic radius ? What is borax bead test ? Aluminium oxide is ——— (neutral oxide, acid oxide, basic oxide, amphoteric oxide) Which one of the following is an ore of aluminium ? (i) Haematite (ii) Bauxite (iii) Dolomite (iv) Cuprite What happens when orthoboric acid is heated ?
12.
Write the formula of orthoboric acid.
B.
Short answer type. (2 marks each)
1.
How is orthoboric acid prepared ?
2.
What happens when orthoboric acid is heated ?
3.
What is borax ? Write two uses of borax.
4.
Write the preparation of orthoboric acid from borax.
5.
To which block of the periodic table B and Al belong ? Explain why.
6.
What happens when aluminium hydroxide is treated with sodium hydroxide solution ?
7.
Name two ores of aluminium.
8.
Complete the equation NaAlO2 + 2H2O
9.
Write the balanced chemical equation for the following:
10.
.... + H2O
On heating, borax first swells forming anhydrous borax and then changes to metaborates and boron trioxide. 10.
Which of the cations Mg2+ and Al3+ is smaller ? Give reason.
11.
Write a note on orthoboric acid.
12.
What happens when borax is heated ?
13.
Give reactions to show that aluminium is an amphoteric element.
14.
Why boron trifluoride is called electron deficient compound ? Explain
BORON FAMILY
551
C.
Short answer type : (3 marks each)
1.
Why boron halides do not exit as dimers while AlCl3 exists as Al2Cl6 ?
2.
BCl3 is trigonal planar while AlCl3 is tetrahedral in dimeric state. Explain.
3.
Why BBr3 is a stronger Lewis acid as compared to BF 3 though fluorine is more electronegative than bromine. (Hint : In BBr3, the size of 4p-orbital of Br containing the lone pair of electrons is much bigger than the empty 2p–orbitals of B and hence donation of lone pair of electrons of Br to B does not occur to any significant extent. So electron deficiency of B in BBr 3 is much higher then that of BF3. Hence it is a stronger Lewis acid.)
D.
Long answer type : (7 marks each)
1.
3.
Write general trends in properties of the group IIIA elements (Boron family). How borax is prepared ? Mention one of its uses. Write notes on (3.5 marks) (i) Aluminothermic process Discuss the similarities and dissimilarities between boron and aluminium.
E.
Explain the following:
1.
Boron has high m.p and b.p.
2.
Boron does not form B3+ ion.
3.
AlCl3 forms a dimer but BCl3 does not form dimer.
4.
Boron and aluminium halides behave as Lewis acids.
5.
Aluminium is stable in air and water.
6.
No visible reaction occurs when aluminium is left in contact with concentrated nitric acid.
7.
Aluminium containers can be used to store conc. HNO3.
8.
Aluminium vessels should not be cleaned with cleaning agent containing washing soda.
9.
Aluminium metal is frequently used as reducing agent for the extraction of metals such as Cr, Mn, Fe etc.
10.
The hydroxides of aluminium and iron are insoluble in water. However, NaOH is used to separate one from the other.
11.
Aluminium hydroxide is amphoteric. Why ?
12.
Why aluminium chloride is not ionic ?
13.
AlCl3 is a Lewis acid, explain.
2.
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MULTIPLE CHOICE TYPE F.
Select the correct answer :
1.
Which of the following is a p-block element ? a) Magnesium b) Aluminium c) Sodium Borax on heating with cobalt oxide forms a bead of
d) Neon
a) Co(BO2)2
d) No reaction
2.
3.
4.
b) CoBO2
c) Co3(BO3)2
The product formed in the reaction, BCl3 + H2O a) H3BO3 + HCl
b) BO3 + HOCl
c) B2H6 + HCl
d) No reaction
is
Which of the following is not an ionic trihalide ? a) AlF3
b) BF3
c) InF3
d) GaF3
5.
In which of the following, +1 oxidation state is stabler than +3 ? a) Ga b) Al c) Tl d) B
6.
Boron carbide, B4C is widely used in ————.
7. 8. 9.
10.
11.
12. 13.
a) Making plaster of paris b) Making boric acid c) As a hardest substance after diamond d) Making acetylene One that marks the paper like lead is ____________. a) Ga b) B c) Ti d) Tl Which is used in high temperature thermometry ? a) Na b) Tl c) Hg d) Ga The two type of bonds present in BH are covalent bond and ________. a) Hydrogen bridge bond b) Co-ordinate bond c) Ionic bond d) None. Borax bead test is responded by a) divalent metals b) metals which form coloured metaborates c) light metals d) heavy metals Alumina is a) Acidic b) Basic c) Amphoteric d) Neutral (CHSE, 1988S) Which metal is protected by layer of its own oxide ? a) Al b) Ag c) Au d) Fe Aluminium forms a) Electrovalent compounds only b) Covalent compounds only c) Electrovalent and covalent compounds both d) Coordinate compounds only.
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14.
Which of the following metals burns in air at high temperature with the evolution of much heat ? a) Cu
15.
17.
18.
19.
21.
b) Sn
c) Ag
d) Al
a) Coating is much smoother
b) It does not tarnish in air
c) It has better shine than silver
d) It does not scratch
The most covalent aluminium halide is a) Aluminium fluoride
b) Aluminium chloride
c) Aluminium bromide
d) Aluminium iodide
Which of the following compounds can make cloth fire proof ? a) Ferrous sulphate
b) Aluminium sulphate
c) Cuprous sulphate
d) Magnesium sulphate
Aqueous solution of borax is b) faintly acidic
c) alkaline
d) strongly acidic
The blue coloured minerals ‘Lapos Lazule’ which is used as a semi precious stone is a mineral of the following class. a) Sodim aluminium silicate
b) Boron trioxide
c) Zinc cobaltite
d) Basic copper carbonate
Which of the following is an ore of aluminium: a) Haematite
22.
d) Al
Aluminium deposited as vapour on glass forms a good mirror, essentially because :
a) neutral 20.
c) Pb
Which metal is powdered, suspended in oil and used as a paint for mirrors ? a) Fe
16.
b) Hg
b) Bauxite
c) Dolomite
d) Cuprite
The structure of diborane (B2H6) contains (a) four 2c–2e bonds and two 3c–2e bonds. (b) two 2c–2e bonds and four 3c–2e bonds. (c) two 2c–2e bonds and two 3c–2e bonds. (d) four 2c–2e bonds and four 3c–2e bonds.
23.
The states of hybricdization of boron and oxygen atoms in boric acid (H 3BO3) are respectively (a) sp2 and sp2
24.
(b) sp3 and sp3
(c) sp3 and sp2 (d) sp2 and sp3 Beryllium and aluminium exhibit many properties which are similar. But the two elements differ in (a) exhibiting maximum covalency in compound. (b) exhibiting amphoteric nature in their oxides (c) forming covalent halies. (d) forming poymeric hydrides.
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ANSWERS A.
D.
1)
H3BO3
2)
Sodium tetraborate decahydrate, Na2B4O7.10H2O. Borax is used in making optical glasses and borosilicate (pyrex) glass which is resistant to heat and shock.
3)
Bauxite, Al2O3.2H2O.
4)
Boron and aluminium.
5)
Na3AlF6 or AlF3.3NaF
6)
Al2O3 + 2NaOH
7) 8) 9) 10) 11) 12)
Aluminium Refer Text. amphoteric oxide Bauxite B2O3 H3BO3
1.
Boron has very high mp. and b.p. because it exists as a giant covalent, polymeric structure both in solid as well as in liquid state. Boron has small size and high ionisation energy for which it is very difficult to
2.
2NaAlO2 + H2O
remove three electrons to form B3+ ion. 3.
AlCl3 forms dimer by completing octet of Al accepting electron pairs. Cl Al Cl
Cl
Cl
Cl
+
Cl
Al
Cl Al
Al
Cl
Cl
Cl
Cl
Cl
4.
Both boron and aluminium atoms in their halides are electron deficient.
5.
Aluminium reacts readily in air and water at ordinary temperature to form protective layer of its oxide which protects it from further action.
6.
Aluminium becomes passive due to coating of Al 2O3 when comes in contact with conc. HNO3 and becomes inert for further action.
7.
Aluminium in contact with conc. HNO 3 becomes passive due to the coating of Al2O3 on it. Therefore aluminium containers can be used to store conc. HNO3.
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BORON FAMILY
8.
9.
Washing soda, Na2CO3 reacts with water to form NaOH and aluminium dissolves in sodium hydroxide. Na2CO3 + H2O
2NaOH + CO2
2Al + 2NaOH + 2HO
2NaAlO2 + 3H2
(sodium meta aluminate) Al has greater affinity for oxygen and thus it reduces a large number of less electropositive metal oxides to metals. Fe2O3 + 2Al
10.
Al2O3 + 2Fe;
H = –ve
Al(OH)3 gets dissolved in sodium hydroxide whereas Fe(OH) 3 is insoluble in NaOH. Al(OH)3 + NaOH
NaAlO2 + 2H2O Sodium meta aluminate (soluble)
E.
11.
Al(OH)3 reacts with acid as well as base to give corresponding salts.
12. 13.
Refer text. Refer text.
1) b 7) d 13) c
2) a 8) d 14) d
3) a 9) a 15) d
4) b 10) b 16) b
5) c 11) c 17) d
6) c 12) a 18) b
19) c
20) a
21) b
22) a
23) d
24)a
qqq
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CHAPTER - 15
THE CARBON FAMILY (GROUP 14 ELEMENTS)
The carbon family comprises of the elements Carbon, Silicon, Germanium, Tin and Lead. These elements belong to group 14 of the periodic table. Carbon and silicon are nonmetals. Germanium is a metalloid since it shows both nonmetallic and metallic properties. Tin and Lead are well known metals. 15.1
ELECTRONIC CONFIGURATIONS : The atomic numbers and electronic configurations of the elements are given below : Table 15.1 Electronic configurations of the elements of carbon family
Element
Symbol
Atomic number
Electronic Clonfiguration Oxidation states
Carbon
C
6
[He] 2s22p2
4
Silicon
Si
14
[Ne] 3s23p2
4
Germanium
Ge
32
[Ar] 4s24p2
2
4
Tin
Sn
50
[Kr] 4d105s25p2
2
4
Lead
Pb
82
[Xe] 4f145d106s26p2
2
4
15.2
OCCURRENCE :
Carbon occurs in nature in free state as diamond, graphite and coal. Coal and graphite are mined in large quantities. Diamond is mined in extremely small quantities. In combined state carbon occurs in large quantities in crude oil and metal carbonates such as lime stone CaCO3, magnesite MgCO3 and dolomite MgCO3. CaCO3. Carbon dioxide occurs in small amounts in the atmosphere and plays the vital role in the carbon cycle in photosynthesis and respiration. Again, we know that carbon is the most essential constituent of all organic compounds. Silicon is the second most abundant element by weight in earth's crust. Silicon occurs very widely as silica SiO2 in the form of sand and quartz. Silicon is also present in large varieties in silicate minerals and clays. Germanium is available in traces in some silver and zinc ores and in some types of coal.
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Tin is present in its important ore tinstone or cassiterite SnO 2. Lead occurs in its ore galena (PbS).
15.3
GENERAL CHARACTERISTICS OF THE ELEMENTS :
Physical properties : Some of the important physical properties of the elements are given in Table 15.2 below. Table 15.2 Physical properties of the elements of group 14 Element
Atomic mass
Atomic radius
Density gm/cc
M.P K
B.P K
Ionisation Electronegativity energy KJ/mol
C
12
0.77
2.22
853
5100
1086
2.5
Si
28
1.15
2.33
1673
2950
787
1.74
Ge
72.6
1.22
5.32
1210
3103
460
2.00
Sn
118.7
1.41
5.75
505
2960
707
1.70
Pb
207.2
1.47
11.35
600
1990
715
1.55
1.
Atomic radius: Atomic radii of elements increase with increase in atomic number as expected. This is due to successive addition of new orbits.
2.
Density : The density of elements gradually increases with increase in atomic number.
3.
Melting and Boiling points : The melting point as well as boiling point of carbon, the first element are exceedingly high. However, within the group the value gradually decreases with increase in atomic number. The melting points of Sn and Pb are relatively low.
4.
Ionisation energy : All these elements have high ionisation energies. Ionisation energy of carbon is very high due to its small size. In case of carbon the valence electrons are strongly held by nucleus and more energy is required to remove a valence electron. Again there is a sharp decrease in ionisation energy from carbon to next element silicon. That is due to an appreciable increase in size of silicon atom. There is however, a very small decrease in ionisation energy as we move from Si to Sn. In case of Ge and Sn, there are the presence of 'd' electrons in the inner configuration which shield the nuclear charge less effectively than is done by 's' and 'p' electrons for silicon atom. In case of Pb there are 'f' electrons in the inner configuration and shield the nuclear charge less effectively than 'd' electrons. And hence the ionisation energy of Pb, instead of decreasing, shows marginal increase.
5.
Electronegativity : Carbon is the most electronegative element of this group. The electronegativity values decrease from carbon to silicon and thereafter no regularity is observed. This appears to be due to the filling of d - orbitals in the case of germanium and tin and also f- orbitals in the case of lead.
558
6.
15.4
+2 CHEMISTRY (VOL. - I)
Metallic and nonmetallic characters : Since the ionisation energy decreases on descending the group, the metallic character increases. Carbon and silicon are nonmetals. Germanium exhibits both nonmetallic as well as metallic character and hence it is a semimetal. Tin and lead are metals. The metallic behaviour of these elements are reflected in their chemical properties also. The increased tendency to form M 2+ ion and decrease in acidic character or increase in basic character of their oxides and hydroxides explains their metallic character. OXIDATION STATES :
All these elements belong to p Block having ns2np2 general electronic configuration in the outermost shells of their atoms. Due to the presence of four electrons (two s electrons and two p electrons) in their valence shells, they all exhibit valency four corresponding to +4 oxidation state. Besides, the elements show +2 oxidation state for the presence of two electrons in their outermost ‘p’ orbital. Silicon exhibits unstable +2 oxidation state. Germanium and tin have well characterised +2 oxidation states. Lead shows stable +2 oxidation state. Chemical properties : Majority of the compounds of group 14 elements are tetracovalent. In this case all four outermost electrons participate in bonding. Promotion of electrons from the ground state to an excited state is followed by sp 3 hybridisation resulting in a tetrahedral structure. Ground state of carbon
1s22s22px12py12pz0
Excited state of carbon
1s22s12px12py12pz1
4 unpaired electrons form 4 covalent bonds. From the ground state one ‘s’ electron is promoted to the ‘p’ orbital and then the four unpaired electrons in the excited state occupy four sp3 hybrid orbitals resulting in four equivalent bonds tetrahedrally directed. The high values of their ionisation potential do not favour M 4+ oxidation state. The elements do not also form M4– ion for their low electronegativity. However, germanium, tin and lead show M4+ and M2+ oxidation states. On descending the group, the +2 oxidation state is more stable than the +4 oxidation state. The stability of M 2+ ion is more than that of M4+ ion and this is due to inert pair effect. The inert pair effect is more prominent in the heavier elements. In such cases only the two outermost unpaired p electrons take part in bonding and the outermost paired s electrons do not. The tendency of non-participation of the paired s electrons in bond formation is known as inert pair effect. The last three elements Ge, Sn and Pb have a tendency to form M2+ ions as well as M4+ ions. However, they prefer to form M2+ ions. Thus, Sn2+ and Pb2+ exist as stable ions and are more common than Sn 4+ and Pb4+ ions. Again, Sn (IV) and Pb (IV) compounds are more covalent than ionic in nature as expected from Fajan's rule. According to Fajan's rule, the smaller the ions, the greater is its
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559
tendency towards covalency. Sn4+ and Pb4+ ions have much smaller radii than Sn2+ and Pb2+ and thus they tend to form covalent compounds. 15.5
GENERAL TRENDS IN THE CHEMICAL PROPERTIES OF THE ELEMENTS :
Hydrides : All the elements form volatile covalent hydrides with hydrogen. The formation of number of hydrides and their stability decrease form carbon to lead. Carbon with hydrogen forms a series of aliphatic hydrocarbons alkane. C nH2n + 2 , alkene CnH2n and alkyne CnH2n – 2. It also forms aromatic hydrocarbons, for example benzene C 6H6 and its derivatives. Carbon forms alkane CnH2n + 2 i.e. CH4, C2H6, C3H8, C4H10 etc. The number of hydrides is less for silicon and they are silanes, SiH4 , Si2H6 etc, the germanium hydrides or germanes are similar to silicon hydrides. Stannane SnH4 and Plumbane PbH4 are only hydrides of tin and lead and they are difficult to prepare. Oxides : All the elements form two types of oxides. They are monoxides and dioxides. The monoxides are CO, SiO, GeO SnO and PbO. All these elements form dioxides of the general formula MO2. The acidic properties of their dioxides decrease down the group. Carbon dioxide CO2 and silica SiO2 are acidic. Germanium dioxide GeO2 is less acidic than SiO2. Tin dioxide SnO2 and lead dioxide PbO2 are amphoteric having both acidic and basic characters. Halides : All elements of this group form tetrahalides of the type MX 4 which are covalent and tetrahedral. The stability of tetrahalides decreases on moving down the group from carbon to lead. The tetrahalides of all these elements except carbon are hydrolysed by water. SiCl4 + 4H2O ® Si(OH)4 + 4HCl
Cl
Cl
Si Cl
Cl
..O ..
+ Cl
H
–HCl H
Si Cl
Cl
OH Si HO
OH
OH
Fig. 15.1 : Structure of Silicic Acid
+3H2O 3HCl
OH
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In the presence of excess acids, the halides of all elements except carbon form complexes. Carbon halides do not form complexes because of absence of vacant d - orbitals in carbon. Complex formation with other halides take place due to the presence of vacant d -orbitals of the group elements. Ex : SiF62–, [GeCl6]2–, [Sn(OH)6]2– etc. Here central atom is sp3d 2 hybridised. All these elements are known to form dihalides of the type MX 2. The stability of halides increase down the group. Carbon dihalides are highly unstable whereas SnCl 2 and PbCl2 are quite stable. 15.6
ANOMALOUS CHARACTER OF CARBON :
Carbon differs considerably from other members of the group due to its small size, high ionisation energy and maximum electronegativity. (i) It is the hardest element and has highest melting and boiling points. (ii) Carbon exhibits maximum covalency of four, whereas other members of the group can extend their covalency beyond four by using their vacant d - orbitals. (iii) Carbon does not form complex compounds due to the absence of the vacant d orbitals. (iv) It has unique ability to form multiple bonds such as C = C, C º C. C = O, C = S and C º N. 15.7
CARBON : CATENATION
Carbon has marked tendency to form long chains. This property is called catenation or self-linkage. Catenation is the tendency of an element to form long chains of identical atoms. Among the group 14 elements carbon has maximum tendency for catenation. The tendency for catenation decreases down the group. Sn and Pb hardly have this tendency. Carbon having this remarkable property forms thousands of organic compounds in which carbon atoms are bonded to each other. Comparison of carbon and silicon : Carbon and silicon, the typical elements of Group 14 are very much similar in properties and compounds. These two elements have some points of similarity in occurrence, electronic configuration and tetracovalency, nonmetallic character, tendency for catenation and formation of similar compounds. There are marked similarities between compounds of carbon and silicon as shown below. Table 15.3
Formation of similar compounds.
Compounds
Carbon
Silicon.
Dioxides Acids Hydrides
CO2is acidic, formed by burning C with oxygen. Carbonic acid, H2CO3 Methane CH4, Ethane C2H6
Halides Other compounds
Carbon tetrachloride CCl4 Chloroform CHCl3
SiO2 is acidic, formed by burning Si with oxygen. Metasilicic acid H2SiO3 Silicomethane SiH4, Silico ethane Si2H6 Silicontetrachloride SiCl4 Silico - chloroform SiHCl3
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561
These two elements also differ from each other in many respects as given below. (table - 15.4) Table.15.4 Property
15.8
Difference between carbon and silicon.
Carbon
Silicon.
Valency
Shows maximum covalency of 4. Therefore, forms more stable compounds.
Shows maximum covalency of six and thus silicon compounds are comparatively unstable.
Conductivity
Some varieties of carbon are good conductors of electricity
Bad conductors of electricity.
Melting point
Does not melt at all. M.P. is very high 3773 K.
Comparatively low M.P. 1673 K
Nature of hydrides
Hydrocarbons e.g. CH4 and C2H6 . These are very stable
SiH4 comparatively less stable
Action with O2
CO and CO2 are gases
SiO2 is a solid
Action with alkalis
No action with alkalis.
Liberates hydrogen when reacts Si+2NaOH + H2O®Na2SiO3+2H2
Nature of hydrolysis
Carbon tetrachloride and
SiCl4 is easily hydrolysed by water
of halides
chloroform are stable liquids and are not hydrolysed easily by water
SiCl4 + 3H2O
® H2SiO3 + 4 HCl
ALLOTROPES OF CARBON :
Different structural forms of the same element show allotropy . The allotropic forms or allotropes have same chemical properties but different physical properties. Carbon exists in different allotropic forms. There are three allotropic forms of carbon. (1) Diamond (2) Graphite (3) Amorphous carbon. Diamond and graphite belong to crystalline variety. Amorphous type of carbon includes charcoal, coke, lamp black, gas carbon etc. Again charcoals are of many types. These are wood charcoal, sugar charcoal and animal charcoal. Blood charcoal and bone charcoal belong to the category of animal charcoal. Table 15.5 Comparison of physical properties among allotropic forms of carbon. Property Diamond Graphite Amorphous carbon Colour Appearance Specific gravity Purity Hardest Electrical conductivity
Colourless Transparent Cystalline 3.5 100% Hardest Bad conductor
Dark grey Lustrous Soapy touch 2.5 95 –99% Soft and marks paper Good conductor
Black Opaque 1 to 1.5 60 – 90% Hard (Lamp black is soft) Bad conductor (Gas carbon is a good conductor)
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15.8.1 Diamond Structure : Ocurrence : Diamond occurs native in nature. In India it is mined from Golconda mines. It is mined from Kimberley mines in South Africa. South Africa supplies about 90% of the world production of diamond. Australia and South America have also diamond mines. Natural diamond varies in size, colour and lustre. The cost of diamond depends upon its lustre, colour and size. The weight of diamond is measured in carats. One carat weighs 0.2054 gram. Cullinan, the largest diamond, weighs about 3026 carats. The following are some well known varieties of diamond. Diamond
Weight in Carat
Cullinan
3026
Excelsier
97.2
Hope
44.6
Kohinoor
11.6
Preparation of artificial diamond : In 1896, Moissan first prepared diamond by heating iron with sugar charcoal (carbon) in a graphite crucible in an electric furnace at a temperature of about 3300 K. At this condition molten iron dissolved most of the carbon. The crucible was then taken out and dipped into molten lead at 600K. During the process the outer layer of iron solidified due to such sudden cooling and expanded. Iron, on expansion, exerted a great presure on the interior liquid mixture and converted it to solid. The carbon at such high pressure was converted to diamond crystals and emerged out of the solution mixture. The diamond crystals and graphite were washed with hydrochloric acid to dissolve iron. Finally diamond was separated from graphite. Structure : Diamond is a three - dimensional polymer consisting of a large number of carbon atoms tetrahedrally arranged. The structure of diamond was established by X - ray diffraction studies. In diamond each carbon atom lies at the centre of a regular tetrahedron and is linked to four surrounding carbon atoms lying at the corners of the tetrahedron. The carbon atoms are bound to each other by sp3 hybridised covalent bonds. Since all the carbon atoms are joined tetrahedrally the resulting structure becomes a giant hexagonal structure containing a series of interlocking hexagons.
· ·
· ·
·
· 1.54A · ·
· ·
· ·
0
·
·
Fig 15.2 : Structure of diamond
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563
In diamond the C - C bond distance is 1.54 A0, each bond angle is 109028'. Due to short bond distance a large amount of energy is required to break the bonds. Therefore diamond is very hard and melts at about 3800 K. During the formation of four equivalent sp 3 hybridised bonds around each carbon atom, all the four outermost valence electrons of carbon are utilised. There are neither free electrons nor any vacant orbital present for the migration of electrons and therefore diamond is a non-conductor of heat and electricity. Chemical properties : Diamond is chemically inert, does not react under ordinary condition. Dut to its compact structure, it is less prone to chemical attack. 1.
When heated in presence of oxygen at 1180 K, diamond forms CO2 without leaving any residue which shows that it is 100% carbon . C + O2 ® CO2
2.
It is oxidised to carbon dioxide when heated with potassium dichromate and concentrated sulphuric acid at 475 K.
3.
C
2O ¾ ¾¾¾¾¾¾® K 2 Cr2 O 7 + H 2SO 4
CO 2
When fused with Na2CO3 diamond is converted to carbon monoxide C + Na2CO3 ® Na2O + 2CO
Uses : 1.
Diamond is used as a valuable decorative gem since it has a high refractive index 2.45 and reflects any amount of light that falls on it. This property attributes to the bright lusture of diamond. It is measured in carats (1 carat = 200 mg)
2.
Black diamonds such as Carbonado and Bort are used for cutting glass, marble stones and other gems. It is also used for cutting, grinding and polishing hard materials.
3.
Diamond is used in making rock drills and diamond dust is used as an abrassive in polishing colourless stones.
4.
Diamond dies are used to draw wires of tungsten filaments for electric lamps.
15.8.2 Graphite : Graphite, also called Black lead or Plumbago is found widely distributed in nature. It occurs as a soft grey or black mineral in countries like USA, Canada, India, Cylone, Italy and Siberia. Manufacture of artificial graphite (Acheson Process) Graphite is obtained from coke or anthracite by heating with sand in heat resistant electric furnace fitted with two carbon electrodes connected by thin carbon rods. The charge is covered with a mixture of carbon and sand. On passing alternating current through the carbon rods a temperature of 3775 K is produced. The temperature is maintained for 30 hours.
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Fig 15.3 Manufacture of artificial graphite.
During the process silicon carbide is formed first, but it decomposes at high temperature of the arc into graphite and silicon. 3C + SiO2 ® SiC + 2CO Coke SiC ® Si + C Structure : Graphite forms a two dimensional sheet like polymeric structure. 1.42A0
3.4A0
Fig 15.4 Structure of graphite.
Chemical properties : 1.
Graphite is chemically inactive and is not affected by dilute acids and alkalies.
2.
It burns in oxygen to form CO2.
3.
It is slowly oxidised to CO2 by chromic acid.
4.
It is oxidised by concentrated HNO3 to give insoluble yellowish green graphitic acid, C11H4O5.
5.
When fused with Na2CO3 it forms CO.
6.
Fluorine attacks graphite forming (CF) n
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565
In graphite each carbon atom is bonded to three other neighbouring carbon atoms through sp2 hybridised sigma(s) covalent bonds. The fourth unhybridised orbital of each carbon atom containing one electron may laterally overlap on the similar orbital of each neighbouring carbon atom to form pi ( p ) bond. These p electrons are delocalised and are free to move under the influence of heat and electricity. Thus, graphite is a good conductor of heat and electricity. The carbon atoms in graphite are arranged in a number of layers composed of hexagonal planar rings. The C - C bond distance 1.42 A0 being short indicates strong bonding. Each layer may be regarded as a fused system of benzene rings. These account for the stability of graphite, with each layer of carbon atoms being firmly bonded by strong covalent bonds. Therefore the melting point of graphite is quite high, 3775 K. The different layers in graphite are held together by relatively weak van der Waals forces and are about 3.4A0 apart (in comparison to short C - C bond distance) for which one layer slides over the other. And therfore graphite is quite soft, cleaves easily between the layers and exhibits lubricating properties. Graphite is chemically inert. It is slightly more reactive than diamond. Due to its more open structure it is more prone to chemical attack as compared to diamond. Uses : Graphite is used in : 1.
Making lead pencils. Here powdered graphite and clay are pressed into sticks.
2.
Making electrodes for electric furnace and refractory crucible.
3.
As a good lubricant for machine parts subjected to high temperature.
4.
Electrotyping i.e making type.
5.
As a pigment in paint.
6.
Nuclear reactors make use of graphite rods as moderators to slow down the high energy neutrons.
7.
Making synthetic diamond when heated at 1880 K by a pressure of 50,000 - 60,000 atmospheres.
15.8.3 Recently discovered allotrope of carbon : Fullerenes : Smalley and Curl of U.S.A and Krotos of U.K. discovered a third crystalline allotrope of carbon in 1985 and named it as Fullerene in honour of Dr. R. Buckminster Fuller, an American Engineer who invented the famous 'Geodesic dome', because this molecule resembles the structure of the geodesic dome-a soccer ball-like structure. Fullerene does not mean a separate molecule or cluster, rather all the members of a special class of carbon clusters. So, it is better to call them together as Fullerenes and term the individual molecules as C 60 fullerene, C70 fullerene etc. depending upon the number of carbon atoms present in each molecule. The most abundant fullerene C60 is also called 'bucky ball' because of its football like structure.
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Preparation : C60 and C70 fullerenes were first prepared in 1985 by Smalley, Curl and Kroto. Carbon was vapourised by directing an intense laser beam on graphite surface and allowed to cool in an inert atmosphere of helium. The carbon cluster, thus formed, was analysed and found to contain two clusters – a dominant cluster with 60 carbon atoms and another with 70 carbon atoms. In 1991, physicists Kratschmer and Huffmann were able to produce isolable quantities of C60 and C70 fullerenes by heating graphite rods in an electric arc in a low pressure helium atmosphere and extracting the soot in an organic solvent. . Structure : Among the .members of the fullerene family, C is the stable one and 60
hence abundant due to its perfect symmetrical structure, the relative position of each and every carbon being identical to that of other 59 carbon atoms. The structures of all the members are made up of combination of pentagons and hexagons. For example, in C60 fullerene 12 pentagons and 20 hexagons have combined to give rise to a perfect football-like structure. (Fig. 15.5)
Fig. 15.5 Bucky ball
Properties : After diamond and graphite, the 'bucky ball' is deemed to be the third crystalline allotrope of carbon. Unlike diamond and graphite, which are actually infinite network solids, the fullerenes are finite with no free bond existing in them. Bucky ball is considered to be the purest form of carbon. In the pure state it is an insulator, but can be suitably converted to semiconductors and even superconductors. Fullerene compounds (fullerides or buckides) can be used for the betterment of mankind and therefore fullerene research has got tremendous potentialities. Uses: 1.
Fullerides or buckides can be used as semiconductors and superconductors.
2.
Buckyball's ability to trap different atoms or molecules has successful application in medical and biomedical fields. Radioactive isotope encapsulated C 60 can be used in cancer therapy and AIDS therapy.
3.
Fullerenes can improve antiwear, antiseize, antifriction properties of lubricating oil.
4.
Catalytic amounts of fullerene can be used in petrochemical refining industries.
THE CARBON FAMILY
15.9
567
SOME IMPORTANT COMPOUNDS OF CARBON
15.9.1. Carbon monoxide CO is a poisonous gas, sparingly soluble in water and neutral in character. It is formed when carbon is burnt in a limited supply of air. Water gas is an equimolar mixture of CO and H2. Producer gas is a mixture of CO and N2. Coal gas is a mixture of CO, H2, CH4 and CO2. All these gaseous mixtures are important industrial fuels. Carbon monoxide is a good reducing agent and can reduce many metal oxides to the metal Fe2O3 + 3CO ® 2Fe + 3CO2 Structure : The structure of carbon monoxide is represented by electron dot formula and resonance hybrid structures. :C::O: C – O « C = O « C– º O+ +
–
Carbon monoxide is found to be an important ligand that can donate lone pair of electron to form co-ordinate bond with many transition metals. And as a result many carbonyl complex compounds are formed, M ¬ C º O. The highly poisonous nature of CO arises because of its ability to form a complex with haemoglobin which is about 300 times more stable than the oxy-haemoglobin complex and destroys the oxygen - carrying property of it resulting in death. While explaining the structure of such carbonyl complexes, the electronic structure of CO has been debated. However, carbon monoxide may be represented as : . .. .. : C :: . .: O : .
or
C¬ =O
15.9.2. Carbon dioxide CO2 is acidic in nature and dissolves in water forming carbonic acid H2CO3. Carbon dioxide is also called the anhydride of carbonic acid. CO2 + H2O ® H2CO3 Carbonic acid is unstable but gives rise to carbonate and bicarbonate salts. Carbon dioxide is present in the atmospheric air about 0.03 % by volume. It is also present in volcanic gases and in supersaturated solution in certain spring waters. Carbon dioxide is obtained by burning carbon in excess of air or by the action of dilute acids on carbonate salts. C + O2 ® CO2 CaCO3 + 2HCl ® CaCl2 + CO2 + H2O The gas when passed through lime water Ca(OH)2 or baryta water Ba(OH)2 , white precipitate of CaCO3 or BaCO3 is formed. By passing exces of carbon dioxide through such solutions the corresponding soluble bicarbonates are formed. Ca(OH)2 + CO2
®
CaCO3 + H2O
CaCO3 + H2O + CO2 (excess)
®
Ca(HCO3)2
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Solid carbon dioxide is known as dry ice. Carbon dioxide gas when solidifies under pressure, dry ice is formed. Dry ice sublimes without melting. It is used as a regrigerant. A mixture of dry ice and chloroform or ether can be used to maintain a very low temperature of about 196K. Structure of carbon dioxide It is a linear molecule with zero dipole moment. C
O
O
2p
2(sp)
2(sp)
2s
2p
2p
2s
2p
2p
There are two oxygen atoms on either p O=C s
2p
2p
2p
side of the carbon atom bound by double bonds. p = O s
Carbon has two sp hybrid orbitals and two unhybridised p-orbitals. Each of the sp orbitals of carbon overlaps with one p orbital of each oxygen atom resulting in two sigma s bonds each disposed on each side of the carbon atom. Further, one unhybrid orbital p overlaps with p orbital of one oxygen. The other p orbital of carbon overlaps with the p orbital of the second oxygen atom. So in addition to two s bonds, two p bonds are formed between C and two 'O' atoms (pp – pp) type.
:
:
:OºC–O : :
:O=C=C=O:
:
:O–Cº O:
:
:
In order to explain the carbon - oxygen bond length 1.15 A0 in carbon dioxide the resonance hybrid structures are also proposed. Resonance structure of CO 2 :
15.9.3. Carbon tetrachloride : It is the most important halide of carbon. Since carbon and chlorine do not combine directly, it is prepared by reacting carbon disulphide with chlorine. Carbon combines with sulphur directly at high temperature to form carbon disulphide. D ® CS C + 2S ¾¾ 2
Carbon disulphide liquid is treated with chlorine gas forming carbon tetrachloride. CS2 + 3Cl2 ® S2Cl2 + CCl4 Carbon tetrachloride is then distilled from sulphur monochloride. Carbon tetrachloride is a colourless, non-inflammable liquid which is an important industrial solvent. It is used as a fire extinguisher and in dry washing of clothes. Carbon tetrachloride is also used to prepare Freons. CCl4 + 2HF ® CCl2F2 + 2HCl (dichloro difluoro methane)
THE CARBON FAMILY
569
Freons are mixed chlorofluro hydrocarbons such as CFCl 3, CF2Cl2, and CF3Cl. These are unreactive and are widely used as refrigeration fluids.
15.9.4. Carbides Carbides are compounds of carbon with another less electronegative element than itself. Carbides have been broadly classified in to three groups depending on the nature of bonding and the consequent properties. (i) Ionic or salt like carbides : These are carbides of strongly electropositive elements. Their structures are comparable to those of ionic compounds. Some ioinic carbides are CaC 2, Al4C3, etc. (ii) Covalent carbides : In these carbides carbon is covalently bound with an electronegative element of close electronegativity ; Examples : CH 4, CO2, CCl4, CS2, SiC etc. CH4, CO2, CCl4, CS2 are small molecules and have low melting and low boiling points, whereas SiC contains giant molecules with very high melting point. (iii) Transition metal carbides or Interstitial carbides : These compounds are formed by transition elements with carbon at high temperature. Often they donot have normal composition. Small carbon atoms penetrate the crystal lattice of the metal. Examples : TiC, ZrC, WC, Mo2C, etc. These carbides have a wide range of stoichiometries. Carbides are usually obtained by heating the element or its oxide with carbon at very high temperature. Carbides find numerous applications in industry. Calcium carbide CaC2 : Calcium carbide is prepared by heating lime, CaO with coke in an electric furnace. CaO + 3C ® CaC2 + CO Calcium carbide reacts with water at room temperature to produce the important gas, acetylene, C2H2. CaC2 + 2H2O ® Ca(OH)2 + C2H2. Acetylene is a very important starting material in the manufacture of a large number of organic chemicals. Acetylene is also a valuable fuel in oxy - acetylene torches. 15.10 COMPOUNDS OF SILICON 15.10.1. Silicon dioxide : Silicon dioxide is a high melting polymeric solid. It is commonly called silica and exists in three different cystalline forms : Quartz, Tridymite and Cristobalite. In nature silica occurs mostly in crystalline quartz form. Though each of these forms has a different structure basically in all these forms each silicon is bonded tetrahdrally to four oxygen atoms and each oxygen atom is common to two tetrahedra. The difference between these structures is the arrangement of the tetrahedral SiO 4 units.
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I I I – Si – O —Si – O — Si – I I I O O O I I I – Si – O —Si – O — Si – I I I O O O I I I – Si – O —Si – O — Si – I I I Fig. 15.6 The nature of bonding in silica. The p bond formation in CO 2 is possible because of small size of carbon atom (9.77 A ). The p - orbitals of carbon atom can approach quite closely the p - orbitals of oxygen 0
for an effective overlap to form p bonds. Thus, carbon dioxide is a linear molecule existing in geseous form. Such a case does not arise in SiO2 because Si atom cannot form double bonds with oxygen due to its large size. Silicon satisfies its tetravalency by forming a three dimensional net work in which each silicon atom is bound to four oxygen atoms in a tetrahedral arrangement and therefore SiO2 remains in solid form.
Silicon dioxide is one of the most abundant natural compounds. Molten SiO 2 when cooled does not cystallise but slowly solidifies to a glass that becomes a rigid solid with random arrangement of its atoms. This substance is the basic ingredient in manufacture of different types of glasses. Different oxides are added to SiO 2 to make varieties of glasses. Glass is amorphous in its structure. SiO2 and glasses are not attacked by acids except hydrofluoric acid. SiO2 + 4HF ® SiF4 + 2H2O. However, basic substances may react with silica at high temperature to form silicates. SiO2 + 2NaOH ® Na2SiO3 + H2O sodium silicate 15.10.2. Silicates and Zeolites : A large number of silicate minerals and aluminosilicate clays occur in earth's crust. These are often extremely complicated in structural pattern. Some simpler types of silicates are : (a)
Orthosilicates containing SiO44– units, tetrahedral and monomeric, example : Mg 2SiO4.
(b)
Pyrosilicates containing Si2O76– units formed by linking two tetrahedra SiO44– units through a common oxygen atom, example : Se2Si2O7.
(c)
Cyclic silicates containing Si3O96– units in which three SiO44– tetrahedra are linked through three common oxygen bridges, example : BaTiSi 3O9.
(d)
Chain silicates containing Si6O1812– units in which six SiO44– tetrahedra are linked with each other through six bridging oxygen atoms, example is beryl Be 3Al2Si6O18.
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571
Besides these above silicates, we have sheet silicates and three dimensional silicates. Sheet silicates are two dimensional in structure. Three dimensional silicates are quatrz or Rock crystal, Zeolite and Ultramarines. Quartz is a transparent colourless cystal, electrically neutral used for making optical and chemical apparatus. Some quartz are opaque or milky. Quartz has framework structure with silicon and oxygen only, having graphite like structure. Zeolites are alluminosilicates with three dimensional structures. Zeolites like NaAlSiO 4, NaAlSi2O6 and NaAlSi3O8 are known. These are complicated silicates which have ion exchange properties. These find use as water softening materials. Zeolites have honeycomb like structure that absorbs or loses certain molecules and acts as molecular sieves. Sodium zeolites can exchange their sodium ions for calcium ions present in hard water. Thus, they act as cation exchangers when used for softening hard water. Sodium silicate : Sodium metasilicate commonly known as sodium silicate Na2SiO3 is a glass. It dissolves in water and hence it is often called water glass. Sodium silicate with excess of sand is used to manufacture water glass for commercial uses. It is used in soaps, cleansers, in fireproofing, as an adhesive and in preserving eggs. Silicates such as asbestos or mica and silicates containing aluminium are heated with lime stone to make portland cement . When soluble silicates are dissolved in dilute acid solutions, a hard jelly is formed. The jelly is then dehydrated to obtain a porous variety of silica known as silica gel. Silica gel is used as dehumidifying agent in commercial air conditioning units and as a deydrating agent in desiccators. It is used to recover industrial vapours since it is able to absorb gases. It is also used in the chemical industry as a catalyst. 15.10.3. Silicon Carbide, SiC Silicon carbide SiC is among the hardest known substances. In industry it is widely used as an abrasive material under the name carborundum. It is prepared by heating to about 3000K, a mixture of five parts silica and three parts powdered coke along with some saw dust. The saw dust keeps the mixture porous. The heating is done in an electric arc furnace between two carbon electrodes. SiO2 + 3C ® SiC + 2CO After completion of the reaction, the carborundum is washed with H 2SO4 and then with NaOH. Pure samples of silicon carbide are pale yellow to colourless. Sometimes it is often dark, dark purple or dark green due to traces of iron and other impurities. It is very hard, as hard as diamond and infusible. Silicon carbide is chemically unreactive. It is unaffected by acids. However it reacts with sodium hydroxide in presence of air. SiC + 2NaOH + 2O2 ® Na2SiO3 + CO2 + H2O
D ¾¾
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It reacts also with chlorine at about 300 K to form silicon tetrachloride. SiC + 2Cl2 ® SiCl4 + C Uses : 1. 2. 3. 4.
Due to its extreme hardness silicon carbide is used as an abrasive for grinding surfaces and sharpening metallic tools. Used as a refractory material. It is often used as semi-conductor in transistor. It finds application as a container for fuel elements used in atomic reactors.
Structure : Silicon carbide has a three dimensional structure of Si and C atoms. Each atom is tetrahedrally surrounded by four of the other kind. Its structure is similar to that of diamond. 15.10.4. Silicon Tetracloride (SiCl4) : Preparation : By heating silicon or silicon carbide with chlorine silicon tetrachloride as formed.
D Si(s) + 2Cl2(g) ¾¾® SiCl4(l) D SiC(s) + 4Cl2(g) ¾¾® SiCl4(l) + CCl4(l)
Properties and Uses : It is a volatile liquid with boiling point 330.57K. Hydrolysis of SiCl 4 gives silicic acid, Si(OH)4 which on further heating gives silica gel. D SiCl4(l) + 2H2O (l) ¾¾® Si(OH)4(aq)
Silicic acid
D ¾¾®
SiO2.xH2O(s) Silica gel.
Silica gel is used as a catalyst in petroleum industry and as an adsorbent in column chromatography. Reduction of SiCl4 with dihydrogen gives ultrapure form of silicon which is useful for making transistors, computer chips and solar cells. D SiCl4(g) + 2H2(g) ¾¾® Si(s) + 4 HCl(g)
15.10.5. Silicones Silicones are long chain organosilicon polymers. Preparation : The preparation of silicones involves three steps. (i) Preparation of alkyl chlorosilanes (ii) Hydrolysis of alkyl chlorosilane to form alkyl hydroxy silane (iii) condensation of alkyl hydroxy silane to get long chain polymers of silicones.
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573
Preparation of alkyl chlorosilane : When silicon reacts directly with chlorinated hydrocarbons in presence of a metal catalyst alkyl chlorosilanes are prepared. For example, silicon reacts with chloromethane (methyl chloride)in presence copper catalyst to yield dimethyl dichlorosilane. Cu
¾¾¾® (CH3)2 SiCl2 Dimethyl dichlorosilane Si + 2CH3Cl ¾catalyst
Alkyl chlorosilanes are also prepared by a Grignard's reaction. SiCl4
+ CH3 MgCl (methyl magnesium chloride)
®
CH3SiCl3
+
MgCl2
CH3SiCl3 + CH3MgCl ® (CH3)2 SiCl2 + MgCl2 (ii)
Hydrolysis of dimethyldichlorosilane results in dimethyl dihydroxysilane (CH3)2SiCl2 + 2H2O ® (CH3)2Si(OH)2 + 2HCl dimethyl dihydroxysilane.
(iii) Two molecules of dimethyl dihydroxysilane undergo condensation with elimination of one water molecule resulting in a unit containing Si – O – Si linkage. CH3 CH3 CH3 CH3 | | | | -H2O HO – Si – O H + HO – Si – OH ¾¾¾ HO – Si – O – Si – OH ¾® | | | | CH3 CH3 CH3 CH3
The condensation reaction can be continued to get silicone polymer. CH3 CH3 `CH3 CH3 CH3 | | | | | – O – Si – O – Si – O – Si – O – Si – O – Si – O – Si | | | | | CH3 CH3 CH3 CH3 CH3
CH3 CH3 | | – O – Si – O – | | CH3 CH3
A silicone polymer. Different types of alkyl or aryl substituted chlorosilanes can be used as starting materials to prepare varieties of silicone polymers. The hydrolysis of alkyl trichlorosilane RSiCl3 results in a complex crossed link polymer.
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| R–
–O –
O R | | Si – O – Si – O – | | O O R | | | Si – O – Si – O – Si – O – | | | R R O | |
In the above structure we notice that from RSiCl 3 on hydrolysis we obtain three. oxygen links that are bonded to other three Si atoms.
Cl OH | | Hydrolysis R – Si – Cl R – Si – OH Condensation H O 2 | | Cl OH
| O | R – Si – O – | O |
Uses 1.
Straight chain silicone polymers are used as silicone oils and rubbers. Viscous polymers are obtained with increased chain length. Silicone rubbers for its elasticity are used as good electrical insulators. Silicone oils are used as greases, varnishes and resins.
2.
Silicones are water-repellants and are good insulators. They are, therefore, used for water proofing and electrical condensers.
3.
These polymers are used in waterproofing textiles, in glassware, as lubricants, as polish and antifoaming agents.
4.
Branched chain silicone polymers are used in paints, varnishes and used as water repellants for treating masonry.
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CHAPTER (15) AT A GLANCE 1.
Group 14 comprises of elements like Carbon, Silicon, Germanium, Tin and Lead.
2.
Electronic configuration of the outermost shell of these elements is ns2np2.
3.
All elements exhibit +4 oxidation state. Besides +4 oxdn. state, +2 oxidation state is also exhibited by germanium, tin and lead.
4.
The stability of M2+ ion is more than M4+ ion. This is due to inert pair effect. The tendency of nonparticipation of the paired s-electrons in bond formation is known as Inert Pair effect.
5.
Carbon differs from other members of the group due to its small size, high ionisation energy and high electronegativity. Carbon has the tendency to combine with other carbon atom thereby forming a long chain. This self-linking property is known as catenation.
6.
CO2 is a gas whereas SiO2 is a solid. CCl4 can be easily hydrolysed.
7.
When different structural forms of the same element have same chemical properties but different physical properties, the phenomenon is known as Allotropy and the different forms are Allotropes. Diamond, graphite, charcoal, lamp black, coke etc. are the different allotropes of carbon. Buckyball or Fullerene (C 60) is a recently discovered crystalline allotrope of carbon.
8.
Diamond is the hardest naturally occurring allotrope of carbon. The carbon atoms are sp 3 hybridised, therefore joined tetrahedrally forming a giant hexagonal structure. C–C bond distance is 1.54 A0 and bond angle is 109028'. It is bad conductor of heat and electricity.
9.
Graphite (Black lead) is another crystalline allotrope of carbon. It forms a two dimensional sheet like polymeric structure. Carbon atom is sp 2 hybridised, arranged in a number of layers composed of hexagonal planar rings. C–C bond distance is 1.42 A 0. Different layers are held by weak van derWaal's force and are about 3.4 A 0 apart. One layer slides over the other. Graphite is quite soft and exhibits lubricating property. It is good conductor of heat and electricity.
10.
Structure of CO :
C+
.. :C: : O :
O
or
C=O
¬ C = O
C
O+
Resonating structures
11. 12.
13. 14. 15.
Solid CO2 is known as dry ice. It is used as a refrigerant. Besides ortho, cyclic and chain silicates there are sheet silicates and three dimensional silicates. Sheet silicates are two dimensional. Three dimensional silicates include quartz or rock crystal, zeolite and ultramarines. Quartz is a transparent colourless crystal, electrically neutral used for making optical and chemical apparatus. Sodium silicate dissolves in water and hence is called waterglass . Freons are mixed chlorofluro carbons, such as CFCl3, CF2Cl2, CF3Cl. and used as refrigeration fluids. Silicones are long chain organosilicon polymers. These are water repellants and are good insulators. These are used for water proofing and electrical condensers .
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QUESTIONS A. Very short answer type questions. (1 mark each) 1.
How many elements constitute 14 Group of periodic table ? Name them.
2.
What is the characteristic electronic contiguration 14 Group of periodic table ?
3.
Is diamond an element or compound ?
4.
What is dry ice ?
5.
What is carborundum ?
6.
Which non metal conducts electricity ?
7.
What are silanes ?
8.
Name the element of group 14 which is a metalloid.
9.
What is the arrangement of atoms in graphite ?
10.
Which allotrope of Carbon is used as a moderator in atomic reactors.
11.
What is the hybridisation of carbon in carbondioxide ?
12.
Whether carbon dioxide is a Lewis acid or base ?
13.
What is the most recently discovered allotrope of carbon ?
14.
What is the commercial name of SiC ?
B. Point out the true or false statements (1 mark each ) 1.
Flint is a natural variety of silica
2.
Silicon is the second most abundant element by weight in earth's crust
3.
Carbon dioxide is an amphoteric oxide
4.
Germanium metal finds extensive applications as semiconductor.
5.
Silanes are strong reducing agents
6.
Opal is an amorphous variety of silica
7.
Carbon dioxide is less acidic than SiO2
8.
Carbon monoxide is a neutral oxide
9.
Fluoro carbons, knows as Freon are used as refrigerant.
10.
Zeolites are cationic exchangers.
11.
Chief constituent of water glass is sodium silicate
12.
Chief constituent of water gas is sodium silicate
13.
A mixture of CO2 and 5% to 10% any gas is called carbogen used in artificial respiration.
14.
Silicon carbide is used in the construction of atomic reactors.
15.
Coke is an amorphous type of carbon.
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577
C. Very short answer objective type questions. ( 1 mark each) 1.
Diamond and graphites are (a) isotopes (b) allotropes (c) isomers (d) polymers
2.
Variety of carbon used as lubricant under trade name oil drag is (a) coke (b) oil carbon (c) graphite (d) diamond
3.
Lead pencil contains (a) lead (b) PbS (c) an alloy of lead (d) graphite
4.
Which of the following gases has the highest percentage in coal gas. (a) CO2 (b) CO (c) N2 (d) H2
5.
Glass is best described as (a) solid (b) a liquid (c) occluded solid (d) a super cooled liquid
6.
Solid carbon dioxide is called dry ice because (a) it does not wet hands easily (b) it looks like ice (c) it sublimes and releases no trace of water (d) all the above
7.
When SiCl4 is hydrolysed, we get (a) SiO2 (b) Si(OH)4 (C) H2SiO2 (d) H2SiO
8.
Water gas is a mixture of (a) CO and H2 (b) CO2 and H2O (C)C2H2 and H2O (d)CO and H2O
9.
Which allotrope of carbon is used as a moderator in atomic reactors ?
D. Short answer type questions. ( 2 marks each) 1.
Explain : Carbon shows tetravalency but does not show divalency.
2.
Describe the structure of diamond. Why diamond is an insulator ?
3.
What is catenation ?
4.
Define allotropy
5.
What is animal charcoal ?
6.
Write the symbols, atomic numbers and electronic configuration of group IV elements
7.
Explain why SiO2 is hard solid but CO2 is a gas.
8. 9.
Explain on the basis of structure the conducting and lubricating properties of graphite Why the melting point of diamond is very high ?
10.
What is sugar charcoal ?
11.
State Fajan's rule.
12.
Explain why diamond is an insulator and graphite is a conductor
13.
What is inert pair effect ?
14.
Explain why SnCl2 is a solid while SnCl4 is a liquid ?
15.
What happens when carbon dioxide passed through lime water ?
16.
Give the allotropic forms of carbon
17.
What happens when magnesium wire burns in carbon dioxide ?
18.
Justify the position of carbon and lead in the periodic table on the basis of electronic configuration
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+2 CHEMISTRY (VOL. - I)
19.
Why dryice does not wet the cloth in which it is placed.
20.
Explain why diamond is hard while graphite is soft.
21.
Mention the hybrid state of carbon in diamond and graphite.
22.
What is allotropy ?
23.
What is water gas ? How is it prepared ?
E.
Long type questions.
1. 2.
Give a comparative account of Group 14 elements (Carbon family) of the periodic table. What are silicones ? How silicones are prepared ? Mention the important uses of silicones.
ANSWERS A.
1. Five Carbon, Silicon, Germanium, Tin and Lead. 2. 3. 4. 5. 6. 7. 8.
ns2np2 Element Crystalline form of carbon Solidified carbon dioxide Silicon carbide graphite carbon Silicon hydrides Germanium.
9. Carbon atoms are arranged in a number of layers composed of hexagonal planar rings. 10. Graphite 11. sp2 12. Lewis acid 13. Fullerene 14. Carborundum B.
C.
1. True
2. True
3. False
4. True
5. True
7. False 13. True
8. True 14. True
9. True 15. True
10. True
11. True 12. True
1. b
2. c
3. d
4. d
6. d
7. b
8. a
5. d
6. True
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579
MULTIPLE CHOICE TYPE QUESTIONS 1.
The general electronic configuration of Group 14 elements is (a) ns2np2
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
(b) ns2
(c) ns2np6 (d) None Which of the following occurs in free state ? (a) C (b) Si (c) Ge (d) Sn Diamond and graphite are (a) isomers (b) allotropes (c) isotopes (d) polymers A gas which burns with blue flame is (a) CO (b) CO2 (c) O2 (d) N2 Poisonous gas present in the exhaust fumes of a car is (a) CH4 (b) C2H6 (c) CO2 (d) CO Sugar of lead is (a) PbSO4 (b) (CH3COO)2Pb (c) PbCl2 (d) PbCO3 Which of the following can attack glass ? (a) H2SO4 (b) HI (c) HCl (d) HF Silicon is an important constituent of (a) Chlorophyll (b) Haemoglobin (c) Rocks (d) Amalgams Which of the following compounds has peroxide linkage ? (a) Pb2O3 (b) CO2 (c) PbO2 (d) SiO2 Graphite is a good conductor because (a) It is crystalline
(b) It has free atoms
(c) It has free electrons
(d) It has sp2 hybridisation
Property of catenation is strongest in (a) C
(b) Si
(c) N
(d) O
Which of the following is an insulator ? (a) Diamond
(b) Graphite
(c) Aluminium
(d) Silicon
580
13.
14.
+2 CHEMISTRY (VOL. - I)
Which of the following is not a crystalline form of silica ? (a) Azurite
(b) Quartz
(c) Tridymite
(d) Crystobalite
The chemical name of Phosgene is (a) Phosphine
(b) Canbonyl chloride
(c) Phosphorus oxychloride (d) Phosphorus trichloride 15.
16.
17.
18.
19.
Pb shows oxidation states (a) + 3, + 4
(b) + 1, + 2
(c) + 2 , +4
(d) + 4
Artificial Carborundum is (a) Silicon carbide
(b) Boron nitride
(c) quartz
(d) Calcium carbide
Red lead is (a) PbO2
(b) PbO2+
(c) PbO2 2PbO
(d) None
The softest form of carbon is (a) Diamond
(b) Graphite
(c) Charcoal
(d) Lamp black
Which of the following is isolectronic with carbon atom ? (a) N2+
20.
21.
(b) Al3+
(c) O2— (d) N+ Which of the following is not used as a refrigerant (a) NH3 (b) CO2 (c) SO2 (d) CO Thermodynamically the most stable form of carbon is (a) diamond (b) graphite (c) fullerence (d) coal
ANSWERS TO MULTIPLE TYPE QUESTIONS 1. a 6. b 11. a 16. a 21. b
2. a 7. d 12. d 17. c
3. b 8. c 13. a 18. d
qqq
4. a 9. c 14. b 19. d
5. d 10. c 15. c 20. d
UNIT - XII
ORGANIC CHEMISTRY CHAPTER - 16
SOME BASIC PRINCIPLES AND TECHNIQUES 16.1
GENERAL INTRODUCTION
The term organic chemistry was coined in 18th century for those compound which were obtained from plants or animals i.e. from living organisms. It was Lavoisier who first proved that all the compounds of plant origin are composed of carbon, hydrogen and oxygen. So also the compounds of animal origin contain the same elements along with nitrogen or sulphur or phosphorus. Since it was assumed that organic compounds could be produced only by living organisms, Swedish chemist Berzelius (1779-1846) put fortth tital force theory with the assumption that a vital force (life force) was responsible for the production of organic compounds. Synthesis of these compounds in the laboratory was impossible due to the absence of this vital force. Vital force theory remained unchallenged till 1828 when a German chemist Friedrich Wohler reported that lead cyanate on treatment with ammonium hydroxide produced urea, a well known organic compound already isolated from human urine by Roulle in 1780. Pb (CNO)2 Lead cyanate
NH 4 OH ¾¾ ¾ ¾ ¾®
D ¾¾® NH4CNO Ammonium Cyanate
NH2CONH2 Urea
Synthesis of urea by Wohler followed by synthesis of acetic acid by Kolbe in 1845 and synthesis of methane by Berthelot by 1856 discarded the baseless Vital force theory. The synthesis of urea marked the beginning of a new era in organic chemistry with a new definition which is not restricted only to the carbon compounds of living system but also includes the synthetic organic compounds under one roof. Organic chemistry is defined as the study of carbon compounds where carbon is covalently bonded to carbon, hydrogen, oxygen, halogens, nitrogen and sulphur. Compounds of carbon containing only carbon and oxygen like carbon dioxide, carbonates, carbon monoxide etc. and compounds containing only carbon and nitrogen are not considered as organic compounds.
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Organic compounds made up of only carbon and hydrogen atoms are called hydrocarbons and all other organic compounds may be considered as the derivatives of hydrocarbons by the replacement of one or more of their hydrogen atoms by other atoms or groups. Thus organic chemistry may be defined more precisely as the chemistry of hydrocarbons and their derivatives.
16.2
BASIC CONCEPTS OF ORGANIC COMPOUNDS
Organic compounds are studied separately because of the following characteristic features of such compounds. 1.
Vast number :
The number of organic compounds is very large compared to inorganic compounds. At present, the number is more than million and it is steadily increasing since new compounds are being synthesised in different laboratories of the world everyday. The vastness of organic compounds is due to the special nature of the carbon atom, a property of the carbon atom to combine with itself through covalent bonds forming long chains or rings of different sizes and shapes. This self-linking property is known as catenation.
C C C C C C
2.
Complex nature :
Organic compounds are large and complex. For example, cane sugar has the molecular formula C12H22O11 and fat has the formula C57H110O6. Fats and proteins have high molecular masses ranging from several thousands to over a million. 3.
Non-ionic and reversible nature of organic reactions :
Organic reactions are mostly non-ionic in nature and involve cleavage and formation of covalent bonds. They are very slow and reversible. 4.
Isomerism :
Organic compounds having the same molecular formula but different physical and chemical properties are called isomers or isomerides while the phenomenon is known as isomerism. This is very common in organic compounds. For example, molecular formula C2H6O represents both ethyl alcohol (CH3CH2OH) and dimethyl ether (CH3-O-CH3) which have entirely different physical and chemical properties.
ORGANIC CHEMISTRY
5.
583
Homologous series : Although the number of organic compounds is very large, they can be classified into
different groups or families known as homologous series. Different members belonging to a homologous series (homologues) are characterised by the presence of a characteristic group, called functional group and can be represented by one general formula having similar methods of preparation and chemical properties. One member of a particular homologous series differs from the consecutive member by a –CH 2– group (molecular mass 14). 6.
Solubility : Organic compounds mostly dissolve in organic solvents like alcohol, ether, benzene
etc. in which inorganic compounds do not dissolve. 7.
Vast application : Organic compounds play a very vital role in our daily life. All our requirements of
life are mostly organic in nature. Our body is made up of tissues which are organic compounds. The following illustration gives an idea about the vast application of organic compounds. : Carbohydrates, fats, proteins, vitamins.
: Coal, wood, petroleum, kerosene, natural gas. : Cotton, wool, silk, nylon, terylene etc. anaesthetics,
: Quinine, penicillin, sulpha drugs, aspiri, antibiotics, hormones etc. : Detergents, disinfectants, soaps, plastics, rubber etc. : Gasoline, artificial rubber, plastic car body, antifreze
etc. : Fertilizers, Insecticides, Explosives, Rubber pipes, Paints, Transparent wrappers, polymers etc. : Dyes, perfumes, cosmetics etc. : Photographic films, developers, plastic decorative items etc.
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16.3 CHARACTERISTICS OF CARBON ATOM
The essential constituent of all organic compounds is carbon. The structure, formation and nature of the organic compounds can be very well understood by studying the characteristics of carbon atom. 1.
Tetracovalency of Carbon atom :
Carbon is placed in the Group-IV of the periodic table. Its high ionisation potential, moderate electronegativity and electron affinity do not permit it to lose or gain electrons. Since it has four electrons in the outermost shell, it can complete its octet by sharing four electrons with other atoms including other carbon atoms and exhibits tetracovalency. Le Bel and van't Hoff proved that in carbon atom the four valencies are directed towards the four corners of a regular tetrahedron making an angle of 109028' or 109.50 between any two bonds. All saturated compounds retain the tetrahedral configuration and hence stable. Thus, methane is represented as shown in Fig. 16.1 Fig 16.1 Tetrahedral carbon atom 2.
Equivalence of four valencies in Carbon atom :
In organic compounds of type CH4 (methane), all the four valencies of carbon are equivalent. Thus, if a hydrogen atom is replaced by a bromine atom, one at a time, then the same methyl bromide will be obtained.
ORGANIC CHEMISTRY
3.
585
Catenation :
Carbon has the unique tendency to combine with other carbon atoms repeatedly to form carbon chains or rings. This self-linking property is called catenation and this occurs due to the formation of covalent bonds between carbon atoms. Thus, carbon atoms can form linear chain, branched chain or cyclic structure as shown below.
C C C
C C C C
C
(Straight chain)
4.
(Branched chain)
(cyclic)
Type of bonds : Three types of bonds are formed in carbon compounds : (a) single (b) double (c)
triple (a)
Single bond :
H
H
X
X
·
·
HX· C ·
X
H
It is formed when two carbon atoms or a carbon atom and an atom of another element share a pair of electrons e.g. ethane or methyl chloride. The bond is a sigma ( s ) bond, which is quite strong and does not easily break during chemical reaction.
·
C·X H ·
or
H
H
H
C
C
X
H
H
CH3CH3 Ethane (Condensed formula)
H Ethane Ethane (electron-dot structure) (Graphic structure, straight line for single bond) (b) Double bond : It is formed when two pairs of electrons are shared by two carbon atoms or by a carbon atom and another atom, e.g. ethylene (ethene). Out of the two bonds of a double bond, one is a strong sigma ( s ) bond and the other is a weak pi (p) bond. The double bonds, also known as ethylenic bonds, though stronger than a single bond, can be ruptured easily during chemical reaction. The double bonds are very reactive and always tend to undergo addition reactions whereby the p bonds are destroyed. H
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+2 CHEMISTRY (VOL. - I)
H
H
X
X
·
C ·
H
H
C
C
H
H
· · ·
· ·
C
or
·
X
X
H
H
Ethylene or Ethene
or
Ethylene or Ethene
Ethylene or Ethene
Ethylene or ethene (c)
Triple bond :
·
·
CH2 = CH2
Ethane
It is formed by sharing of three pairs of electrons between two carbon atoms or between a carbon atom and another atom. e.g. as in acetylene (ethyne) molecule.
H X · H ·· ·· H · X H Acetylene
or
or Ethyne
H – C º C – H Acetylene Acetylene or Ethyne
or
CH º CH or Ethyne
A triple bond, also known as an acetylenic bond consists of one s bond and two p bonds. Triple bonds, though stronger than other two types of bonds are very reactive and are easily during chemical reaction. The triple bonds being very reactive undergo addition reactions easily.
Acetylene or Ethyne
Ethane
16.4 HYBRIDISATION AND TETRACOVALENCY OF CARBON The electronic configuration of carbon atom is 1s 2, 2s2, 2px1, 2py1. Since there are two half-filled p-orbitals carbon is expected to be divalent. But carbon is tetracovalent. To explain this, it is asumed that one of the 2s electrons is promoted to the vacant 2p z orbital. The
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electronic configuration of carbon in this excited state would be 1s 2, 2s1, 2px1, 2py1 2pz1. Now, the four half-filled orbitals would explain the tetracovalency of carbon. However, the energy of 2s orbital is less than 2p orbitals, so one of the bonds of carbon should be different from the other three. But it has been established that all the four bonds of methane are equivalent. To explain this discrepancy, the concept of hybridisation was introduced. According to this concept, the atomic orbitals of an atom having equal or nearly equal energies mix and redistribute their energies to form an equal number of equivalent or hybrid orbitals. Three different types of hybrid orbitals – sp 3, sp2 and sp may be formed by the mixing of s and p orbitals. The hybrid orbitals are similar to atomic orbitals except that (i) they do not overlap laterally (ii) they have directional character Because of the directional character of the hybrid orbitals, the compounds formed by the electrons of such orbitals have a definite geometrical shape. sp3-hybridisation :
In this case, one 2s orbital and three 2p orbitals mix to form four equivalent sp3 hybrid orbitals, each having 25% s character and 75% p character. They are directed towards the four corners of a regular tetrahedron, the angle being 109 028' between any two bonds. Each of these hybrid orbitals overlap linearly with the 1s orbitals of four hydrogen atoms to form four equivalent C-H s bonds in methane (Fig. 16.2). In saturated compounds carbon is always sp3 hybridised.
sp3 hybridisation
Methane molecule
Fig.16.2 sp3-hybridisation in methane molecule sp2-hybridisation : This type of hybridisation occurs in compunds containing double bonds. One 2s and two 2p orbitals mix to produce three equivalent sp 2 hybrid orbitals. sp2 hybrid orbitals are
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planar making an angle 1200 with each other (Fig 16.3) sp2 hybrid orbitals have 33.33% s character and 66.66% p character. The third 2p orbital lies perpendicular to the plane containing the hybridised orbitals.
1s Orbital
p Orbital x
p Orbital y
Three sp2 hybrid orbitals
Fig 16.3 : Formation of sp 2 hybridised orbitals
In ethylene (ethene) molecule the two sp 2 hybrid orbitals of each carbon atom overlap linearly forming a C-C s bond. The two other sp2 hybrid orbitals of each carbon atom overlap with 1s orbital of four hydrogen atoms to form four C–H s bonds. The unhybridised p orbitals overlap laterally, partly above and partly below the nuclear plane to form C–C p bond (Fig 16.4).
Fig 16.4 : Orbital structure of ethylene molecule
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Since lateral overlapping is less significant than linear overlapping, the p bond is weaker than the s bond and can be easily broken. sp-hybridisation : In compounds containing triple bonds, one 2s and one 2p orbital mix to form two equivalent sp hybrid orbitals. They are linear (i.e. they make 180 0 with each other (Fig 16.5) The sp hybrid orbitals possess 50% s and 50% p character. This hybridisation leaves two pure 2p orbital unaffected.
s-Orbital
p - Orbital
two sp hybrid orbitals
Fig 16.5 : Formation of sp hybrid orbitals
In acetylene (ethyne) molecule, sp hybrid orbital of each carbon atom overlap linearly to form C–C s bond. The other sp hybrid orbital of each carbon atom overlaps with 1s orbitals of two hydrogen atoms to form two C–H s bonds. The two pure p orbitals which lie perpendicular to the plane containing hybrid orbitals and with each other overlap laterally with similar p orbitals of the other carbon atom forming two C–C p bonds (Fig 16.6). Therefore, the triple bond between two carbon atoms consists of one s bond and two p bonds.
Fig 16.6 : Orbital structure of acetylene molecule
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The simplest way to write the structural formula of a compound is to represent each covalent bond between the combined atom by a line (—), Thus, methane can be represented H
as
H
C
But this does not give the idea about the spatial configuration. For
H
H
example, methane is tetrahedral and not planar. In case of larger molecules, since the detailed structural formulae, require a lot of space, condensed structural formulae are often written without showing their bonds (except double and triple bonds). Atoms of elements other than hydrogen are written after the hydrogen atom. For example.
H
H
H
H
H
C
C
C
C
H
H
H
H
H
or
CH3CH2CH2CH3 Condensed formula of n-butane.
Structural formula of n-butane It may be noted that the carbon is always tetravalent and in condensed formula when carbon atom is written near another carbon atom, it should be understood carbon is linked with carbon and hydrogen atoms are attached to the carbon atoms as shown above. Less often ethane is represented as H3CCH3, rather CH3CH3 is more commonly used. 16.5 METHODS OF PURIFICATION OF ORGANIC COMPOUNDS Organic compounds whether extracted from a natural source o prepared in the laboratory are never found free from impurities. Various methods used for the purification of organic compound depend on (i)
The nature of the organic compound and
(ii)
The nature of the impurity present in it.
The common techniques applied for purification are : A. CRYSTALLISATION : Crystallisation is the most commonly used method for purification of solid substances which is based on the difference in the solubilities of the compound and the impurities in a suitable solvent.
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The process of crystallisation involves the following steps : (a)
Choice of the solvent : Solvent must satisfy the following characteristics (i)
It does not dissolve the compound completely at room temperature, but appreciably soluble at higher temparature.
(ii)
It dissolves the impurities to a small extent.
(iii) It possesses a relatively low boiling point in order to effect its easy removal from the crystals of the pure compound. (iv) The solvent does not react chemically with the substance. (b)
Making the solution :
The crude sample is dissolved on heating in the minimum volume of suitable solvent to get nearly a saturated solution. A pinch of animal charcoal may be added to the solution during boiling to remove colouring materials by the process of adsorption. (c)
Hot filteration : The saturated solution is filtered hot to remove the insoluble impurites.
(d)
Cooling and separation of crystals :
The hot solution is then cooled in a container surrounded by ice cold water. The pure solid separates out as crystalline precipitate which is separated by filtration (Fig 16.7) (e)
Drying :
The precipitate is then dried in a steam oven or hot air oven at a suitable temparature or by means of a vacuum desiccator (Fig 16.8)
Filter paper
Fig. 16.7 Filtration under reduced pressure using Buchner funnel.
Fig. 16.8 Vacuum desiccator
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SUBLIMATION :
Sublimation may be defined as the process in which solid substances when heated, pass directly from the solid to the vapour state without melting and the vapour when cooled give back the solid substances. Solid
Heat Cool
Vapour
This process is very useful in separating volatile from non-volatile solid but is of limited application as only a few substances like naphthalene, camphor and benzoic acid can be purified by this process. The process of sublimation can be shown in Fig. 16.9 where the impure substance is heated in a dish covered with a perforated asbestos sheet on which an inverted funnel is placed. The funnel is kept cool by wrapping it with a wet filter paper or wet cloth. Vapours of the substance rise up and condense on the cooler walls of the funnel.
Fig. 16.9 Sublimation
C.
DISTILLATION :
The operation of distillation is employed for (i) the separation of volatile liquids from non-volatile impurties and (ii) the liquids having sufficient difference in their boiling points. Various methods employed for the purification of liquids depend on the nature of the liquid and the nature of impurities present. These are : (a)
Simple distillation :
Liquids which boil under ordinary pressure without decomposition and are associated with non-volatile impurities are generally purified by simple distillation. It is a process in which vaporization and condensation are going on side by side. The impure liquid is taken in a distillation flask fitted with a thermometer and a condenser and heated on a wire gauze. The liquid vaporizes and vapours are condensed as they pass through the air or water
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condenser. The pure liquid is collected in the receiver while the non-volatile impurity is left behind as shown in Fig. 16.10. Some glass beads are generally added to the distillation flask to avoid bumping.
Fig. 16.10 Simple distillation apparatus
(b)
Fractional distillation :
When the liquids present in the mixture have their boiling points close to each other, the separation is best effected by fitting the distillation flask with a fractionating column which in turn is connected to the condenser as shown in Fig. 16.11. Vapour of the liquid with higher boiling point condense before the vapour of the liquid with lower boiling point and consequently the vapours rising up in the factionating column will be richer in more volatile component. The role of fractionating column is to provide many Fig. 16.11 Fractional distillation surfaces for heat exchange between the ascending vapours and the descending condensed liquid. So the vapour of low boiling point on reaching the top become pure and pass through the condenser and the pure liquid is collected in a receiver. When series of successive distillations are carried out, the remaining liquid in the distillation flask get enriched in high boiling component. Each successive condensation and vaporization unit in the fractionating column is called a Theoretical Plate.
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The use of fractionating column has found a remarkable application in modern industry, especially in the distillation of petroleum, coal-tar and crude alcohol. (c)
Distillation under reduced pressure :
This method is applicable to purify liquids having very high boiling points and those, which decompose before their boiling point is reached. A liquid boils when its vapour pressure is equal to the atmospheric pressure. When the pressure is reduced by a water pump or a vacuum pump, the liquid boils at a lower temperature and distils over undecomposed. Distillation under reduced pressure is carried out in a specially designed apparatus (Fig. 16.12). The receiver is attached to a vacuum pump to lower the pressure registered by the manometer provided and the distillation is carried out as usual.
Fig. 16.12 Distrillation under reduced pressure
An important application of this process is the recovery of glycerol form spent-lye in soap industry. Glycerol decomposes at its boiling point (298 OC) but can be distilled unchanged at 12mm pressure when it boils at 180OC. Another application of vacuum distillation is the concentration of sugar juice under reduced pressure where it serves to economise fuel. (d)
Steam distillation :
Liquid practically immiscible with water, volatile in steam and possessing a fairly high vapour pressure are purified by steam distillation. Steam is bubbled through the impune liquid in a flask heated on a sand bath (Fig. 16.13). On vigorous boiling the vapours of the organic substance mixed with steam rise up and condense as they pass through the water condenser. Thus the condensate is a mixture of the organic substance and water. The two, being immiscible are separated in a separating funnel.
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Fig. 16.13 Steam distillation
Theory of steam distillation : Let p1 and p2 be the vapour pressures of water vapour and liquid at the distillation temparature. The liquid boils when their sum is equal to the atmospheric pressure P i.e. p1 + p2 = P or p2 = P – p1 So when a liquid boils in the presence of steam p 2, the partial pressure of organic liquid in less than the atmospheric pressure. That implies the substances boils at a temperature lower than its boiling points. The actual amount of the substance which distils over is given as w 1 p1 ´18 = w 2 p2 ´ M
where
w1= weight of water which distils over w2 = weight of compound which is carried over p1 = vapour pressure of water p2 = vapour pressure of the compound Molecular mass of water = 18 Molecular mass of the compound = M
Steam distillation is employed in industry for the recovery of various essential oils from plants and flowers. It is also used in the manufacture of aniline and turpentine oils. D.
EXTRACTION WITH A SOLVENT ;
The process of removing a substance from its aqueous solution by shaking with a suitable organic solvent is termed extraction. When an organic substance is present as solution in water, it can be recovered from the solution by the following steps :
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(i)
The aqueous solution is shaken well with an immiscible organic solvent in which solute is more soluble.
(ii)
The solvent layer is separated by means of a seprating funnel as shown in Fig.16.14
(iii) The organic substance is then recovered from it by distilling off the solvent. According to Partition Law, it is always better to extract two or three times with smaller quantities of the solvent than once with the bulk of the solvent provided.
Fig. 16.14 Extraction with a solvent
E.
CHROMATOGRAPHY :
Chromatography is an important modern techniques for the separation, isolation, purification and identification of the components present in a mixture. The name derived from Greek words Chroma meaning colour and Graphe meaning writing, suggests that this technique was originally confined to the separation of the coloured substances like plant pigments or dyestuffs, But now this technique is equally applied to colourless substances. (a) Principle : Chromatography is based on the principle of selective distribution of the components of a mixture between two phases, a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. When the stationary phase is solid, the basis is adsorption, and when it is a liquid, the basis is partition. (b) Types of Chromatography : Basing on the principle involved, chromatography can be classified into two different types. These are : 1. Adsorption chromatography : In this technique, the fixed phase is a solid such as alumina, silica gel, magnesium oxide where the components of a mixture are adsorbed to different degrees. When a mobile phase is allowed to more over the stationary phase (adsorbent), the components move with different mobility.
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Again it can be of two types. (i) Column chromatography : Column chromatography is named after the container of the stationary phase in which the adsorbents like alumina, silica get etc. are packed in a glass tube filted with a stopcock at its lower end. (Fig 16.15)
Fig. 16.15 Separation of a mixture by column chromatography (a) mixture added to top of the column; (b) appearance of bands; (c) collection of fraction.
A small amount of the concentrated solution of the sample is placed on the top of the column. A suitable solvent called eluant or a mixture of liquids is allowed to flow down the column slowly. As the components move down the column, their rates are determined by the distribution behaviour between the fixed adsobent and the moving liquid. The most readily adsorbed substances are retained near the top and others come town to various distances. As the process of elution continues, the zones of enhanced concentration gradually separate and become distinct. Now each zone contains only a single component. The process of separating the components of a mixture into zones or bands of pure substances, each located at a different place on the column is called the development of the chromatogram and the process of removing each component from column and collecting them one by one is called elution. (ii) Thin Layer Chromatography (TLC) : With thin layer chromatography, it is possible to separate even minute quantities of mixtures. A sheet of glass is coated with a thin layer of absorbent which acts as the fixed or stationary phase. After drying the plate, a drop of the solution of the mixture is placed just above one edge, which is then dipped in a pool of solvent. The solvent is drawn up the adsorbent layer by capillary action and consequently the components of the mixture move up along with the eluent to different distances depending on their degree of adsorption and separation taking place (Fig. 16.16).
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The relative adsorption of each component of the mixture is expressed in terms of its retardation factor i.e. Rf value given as Rf =
Distance moved from the base line by the substance Distance moved from the base line by the solvent front
Fig. 16.16 Thin Layer Chromatography
The spots of coloured compounds are visible on TLC plate due to their original colour. But the spots of colourless compounds can be detected by spraying the plate with developing or colouring agents like ninhydrin reagent as in case of aminoacids. 2.
Partition chromatography :
It operates by a mechanism analogous to counter-current distribution. The fixed phase may be a liquid strongly adsorbed on solid which acts as a support. In this case the solute gets distributed between the fixed liquid and the moving liquid (solvent). Paper chromatography is a special type of partition chromatography in which the water trapped in the cellulose molecules of the paper acts as the stationary phase. The paper chromatographic technique is partly partition and partly adsoption. A mixture of components to be separated is generally dissolved in water and is then placed on a strip of chromatographic paper. The second solvent (mobile phase) placed in the vessel is allowed to travel along the strip. The cellulose in the filter paper acts as a support for the stationary phase with water adsorbed on it. The mixture of the components is thus subjected to partition between the stationary phase and moving phase. Due to continuous and repeated partitioning, the rate of migration of each component in different causing the separation of the components (Fig. 16.17).
Fig. 16.17 Paper Chromatography
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Rf value which is characteristic of a particular component can be determined as : Rf =
Distance travelled by the solute from the base line Distance travelled by the solvent from the base line
16.6 QUALITATIVE ANALYSIS OF ORGANIC COMPOUNDS For the complete diagnosis and characterisation of an organic compound, the first and foremost step is to detect the elements present in it. The principal elements present in organic compounds are : carbon, hydrogen and oxygen. In addition to these, they may contain nitrogen., sulphur, halogens and phosphorus. (A) Detection of Carbon and Hydrogen : A mixture of organic compound and dry copper (II) oxide is heated in a dry test tube fitted with a delivery tube provided with a bulb in middle containing anhydrous CuSO 4. The other end of the delivery tube is dipped into lime water in another test tube.
Fig. 16.18 Detection of carbon and hydrogen
On heating the mixture, carbon and hydrogen present in the organic compound get oxidized to CO2 and H2O respectively. Δ C + 2CuO ¾¾® 2 Cu + CO2 Δ 2H + CuO ¾¾® Cu + H2O
(i)
Water formed converts anhydrous CuSO4 (white) to CuSO4. 5H2O (blue) indicating the presence of hydrogen in the organic compound. CuSO4 + 5H2O ® CuSO4 . 5H2O Anhydrous copper sulphate (white) Copper sulphate pentahydrate (blue)
(ii)
CO2 formed turns lime water milky indicating the presence of carbon in the organic compound. Ca(OH)2 + CO2 ® CaCO3 ¯ + H2O lime water
(B)
Detection of Oxygen :
Though there is no conclusive test for oxygen, its presence in organic compounds is indicated by indirect methods.
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(i)
(ii)
When the substance is heated alone in a dry test tube, in an atmosphere of nitrogen, formation of droplets of water on cooler parts of the tube shows the presence of oxygen. A negative result, however, does not necessarily show the absence of oxygen. O O If any of the various oxygen containing groups like – OH, – C –, – C – OH, – NO2 etc is detected, the presence of oxygen is confirmed.
(iii) The sure test for oxygen depends on the determination of the percentage of all other elements present in the compound. If the sum of these precentages falls short of hundred, the remainder gives the percentage of oxygen and thus confirms its presence. (C) Detection of Nitrogen : The presence of nitrogen in an organic compound is shown by the following tests: (i)
Ignition test : When heated strongly, smell of burning feathers indicates the presence of nitrogen. However, a negative result is not a proof of the absence of nitrogen.
(ii)
Soda-lime test : Liberation of ammonia gas on heating the given organic compound with soda-lime confirms the presence of nitrogent in it. CH3CONH2 + NaOH ¾CaO ¾¾® CH3COONa + NH3 Acetamide But nitrogenous compounds including nitro and diazo derivatives do not respond to this test, so a negative result is not a proof of the absence of nitrogen.
(iii) Lassaigne's test : This is a confirmatory test for the detection of nitrogen in all cases of nitrogenous compounds. It involves the following steps. (a)
A freshly cut small piece of sodium metal is melted in a small fusion tube and fused with the organic compound. The red hot tube is broken by immersing in distilled water in a china dish. The contents are boiled in water to extract the fused mass and filtered. This extract is known as sodium fusion extract. Nitrogen if present, combines with sodium and carbon of the organic compound to form sodium cyanide. Na + C + N ® NaCN From organic compound
(b)
The sodium fusion extract component is boiled with rerrous sulphate solution and cooled. FeSO4 + 2NaOH ® Fe (OH)2 + Na2SO4 from excess of sodium 6NaCN + Fe(OH)2 ® Na4 [Fe (CN)6] + 2NaOH Sodium ferrocyanide.
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601
To the cooled solution a little ferric chloride solution is added followed by excess of conc.hydrochloric acid. 3Na4 [Fe(CN)6] + 4FeCl3 ® Fe4[Fe(CN)6]3 + 12 NaCl Prussian Blue The formation of Prussian blue or green colouration confirms the presence of nitrogen. In case sulphur is present along with nitrogen in the organic compound, a blood red colouration may appear while performing the test for nitrogen. Na + C + N + S ® NaCNS Sodium sulphocyanide. 3NaCNS + FeCl3 ® Fe (CNS)3 + NaCl Ferric Sulphocyanide (Blood red)
D.
Detection of Halogens : (i)
Lassaigne's test : If halogens present in the organic compound, the sodium fusion extract contains sodium halides. Na + X ® NaX A little amount of sodium extract is acidified with dilute nitric acid and silver nitrate solution is added to it. A precipitate proves the presence of halogen. NaX + AgNO3 ® Ag X ¯ + NaNO3 White precipitate soluble in dil. NH4OH indicates chlorine. Yellowish white precipitate insoluble in dil. NH4OH, but soluble in conc. NH4OH indicates bromine. Yellow precipitate insoluble even in conc. NH 4OH indicates iodine. However, when nitrogen or sulphur is also present in the compound. NaCN and /or Na2S present in the extract interferes in the test of halogens. They can be removed by boiling with strong nitric acid to decompose the cyanide and sulphide. If not removed they will form a white and black precipitate respectively on the addition of silver nitrate. Na + C + N ® NaCN NaCN + AgNO3 ® AgCN ¯ + NaNO3 white ppt. 2Na + S ® Na2S Na2S + 2AgNO3 ® Ag2S ¯ + 2NaNO3 Black ppt. NaCN + HNO3 ® NaNO3 + HCN Na2S + 2HNO3 ® NaNO3 + H2S -
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(ii)
E.
Beilstein test : The copper wire flattened at one end is heated in an oxidizing Bunsen flame till it ceases to impart any green colour to the flame. Now a small amount of organic compound is taken on the copper wire and heated in the flame. A green flame, if imparted to the flame, is due to the formation of a volatile copper halide and proves the presence of a halogen. Though sensitive, this test is not always reliable for example, urea imparts green colour though it does not contain halogen.
Detection of Sulphur : Sulphur may be detected in an organic compound by the following tests : (i)
Lassaigne's test : If sulphur present then the sodium fusion extract contain Na 2S. 2Na + S ® Na2S Presence of sodium sulphide in the extract can be confirmed as follows :
(a)
A purple colour is obtained when a portion of the extract is treated with sodium nitroprusside solution. Na2S + Na2 [Fe(CN)5NO] ® Na4 [Fe(CN)5 NOS] Purple A black precipitate of lead sulphide is produced when lead acetate solution is added to the sodium fusion extract. Na2S + (CH3COO)2Pb ® PbS + 2CH3COONa Lead acetate Lead suphide Black ppt
(b)
(ii)
F.
Oxidation test : The organic compound is fused with a mixture of potassium nitrate and sodium carbonate. Sulphur if present is oxidised to sulphate. Na2CO3 + S + 3O ® Na2SO4 + CO2 The fused mass in extracted with water, acidfied with hydrochloric acid and then barium chloride solution is added to it. A white precipitate of BaSO 4 indicates the presence of sulphur. BaCl2 + Na2SO4 ® BaSO4 + 2Nacl Barium sulphate (white ppt)
Detection of Phosphorus :
The compound containing phosphorus when heated with an oxidising agent like hydrogen peroxide, phosphate is formed. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus. Na3PO4 + 3HNO3 ® H3PO4 + 3NaNO3 H3PO4 + 12 (NH4)2 MoO4 + 21HNO3 Ammonium molybdate ® (NH4)3PO4.12MoO3 Ammonium phosphomolybdate
+ 21 NH4NO3 + 12H2O
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16.7 QUANTITATIVE ANALYSIS After detection of elements, the next step in the molecular diagnosis of an organic compound is its quantitative analysis i.e. finding the prercentage composition of the elements by weight which can be done by the following methods. A.
Estimation of Carbon and Hydrogen (Liebig Method) : (a)
Principle : Both carbon and hydrogen are estimated together in one operation. When a known weight of the organic substance is burnt in excess of oxygen, carbon and hydrogen present in it are oxidised to carbon dioxide and water respectively. yö æ y CxHy + ç x + ÷ O2 ® xCo2 + HO 4ø 2 2 è
The weights of CO2 and H2O thus formed are determined and the amounts of carbon and hydrogen in the original substance calculated. (b)
Apparatus : The apparatus for the Liebig method consists of three units as shown in fig. 16.19 (i)
Oxygen supply
(ii)
Combustion tube
(iii) Absorption apparatus.
Fig.16.19 Apparatus for the esimation of C and H The products of combustion containing moisture and carbon dioxide are passed through the absorption apparatus which consists of weighed U-tube packed with anhydrous CaCl 2 or pumice soaked in conc. sulphuric acid to absorb water and another U-tube containing conc. solution of potassium hydroxide to absorb carbon dioxide. These tubes are connected in series. The U-tube and the potash bulb are then detached and the increase in weight of each of them determined. (c)
Calculation :
Let the weight of the organic compound = w gm Increase in weight of U-tube (H2O) = x gm Increase in weight of potash bulb (CO2) = y gm Percentage of carbon =
12y 100 ´ 44 w
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Percentage of hydrogen =
2x 100 ´ 18 w
Example : 0.2 gm of an organic compound produces 0.5764 gm of CO2 and 0.1512 gm of water on combution. Calculate the percentage of carbon and hydrogen in the compound. Solution : Weight of the organic compound = 0.2 gm Weight of CO2 produced = 0.5764 gm Weight of H2O produced = 0.1512 gm Percentage of carbon =
12 0.5764 ´ ´ 100 = 78.5 44 0 .2
Percentage of hydrogen = B.
2 0.1512 ´ ´ 100 = 8.4 18 0.2
Estimation of Nitrogen : The two main method for the estimation of nitrogen in an organic compound are :
(i)
Dumas Method : (a)
Principles : Duma's method is based on the fact that nitrogenous compounds when heated with copper oxide in an atmosphere of carbon dioxide yield free nitrogen. C + 2CuO ® CO2 + 2Cu 2H + CuO ® H2O + Cu 2N + CuO ® N2 + Oxides of nitrogen or
CxHyNz + (2x +
y
/2)
CuO ® xCO2 +
y
/2
H2O +
z
/2
N2 + (2x +
y
/2)
Cu
Traces of oxides of nitrogen forrmed, if any are reduced to nitrogen by passing the gaseous mixture over a heated copper spiral. Cu + Oxides of nitrogen ® CuO + Nz The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the uppar part of the graduated tube (Fig. 16.20)
Fig. 16.20 Dumas method for estimation of nitrogen
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605
Calculation :
Let the weight of the organic compound = W g Volume of nitrogen collected = VIml Room temparature = TIK
V2 = volume of nitrogen at STP =
P1V1 273 ´ T1 760
by applying the ided gas equation
P1V1 P2 V2 = T2 T1
Here P1 = Atomospheric pressure – aqueous tension = (P–p) mm of Hg. Thus V2 =
(P - p)V1 273 ´ = Vml (say) T1 760
How 22,400 ml of N2 at STP weighs 28 g Vml of at STP weighs =
28 × V g 22400
Percentage of nitrogen = Example : 0.2 gm of an organic compound yielded 20.7 ml of nitrogen at 15 OC and 758 mm of Hg by Duma's method. Calculate the percentage of nitrogen in the compound. (Aqueous tension at 15OC = 12mm) Solution : Volume of N2 at STP = =
( P - p) V1 273 ´ T1 760
(758 - 12) 20.7 273 ´ = 19.2604 ml 288 760
Weight of ml of Nitrogen =
Percentage of Nitrogen = (ii)
28
28 ´ 22400 19.2604 g
19.2604 22400
100 =12.04 0.2
Kjeldahl's Method : (a) Principle : Kjeldahl's method is based on the fact that when an organic compound containing nitrogen is heated with concentrated sulphuric acid, the nitrogen present in it is quantitavely converted into ammonium sulphate. D CO + H O + (NH ) SO C + H + N + H2SO4 ¾¾® 2 2 4 2 4 from organic (Conc.) compound
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The resultant liquid is heated with concentrated alkali and the liberated ammonia gas is passed through a known excess of a standard acid solution. The volume of the unreacted acid is determined by titration with a standard alkali solution. The amount of ammonia and of nitrogen is determined by knowing the amount of acid reacted with ammonia. (NH4)2 SO4 + 2NaOH ® Na2SO4 + 2H2O + 2NH3(b)
Procedure : The experiment is carried out in three steps. (i)
Formation of ammonium sulphate in Kjeldahl's flask (Fig. 16.21)
(ii)
Distillation with alkali to get ammonia (Fig. 16.22)
(iii) Titration of excess acid.
Fig. 16.21 Kjeldahl's flask
(c)
Fig. 16.22 Kjeldahl's method for estimation of nitrogen
Calculation : Let the organic comound = Wg Volume of H2SO4 of molarity M taken = Vml Volume of NaOH of molarity M used for titration of excess of H 2SO4 = V1ml V1ml of NaOH of molarity M =
V1 ml of H2SO4 of molarity M 2
V1 ö æ Volume of H2SO4 of molarity M unused = ç V - ÷ ml 2 ø è V ö æ ç V - 1 ÷ ml of H2SO4 of molarity M 2 ø è
V1 ö æ = 2 ç V - ÷ ml of NH3 solution of molarity M 2ø è
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1000 ml of 1M NH3 solution contains 17g of NH3 or 14 g of N. V1 ö æ Hence, 2 ç V - ÷ ml of NH3 solution of molarity M contains 2ø è
14 × M × 2
(V
V1 2
)
g of Nitrogen
1000
14 M 2 V Percentage of Nitrogen
=
= (d)
V1 2
100 M
1000
(
1.4 M 2 V
V1 2
)
M
Limitation : This method is not applicable to organic compound containing nitro and azo groups and also to pyridine where nitrogen atom is present in the ring. Problem : 0.4g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralise 20 ml of 0.1 M H 2SO4. Calculate the percentage of nitrogen in the compound. Solution : Weight of the organic compound = 0.4g Volume of 0.1 M H2SO4 required for neutralisation of ammonia = 20 ml 20 ml of 0.1 M H2SO4 = 2 × 20 ml of 0.1 M NH3 = 40 ml of 0.1 M NH3 = 4 ml of 1 M NH3 Now 1000 ml 1M NH3 = 17g NH3 = 14g of Nitrogen 4 ml 1M NH3 =
14 × 4g = 0.056g of Nitrogen. 1000
Percentage of nitrogen = C.
0.056 × 100 = 14.0 0 .4
Estimation of Halogens by Carius method : (a)
Principle : Carius method for estimation of halogens is dependent on the fact that when a known weight of the organic compound containing halogens is heated in the presence of silver nitrate in a hard glass tube known as Carius tube (Fig. 16.23) in a furnace, silver halide is formed. It is filtered, washed, dried and weighed. From the amount of silver halide produced, the percentage of halogen is calculated. D CO2 + H2O + AgX C + H + X + fuming HNO3 + AgNO3 ¾¾®
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Organic compound
Fig. 16.23 Heating the bomb tube in a furnace
(b)
Calculation : Let the weight of the organic compound = wg Weight of silver halide formed = xg Since 1 molecule of Agx º 1 atom of x (where X = Cl, Br or I) Percentage of halogen =
Atomic weight of X x ´ ´ 100 Molecular weight of AgX w
Problem : 0.1890 g of an organic compound gave 0.2870g of silver chloride in Carius method of estimation of halogen. Find out the percentage of chlorine in the compound. Solution : Molecular mass of AgCl = 108 + 35.5 = 143.5 g mol–1 143.5 g of AgCl contains 35.5 g chlorine. 0.2870 g of AgCl contains
35.5 × 0.2870 g of chlorine. 143.5
Weight of the organic compound = 0.1890 g \ Percentage of chlorine D.
35.5 ´ 0.2870 ´ 100 = 37.57 143.5 ´ 0.1890
Estimation of Sulphur : (a)
Principle : A known weight of the organic compound is heated with sodium peroxide or fuming nitric acid in the Carius tube at 300 OC for six hours. Sulphur present is oxidised to sulphuric acid which is precipitated as barium sulphate by the addition of excess of barium chloride. Barium sulphate so precipitated is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the weight of barium sulphate. S + (from organic compound)
6HNO3 ® H2SO4 + 6NO2 + 2H2O
H2SO4 + BaCl2 ® BaSO4¯ + 2HCl (white ppt.)
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609
Calculation :
Let the weight of the organic compound = wg Wt. of BaSO4 formed = xg 1 mol of BaSO4 = 233g BaSO4 = 32 g sulphur xg BaSO4 contains
Hence,
32 ´ x g of sulphur 233
Percentage of sulphur =
32 ´ x ´ 100 233 ´ w
Problem : 0.395 g of an organic compound by Carius method for the estimation of sulphur gave 0.582g of BaSO4. Find the precentage of sulphur in the substance. Solution : Weight of the organic compound = 0.395g Weight of BaSO4 formed = 0.582g 233g of BaSO4 contains 32g of sulphur \ 0.582g of BaSO4 contains
32 × 0.582g of sulphur 233
Hence, percentage of sulphur in the organic compound = E.
32 100 × 0.582 × = 20.24 233 0.395
Estimation of Phosphorus : (a)
Principle : A known weight of an organic compound is heated with fuming nitric acid in a Carius tube where upon phosphorus present in the compound is oxidised to phosphoric acid (H3PO4). The phosphoric acid is precipitated either as (i) ammonium phosphomolybdate, (NH4)3 PO4.12 MoO3 by the addition of ammonia and ammonium molybdate solution or (ii) as magnesium ammonium phosphate by adding magnesia mixture (MgCl2 + NH4Cl + NH4OH) which on ignition yeilds magnesium pyrophosphate. 2MgNH4PO4 ® Magnesium ammonium phosphate
Mg2P2O7 + 2NH3 + H2O Magnesium pyrophosphate
Precipitate of ammonium phosphomolybdate or magnesium pyrophosphate is separated, dried and weighed from which the percentage of phosphorus can be calculated. (b)
Calculation : Let the weight of the organic compound = wg (i)
weight of ammonium phophomolybdate = xg molecular wt. of (NH4)3 PO4.12 MoO3 = 1877 g Hence, percentage of phosphorus =
31 ´ x ´ 100 1877 ´ w
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(ii)
weight of magnesium pyrophosphate = yg molecular weight of Mg2P2O7 = 222 Hence, percentage of phosphorous =
62 ´ y ´ 100 222 ´ w
Problem : The phosphoric acid formed by heating 0.31g of an organic compound with fuming nitric acid in carius tube was precipitated as MgNH 4 PO4 which on ignition left 0.148g of Mg2P2O7. find out the percentage of phosphorus in the organic compound. Solution : wt. of the organic compound = 0.31g wt. of Mg2P2O7 = 0.148g 222g of Mg2P2O7 contain 62g of phosphorus. 0.148 g of Mg2P2O7 contain phosphorus
62 × 0.148g 222
Hence, percentage of phosphorus is the organic compound = F.
62 × 0.148 × 100 = 13.33 222 × 0.31
Estimation of Oxygen :
Though the percentage of oxygen in an organic compound is obtained by substracting the sum of the percentages of other elements from 100, it can be estimated by a direct method. (a)
Principle : A known weight of the organic compound is pyrolysed in a stream of nitrogen gas. The mixture of gaseous products along with oxygen is passed over red hot coke (1373K) when all the oxygen is converted to carbon monoxide. This when passed through warm (448K) iodine pentoxide (I2O5) is oxidised to carbon dioxide producing iodine (Fig. 16.24) D Organic compound ¾¾® O2 + Gaseous products K ® 2CO(g) 2C + O2(g) ¾1373 ¾ ¾¾
5CO + I2O5 448K
5CO2 + I2
By determining the amount of CO2 (or I2) produced, the percentage of oxygen in the original organic compound can be calculated, which can be done by passing the mixture of CO2 and I2 over a bed of potassium iodide solution, thus removing iodine. The residual gases containing CO2 are finally passed through a sodaasbestos U-tube where CO2 is retained. This U-tube is weighed before and after pyrolysis, the difference of which gives the amount of CO 2 obtained from the given sample of the organic compound (Fig. 16.25)
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Fig. 16.24 Pyrolysis unit of apparatus for the estimation of oxygen.
Fig. 16.25 Absorption Unit (b)
Calculation :
Let the weight of the organic compound = wg The weight of CO2 produced = xg 44g of CO2 contain 32g of oxygen. xg of CO2 contain oxygen =
32 × x g 44
Hence, percentage of oxygen = G.
32 100 × x × 44 w
CHN Analyser :
Recently it has become possible to carry out the estimation of elements like carbon, hydrogen and nitorgen present in an organic compound by using only micro quantities (1– 3 mg) of the compound and automatic experimental techniques like CHN Analyser. This instrument displays the results on a screen within a short time. 16..8 CLASSIFICATION OF ORGANIC COMPOUNDS Classification of organic compounds has been made essentially to systematise and simplify the study of organic compounds. Two main systems of classification are
A.
A.
Classification based on carbon skeleton.
B.
Classification based on functional group.
Classification based on Carbon skeleton : Mainly there are two important classes of organic compounds. 1.
Acyclic, open-chain or aliphatic compounds : Compounds which have an open chain of carbon atoms (branched or straight chained) are called acyclic, open chain or aliphatic compounds. Examples are
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OH CH3
CH3CH2CH3
(Propane) 2.
CH
CH3
(Isopropyl alcohol)
CH3CH2COOH
(Propionic acid)
Cyclic or closed chain compounds : Compounds with a closed chain of atoms (ring structure) are termed as cyclic or closed chain compounds. Examples are cyclohexane, pyridine, benzene etc.
(Cyclohexane)
(Pyridine)
(Benzene)
There are two types of cyclic compounds. (a)
Homocyclic or Carbocyclic compounds : They have one or more rings composed of carbon atoms only. They may be alicyclic or aromatic. (i)
Alicyclic compounds : Compounds like cyclopropane, cyclobutane, cyclohexane though cyclic resemble aliphatic compounds. They are called alicyclic compounds. CH 2
CH2
H2C
CH2 (Cyclopropane)
(ii)
CH2
CH2
CH2
CH2
(Cyclobutane)
CH2
CH2
CH 2
CH2
CH2 (Cyclohexane)
Aromatic compounds : Aromatic compounds can be classified into two categories. These compounds contain one or more benzene rings in their structure. For example,
ORGANIC CHEMISTRY
613
(a)
Benzenoid compounds
(b)
Non-benzenoid compounds : These are aromatic compounds having no benzene ring. Example- Tropolone Tropolone
(b)
Heterocyclic compounds : Heterocyclic compounds contain one or more atoms or elements other than carbon and hydrogen alongwith carbon atoms in the ring. The atoms of elements other than carbon are termed as hetero atoms which may be N,O,S etc. Example are furan, pyrrole, thiophene, pyridine etc.
(Furan)
(Pyrrole)
(Thiophene)
The scheme of classification may be represented as follows :
(Pyridine)
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+2 CHEMISTRY (VOL. - I)
Classification based on Functional Group :
The skeletal classification of organic compounds is very wide and therefore needs further systematisation for study. This is achieved by dividing the above classes of compounds on the basis of functional group or groups present. Functional group : The simplest compounds are hydrocarbons consisting of carbon and hydrogen only. All other compounds are treated as the derivatives of the hydrocarbons and formed by replacing one or more hydrogen atoms with other atoms or groups.
R — H (Hydrocarbon)
-H
¾¾ ¾® +X
R — X (Organic compound)
where R is the hydrocarbon residue and X is a group. Thus, an organic compound can generally be splitted up into two or more groups or radicals. For example, methyl alcohol can be splitted up into methyl (CH3–) and hydroxyl (–OH) group. A group contains heteroatoms like O,N,S, halogens etc. The groups determine the properties of organic compounds to a great extent. The reactive groups which determine the chemical characteristics of the organic compounds are termed as functional or characteristic groups. In fact, all the alcohols are characterised by the pressence of the hydroxyl group (–OH) in their molecules. Hence, hydroxyl group (–OH) is functional group. Similarly, all the carboxylic acids are characterised by the presence of the carboxyl group (–COOH) and therefore carboxyl group (–COOH) is a functional group of a class of compounds called carboxylic acids. In unsaturated compounds, the double bond or triple bond is the reaction site and hence they are considered as functional group(s). Only saturated hydrocarbons do not contain a functional group. The classification of organic compounds on the basis of functional groups present makes the study simple and systematic. Some important functional groups and the class alongwith common examples are listed in Table - 16.1
ORGANIC CHEMISTRY
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TABLE - 16.1 Class of
Functional group
compound
symbol and name
1.
Alkanes
2.
Alkenes
3. 4.
Alkynes Alkyl halides
Nil C = C (double bond)
–C º C – (triple bond) –X (F,Cl,Br,I) (halo)
Common examples
Methane(CH4),Ethane (CH3CH3) Propane (CH3CH2CH3) Ethylene (CH2=CH2) Propylene (CH3–CH=CH2) Acetylene (CH º CH) Methyl chloride (CH3Cl), Ethyl bromide (CH3CH2Br), Isopropyl
(CH
5.
Alcohols or
6.
Alkanols Ethers
– O – R (Alkoxy)
7.
Aldehydes
–CHO (formyl
or 8.
or
or
Alkanals
O – C aldehyde) H
Ketones
=CO (carbonyl
or
or
Alkanones 9.
–OH (hydroxy)
Carboxylic acids or Alkanoic acids
10. Esters
12. Acid amides
Ethyl alcohol (CH3CH2OH) Dimethyl ether (CH3 – O – CH3) Diethyl ether(CH3CH2-O-CH2CH3) Formaldehyde (H–CHO) Acetaldehyde (CH3–CHO) Acetone (CH3COCH3)
or
C = O oxo) –COOH
Formic acid (H–COOH)
or (carboxyl) –C –C
O OH
Acetic acid (CH3–COOH)
O Ester or OR
Ethyl formate (H – COOC2H5)
alkoxycarbonyl)
11. Acid chlorides
)
iodide CH 3 CH I 3 Methyl alcohol (CH3OH)
O (chloroformyl) Cl O (amide) –C NH2 –C
Methyl acetate (CH3–COOCH3) Acetyl chloride (CH3COCl) Acetamide (CH3CONH2)
13. Nitriles
–CºN
(cyano)
Acetonitrile (CH3CN)
14. Amines
–NH2
(amino)
Methylamine (CH3NH2)
15. Nitroalkanes
–NO2
(nitro)
Nitromethane (CH3–NO2)
(benzene)
Chlorobenzene (Cl
16. Arenes
O
Toluene (CH3
O )
O )
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HOMOLOGOUS SERIES A class of compounds containing the same functional group when arranged in the order of increasing molecular mass, each compound differing from the adjacent member by a –CH2 group forms a homologous series and the phenomenon is called homology. Each member of a homologous series is the homologue of its immediate neighbour. A homologous series has the following characteristics. 1.
All members of a homologous series can be represented by a general formula R–X where R is the hydrocarbon residue and X is the functional group. The homologous series of alcohols can be represented by the general formula CnH2n+1OH and the formula of various homologues can be obtained by putting 1,2,3....etc. for n.
2.
All members of a series can be prepared by the same general method.
3.
The physical properties of the members of a series show a regular gradation with increase in molecular mass though the chemical properties of all the members are similar.
A few important homologous series are mentioned in Table 16.2 TABLE – 16.2 Some Important Homologous Series Name of the series
General
Examples
Formula
1.
Alkane
CnH2n+2
CH4,C2H6, C3H8 C4H10....
2.
Alkene
CnH2n
C2H4, C3H6, C4H8......
3.
Alkyne
CnH2n–2
C2H2, C3H4, C4H6......
4.
Arene
CnH2n–6
C6H6, C7H8, C8H10......
5.
Alcohol
CnH2n+1OH
CH3OH, C2H5OH, C3H7OH.....
6.
Aldehyde
CnH2n+1CHO
HCHO, CH3CHO, C2H5CHO......
7.
Ketone
RCOR'
CH3COCH3, CH3CH2COCH3......
8.
Carboxylic acid
CnH2n+1COOH
HCOOH, CH3COOH, C2H5COOH......
9.
Alkyl halide
CnH2n+1X
CH3Cl, C2H5Cl, C3H7Cl....
10.
Amine
CnH2n+1NH2
CH3NH2, C2H5NH2, C3H7NH2....
11.
Ester
CnH2n+1COOR
HCOOCH3 ,CH3COOCH3,C2H5COOCH3.....
12.
Ether
R – O – R'
CH3-O-CH3, CH3OCH2CH3, CH3CH2OCH2CH3...
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16.9 NOMENCLATURE OF ORGANIC COMPOUNDS Early system of nomenclature of organic compounds was based on the sources from which they were obtained. For example, methyl alcohol was named after its source (Greek word : methu means wine and hule means wood), since it was obtained by destructive distillation of wood. Thus, the names citric acid (from citrus fruits) formic acid (Latin : formica means red ant), uric acid (from urine), acetic acid (Latin : acetum means vinegar) etc. have been derived from their sources. These are known as common or trivial names. Due to the unique property of catenation carbon forms a large number of compounds. Problems were raised in naming them by this system. The systematic nomenclature was first developed in an international conference of chemists in 1928 in Geneva. This system was named as Geneva System which was subsequently modified in 1930 in Belgium at the International. Union of Chemistry (IUC). The most systematic nomenclature was developed by the International Union of Pure and Applied Chemistry in 1957 referred to as IUPAC system of nomenclature. This system is modified from time to time. The latest rules are mentioned here following "A guide to IUPAC nomenclature of organic compounds" (1994 edition) by R. Panico, W.H. Powell and Jean Claude Richer (Sr.editor), Blackwell Scientific Publications, Oxford. The IUPAC system provides a set of rules for naming the complex organic compounds. Some of the organic compounds have still retained their common names because of familarity and long usage. A. GENERAL RULES FOR IUPAC NOMENCLATURE
According to the latest rules, the name of an organic compound is derived from the name of the parent hydrocarbon by using suitable prefixes, infixes and suffixes. The name of the organic compound consists of three parts. Prefix–Word root – Suffix Word root is the basic unit of the name. The total number of carbon atoms in the principal chain usually dermines the word root. They are fixed as follows : Word root
Chain length (system)
Meth
–
C1 (for one carbon system)
Eth
–
C2 (for two carbon system)
Prop
–
C3 (for three carbon system)
But
–
C4 (for four carbon system)
Pent
–
C5 (for five carbon system)
Hex
–
C6 (for six carbon system)
Hept
–
C7 (for seven carbon system)
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Oct
–
C8 (for eight carbon system)
Non
–
C9 (for nine carbon system)
Dec
–
C10 (for ten carbon system) and so on
The suffix refers to the functional group of a particular class of compound. Suffix may be of two types. (i) Primary suffix :This suggests the nature of linkage between the carbon atom. Linkage
Suffix
Carbon-Carbon single bond
C
Carbon-Carbon double bond
C
C
– – ene
Carbon-Carbon triple bond
C
ºC
– – yne
C
– – ane
(ii) Secondary suffix : This suggests the characteristic functional groups present in the compound and is added to the name after the primary suffix. Some of the secondary suffixes are : Functional group
Suffix
Functional group
Suffix
Alcohol (–OH)
–ol
Amide (–CONH2)
–amide
Aldehyde (–CHO)
–al
Acyl group (R–CO–)
–oyl
Ketone ( C
–one
Cyanide (–C º N)
–nitrile
Carboxylic acid (–COOH)
–oic acid
Amine (–NH2)
–amine
Ester (–COOR)
R.....–oate
Sulphonic acid (–SO3H) –suphonic acid
Acid halide (–COX)
–oyl chloride
O)
Note : When using secondary suffixes, the carbon of the functional group (if any) is normally included in the principal chain. Prefixes indicate the substitution of other atom or group in place of hydrogen atom(s) in the main chain of carbon atoms under consideration. They are not regarded as principal functional groups. Commonly used prefixes are :
ORGANIC CHEMISTRY
619
Substitutent
Prefix
Substituent
Prefix
CH3–
Methyl
Br –
Bromo
CH3CH2–
Ethyl
I–
Iodo
CH3CH2CH2–
Propyl
– NO2
Nitro
CH3CH2CH2CH2 –
Butyl
– OR
Alkoxy
F –
Fluoro
+ -N º N
Diazo
Cl –
Chloro
– N = O
Nitroso
B. COMMON CONVENTIONS IN IUPAC SYSTEM 1. Position of locants : A locant is a number (1, 2, 3...) or a letter (o –, m –, p – etc.) used to locate the position of the substituent or a carbon-carbon multiple bond. These locants are placed immediately before the part of the name to which they relate. For example : CH3
(i)
CH3– CH2– CH– CH2– CH3
3 – Methylpentane (ii)
CH3–CH2–CH = CH2 But–1–ene
(According to the latest rule, names like 1- Butene or Butene - 1 are not correct)
2. Numerical prefixes : Numerical prefixes are used to describe a multiplicity of identical features of a structure in nomenclature. Some common numerical prefixes are : Number
Numerical term
Number
Numerical term
1
mono–
6
hexa –
2
di –
7
hepta –
3.
tri –
8
octa–
4.
tetra –
9
nona –
5.
penta–
10
deca –
For example :(i) CCl4
(ii)
CH2 – OH
Tetrachloromethane CH2 – OH Ethane–1,2–diol
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In example (i) above, the prefix 'tetra' before chloro indicates the presence of four chlorine atoms in the molecule. Similarly, in the example (ii) above, the prefix 'di' before the suffix 'ol' indicates the presence of two –OH groups in the molecule. When the substituent itself is substituted (substituted substituent), the multiplicity is indicated by numerical prefixes such as 'bis' (2), 'tris' (3), tetrakis (4) and so on. CH2 – CH2 – Br CH3 – CH2 – C – CH2 – CH2 – CH3 CH2 – CH2 – Br
3,3 – Bis (2–bromoethyl) hexane no hyphen Here, 2 bromoethyl is a substituted substituent because the ethyl group is substituted at the second carbon atom and two such groups are present at position 3. Hence, numerical prefix used is 'bis' and not 'di' as in simple substituents. 3. Brackets or Parentheses : Brackets or parentheses are used around prefixes defining substituted substituents after the numerical prefix if any. For example : 3,3 – Bis (2–bromoethyl)hexane. 4. Use of hyphen : A hyphen separates a locant from the syllable of the name. CH3 CH3 – CH – CH3 OH
(2–Methylpropane)
OH OH
CH2 – CH – CH2
(Propane – 1,2,3 – triol)
When a substituent is put in the bracket, a hyphen is given after the end of the bracket only if the final bracket is followed by a locant. CH3 CH2 – CH2 – Br 1CH 3
– 2CH – 3C – 4CH2 – 5CH2 – 6CH3 CH2 – CH2 – Br
3,3 - Bis (2- bromoethyl) - 2 - methylhexane hyphen
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CH2 – CH2 – Br CH3 – CH2 – C – CH2 – CH2 – CH3 CH2 – CH2 – Br
3,3 - Bis (2-bromoethyl) hexane
no hyphen
5. Omission of 'e' : The terminal 'e' in the primary suffix is omitted when followed by a second suffix beginning with 'a', 'i', 'o', 'u', or 'y' OH 1CH – 2CH– 3CH – 4CH 3 2 3
Butan – 2 – ol
The terminal 'e' of the primary suffix 'ane' is omitted because it is followed by 'o' of the secondary suffix – 'ol'. OH
OH
1CH – 2CH– 3CH– 4CH 3 2
Butane – 2,3 – diol.
Here, 'e' of the primary suffix 'ane' is not omitted because it is followed by 'd' of secondary suffix 'diol'. 6. Alphabetical order of the prefixes : When two or more prefixes are attached to the principal chain, then the prefixes are arranged in the alphabetical order. While arranging prefixes in alphabetical order, di–, tri, tetra – etc (if any) are not considered except for substituted substituents.
Br
Cl 1
CH2 – 2 CH– 3 CH2 – 4 CH2 – 5 CH3
2–Bromo – 1 – chloropentane.
The prefix 'bromo' is placed before 'chloro' because 'b' comes before 'c' in alphabetical order. CH3 I 1CH
3
– 2C – 3CH – 4CH2 – 5CH3 CH3
3 – Iodo – 2, 2 – dimethylpentane.
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Since 'methyl' is a simple substituent, di– is not considered while arranging the prefixes in alphabetical order. 'I' comes before 'm' and therefore 'Iodo' comes first. In case of substituted substituents, the numerical prefixes 'di', 'tri' etc. are considered while arranging the prefixes. The following example will clarify the rule. Br CH3 1CH
3
1CH
– 2CH2 – Br
– 2CH2 – 3CH – 4CH – 5CH2 – 6CH2 – 7CH3
4 – (1,2 – Dibromoethyl) – 3 – methylheptane.
Here, '1,2 – dibromethyl' is a substituted substituent. Hence, numerical prefix 'di' shall be considered while arranging prefixes. Obviously, 'd' of dibromoethyl comes before 'm' of methyl is alphabetical order and written accordingly. When two or more prefixes consist of identical words priority for citation is given to that group which contains the lowest locant at the first point of difference. If the locants at the first point are same, the second point of difference is considered and so on and the lowest locant is preferred. This rule can be explained by the following examples : Cl Cl – 2CH2 – 1 CH2 8
1
CH – 2 CH3
CH3 – 7CH2 – 6CH2 – 5 CH – 4CH – 3 CH2 – 2CH2 – 1CH3
4 – (1–Chloroethyl) – 5 – (2–chloroethyl)octane
Here, the two prefixes have identical words 'chloroethyl'. Priority for citation is thus given to that group which contains the lowest locant at the first point of difference. Therefore, 1– chloroethyl is written first. Br – 1CH – 2CHBr2 7CH 2
– 6CH2 – 5CH2 – 4CH – 3CH – 2CH2 – 1CH3 1CBr 2
– 2CH2Br
3 – (1,1,2–Tribromoethyl)–4 –(1, 2, 2 –tribromoethyl)heptane Here, the prefixes have identical words and the locants at the first place in both the prefixes are same. Therefore, the locants at the second place is considered and (1,1,2) set being lower than (1,2,2) set is given priority.
ORGANIC CHEMISTRY
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7. The name of an organic compound should start with a capital letter. There are certain exceptions which are given below. CH 3 CH 3 – C – CH 3 1 CH
3–
2 CH
2–
3 CH
2–
4 CH
– 5 CH 2 – 6 CH 2 – 7 CH 2 – 8 CH 3
4 – tert – Butyloctane
Cl
Cl
Cl
Cl
Cl
Cl o – Dichlorobenzene
8.
m – Dichlorobenzene
p – Dichlorobenzene
Lowest set of locant rule : (a)
If there is only one substituent, the carbon atoms of the principal chain is numbered from such an end so as to give the lowest number to the carbon atom carrying the substituent.
CH 3 5 CH
3–
4 CH
2–
3 CH
2–
2 CH
2–Methylpentane (Correct)
– 1 CH 3
CH 3 1 CH
3–
2 CH
2–
3 CH
2–
4 CH
– 5 CH 3
4–Methylpentane (Incorrect)
In the above example, the correct locant is 2 and not 4, because 2 is lowere than 4. (b)
If two or more number of substituents are present, the carbon atoms of the parent chain is numbered from such an end so as to give the lowest set of locants possible.
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The lowest set of locants is decided as follows : (i)
When series of locants containing the same number of terms are compared term by term, then the lowest set is one which contains the lowest number on the occasion of the first difference. (In case the first locant in both the sets is identical, the second set is compared and so on till the lowest locant in a set is obtained and that set shall be the lowest set eventhough the lowest sum rule applied previously is violated.) The following examples will clarify the situation. CH3
Correct Incorrect
1 2 3 4 5 6 7 8 9 10 CH3 – CH – CH2 – CH2 – CH2 – CH2 – CH – CH – CH2 – CH 10 9 8 7 6 5 4 3 2 1 CH3
CH3
Correct name : 2,7,8 – Trimethyldecane (Sum of locants : 2+7+8 = 17) Incorrect name : 3,4,9 – Trimethyldecane (Sum of locants : 3+4+9 = 16) Here, the two sets of locants are 2,7,8 and 3,4,9. The first points of difference is the first locant which is 2 in the first set and 3 in the second set. As 2 is the lowere locant than 3, the first set of locants i.e. 2,7,8 is the correct set eventhough the sum of the locants is 17, higher than the second set which is 16 (3+4+9). Therefore, the correct name is 2,7,8– Trimethyldecane. [Note : Commonly used lowest sum rule in the case will lead to incorrect name] Another example may be taken. 6 5 4 3 2 1 Br – 1CH2 – 2CH – 3CH2 – 4CH – 5CH – 6CH2 – Br Br
Br
Br
Correct name : 1,2,3,5,6 – Pentabromohexane (Sum of locants : 1+2+3+5+6 = 17) Incorrect name :1,2,4,5,6 – Pentabromohexane (Sum of locants : 1+2+4+5+6 = 18)
Correct Incorrect
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Here, out of the two sets of locants i.e. 1,2,3,5,6 and 1,2,4,5,6 the first set 1,2,3,5,6 is lower. The first two locants in both the sets are identical and the difference comes in the third locant. 3 being lower than 4 the first set 1,2,3,5,6 is the lowest set. Therefore, the correct name is 1,2,3,5,6 – Pentabromohexane. [Note : In this case, the lowest set of locant rule and the lowest sum of locant rule give the same set of locants.] Important : It is important to note that both the lowest set of locant rule and the lowest sum of locant rule lead to the correct name when applied to smaller carbon chain (up to 7 carbon atoms). For the longer carbon chains containing more than 7 carbon atoms. lowest sum of locant rule will lead to incorrect name. Therefore, it is better to use the lowest set of locant rule irrespective of the length of the chain. 9.
When two or more different substituents are present to equivalent positions from the two ends of the principal chain (i.e. same set of locants is obtained when numbered from either end), then the numbering is done is such a way that the substituent which comes first in the alphabetical order gets the lower locant.
6CH
3
– 5CH2 – 4CH – 3CH – 2CH2 – 1CH3 NO2 CH3
3–Methyl – 4– nitrohexane.
Both the substituents 'nitro' and 'methyl' are present at equivalent positions giving same set of locants (3,4). Since 'm' of 'methyl' comes before 'n' of 'nitro', the numbering is done from right end to the left giving the lowest locant to the methyl group.
C. NAMES OF SIMPLER MEMBERS OF ALIPHATIC HOMOLOGOUS SERIES The names of the aliphatic homologous series and their simpler members are discussed below which will be helpful for naming complex organic compounds. 1.
Hydrocarbons
Saturated hydrocarbons are called paraffins or alkanes. Unsaturated hydrocarbons containing double bond are called olefins or alkenes and those containing triple bond are called acetylenes or alkynes. Thus, there are three homologous series in hydrocarbons.
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Saturated hydrocarbons or Alkanes or Paraffins (Suffix – ane)
Alkanes are the parent compounds from which all other compounds are structurally derived. The carbon atoms in the chain are linked by single covalent bonds. Some typical members with their common and IUPAC names are listed below. TABLE : 16.3 ALKANES General formula : CnH2n+2 where n = 1,2,3,4,.....etc.
Formula
Alkane
Common name
IUPAC name
CH4(n=1)
CH4
Methane
Methane
C2H6(n=2)
CH3–CH3
Ethane
Ethane
C3H8(n=3)
CH3–CH2–CH3
Propane
Propane
C4H10(n=4)
CH3–CH2–CH2–CH3
n-Butane
Butane
C4H10(n=4)
CH 3 – CH – CH 3
Isobutane
2–Methylpropane
n-Pentane
Pentane
Isopentane
2–Methylbutane
Neopentane
2,2–Dimethylpropane
CH 3
C5H12 (n=5) C5H12 (n=5)
CH3–CH2–CH2–CH2–CH3 CH3– CH – CH2 – CH3 CH3 CH3
C5H12(n = 5)
CH3– C – CH3 CH3
Alkyl groups or radicals : The monovalent hydrocarbon residue left after removing one hydrogen atom from the alkane is called an alkyl group or radical, often represented by R –. The alkyl group is named by removing – ane from alkane and adding 'yl'. Thus, CH3– group is called methyl group. CH3–CH2–group is called ethyl group and so on. The list of a few alkyl groups is given below.
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TABLE : 16.4 SOME ALKYL GROUPS OR RADICALS. Alkyl radical
Structure
Common name
Methyl (CH3–)
CH3–
methyl
methyl
Ethyl (C2H5–)
CH3–CH2–
ethyl
ethyl
Propyl (C3H7–)
CH3CH2CH2–
n-propyl
CH 3
CH –
1–methylethyl or Isopropyl
CH3CH2CH2CH2– n– butyl CH3 CH3
CH – CH – Isobutyl 2
CH 3 CH 2
CH –
CH 3 CH 3 CH 3 CH 3
(ii)
propyl
Isopropyl
CH 3
Butyl (C4H9–)
IUPAC name
C–
butyl
2–methylpropyl or Isobutyl
sec–Butyl
tert – Butyl
1– methylpropyl or sec – Butyl
1,1 – dimethylethyl or tert – Butyl
Olefins or Alkenes (Suffix–ene) General formula – CnH2n where n = 2,3,4 etc. Characteristic group –
C=C
The IUPAC name of a member is derived from the alkane having the same number of carbon atoms by replacing 'ane' of alkane by 'ene'. The first member, obtained by putting n = 2, is ethene C2H4 corresponding to the alkane, ethane. The common name of alkene is alkylene. Thus, ethene is named as ethylene in common system.
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TABLE : 16.5 Alkane
Alkene
Common name
IUPAC name
CH4 (Methane)
Does not exist
—
—
C2H6(Ethane)
C2H4(CH2=CH2)
Ethylene
Ethene
C3H8(Propane)
C3H6(CH3–CH=CH2)
Propylene
Propene
C4H10(Butane)
C4H8(CH3–CH2–CH=CH2) a – Butylene C4H8(CH3–CH=CH–CH3) b – Butylene
But–1–ene But–2–ene
C4H 8(CH 3 – C = CH 2)
Iso–butylene
CH3
2-Methylprop–1–ene
Alkenyl group : The monovalent hydrocarbon residue left after removing one hydrogen from the alkene is an alkenyl group. The common names are more prevalent. TABLE 16.6 Alkenyl group
Formula
Common name
C2H3 –
CH2 = CH –
C3H5 –
CH3 – CH = CH –
vinyl
CH2 = CH – CH2–
IUPAC name Ethenyl
propenyl allyl
1–propenyl 2–propenyl
In certain cases, one or more multiple bonds are considered to be present in sidechains. For this purpose, the following prefixes are mostly used. CH2 = (Methylidene) (iii)
CH3 – CH = (Ethylidene)
Acetylenes or Alkynes (Suffix–yne) General formula
:
Characteristic group
CnH2n-2 where n = 2,3,4 etc. :
– C º C –
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The IUPAC name of a member is derived from the alkane containing the same number of carbon atoms by replacing the suffix 'ane' by 'yne'. The first member is one containing two carbon atoms, C2H2 and is commonly known as acetylene. TABLE : 16.7 Alkane
Alkyne
Common name
IUPAC name
C2H6(Ethane)
C2H2(CHºCH)
Acetylene
Ethyne
C3H8(Propane)
C3H4(CH3–C º CH)
Allylene
Propyne
C4H10(Butane)
C4H6(CH3CH2 – C º CH) a – Crotonylene
But–1–yne
C4H6(CH3–C º C – CH3)
But – 2 – yne
b – Crotonylene
Alkynyl group : The monovalent hydrocarbon residue left after removing one hydrogen atom from an alkyne is an alkynyl group. TABLE : 16.8 Alkynyl group
Formula
Common name
IUPAC name
C2H–
HC º C –
Acetylide
Ethynyl
C3H3–
HC º C – CH2–
Propargyl
2 – Propynyl
CH3 – C º C –—
1 – Propynyl
2. Halogen Derivatives of Hydrocarbons The halogen derivatives of the hydrocarbons are obtained by replacing one or more number of hydrogen atoms by halogen atom or atoms. Thus, a monohalogen compound contains one halogen atom in its molecule. Di –, tri, or polyhalogen compounds contain two, three or more number of halogen atoms in their molecules. (i) Monohalogen derivatives or Alkyl halides : Alkyl halides are represented by the general formula R – X, where R is an alkyl group and X is a halogen. The common names for these compounds are alkyl halides whereas in IUPAC system, they are known as haloalkanes.
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TABLE : 16.9 General formula CnH2n+1X or R – X Formula
Common name
IUPAC name
CH3 – Cl
Methyl chloride
Chloromethane
C2H5 – Cl
Ethyl chloride
Chloroethane
C3H7 – Br
Propyl bromide
Bromopropane
C4H9 – I
Butyl iodide
Iodobutane
3. Alcohols or Alkanols (Suffix – ol) Alcohols are the hydroxy derivatives of hydrocarbons. When one hydrogen atom is replaced by a hydroxyl (–OH) group, the alcohol formed is called a monohydric alcohol. In trivial system, monohydric alcohols are named as alkyl alcohols, whereas in IUPAC system the alcohols are called alkanols, obtained by replacing 'e' of alkane by '–ol'. - e( - H )
R – H (Alkane) ¾ ¾ ¾¾ ® + o l( + O H )
R – OH (Alkanol)
General formula : CnH2n+1OH or R – OH TABLE : 16.10 Formula
Common name
IUPAC name
Alkyl alcohol
Alkanol
CH3 – OH
Methyl alcohol
Methanol
CH3CH2 – OH
Ethyl alcohol
Ethanol
CH3CH3CH2OH
n – Propyl alcohol
Propan–1–ol
Isopropyl alcohol
Propan–2–ol
tert – Butyl alcohol
2-Methylpropan–2–ol
General : R – OH
CH3 CH3
CH3 CH3
CH – OH
CH3 C – OH
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When two hydrogen atoms of a hydrocarbon attached to two different carbon atoms are replaced by two – OH groups, the alcohol formed is called a dihydric alcohol or glycol or alkanediol. Similarly, a compound with three – OH groups attached to three different carbon atoms is called a trihydric alcohol or glycerol or alkanetriol. 4.
Ethers or Alkoxyalkanes : (Prefix – alkoxy) When two same or different alkyl groups are linked through oxygen atom, an ether
results. This is represented by R – O – R1 where R and R1 may be same or different alkyl groups. In common system, ethers are called alkyl ethers (smaller alkyl group goes first) and according to IUPAC system, ethers are treated as alkoxy (alkyl + oxy) alkanes in which one hydrogen atom of the hydrocarbon is replaced by an alkoxy group (RO–). The parent alkane will be the one containing maximum number of carbon atoms. General formula : CnH2n+2O represented by R – O – R1 TABLE : 16.11 Formula
Common name
General : R – O – R1
5.
Dialkyl ether
IUPAC name Alkoxyalkane
CH3 – O – CH3
Dimethyl ether
Methoxymethane
CH3 – O – CH2CH3
Methylethyl ether
Methoxyethane
CH3CH2 – O – CH2CH3
Diethyl ether
Ethoxyethane
Aldehydes or Alkanals : (Suffix – al) The functional group of this class of compounds is an aldehydic group of formyl O
group, –CHO, structurally represented as
– C
H
. Aldehydes have the general formula
R – CHO where R may be hydrogen or any alkyl group. The common names are derived from the acids they produce on oxidation. But the IUPAC name is obtained by replacing 'e' of the hydrocarbon with 'al'. (The number of carbon atom (s) determines the parent alkane). O
CH4 ® H – C
- ic acid
O
¬¾ ¾¾ ¾ H – C H + aldehyde (Methanal (Formic acid) H
IUPAC name (Formaldehyde) Common name
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O Thus, HCHO, structurally represented as H – C – H is called formaldehyde by common system and methanal by IUPAC system of nomenclature. General formula : CnH2nO represented by R – CHO (R = H or alkyl group)
TABLE : 16.12 Formula General : R – CHO
6.
Common name
IUPAC name
Aldehyde
Alkanal
H – CHO
Formaldehyde
Methanal
CH3–CHO
Acetaldehyde
Ethanal
CH3CH2 – CHO
Propionaldehyde
Propanal
CH3CH2CH2 – CHO
Butyraldehyde
Butanal
Ketones or Alkanones (Suffix – one)
This class of compounds contain a bivalent keto group or carbonyl group ( C = O) and both the valencies are satisfied by two same or different alkyl groups. Ketones are O
represented by the general formula R – CO – R 1 (R–C–R 1 ) . According to the common system, the name ketone is preceded by the alkyl groups arranged in alphabetical order. In IUPAC system, the name of a ketone is obtained by replacing 'e' of the parent hydrocarbon by 'one', O
resulting in alkanone. Thus, CH3COCH3 (CH 3– C – CH 3 ) is named as dimethyl ketone or acetone in common system and as propanone in IUPAC system of nomenclature.
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General formula : CnH2nO O
represented by R– C –R 1 (R & R1 may be same or different) TABLE : 16.13 Formula
Common name
IUPAC name
Dialkylketone
Alkanone
CH3 – CO – CH3
Dimethyl ketone or Acetone
Propanone
CH3CH2 – CO – CH3
Ethylmethylketone
Butanone
CH3CH2COCH2CH3
Diethylketone
Pentan–3–one
General : R – CO – R1
7. Carboxylic Acids (Fatty acids) or Alkanoic acids (Suffix – oic acid) O
Organic acids are characterised by the presence of carboxyl groups (– C – OH) formed by CARBOnyl ( C = O) and hydroXYL (–OH) groups. Compounds containing carboxyl group (–COOH) are called carboxylic acids. The general formula of the carboxylic acid is R – COOH (one hydrogen atom of the hydrogen atom is replaced by – COOH), where R may be H or any alkyl group. Compounds containing one, two or three carboxyl groups are called mono, di–or tricarboxylic acids respectively. The common system of naming the monocarboxylic acids is based on the sources from which they are obtained whereas in IUPAC system, 'e' of the parent hydrocarbon (taking C atom of –COOH group alongwith C atoms of alkyl group) is replaced by 'oic acid'. General Formula : CnH2n+1 COOH or CnH2nO2 represnted by : R – COOH (R may be H) TABLE : 16.14 Formula
Common name
IUPAC name
Monocarboxylic acid
Alkanoic acid
H – COOH
Formic acid
Methanoic acid
CH3 – COOH
Acetic acid
Ethanoic acid
CH3CH2–COOH
Propionic acid
Propanoic acid
CH3CH2CH2–COOH
Butyric acid
Butanoic acid
CH3CH2CH2CH2–COOH
Valeric acid
Pentanoic acid
General : R – COOH
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Derivatives of Carboxylic acids : There are four main derivatives of carboxylic acid. (i)
Acid halides or Acyl halides or Alkanoyl halides The chloride is most important among all the acid halides. Here, –OH group of –COOH is replaced by chlorine (halogen) to give acid chloride or acylchloride. The IUPAC names are obtained by substituting 'e' of the corresponding alkane by 'oyl chloride'. More systematically, 'oic acid' of corresponding acid is replaced by 'oyl chloride'. General formula : R – CO – X
TABLE : 16.15 Formula
Acid
Common name
IUPAC name
Alkanoic acid
Acyl halide
Alkanoyl halide
Formic acid
Formyl chloride Methanoyl chloride
O General :
R–C–X O H – C – Cl
(Methanoic acid) CH3COCl
Acetic acid
Acetyl chloride
Ethanoyl chloride
(Ethanoic acid) CH3CH2COCl
Propionic acid
Propionyl chloride Propanoyl chloride
(Propanoic acid)
(ii)
Acid anhydride or Alkanoic anhydride : Structurally, in these compounds two acyl groups are connected through an oxygen atom. They are considered as acid derivatives formed by elimination of a molecule of water from two carboxyl groups. Hence, they are known as acid anhydrides in common system. In IUPAC system their names are obtained by replacing 'acid' from the name of the alkanoic acid by anhydride.
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General formula : R – CO – O – CO – R
TABLE : 16.16 Corresponding Acid
Formula R – CO – O – CO – R
Carboxylic acid (Alkanoic acid)
CH3CO – O – COCH3
Acetic acid (Ethanoic acid)
Common name Acid anhydride
Acetic anhydride
C2H5CO – O – COC2H5 Propionic acid Propionic anhydride (Propanoic acid)
IUPAC name Alkanoic anhydride Ethanoic anhydride
Propanoic anhydride
(iii) Esters or alkyl alkanoate : When the hydrogen atom of the carboxyl group is replaced by an alkyl group, the compound formed is called an ester. Esters are represented by the general formula R – COOR1 (R and R1 may be same or different) and are named as alkyl carboxylate in common system. The IUPAC name for an ester is obtained by replacing 'ic acid' by 'ate' and the name of the alkyl group (R1) attached to oxygen is prefixed to the word. General formula : R – COOR1
(where R may be H or R & R1 may be same or different alkyl groups.)
TABLE : 16.17 Corresponding Acid
Common name
Carboxylic acid
Alkyl carboxylate Alkyl alkanoate
H – COOCH3
Formic acid (Methanoic acid)
Methyl formate
Methyl methanoate
CH3 – COOCH3
Acetic acid (Ethanoic acid)
Methyl acetate
Methyl ethanoate
CH3 – COOC2H5
Acetic acid (Ethanoic acid)
Ethyl acetate
Ethylethanoate
C2H5 – COOCH3
Propionic acid (Propanoic acid)
Methyl propionate Methyl propanoate
Formula
General : R – COOR1
IUPAC name
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(iv) Acid amides or Alkanamides Acid amides are obtained by replacing –OH of the carboxyl group by –NH 2 group and can be represented by the general formula R – CO – NH2 where R may be H or any alkyl group. In common system, these are called acidamides, whereas in IUPAC system the name of an amide is obtained by replacing 'oic acid' of the corresponding alkanoic acid with 'amide'. O
General formula : R – C – NH 2 (R may be H or any alkyl group) TABLE : 16.18
Formula
Corresponding Acid
Common name
IUPAC name
Carboxylic acid (Alkanoic acid)
Acid amide
Alkanamide
H – CO – NH2
Formic acid (Methanoic acid)
Formamide
Methanamide
CH3 – CO – NH2
Acetic acid (Ethanoic acid)
Acetamide
Ethanamide
C2H5 – CO – NH2
Propionic acid (Propanoic acid)
Propionamide
Propanamide
General : R – CO – NH2
9. Amines : Amines are alkyl derivatives of ammonia, obtained by replacing hydrogen atom(s) from NH3 by the alkyl group(s). According to the common system, monoalkyl derivatives are known as alkyl amines, dialkyl derivatives are known as dialkylamines, trialkyl derivatives, are known as trialkylamines. In IUPAC system, alkylamines are called alkanamines, dialkylamines are called alkylalkanamines and trialkylamines are called dialkylalkanamine. The longer alkyl group present is considered as the parent chain. The remaining alkyl group(s) are named as substituents attached to the nitrogen and a prefix N – is used with the name of the alkyl group.
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R N
General formula : R – NH2 , Where R, R1
R1
& R11
,
R R1 R11
N
are same or different alkyl groups.
TABLE : 16.19 Formula
Common name
IUPAC name
Alkylamine
Alkanamine
NH
Dialkylamine
N–Alkylalkanamine
N
Trialkylamine
N–Alkyl–N–alkylalkanamine
Methylamine
Methanamine
Isopentylamine
3–Methyl–1–butanamine
General : R – NH2 R R1
R R1 R11
CH3–NH2 CH3 – CH– CH2CH2 – NH2 CH3
CH3 – NH – CH2 – CH3 Ethylmethylamine
N–Methyl ethanamine
CH 3 CH 3 – CH 2 – N – CH 3
Ethyldimethylamine
N,N–Dimethylethanamine
Ethyl methylpropylamine
N–Ethyl–N–methylpropanamine
CH3 CH2 CH2 – N – CH3 CH2 – CH3
NH2–CH2–CH2–CH2–CH2–NH2
—
1,4 – Butanediamine
Note : When used as the substituent, the –NH2 group is named as amino and is prefixed with a number indicating the carbon atom to which it is attached. 10.
Nitroalkanes : The nitroalkanes are nitro derivatives of the corresponding alkanes and represented
by the general formula, R – NO2 (R
+ O –N – OO
, Nitro group is always prefixed to the alkane.
The common and IUPAC names are generally same except that is IUPAC system, the position of the nitrogroup is indicated.
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General formula : CnH2n+1NO2 or R – NO2 TABLE : 16.20 Formula General :R – NO2
Common name Nitroalkane
Nitroalkane
CH3 – NO2
Nitromethane
Nitromethane
CH3CH2CH2 – NO2
Nitropropane
1- Nitropropane
CH3– CH – CH3
—
NO2
11.
IUPAC name
2 – Nitropropane
Alkyl cyanides or Alkane nitriles :
The functional group of alkyl cyanides is the –C º N group. The general formula is R–CN. These are commonly named either as alkyl cyanides or the nitrile of the acid which is formed on their hydrolysis by changing the suffix – ic acid by – onitrile. Their IUPAC names, however, follow the parent alkane as illustrated below. General formula : CnH2n+1 CN or R – CN TABLE : 16.21 Formula General :R – CN
Common name
IUPAC name
Alkyl cyanide or Acid nitrile
Alkane nitrile
CH3 – CN
Methyl cyanide or Acetonitrile
Ethane nitrile
CH3 – CH2 – CN
Ethyl cyanide or Propiononitrile
Propane nitrile
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639
Nomenclature of branched-chain alkanes and univalent radicals :
The following rules are adopted while naming the higher branched-chain saturated hydrocarbons or alkanes. (1) Longest chain rule : (a)
Select the longest continuous chain of carbon atoms which constitues the principal chain or parent chain. Other carbon atoms which are not included in the chain are regarded as substituents and denoted as prefixes. The given alkane is named as a derivative of the alkane representing the parent chain. The longest continuous chain need not be horizontal. CH2 – CH 3
CH2 – CH3
CH3 – CH – CH – CH 3
CH3 – CH – CH – CH 3
CH2 – CH 3
CH2 – CH3
Incorrect chain (contains only 4 carbon atoms)
Correct chain (contains 6 carbon atoms)
In the above example, the longest continuous chain of carbon atoms contains 6 carbon atoms though the chain is not horizontal (straight). So, it is a derivative of the parent hydrocarbon hexane. (b) If the molecule contains two or more chains containing same number of carbon atoms, select the parent chain with greater number of side chains as substituents. The compound contains two carbon chains containing 7 carbon atoms. The horizontal chain is selected because it contains 3 substituents while the other chain contains only 1 substituent. CH32CH2 – 1CH3 1CH3
3
4
5
6
7
– 2C – 3CH – 4CH2 – 5CH2 – 6CH2 – 7CH3 CH3
(2) Numbering the principal chain : (a)
Number the carbon atoms of the parent hydrocarbon by Arabic numerals 1,2,3.....etc. from one end to the other such that the carbon atom carrying the substituent gets the lowest numeral (lowest locant).
640
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3
– 2CH – 3CH2 – 4CH2 – 5CH3
5CH
3
– 4CH – 3CH2 – 2CH2 – 1CH3
CH3
CH3
Correct numbering
Incorrect numbering
(Substituent has the locant 2)
(Substituent has the locant 4)
(b) If more number of substituents are present, number the carbon atoms of the parent chain from such an end so as to give the lowest set of locants possible to the side chains (Lowest set of locant rule may be referred to). CH3 1C H
3
CH2 – CH3
– 2C H – 3C H – 4C H 2 – 5C H 3
3 – Ethyl – 2 – Methylpentane Here, numbering from left to right gives the lowest set of locants to the substituent (2,3) and hence, the numbering is done as shown. (c)
Prefix the substituent preceded by the locant to the name of parent hydrocarbon. Separate the locant from the name of the substituent by a hyphen.
The following example will clarify the statement mentioned above. CH 3 1CH
3
– 2CH 2 – 3CH – 4CH2 – 5CH 3
3 – Methylpentane The parent hydrocarbon contains 5 carbon atoms and a methyl group is attached having the locant 3, Hence, the name of the compound is 3–Methylpentane. (3) Alphabetical order of substituents : If the number of substituent is more than one, arrange them in alphabetical order prefixed by its positional number. CH 3 1CH
3
CH 2 – CH3
– 2CH – 3CH – 4CH 2 – 5CH3
In this example, 'ethyl' group is alphabetised under 'e' and is written before 'm' of methyl group. Hence, the name of the compound is 3–Ethyl – 2 – methylpentane.
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If a particular substituent is present two or three times, attach the prefix di – or tri – respectively to the name of substituent. Write the locants in the increasing order separated by commas amongst themselves and by a hyphen from the name of the substituent. Do not consider di – or tri – while deciding the alphabetical order of the simple substituents. * Thus, CH3 1CH 3
CH2 – CH3
– 2C – 3CH2 – 4CH – 5CH2 – 6CH3
is named as
CH3
4 – Ethyl – 2, 2 – dimethylhexane and NOT
2,2 – Dimethyl – 4 – ethylhexane
Note : If two or more different substituents are present at equivalent positions from the two ends of the parent chain, numbering is done from such an end so as to give the lowest locant to the substituent which comes first in the alphabetical order. (Rule – 9 of the common conventions in IUPAC system). Thus, CH 3 1 CH
3
– 2 CH 2 – 3 CH – 4 CH – 5 CH 2 – 6 CH 3
is named as
CH 2 – CH 3
3 – Ethyl – 4 – methylhexane and NOT
4–Ethyl – 3– methylhexane (numbering being done from right to left)
(4)
Rule for branched substituents : If the substituent is itself branched, name it as the substituted alkyl group. Number the carbon chain of the substituent from the carbon atom attached to the parent chain.
The following compound is a derivative of the parent hydrocarbon containing 9 carbon atoms i.e. nonane. Ethyl group is present at position 3 and another complex substituent (substituted alkyl group) is present at position 5. This branched side chain is now numbered in a way so that the carbon atom of the side chain attached to main parent chain gets 1 (longest chain rule is also applicable to such side chains). The compound is named as *
Where names of the complex radicals are composed of identical words, priority is decided by considering the locants in the complex radicals. (Rule – 6 of the common convention in IUPAC system)
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CH 2 – CH 1 CH
3
– 2 CH 2 – 3 CH –
4 CH
3 2
– 5 CH – 6 CH 2 – 7 CH 2 – 8 CH 2 – 9 CH 3 1 CH
n o te
– CH3
2 CH
– CH3
3 CH
3
5 – (1,2 Dimethylpropyl) – 3 – ethylnonane.
Important :Complex radicals are alphabetised under the first letter of their complete name i.e. including the numberical affix (if any). Here, 'dimethylpropyl' as a complex single substituent, is alphabetised under 'd' and written before 'e' of ethyl group. The following trivial, common or semi-systematic names may be used for the unsubstituted radicals. CH 3
CH –
CH3 – CH – CH2 –
CH3
CH3
CH3
CH 3
Isopropyl
Isobutyl
CH3 – CH2 – CH – CH3
sec – Butyl
CH – CH2 – CH2 –
Isopentyl
CH3 CH3
CH 3
–C–
CH 3
– CH – C – 2
CH3
CH 3
tert – Butyl
tert – Pentyl
CH3
– C – CH –
CH3
2
CH3
Neopentyl These trivial or common or semi-systematic prefixes can only be used when they are unsubstitued. This will be clear from the following example. Br H 3C – C 1 – 2CH 3 8
Substituted CH3
1 CH 3 – 7CH 2 – 6CH2– 5CH – 4CH2– 3 CH – 2CH – CH 3
H 3C – CH – CH
3
Unsubstituted
5–(1– Bromo–1–methylethyl)–3–isopropyl–2–methyloctane.
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E. NOMENCLATURE OF UNSATURATED ACYCLIC HYDROCARBONS : When carbon–carbon multiple bonds are present in an open chain hydrocarbon, the following additional rules are followed while naming these compounds. (1)
Selecting the principal chain : Select the longest continuous chain of carbon atoms containing maximum number of multiple bonds as the principal chain. CH 2
CH2
CH 3 – CH 2 – C – CH 2 – CH 3
C H 3– C H 2 – C – C H 2 – C H 3
(selected)
(rejected)
Here, the longest continuous chain containing the double bond has four carbon atoms and is selected though the longest continuous chain without the double bond contains 5 carbon atoms which is rejected. CH2
CH2
CH 3 – CH 2 – C – C – CH 2 – CH 3
C H 3– C H 2 – C – C – C H 2 – C H 3 CH – CH3
CH – CH 3
(selected)
(rejected)
Though the longest continuous chain of carbon atoms contains 6 carbon atoms, it does not contain double bond whereas two double bonds are present in a chain which contains 5 carbon atoms. Therefore, the chain containing two double bonds is selected. (2)
Numbering the principal chain :
(i)
If there is one multiple bond, number the principal chain from such an end that the carbon atom containing the multiple bond gets the lowest numeral. For example : 1
CH3 – 2CH = 3CH – 4CH2 – 5CH3 (Correct)
5
CH3 – 4CH = 3CH – 2CH2 – 1CH3 (Incorrect)
Numbering from left end gives the lowest numeral to the carbon carrying double bond. Therefore, numbering from left to right will lead to correct nomenclature. (ii)
If the principal chain contains two or more multiple bonds, number the principal chain from such an end so that the multiple bonds get the lowest set of locants.
The following illustration will explain the rule mentioned above. 1 5 CH 3
2
– 4 CH =
3 3 CH
–
4 2C
5 1 CH
Incorrect (Set 2,4) Correct (Set 1,3)
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In the above case, numbering from two different ends gives two different sets of locants i.e. 2,4 and 1,3. Out of these two sets of locants set 1, 3 is lower. Hence, numbering is done from right to left. (iii) If numbering of the principal chain from both ends give the same set of locants to the multiple bond, then select the set which gives the lower locant to the double bond. This means that if choice exists the double bond gets preference over triple bond in numbering. 1 5 CH
2
3
4
5
ºº 4 C – 3CH 2 – 2CH = 1CH2
Incorrect Correct
Numbering the carbon atoms of the principal chain from both the ends gives the same set of locants i.e. 1,4. Therefore, the numbering is done from right to left so as to give the lowest numeral to the carbon carrying the double bond (1). (3)
(i)
The primary suffix for double bond is 'ene' and for triple bond, it is 'yne'. For example : 1
CH3 – 2CH = 3CH – 4CH2 – 5CH3 is named as Pent – 2 – ene CH3
and
5CH
3
– 4CH – 3C º 2C – 1CH3
is named as 4 – Methylpent – 2 – yne
[Note : Old conventional names like 2 – Pentene or Pentene – 2 and 4 – Methyl – 2 – pentyne or 4 – Methylpentyne – 2 are no longer correct] (ii)
When both double and triple bonds are present in the principal chain, the compound is named as a derivative of alkyne i.e. the suffix – 'ene' always comes before –yne. In such cases, 'e' of –ene is deleted if it is followed by suffix starting with 'a', 'i', 'o', 'u' or 'y'. For example : 5
CH3 – 4CH = 3CH – 2C º 1CH Pent – 3 – en – 1 – yne
is named as
(Numbering as shown above gives the lowest set of locants i.e. 1,3 and 'e' of '–ene' is deleted since it is followed by 'y' of –yne) 5
CH º 4C – 3CH2 – 2CH = 1CH2 Pent – 1 – en – 4 – yne
is named as
(Since numbering from both the ends gives the same set of locants i.e. 1,4, double bond is given preference over the triple bond) (iii) If the principal chain contains two double bonds or two triple bonds, then the suffix is – diene or diyne respectively. In such cases, as per the rule, 'a' is added to the word root. For example – 6CH
3
– 5CH = 4CH – 3CH – CH2 – CH2 – CH3 2CH
= 1CH2
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Considering the longest continuous carbon chain containing both the double bonds and lowest set of locants rule, the compound is named as 3 – Propylhexa – 1, 4 – diene Similarly, 1CH º 2C – 3C º 4C – 5C º 6C – 7CH3 is named as Hepta – 1, 3, 5 – triyne (4)
If the unsaturated acyclic hydrocarbon contains side chains alongwith the multiple bonds, number the principal chain in such a way so as to give the lowest set of locants to the multiple bonds. (The side chain may get the higher locant) 5CH 2 1CH 2
– 6CH2
= 2CH – 3CH = C4 – CH3
is named as
4 – Methylhexa – 1, 3 – diene But if numbering from both ends gives the same set of locants to the multiple bonds, number the principal chain in a manner so as to give the lowest set of locants to the side chain (s). For example : CH 3 C5H 2 = C4H – 3CH – C2 = C1H2 is named as CH 2 – CH 3
3 – Ethyl – 2 – methylpenta – 1, 4 – diene Numbering of the parent chain as mentioned above gives the lowest set of locants to the side chains (2&3). (5)
Sometimes more than two multiple bonds may be present in the unsaturated hydrocarbon and it is not possible to accomodate all the multiple bonds in the principal chain. In such cases, the multiple bonds are considered as side chains. For this purpose, the following prefixes are commonly used. CH2 =
CH3 – CH =
CH2 = CH –
Methylidene Ethylidene
Vinyl
CH2 = CH – CH2 – Allyl
Such a case is illustrated in the following example. CH 2 1 CH
2
= 2CH – 3C – 4CH = 5 CH – 6CH 3
3 – Methylidenehexa – 1, 4 – diene (Longest continuous chain of carbon atoms with maximum number of double bonds)
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NOMENCLATURE OF ORGANIC COMPOUNDS CONTAINING ONE PRINCIPAL FUNCTIONAL GROUP (1)
Selection of principal chain : Select the longest continuous chain of carbon atoms which will include the principal functional group and that will constitute the parent chain. If multiple bonds are present, the principal or parent chain should include maximum number of multiple bonds. If there are two or more chains containing equal number of carbon atoms, select the one with maximum number of substituents. 1CH
3
– 2CH = 3C – 4CH 2 – 5CH 2 – 6CH 3 CH 2 – OH
(Incorrect) CH 3 – CH = 2 C – 3 CH 2 – 4 CH 2 – 5 CH 3 1 CH
2
– OH
(Incorrect) 4 CH
3
– 3CH = 2C – CH 2 – CH 2 – CH 3 1 CH
2
– OH
(Correct) In the above example, the correct principal chain contain the functional group and the double bond and contains only 4 carbon atoms. The other two chains, though contain 6 carbon atoms, do not contain either the functional group or the double bond and hence, they are not selected. (2)
Numbering the principal chain : (Lowest locant to functional group) Number the principal chain in such a way that the principal functional group gets the lowest locant even if it violates the lowest set of locant rule. O 1 CH
3
5 CH
= 6 CH 2
– 2 C – 3 CH 2 – 4 CH – CH 2 – CH 2 – CH 3
The principal chain, in this case contains six carbon atoms as it must include the double bond. The numbering is done so as to give the lowest numeral to the principal functional group ( C = O)
i.e.2.
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If characteristic functional group such as – CHO, – COOH, – COOR, – COCI, – C º N etc. is present in the organic compound, then the numbering should be done in such a way that the characteristic functional group gets the lowest locant 1. This locant for the principal functional group is generally omitted if there is no ambiguity. (3)
Prefix the substituents to the parent hydrocarbon as per IUPAC rules without considering the functional group.
(4)
Add the suffix representing the functional group in place of 'e' of the hydrocarbon.
(5)
Insert the number denoting the position of the functional group before the related suffix and separate on either side by hyphens.
(6)
Halo and nitro groups are always taken as substituents and are prefixed.
A few examples are given below to understand the above rules. Example - 1 :
CH3 – CH2 – 2CH – 3CH2 – 4CH2 – 5CH3 is named as 1CH
2
– OH
2 – Ethylpentan – 1 – ol 5
O
Example - 2 :
1
CH = 6 CH 2
CH3 – 2 C – 3 CH2 – 4 CH – CH 2 – CH 2 – CH 3
4 – Propylhex – 5 – en – 2 – one Example - 3 :
CH3 – 4C º 3C – 2CH2 – 1CHO Pent – 3 – yn – 1 – al
5
3 CH
Example - 4 :
2
O
CH 3 – CH 2 – 2C – 1C – OCH 3
Methyl – 2 – ethylprop – 2 – enoate O
Example - 5 :
5CH
3
– 4CH – 3C – 2CH – O – CH 2 – CH 3 OCH 3
1CH
3
2 – Ethoxy – 4 – methoxypentan – 3 – one (When the position of the principal functional group remains same while numbering from both ends of the parent chain, Ethoxy group gets precedence over Methoxy group)
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Example - 6 :
5 CH
3
– 4 CH – 3 CH – 2 CH – 1 CH 3 Cl
Br
(CBSE, 1998C)
OH
3 – Bromo – 4 – chloropentan – 2 – ol Example - 7 :
Example - 8 :
CH2 = 2CH – 1CN Prop – 2 – enenitrile
(AISB, 1997)
3
CH3 – CH2 – CH2 – 2C = 3CH – 4CH3 1COOH
2 – Propylbut – 2 – en – 1 – oic acid G.
NOMENCLATURE OF POLYFUNCTIONAL ORGANIC COMPOUNDS (1)
When a compound has more than one functional group, the group which determines the class of the compound is referred to as the principal group and forms the suffix. The remaining groups are treated as substituents. The order of seniority of some common groups is listed below with refix and suffix names. The functional group placed higher in the table takes precedence over the others below. The longest continuous chain should contain maximum number of functional groups. TABLE : 16.22 PRIORITY OF FUNCTIONAL GROUPS
Order of Seniority 1 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.
Functional Group – COOH – COOR – COCl – CONH2 –CºN – CHO ( C = O) – OH –NH2 C=C CºC –NO2 –X (X = F, Cl, Br, I)
Prefix
Suffix
Carboxy– Alkoxycarbonyl– Chloroformyl– Carbamoyl– Cyano – Formyl – Oxo Hydroxy– Amino– — — Nitro– Halo–
–oic acid Alkyl....oate –oyl chloride –amide – nitrile – al –one –ol –amine –ene –yne — —
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Important : When –CHO, –CN, –COOH, –COCl etc. are used as substituents, their carbon atoms are not counted in the principal chain. However, when prefix 'oxo' is used for the keto group, its carbon is counted in the principal chain because of its non-terminal nature. If oxo ( C = O) group occurs at the end of a chain, it is known as an aldehyde group. Therefore, aldehyde group can be represented by two types of prefixes : 'oxo', when carbon of –CHO is counted in the chain and 'formyl', when it is not counted in the chain. (2)
The principal chain of the polyfunctional compound must be numbered in such a way that the principal functional group gets lowest possible number followed by double bond, triple bond and substituents.
(3)
The substituents must be placed in an alphabetical order before the word root. When two substituents occupy identical positions from either end of the principal chain, then the lowest number must be given to the substituent which comes first in the alphabetical order.
(4)
When the substituent is itself substituted, the nomenclature of the whole substituent follows the rules mentioned earlier. There are a few exceptions. (i)
No characteristic group is expressed as suffix. Suffixes such as 'yl', ylidene and ylidyne are used. – CH2 – CHO
–CH2Cl
CH3 – CH2 – C º
CH2 =
Formylmethyl
Chloromethyl
Propylidyne
Methylidene
–CH2 – CH2 – CO – NH2 Carbamoylethyl (ii)
–CH2– CN Cyanomethyl
The point of attachment of complex substituent has the lowest possible locant. For example, CH 3
–
1 CH
–
OH 2 CH
2 Br
2 – Bromo – 1 – methylethyl
3 CH
3
–
2 CH
–
1 CH
=
2 – Hydroxypropylidene
(iii) Numerical terms used for similar substituted substituent are bis (for two), tris (for three), tetrakis (for four) etc. For simple substituents di – , tri, tetra – etc. are used. The following examples illustrate the above principle of nomenclature of polyfunctional organic compounds.
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Example : 1 :
6CH
3
– 5CH2 – 4CH – 3CH2 – 2CO – 1CH3 OH
The principal chain contains 6 carbon atoms and the two functional groups C = O and – OH. Keto group being senior takes precedence over – OH group Therefore – OH group appears as a substituent. The name of the compound will be 4 – Hydroxyhexan – 2 – one Example : 2 :
6
CH3 – 5CH – 4CH = 3C – 2CH2 – 1CHO OH
CH3
The name of the compound is 5 – Hydroxy – 3 – methylhex – 3 –en – 1 – al. CN
Example : 3 :
7CH
3
O
– 6CH – 5CH = 4CH – 3C – 2CH2 – 1COOH
The principal chain containing maximum number of functional groups contains 7 carbon atoms. – COOH being the senior most group, the other groups appear as substituents. Further, a double bond is present. Carbon atom of –COOH group is given the number –1. Thus, the name of the compound is 6 – Cyano – 3 – oxohept – 4 – enoic acid O
Example : 4 :
CH 2 – CH 3
H – 4 C – 3CH 2 – 2 CH – 1COOH
When – CHO group is taken in the principal chain, the compound is 2 – Ethyl – 4 – oxobutanoic acid. When – CHO group is taken as a substituent, 2 – Ethyl – 3 – formylpropanoic acid. Cl
Example : 5 :
OH
5CH – 4CH – 3CH 3
NH2 – 2CH – 1COOH
2 – Amino – 4 – chloro – 3 – hydroxypentanoic acid COOC2H 5
Example : 6 :
5CH
3–
4C
= 3CH – 2CH 2 – 1COOH
4 – Ethoxycarbonylpent – 3 – enoic acid
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IV (5) Nomenclature of polyfunctional organic compounds containing two or more similar terminal functional groups. Carbon containing functional groups which occur at the end of a chain are termed as terminal functional groups. They are – COOH, –COOR, –COCl, –CO – O – CO – (anhydride), – CHO and –CN groups. When two or more such groups are present in the organic compound, special rules are followed for their nomenclature. (i) For organic compounds containing only two similar terminal groups : In such compounds, the carbon atoms of both the terminal groups are counted in the principal chain and the groups are indicated by a suitable suffix alongwith prefix 'di'. The following examples will clarifiy the rule CH 3
Example - 1 :
HOOC 4 – 3 CH 2 – 2CH – 1 COOH
is named as
2 – Methylbutanedioic acid C 2H 5
Example - 2 :
NH 2 5CO– 4CH 2 – 3CH 2 – 2CH – 1CONH2
2 – Ethylpentanediamide OH
Example - 3 :
C2H5OOC5 – 4CH2 – 3CH2 – 2CH – 1COOC2H5
Diethyl – 2 – hydroxypentanedioate Example - 4 :
N º 1C – 2CH2 – 3CH2 – 4CH2 – 5C º N Pentanedinitrile
Example - 5 :
OHC1 – 2CH2 – 3CH2 – 4CHO Butanedial
(ii)
For organic compounds containing more than two similar terminal groups : In such cases, two situations may arise.
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(a)
All the similar terminal groups are directly attached to the principal chain In such cases none of these groups forms a part of the principal chain. Special suffixes are used in these cases and when using special suffixes, the carbon atoms of the terminal groups are not counted in the principal chain.
The special suffixes are given below. Functional group
Suffix
– COOH
–carboxylic acid
– COOR
alkyl.......carboxylate
– COCl
– carbonylhalide
– CONH2
–carboxamide
– CN
– carbonitrile
– CHO
– carbaldehyde
This rule is illustrated in the following examples. CHO
Example - 1 :
OHC – 1CH 2 – 2CH – 3CH 2 – 4CH 2 – CHO
Butane – 1, 2, 4 – tricarbaldehyde CN
Example - 2 :
NC – 1CH 2 – 2CH – 3CH 2 – CN
Propane – 1, 2, 3 – tricarbonitrile COOH
Example - 3 :
HOOC – 1CH – 2CH – 3CH2 – COOH COOH
Propane – 1, 1, 2, 3 – tetracarboxylic acid (Numbering from right to left will violate lowest set of locants rule)
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COOCH3 Example - 4 :
CH3OOC – 1CH2 – 2CH2 – 3CH – 4CH2 – 5CH2 – COOCH3
Trimethylpentane – 1, 3, 5 – tricarboxylate CONH 2
Example - 5 :
H 2NCO – C 1H 2 – 2CH – 3CH 2 – CONH 2
Propane – 1, 2, 3 – tricarboxamide. Example - 6 :
ClO C – 4 CH 2 – 3 CH 2 – 2 CH – 1 CH 2 – COCl CO Cl
Butane – 1, 2, 4 – tricarbonylchloride (b) All the similar terminal groups are not directly attached to the principal chain. In such cases, the longest chain with two such similar terminal groups forms the principal chain and the other similar terminal group is treated as substituent and indicated by suitable prefixes. CH2 – COOH
Example - 1 :
HOOC1 – 2CH2 – 3CH – 4CH2 – 5COOH
3 – (Carboxymethyl) pentanedioic acid (Here, note that the carbon atoms of the terminal groups are counted while numbering the principal chain) CH2 CHO
Example - 2 :
OHC6 – 5CH2 – 4CH2 – 3C – 2CH2 – 1CHO CH2 CHO
3,3 – Bis (formylmethyl) hexanedial
Example - 3 :
CH2COOC2H5 C2H5OO1C – 2CH2 – 3CH – 4CH2 – 5COOC2H5
Diethyl 3 – (ethoxycarbonylmethyl) pentanedioate
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NOMENCLATURE OF ALICYCLIC COMPOUNDS While naming the alicyclic compounds, some additional rules are followed in addition to the rules
adopted for acyclic compounds. (1)
The names of the alicyclic compounds are obtained by adding the prefix 'cyclo' to the word root. For example,
Cyclopropane (2)
Cyclopentane
Cyclohexane
The substituents present is indicated by Arabic numerals 1,2,3, etc. If only one substituent is present, its position is not mentioned. But if more than one substituent is present, the numbering is done in a manner (i.e. clockwise or anticlockwise) to give the lowest set of locants to the substituents. While writing the substituents, alphabetical order is maintained like the acyclic compounds. The following examples will illustrate the above rules.
CH 3
CH3
2 1
3
6
4
CH 2 CH 3
5
Cyclopentene
Methylcyclohexane
4– Ethyl – 2 – methylcyclohex – 1 – ene (alphabetical order when writing the name)
CH3
CH 3
1
3
6
2
4
5
3
5
4
(Incorrect)
CH2 CH3
2 6
1
CH 2 CH 3
(Correct) 1 – Ethyl – 3 – methylcyclohexane (alphabetical order when same set of locants are obtained)
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655
When the number of carbon atoms is more in the side chain alkyl group than the ring, the the compound is treated as the cycloalkyl derivative of the alkane. For example, CH3 1CH 3
–
2CH
–
3CH 2
–
4CH 3
CH CH3
2 – Cyclopropylbutane (Side chain contains more number of carbon atoms than the ring) (4)
If the side chain contain multiple bond of functional group, the alicyclic ring is treated as the substituent irrespective of the size of the ring. 3 CH
3CH
3 1 CHO
2 CH
2 – Cyclopentylpropanal (5)
Isopropylcyclohexane (ring contains more number of carbon atoms than the side chain)
2
– 2CH = 1CH2
3 – Cyclohexylprop – 1 – ene
If functional group alongwith other substituent group are present in the ring, it is indicated by suitable prefixes and suffixes as in case of acyclic compounds. If the functional group is directly attached to the ring, its presence is indicated by using suitable suffix, but the carbon atom of such group is not counted in the word root. Suffixes for such groups are as given in III B (ii) (a). For exampleCH2
–
O
CH3
2
3 4
2
3
5
1
4
6
OH
3 – Ethylcyclohexan – 1 – ol
1 5
CHO
6
2 – Oxocyclohexane – 1 – carbaldehyde (–CHO group gets the lowest locant for seniority and suffix 'carbaldehyde' is used instead of 'al' because its carbon is not counted in the ring)
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If the ring as well as the side chain contain functional group, then the parent hydrocarbon is decided as in acylic compounds. For example, 3
2 1 — 3CH2 – 2CH2 – 1COOH
H3C — 4 5
6
3 – (4–Methylcyclohex – 3 – enyl) propanoic acid (Side chain contains only functional group)
5
6 1 — 3CH2 – 2CH2 – 1COOH
HO — 4 3
2
3 – (4–Hydroxycyclohexyl) propanoic acid (Both the ring & side chain contain functional groups) The principal functional group is present in the side chain. OH 6
1 2 — CH2 – CH2 – CH – CH3
5 4
3
OH
2 – (3–Hydroxybutyl) cyclohexan – 1 – ol. (Here the ring contains more number of carbon atoms than the side chain) (6) If an alicyclic ring is attached to a benzene ring, the compound is named as a derivative of benzene.
Cyclopentylbenzene
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I. NOMENCLATURE OF AROMATIC COMPOUNDS Compounds containing one or more benzene rings are known as aromatic compounds. The parent hydrocarbon in the aromatic series is represented by the molecular formula, C 6H6. The IUPAC name for the compound is benzene and is represented as a regular hexagon with alternate double bonds or a regular hexagon with a circle inside.
Or
(1)
If only one hydrogen atom of the benzene ring is replaced by another monvalent atom or group of atoms, a monosubstituted derivative results. The substituents present in the ring are prefixed to the word benzene. A monosubstituted derivative can exist in one form only, since all the six hydrogen atoms in benzene are equivalent. For example, CH 3
Cl
Methylbenzene (2)
Or
Chlorobenzene
NO2
CH = CH2
Vinylbenzene
Nitrobenzene
A number of substituted benzenes have been given special names and have been approved by IUPAC. For example, CH3
OH
Toluene
Phenol COCH 3
CHO
COOH
Benzaldehyde
Benzoic acid
O C
Acetophenone
Benzophenone
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OH
CH3
OH
NH2
Aniline
Salicylic acid
CH 3
o – Xylene OH
O 2N
NH 2
CH3
CH3
COOH
o – Cresol NO2
NO2
o – Toluidine (3)
Picric acid
While naming the disubstituted derivatives of benzene, the positions are indicated by numbers. The carbon atom carrying the principal substituent is given the number 1 and the relative position of other substituent is denoted by numbering the carbon atoms of the ring from 2 to 6 either clockwise or anticlockwise so that the lowest number is given to the substituent. Only three disubstituted products are possible – 1,2;1,3 and 1,4. They are referred to as ortho (o), meta (m) and para (p) deriviatives respectively. When different substituents are present, they are written in alphabetical order . CH 3
OH
Br CH 3
Br 1,2 – Dimethylbenzene (o– Xylene) NH 2
1,3 – Dibromobenzene (m – Dibromobenzene)
Cl 4 – Chlorophenol (p–Chlorophenol)
CH3
OH COOH
NO 2
(4 – Nitroaniline) (p – Nitroaniline)
2 – Hydroxybenzoic acid or o – Hydroxybenzoic acid (Salicylic acid)
OH
2 – Methylphenol (o – Cresol)
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659
For polysubstituted benzene derivatives, the positions of the substituents must be numbered in alphabetical order. However, if one of the groups is associated with a special name, the compound is named as a derivative of the monosubstituted compound, numbered from the group designated in the special name. OH
Br
Cl
I
Br
O 2N
Cl
(1 – Bromo – 4 – chloro – 2 – iodobenzene) nitrophenol) Br
(3 – Bromo – 2 – chloro – 5 –
NH 2
Cl
(3 – Bromo – 5 – chloroaniline) A few aryl groups are mentioned below which will be helpful for nomenclature.
Phenyl CH3
o – Tolyl or 2 – Tolyl
CH 2
CH
C
Benzyl
Benzal
Benzo
CH 3
m – Tolyl or 3 – Tolyl
CH 3
p – Tolyl or 4 – Tolyl
It may be noted that (a) in case of 'tolyl' groups, no substitution is allowed (b) in case of 'phenyl' group, unlimited substitution is allowed (c) in case of 'benzyl' group, substitution is allowed in the ring.
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J. WRITING STRUCTURAL FORMULA FROM IUPAC NAME While writing the structural formula from the IUPAC name, the carbon skeleton of the principal chain is first written. The functional group and the substituents are then incorporated in proper positions as described in IUPAC name. Finally, the residual valencies are filled up with hydrogen atoms. The following illustration may be taken. Let us write the structural formula of the organic compound, 3 – Bromo – 3 – ethyl – 2 – methylpentan – 1 – ol. The principal chain consists of 5 carbon atoms and the skeleton is first written and numbered from one end to the other. 1
C – 2C – 3C – 4C – 5C
The functional group in the chain, (–OH group) is placed on C1 and substituents bromine at C3, methyl group at C2 and ethyl group at C3 are placed. Now, the skeleton takes the following shape. CH3 CH 3 CH 2 HO – 1C – 2 C – 3C — 4 C – 5C – Br
Lastly, the remaining valencies of carbon atoms are satisfied by hydrogen atoms. CH3 H
CH3
HO – 1C – 2C H
H
CH2 3C
Br
H 4C
H – 5C – H
H
H
In the condensed form, the compound is written as CH3 CH 3 CH2 HO – CH 2 – CH – C – CH2 – CH 3 Br
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16.10 ISOMERISM * Compounds having the same molecular formula, but differing in physical and chemical properties are called isomers or isomerides and the phenomenon is known as isomerism (Greek : isos = equal ; meros = part). The difference in properties is due to the difference in the arrangement of atoms within the molecule. Isomerism can be classified as follows : Isomerism
Stereoisomerism
Structural isomerism
Geometrical isomerism
Optical isomerism
Chain or Nuclear isomerism
Position isomerism
Functional isomerism
Metamerism
A. STRUCTURAL ISOMERISM Isomerism due to the arrangement of atoms within the molecule resulting in several structural formulae is termed as structural isomerism. Compounds exhibiting such type of isomerism are known as structural isomers. The structural isomerism is of the folllowing types : (a)
Chain or Nuclear Isomerism : This type of isomerism arises due to the difference in the structure of the carbon chain which forms the nucleus of the molecule. Therefore, the isomerism is called chain or nuclear isomerism. For example, butane has got the molecular formula C4H10 and two structures can be written, one having a straight chain and another having a branched chain.
* Not in the syllabus
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CH 3
CH3 – CH2 – CH2 – CH3
CH 3 – CH – CH 3
(n – butane)
Isobutane or 2 – Methylpropane
Similarly, three chain isomers are possible for the same molecular formula C 5H12 (pentane) CH3 – CH2 – CH2 – CH2 – CH3
CH 3 – CH – CH 2 – CH 3
(n – pentane)
CH 3
(isopentane or 2 – Methylbutane) CH3 CH3
C
CH3
CH3
(neo – pentane or 2,2 – Dimethylpropane) Chain isomerism is also exhibited by alkenes, alkynes, alcohols, ethers, aldehydes, ketones, acids, esters and amines. (b)
Position Isomerism : Isomerism arising out of the difference in the position of the same functional group or substituent in the same carbon chain is called position isomerism. For example, CH3 – CH2 – CH2 – Cl
and
CH 3 – CH – CH 3 Cl
(1 – Chloropropane)
(2 – Chloropropane)
are position isomers. Similarly, CH2 = CH – CH2 – CH3 But – 1 – ene
and
CH3 – CH = CH – CH3 are also But – 2 – ene
position isomers. Disubstituted aromatic compounds exhibit position isomerism due to the different relative positions occupied by the two substituents on the benzene ring. For example, xylene exhibits the following position isomers.
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663
CH3
CH3
CH 3 CH3
CH 3
CH3 (o – xylene)
(m – xylene)
(p – xylene)
Note : Aldehydes and carboxylic acids do not exhibit position isomerism. (c)
Functional Isomerism : Isomerism arising out of the presence of different functional groups is termed as functional isomerism. Alcohols and ethers, aldehydes and ketones, acids and esters, different classes of amines having same molecular formula are functional isomers of each other.
The following examples will justify the above statement. With molecular formula C 2H6O, two compounds with different functional groups can be written – one is an ether and the other is an alcohol. CH3 – O – CH3
and
(Dimethyl ether)
CH3 – CH2 – OH
are
(Ethyl alcohol)
thus, functional isomers. Similarly, CH3COCH3
and
(Acetone)
CH3CH2CHO are functional (Propionaldehyde)
isomers. One is a ketone and the other is an aldehyde. CH3COOH (acetic acid) and HCOOCH3 (methyl formate) are functional isomers. One is an acid and the other, an ester. CH3CH2CH2NH2 (n–propylamine), CH3 – NH – CH2 – CH3 (ethylmethylamine) and CH3–N– CH3 (trimethylamine) are functional isomers. CH3
(d)
Metamerism : Isomerism arising out of the unequal distribution of carbon atoms on either side of the same functional group in molecules of compounds belonging to the same class is known as metamerism. In other words, metamerism occurs due to the difference in the nature of alkyl groups linked to the polyvalent atom of the functional group. For example, ethers, ketones and secondary amines exhibit metamerism.
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Ethers :
CH3 CH2 – O – CH2 CH3
and
(Diethyl ether or Ethoxy ethane
CH3 – O – CH2 CH2 CH3 (Methylpropyl ether or Methoxypropane)
O
Ketones :
O
CH3 CH2 – C– CH2 – CH3
CH3 – C – CH2 CH2 CH3
(Diethyl ketone or Pentan – 3 – one)
Methyl propylketone or Pentan – 2 – one
Secondary amines : CH3CH2NHCH2CH3
and
(Diethylamine or N – ethyl ethanamine)
CH3NHCH2CH2CH3 (Methyl propylamine or N – Methyl propanamine)
Tautomerism : Two compounds with the same molecular formula, existing in dynamic equilibrium with each other are said to be tautomers and the phenomenon is known as tautomerism. When a ketonic substance remains in equilibrium with its enolic form, then it called keto – enol tautomerism. Acetoacetic ester exhibits keto – enol tautomerism. It gives the reactions of both keto compound and unsaturated hydroxy compound. This can only be explained in terms of the dynamic equilibrium existing between the two isomeric forms.
OH
O
CH3 – C = CH – COOC2 H 5
CH3 – C – CH2 – COOC2 H5
(Keto form I)
(Enol form II) (Acetoacetic ester)
The keto and enol tautomers of acetone can be represented as : OH
O
CH 3 – C = CH 2
CH 3 – C – CH 3
Keto form
(enol form) (Acetone)
Tautomerism is considered as a special case of functional isomerism.
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665
B. STEREOISOMERISM Isomerism exhibited by compounds having the same structural formula, but differing in the relative arrangement of atoms or groups in space within the molecules is termed as stereoisomerism. Compounds showing stereoisomerism are called stereoisomers. There are two types of stereoisomerism. (a) Geometrical isomerism (b) Optical isomerism (a)
Geometrical Isomerism : When two compounds have the same molecular formula, but different arrangement of atoms or groups in space around the double bond, they are called geometrical isomers and the phenomenon is known as geometrical isomerism.
Simple examples of this type of isomerism are maleic acid and fumaric acid. Both the acids have the same molecular formula, but the arrangement of atoms and groups in space around C = C is different in both the acid molecules. When two similar atoms or groups lie on the same side, the isomer is a cis – isomer and when they lie on the opposite side, the isomer is a trans – isomer.
H
COOH
H
COOH
H C
C
C
C H
HOOC
COOH
(Maleic acid) Cis
(Fumaric acid) Trans
Therefore, geometrical isomerism is also called cis–trans isomerism, which arises due to the restricted rotation about a carbon to carbon double bond. Another example of geometrical isomerism is but – 2 – ene which exists as cis but – 2 – ene and trans but – 2 – ene. H
H 3C C
C
and
are geometrical isomers.
C
C H 3C
H
H3C
H
(cis – But – 2 – ene)
H
CH3
(trans – But – 2 – ene)
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E – Z Nomenclature ‘E’ stands for Entgegen (German) meaning opposite and ‘Z’ stands for Zussamen (German) meaning together. Consider the isomers is which all the four atoms attached to C = C bond are different. abC = Cxy For naming such compounds a system of nomenclature has been introduced. This is called E-Z nomenclature. Step I - Following the set of sequence rule, priority is given to both the atoms or groups attached to each carbon atom separately. The atom of group of higher priority is numbered 1 and that of lower priority is numbered 2. Step II - If the two groups numbered 1 are on same side, the configuration is labelled ‘Z’. If they are on the opposite side the configuration is labelled ‘E’. Example Assign E or Z configuration to the following compounds 2 (1) H3C2 H C = C 1 CH3C1H2 CH2 CHO (Z) (2)
2
Br
Cl
1
C=C 1
I
H2 (E)
(3)
2
Br
F2 C=C
1
(4)
I (Z) H3C2 1
Cl1 1
OH C= C
H3C H2C
2
CH3 (E)
(b)
Optical Isomerism : When four different atoms or groups are attached to a carbon atom, two structures can be written for the same compound as in case of lactic acid. This is due to the tetrahedral geometry of the carbon atom. The two structures of lactic acid are represented as follows (Fig. 16.26)
The two structures are related as object and its mirror image and are not superimposable. They represent two different compounds. When plane polarised light is passed through these compounds separately, one rotates the plane of polarised light towards left and the other rotates
ORGANIC CHEMISTRY
towards right. But the magnitude of rotation is same. However, these two compounds have similar chemical and almost similar physical properties due to identical structures except that they rotate the plane of polarised light in different directions due to the arrangement of atoms or groups in space around the central carbon atoms to which they are attached.
667
H3 C
CH3
C
C OH
H
COOH (d–lactic acid)
HO
H
HOOC (I–lactic acid)
Fig.16.26 Optical isomers of lactic acid
Compounds having the same molecular formula, but different spatial arrangement of atoms or groups within the molecule and possessing the property of rotating the plane of polarised light are known as optical isomers or enantiomers or enantiomorphs or antipodes and the phenomenon is called optical isomerism or enantiomerism. The enantiomers have similar chemical and almost similar physical properties except that they behave differently towards the plane polarised light. The enantiomer rotating the plane of polarised light towards right is called dextrorotatory and is represented by the letter d – or a (+) sign. Its mirror image enantiomer which rotates the plane of polarised light towards left is called laevo–rotatory and is represented by the letter l – or a (–) sign. The d – and 1 – forms of lactic acid are shown in Fig. 15.7. Since the d – and 1 – isomers rotate the plane of polarised light to the same extent either to right or to left, a mixture containing equal amount of d – and 1 – isomers of an optically active compound will not rotate the plane of polarised light at all and such a mixture is called a racemic mixture or racemic modification represented by dl – or (±) sign. A racemic mixutre is optically inactive due to external compensation and can be resolved (separated) into its optically active antipodes.
CONDITIONS FOR OPTICAL ACTIVITY A compound to show optical activity should possess in its molecule atleast one asymmetric carbon atom or chiral carbon i.e. a carbon atom to which four different atoms or groups are attached. Secondly, the molecule should not have a plane of symmetry, so that the molecule and its image are non-superimposable (dissymmetric nature of the molecule). These are the two condition for a compound to show optical isomerism. OH
Compounds like 2 – chlorobutane, butan – 2 – ol, glyceraldehyde (HC – CH2OH) , malic OH
NH2
CHO
acid (HC – CH2 – COOH) , alanine (CH3CH– COOH) etc. containing one chiral centre show COOH
optical isomerism.
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Specification of Configuration about a Chiral atom R,S Nomenclature : To assign a particular configuration to an optically active compound in terms of prefixes R and S, the following procedure is to be followed, ‘R’ stands for rectus (latin) meaning right. ‘S’ stands for sinister (latin) meaning left. The following steps are involved. Step - I (Fixing of Priorities)- A set of rules known as sequence rules are to be obeyed while, assigning the order of priority to the four groups attached to chiral or asymmetric carbon atom. Rule - 1 The atom with highest atomic number gets the highest priority and the atom with lowest atomic number gets the least priority. Atom or group with highest priority as given number 1 and that with least priority is given number 4. e.g. 1 - chloroethanal CH3 – CH – OH | Cl Order of priority is Cl (1), OH(2), CH 3 (3), H (4) Rule - 2 If the first atom of two or more groups is the same, then priority is decided taking the second or even the third atom of the group as the case may be. e.g. Between CH3– and CH3 – CH2 – group CH3 – CH2 – group is given priority since in CH3– the second atom is Hydrogen whereas is CH 3 – CH2 the second atom is Carbon having higher atomic number. In secondary butyl alcohol CH3 CH – CH2 CH3 | OH Order of priority is OH(1), CH 3– CH2– (2), CH3– (3), H (4) Rule - 3 If the group or atom linked to chiral carbon contains = bond or becomes equivalent to two and three single bonds respectively. N | –C = O ≡ C O C≡ N ≡ C N | | N O e.g. in case of glyceraldehyde CHO | *CHOH , order of priority –OH, –CHO, CH2 OH, H | CH 2OH
º
bond, then that
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669
Step II (Assigning Configuration) The molecule is to be viewed in such a way that the atom or group of the lowest priority is directed away from the eye. If the atoms or groups attached to chiral carbon be 1, 2, 3, 4 and order of priority is also 1, 2, 3, 4, then orientation of the molecule is made in such a way that atom of lowest priority (4) is pointed to the back. If the eye traces a sequence 1®2®3 in a clockwise direction, the configuration is ‘R’. If it traces the sequence 1®2®3 in an anticlockwise direction, the configuration is ‘S’ (3)
(3)
(4)
Viewer (2)
(1)
Viewer
(4)
(1)
(2)
Clockwise ‘R’ Configuration
Anti clockwise ‘S’ Configuration
Examples
(1) Configurations assigned to two enantiomers of secondary butyl alcobol CH3 – CH – CH2 – CH3 | OH Sequence, –OH (1), – CH2 – CH3 (2), –CH3 (3),– H (4) 2
2
C2H5
C2H5 C 1
3
3
OH
H4
C
H4
1
CH3
CH3
R- Configuration (clockwise)
HO
S- Configuration (anticlockwise)
* (2) Configuration assigned to two i somers of lactic acid [ CH 3 - CH (OH ) - COOH ]
Priority,
(1)
OH,
(2)
COOH,
(3)
(4)
CH 3 , H
2
2
COOH
COOH H4
C 1
HO
3
CH3
R- Configuration
C 3
CH3
H4 1
OH
S- Configuration
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Configuration of the molecule drawn on the plane of the paper by Fischer Projection. Consider the configuration of 1–chloroethanol CH 3
* CH |
OH
Cl Priority is – Cl (1), – OH (2), – CH3 (3), – H (4) Case -1 Bring the atom of lowest priority to the bottom. Rotate the eye in order of decreasing priority. If it is clockwise ® ‘R’ configuration. If, anticlockwise ® ‘S’ configuration. 2
OH 3
H2C
1
Cl
C
H4 S- Configuration
Case - 2 If the atom of lowest priority is at the top, rotate the molecule by 180 0 so as to bring it to the bottom. Rotate the eye in order of decreasing priority and assign the configuration. H
2
OH
1
3
H3C
C
Cl
1
3
CH3
C
Cl OH 2
H
R-Configuration Case - 3 If the atom of lowest priority is at the left side of horizontal line, keep the top
position fixed and change the position of remaining atoms in the clockwise direction so as to bring the atom of lowest priority to the bottom. Rotate the eye in order of decreasing priority and assign configuration.
l
H C
H
O
HC
3
C
H
OH C
2
OH
Cl
1
Cl
C
CH3
3
CH3
H
R-Configuration
Case - 4 If the atom of lowest priority is at the right side of the horizontal line, keep the top position fixed and change the position of other atoms or groups is the clockwise direction so as to bring the atom of lowest priority to the bottom. Rotate the eye in the order of decreasing priority and assign configuration.
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671
Assignment of configuration to the molecule containing more than one chiral carbon. Example - (1) Consider tartaric acid HOOC – CH (OH) CH (OH) (COOH) 1
COOH 2 * HO- C - H 3 * H-C-OH 4
COOH
The sequence of priority is –OH (1), –COOH (2), CH (OH) COOH (3), H (4) Consider the asymmetric carbon No (3). 3
3
CH(OH)COOH 4
CH(OH)COOH
1
1
H- C*- OH
HO
2
COOH
2
COOH
C H
4
3S
Here atom of lowest priority is at the left so keep the top group fixed and change the position of other atom in the clockwise direction so as to bring the atom of lowest priority to the bottom. Rotate the eye in order of decreasing priority and assign configuration. Consider the asymmetric carbon No (2). 2
2
COOH 1
COOH
3
4
1
C OH
HOOC (OH) HC
OH- C*- H 3
4
H
CH(OH)COOH
Here the atom of lowest priority is at the right side of horizontal line. Keeping the top position fixed the change of position of other groups is made is the clockwise direction so as to bring the atom of lowest priority to the bottom. Rotate the eye in the order of decreasing priority. Thus, configuration of the compound is (2S, 3S) CH3
(2) Consider H
C
2
Cl
H
C3
Cl
CH3
Sequence is – Cl (1), – CH (Cl) CH3 (2), – CH3 (3), – H(4) Consider Chiral ‘C’* (2) 3
CH3 4
H
C
CH3 1
Cl
..
CH(Cl)CH3 2
3
CH
3
1
Cl
C H
2
CH(Cl)CH3
1
Cl
C
3
2
CH(Cl)CH
H 2S
3
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Consider chiral ‘C*’ (3) 2
2
CH(Cl)CH3
CH(Cl)CH3 4
H
1
C
1
Cl
Cl
CH3
3
C
CH3
H
3
3R
The configuration is (2S, 3R).
16.11
ELECTRONIC DISPLACEMENTS IN A COVALENT BOND
The covalent bond which is formed by the mutual sharing of two electrons between the participating atoms may not be an ideal one, because an attacking agent always bears either a positive or a negative charge and for a reaction to occur on the covalent bond, it must possess oppositely charged centres. In other words, the substrate molecule although as a whole electrically neutral, must develop polarity on some of its carbon atoms and substituents linked together which is made possible by the partial or complete displacement of the bonding electrons. These displacements may be permanent (e.g. inductive, resonance and hyperconjugative effects) otherwise called as polarisation effects or temporary (e.g. electromeric and inductomeric effect) known as polarisability effects. A. INDUCTIVE EFFECT When two electrically dissimilar (having different electronegativities) atoms are connected by a single bond, the more electronegative atom will attract the shared electron pair of the bond towards it than the other atom. As a result, a fractional negative charge is developed (d–) on the more electronegative atom and a fractional positive charge (d+) is developed on the less electronegative atom. The extent of fractional charge created on atom depends on the electronegativity difference between the atoms. d+ C
d Cl
d+ C
d C
d
N
d+
Hg
The polarisation of electrons (or electron density) towards more electronegative atom is a permanenet effect and therefore, called permanent polarisation effect or inductive effect. Electron repelling groups (alkyl groups, metal atoms) have +I effect whereas electron withdrawing groups (NO2, F, Cl, Br, I, OCH3, OH, CN, COOH etc.) have –I effect. Consider a chain of carbon atoms with a chlorine atom joined to the end carbon atom. d+
d
– C3 – C 2 – C1 – Cl
As stated earlier, chlorine being more electronegative than carbon acquires a small negative charge and C1 acquires a small positive charge. As C1 is positively charged, it attracts the electron pair shared between C1 and C2 towards itself. This causes C2 to acquire a small positive charge, but this charge is smaller than the charge on C 1, which is now almost neutralised. Similarly, C 3 acquires a positive charge which is less than that on C 2. This effect is the inductive effect and
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673
decreases rapidly as the distance from Cl increases. Practically, the inductive effect is ignored after the second carbon atom. d+ d
– C3 – C2 – C1 – Cl
Inductive effects are of two types. These are : (i) + I effect : If the substituent attached to the end of the carbon chain is electron donating, the effect is called +I effect. For example, ddd–
dd–
C
d–
C
C
d+
CH 3
+I
The +I effect of some of the atoms or groups in the decreasing order is : (CH3)3C–
>
t-Butyl
(CH3)2CH
>
Isopropyl
CH3– CH2– Ethyl
>
CH3–
>
D>H
Methyl
(ii) –I effect : If the substituent attached to the end of the carbon chain is electron withdrawing, the effect is called –I effect. For example, C
+
C
+
+
C
NO2
(–I effect)
The –I effect of some of the atoms and groups in the decreasing order is : –NO2 > – CN > – COOH – > – F > –Cl > –Br > –1 Inductive effect is reflected in the acidity of the various chloroacetic acids. If we consider monochloroacetic acid (Cl – CH2 – COOH) and trichloroacetic acid (Cl3C – COOH), we find that the trichloroacetic acid is the strongest of all these three substituted acids. That is because the negative charge of the carboxylate ion is attracted by the chlorine atoms, thus stabilizing the anion very strongly. As the number of chlorine atom increases, the anion gets more and more stabilized. CH3COOH
CH3 – COO + HÅ
Ka = 1.8 x 10–5
Cl – CH2COOH
Cl
Ka = 1.4 x 10–3
Cl CH – COOH Cl
Cl Cl
Cl Cl Cl
Cl Cl Cl
CH – COOH
CH2 – COO + HÅ
CH – COO + H
Å
CH – COO + H
Ka = 5.53 x 10–2
Å
Ka = 0.3
(most stable)
Inductive effect can also be used to explain the stability of carbocations. Carbocations are planar and positively charged.
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R R R
C+
H H H
+ CH 2
CH 3
C+
Alkyl groups release electrons and disperse the positive charge by inductive effect. Electrons in a C – C bond are more polarisable than those in a C – H bond. Hence, replacing hydrogen by alkyl group reduces the net charge on positively charged carbon. Thus, the stability of carbocations follows the order, Å Å Å Å C (CH ) ( CH3) 3 C > H 2 > C H3 3 2 C H > CH 3 B. INDUCTOMERIC EFFECT The temporary effect which increases the permanent inductive effect present in a molecule due to the approach of a charged ion (attacking reagent) is termed as inductomeric or inductometric effect. For example, in nitromethane, the –I effect of –NO 2 group is further increased temporarily by the approach of hydroxyl ion to the hydrogen atom which is ultimately removed as water.
+ H O2 N
C–H
+
+
OH
ON – CH 2 + H2 O 2
H + Since inductomeric effect is time dependent, it always favours the reaction and never inhibits it. C. ELECTROMERIC EFFECT When multiple bonds such as , C = C , C = O , - C = N , - C = C - are present in an organic molecule, the shared pair of electrons of the p-bond may be completely displaced towards one of the bonded atoms depending upon its electronegativity when a charged reagent is brought near the molecule. This effect is temporary, because the polarisation of the molecule to positive or negative poles and the displaced electron pair reverts back to its original position when the charge is removed from the vicinity of the molecule. This effect is known as ‘‘Electromeric effect’’ and is represented by E. This effect is more prominent in conjugated systems. When two atoms with different electronegativities are involved in displacement process, the displacement occurs towards the more electronegative atom. Electromeric effect may be represented as given below. + C - O. ...
. C=O .. . C=C
C=C
C=N
..
+ C-C
+ C N
.
C-C
(curved arrow indicates the electron displacement with direction)
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675
When a multiple bond is conjugated to an atom with non-bonded pair of electrons, then that non-bonded pair can also participate to enhance the electromeric effect, Thus,
.. ..
- O - CH = CH2
..
+
..
+
- O = CH - CH2 , - N - CH = CH2
- N = CH - CH2
The ease of of electron release will depend upon the ease of polarisation of the non-bonded electron pair. Thus, one of the sequence is
.C.
>
.N.
.O.. . > .
D. RESONANCE : Occasionally no reasonable single electronic structure can be written for a molecule which can satisfactorily explain all its observed properties. For example, carbon dioxide can be represented by atleast three possible structures. O=C=O I
or
O – C º OÅ
OÅ º C – O
or
II
III
The calculated heat of formation of carbon dioxide for one formula is 350kcals/mol. and the O – O distance should be 2.40A0. However, the observed heat of formation is 380kcals/mol. and O – O distance is 2.30A0 indicating that none of these three structures satisfactorily accounts for its observed properties. Thus the actual structure is one which can not be represented on the paper and all these structures make some contribution to the actual structure of carbondioxide molecule. The actual molecule is a hybrid of all these individual structures, which differ in the location of electrons around the constituent atoms. They are called the contributing structures or canonical structures or resonating structures which are purely hypothetical and individually do not represent any real molecule. This phenomenon is called resonance, a term coined by Heisenberg. Ingold had called this phenomenon mesomerism. The true structure of the molecule is called resonance hybrid. Benzene can be represented by the following structures :
I
II
III
Fig 16.27 I & II are the resonating Structures of Benzene whereas III is the resonance hybrid
None of these structures truly represents the actual structure of benzene. Both the structures contribute significantly (nearly 93%) to the actual structure of benzene. Hence, benzene is said to be a resonance hybrid of these contributing structures.
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When several structures may be assumed to contribute to the actual structure of a molecule, but no one of them can be said to represent it uniquely, then the true structure of the molecule is referred to as a resonance hybrid and the phenomenon is termed as resonance. The situation of resonance is represented by a double headed arrow (4) connecting the resonating structures. A resonance hybrid exhibits unusal stability. The difference between the energy of the hybrid and the energy of the most stable resonating structure is called the resonance energy. The resonance energy for carbon dioxide is 30 kcals/mol. whereas for benzene, it is 36 kcals/mol. Chief Conditions of Resonance : (i)
The various resonating structures should have the same positions of nuclei differing only in the position of electrons.
(ii)
The energy content of all the canonical structures must nearly be same.
(iii) Each canonical form must have the same number of unpaired electrons (including zero). (iv) Resonance occurs in a molecule only when all the atoms lie in the same plane. (v)
All resonating structures do not contribute to the same extent to the hybrid. Stable structures having less energy contribute most and the hybrid behaves more like that form. Among the resonating structures, the one which has (a) more number of covalent bonds, (b) all the atoms with octet of electrons excepting hydrogen which has a duplet, (c) less separation of opposite charges and (d) more dispersal of charge, is more stable than others.
(vi) Resonating structures are interconvertible by one or a series of short electron shifts. For instance O
O CH3 – C – O (Benzene)
CH 3 – C = O
(Acetate ion)
Characteristics of Rosonance hybrid : (i)
Stability : The resonance hybrid is always more stable than any of the contributing structures.
(ii)
Resonance energy : Resonance energy is the measure of the stability of the resonance hybrid. Higher its value, greater is its stability.
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677
(iii) Unusual bond lengths : In a resonance hybrid, the bond lengths are totally different from those of any of the contributing structures. For example, all the (C – C) bond lengths are equal in benzene having value 1.39Ao, which is in between the bond length of 1.54Ao for C–C single bond and 1.34Ao for C = C double bond. Resonance effect or Mesomeric effect : The flow of p electrons from on part of conjugated system to the other creating centres of low and high electron density due to the phenomenon of resonance is termed as resonance effect (R effect) or mesomeric effect (M effect) which is of two types : (i)
Positive resonance (+R) or positive mesomeric (+M) effect: Groups which donate electrons to a conjugated system are said to have +R or +M effect.
: :
– OH, – OR, – NH2, – NHR, – NR2, – Cl, – Br, – I etc. : CH2 – CH = Cl : :
CH2 = CH – Cl :
: Cl
: Cl
:
:
: Cl
:
: Cl :
:
:
:
Vinyl chloride
: Cl :
:
:
Chlorobenzene Negative resonance (–R) or negative mesomeric (–M) effect : Groups which withdraw electrons from the conjugated system towards themselves are said to exhibit – R or – M effect. Examples are – NO2, – CN, – COOH – CHO, – COR etc.
O
: O: :
:
:
(ii)
CH2 = CH – C – H
H2C – CH = C – H
O
O
O
O
O
N=O
N–O
N–O
N–O
N
O
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Effects of resonance : Resonance helps to explan the acidity and basicity of organic compounds. A few common examples are given below. (a) Acidity of organic acids : The acidity of an acid depends upon its ability to donate proton. The equilibrium can be written as HA
+
(acid)
H3O+ +
H2O (base)
(conjugate acid)
A– (conjugate base)
HA is an acid and it donates a porton to H2O (base) and forms a conjugate base (A–). If the conjugate base is more stable than the acid, then the equilibrium shifts towards the right and HA behaves as an acid. Acetic acid is an acid and ethanol is not an acid. CH3 – C
O + H2 O
OH
(acetic acid)
CH3 – C
CH3 – C
O O
+ H3O Å
(acetate ion)
CH3 – C
O Å
OH (Charge separation)
O O
(Equivalent contributing structures)
The conjugate base (acetate ion) is more stable than the acid because of the equivalence of the contributing structures. Though resonance occurs in the acid, it does not contribute to its stability due to charge separation which acts as a restoring force. Hence, the equilibrium shifts towards right and acetic acid is an acid Consider the case of ethanol C2H5OH. The equilibrium can be represented as C 2 H 5 O + H3 OÅ
C 2 H 5 OH + H 2O
(Ethanol) (Ethoxide ion) Resonance does not occur in ethoxide ion. Ethyl group being electron repelling does not accomodate the negative charge. Hence the equilibrium does not shift to the right since the ethoxide ion is not stabilised by resonance. Phenol is acidic Consider the following equilibrium for phenol. O
OH
+ H2O
(Phenol)
+ H3O+
(Phenoxide ion)
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Phenol will be acidic only if the phenoxide ion will be more stable than phenol, so that the equilibrium will shift towards right hand side. Let us write the resonating structures for both the phenol and phenoxide ion.
I
+ OH
+ OH
:OH
II
OH
+ OH
III
IV
V
O
O
Resonating structures of phenol
O
I
O
O
II
III
IV
V
Resonating structures of phenoxide ion For both phenol and phenoxide ion five resonating structures can be written. In case of phenol, besides the two Kekule structures (I & V), other structures contribute negligibly towards the stability because of charge separation. On the other hand, in case of phenoxide ion, the other structures (II , III & IV) do not involve charge separation, but instead delocalise the charge leading to greater stability of the ion. Hence, phenoxide ion is more stable than phenol due to resonance and phenol is acidic. p- Nitrophenol is more acidic than phenol
C 6 H 5 – O + H 3 OÅ
C 6H 5 – OH + H 2O (Phenol)
O
Å N
OH + H2O
O (p–Nitrophenol)
(Phenoxide ion)
O O
N
O + H3O (p–nitrophenate ion)
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p- Nitrophenate is more stable than phenoxide because of resonance. More number of resonating can be written for p- nitrophenate leading to greater stability. Therefore, p- nitrophenol is more acidic (pKa = 7.16) than phenol (pKa = 10). (b)
Basicity of organic bases : All organic compounds containg unshared pair of electrons or acquiring unshared pair of electrons through electron shifts can be considered as bases. The unshared pair of electrons accepts a porton, thus behaving as a base. The equilibrium for basicity is expressed as B + H3O+
®
(base)
BH+
+H2O
(conjugate acid)
The conjugate acid is weak for a strong base and strong for weak base. Basicity of methylamine and aniline CH3NH2
Å
+ H3O Å
CH 3 N H 3 + H 2 O
+ H3O Å
C6 H 5 N H 3 + H 2 O
(Methylamine) C6H5NH2
Å
(Methylamine)
(i)
In case of methylamine, methyl group is electron repelling (+ I effect) whereas phenyl group is electron withdrawing (- I effect) in aniline.
..
Å
– NH 2
(ii)
NH2
The conjugative effect of phenyl group withdraws electron density from nitrogen. This is a consequence of resonance. As a result the basicity of aniline decreases. No such effect is observed in methylamine and hence methylamine is quite basic in nature. Basicity of aniline and p- nitroaniline When nitro group is present at para position, there is greater resonance interaction of the lone pair of electron of nitrogen with p- nitrophenyl system than with phenyl group, thus decreasing its basicity. O
+ N
O
O NH2
O
Å N
Å NH2
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E.
HYPERCONJUGATION (NO BOND RESONANCE) A special type of resonance involving s - electrons and p- electrons is called hyperconjugation. When alkyl groups like H3C–, C2H5 – are directly attached to multiple bonds like C = C, C ≡ C, C = O, C ≡ N, the electron repelling property of the former groups are considerably enhanced. The +I effect of the alkyl groups can not explain this phenomenon properly. To account for this, Baker and Nathan suggested that when C – H bond is attached to an unsaturated carbon atom, the s - electrons of H – C bond become delocalised by entering into partial conjugation with the adjacent p-electrons of unsaturated system, which means s – p conjugation occurs. This type of conjugation between the electrons of single bond and those of multiple bond is termed hyperconjugation (Mulliken, 1941). Hyperconjugation causes a permanent polarity in the molecule and that polar effect is called hyperconjugative effect, Hyperconjugation is represented as follows: H H - C - CH = CH2 H
H H - C = CH - CH2 H
H H
C = CH - CH2 H
H H
C = CH - CH2 H
In the above resonating structures, one of the H – C bond is considered to be cleaved. From this point of view, hyperconjugation is often called ‘‘no bond resonance’’. Though H – C bond is shown to be broken, H+ is never free from the system or changes its position in the molecule. This effect is also called Baker-Nathan effect or Anchimeric assistance. (Please note that the above resonating structures have been written taking one H – C bond only. Other two H – C bonds can also result in different resonating or hyperconjugative structures) Thus it is evident that hydrogen atoms attached to the carbon atom (a-carbon atom) next to the multiple bond (called the a - hydrogen atoms) are responsible for this hyperconjugation. Therefore, with the increased number of a-hydrogen atoms, more number of hyperconjugative structures can be written leading to greater inductive effect. Thus, the order of hyperconjugative effect is CH3 – > CH3 – CH2 – > (CH3)2 CH – > (CH3)3 C – (3a - H) (2a - H) (1a - H) (no a-H) This order is the reverse of inductive effect i.e. (CH3)3C – > (CH3)2 CH – > CH3 – CH2 – > CH3– Effect of Hyperconjugation Hyperconjugation effect is certainly weaker than resonance effect. Due to this effect, the electron density of the unsaturated part of the molecule increases and therefore, the multiple bonds in these systems are found to be more reactive than what is expected from +I effect only. Various physical and chemical properties have been explained through hyperconjugation. Physical : Bond length – In ethane (CH3 – CH3) , C – C bond length is 1.54A0. But in CH3 – C ≡ CH, CH3 – C single bond distance is found to be 1.46A0. This is due to hyperconjugation and the CH3 – C bond develops a partial double bond character resulting in the shortening of bond distance.
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Chemical (i) Stability of alkenes : The stabilities of alkenes can be known from their heat of hydrogenation values. Lower the heat of hydrogenation, greater will be the stability. Greater the number of alkyl groups attached to C = C, greater is the stability. This is evident from the following table. Compound No. of a-H Heat of hydrogenation in kcals mol –1 I. CH 2 = CH2 nil 32.8 (Ethene) II. CH3 – CH = CH2 3 30.1 (Propene) III. CH3 – CH = CH – CH3 6 cis 28.6 But - 2 - ene trans 27.6 CH3 IV. 9 26.9 CH 3 C = CH - CH3 2-methylbut -2-ene CH3 CH3 C=C V. 12 26.6 CH3 CH3 2,3-Dimethylbut-2-ene On the basis of heat of hydrogenation, the stability of alkenes follow the order, V > IV > III > II > I. This is in perfect agreement with the number of a-hydrogen atoms of alkyl groups attached to carbon-carbon double bond (ii) Directive Influence of Alkyl groups
H
H
H-C-H
H-C H
H
+
H
+
H-C H
+
H-C H
:
.. (For other two a-dydrogens, similar structures can be written) CH3– and other alkyl groups have o– and p–directive influence in electrophilic substitution reaction. In case of toluene, the electron density at o– and p-positions with respect to methyl group increases due to hyperconjugation (3-a hydrogens). Hence, electrophilic substitution will preferrably be at o– and p– positions in case of toluene and other alkyl benzenes. (iii) Stability of carbocations and free radicals. The order of relative stability of the carbocations and free radicals is tertiary > secondary > primary. This can be explained on the basis of hyperconjugation. Greater the number of alkyl groups attached to carbocation, greater is the hyperconjugative interaction leading to the stability of the carbocation. Similar explanation can hold good for free radicals also.
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16.12 TYPES OF ORGANIC REACTION Organic reactions may be classified into four fundamental types basing on the nature of the products formed. These are : (i) Substitution reaction (ii) Addition reaction (iii) Elimination reaction (iv) Rearrangement reaction (i)
Substitution reaction : When one or more atoms or groups of an organic compound are replaced by an equivalent number of atoms or groups, without any change in the remaining part of the molecule the reaction is termed as a substitution reaction. In substitution reaction, one or more covalent bonds are broken and formed at the same carbon atom of the reactant. The new atom or group which enters the molecule is called the substituent and the product is called a substitution product. Substitution reactions are characteristic reactions of saturated compounds. For example, when methane reacts with chlorine in the presence of diffused sunlight, all the four hydrogen atoms are substituted by chlorine atoms, subsequently forming a mixture of chloromethanes. CH4 + Cl2
hn ®
CH3Cl
+ HCl
(Chloromethane) Methyl chloride ®
CH2Cl2 (Dichloromethane) Methylene chloride
+ HCl
CH2Cl2 + Cl2 ®
CHCl3 (Trichloromethane) Chloroform
+ HCl
CH3Cl + Cl2
CHCl3 + Cl2
®
CCl4 + HCl (Tetrachloromethane) Carbon tetrachloride
When methyl bromide is treated with aqueous KOH solution, the bromine atom is substituted by the hydroxy group and methyl alcohol is formed. CH3 – Br Methyl bromide
+ KOH (aq)
CH3OH + KBr (aq) Methyl alcohol
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When benzene is treated with a mixture of conc. HNO3 and conc. H2SO4 at 600C, the hydrogen atom of the benzene ring is substituted by a nitro group, forming nitrobenzene. C6H6
Conc . H2 SO4
+ HNO3
60 o C
(Benzene)
(ii)
C6H5NO2
+ H2O
(Nitrobenzene)
Addition reaction : A reaction in which two reacting molecules combine to give a single molecule as the product is called addition reaction. Addition reactions are characteristic reaction of unsaturated compounds. For example : acetylene adds on hydrogen in two steps forming ethylene and ethane respectively. Ni
CH º CH + H2
CH2 = CH2
473K Ni
CH2 = CH2 + H2
CH3 – CH3
473K
Similarly, NaHSO3 adds to C = O group of acetaldehyde forming a bisulphite compound. H
H
CH3 – C – OH
CH3 – C = O + NaHSO3
SO3Na (Acetaldehyde)
Addition of HCl to propene results in 2–chloropropane as the addition product. CH3 – CH – CH3
CH3CH = CH2 + HCl Propane
| Cl
2-Choloropropane
(iii) Elimination reaction : Reaction in which two or more atoms or groups attached to the adjacent carbon atoms of the reactant molecule are eliminated in form of simple molecules like H2O, H2, NH3, HCl etc. leading to the formation of a multiple bond are called elimination reactions. Due to such elimination, a multiple bond (double or triple) is formed. For example, when vapour of ethanol is passed over heated Al2O3, ethylene is formed. A molecule of water is eliminated.
H
OH
H– C – C – H H
H
(Ethanol)
Al2O3 573K
H – C = C – H + H 2O H
H
(Ethylene)
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Similarly, when ethyl bromide is treated with alcoholic KOH, ethylene is formed. H
Br
H– C – C – H H
H – C = C – H + HBr
Alc. KOH
D
H
H
H
(Ethyl bromide)
(Ethylene)
When ethylene dibromide is heated with Zn, ethylene is formed. CH 2 – Br
+ Zn
CH2
heat
CH2
CH 2 – Br
+ ZnBr2
(iv) Rearrangement reaction : These types of reactions involve migration of an atom or group from one site to another within the same molecule. The product is always a structural isomer of the original compound. For example, when n-butane is heated with AlCl 3 at 573K, 2 – methylpropane is formed. AlCl3
CH3 – CH2 – CH2 – CH3
573K
CH3 CH3 – CH – CH3
Similarly, 1 – bromopropane rearranges to form 2 – bromopropane when heated with AlBr3 and HBr.
CH3 – CH2 – CH2 – Br
D AlBr3 – HBr
(1 – Bromopropane)
Br CH3 – CH – CH3
(2 – Bromopropane)
Ammonium cyanate on heating forms urea involving a rearrangement reaction. NH4CNO Ammonium Cyanate
D ¾¾®
NH2 – CO – NH2 Urea
16.13 FISSION OF A COVALENT BOND : Irrespective of the organic reaction each reaction involves the breaking of a covalent bond in the substrate in the presence of a reagent and formation of a new covalent bond resulting with the product. For example, the substrate R – X reacts with an attacking agent Y to give product R – Y. R–X+Y
R–Y+X
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In this reaction the covalent bond R – X is broken and a new covalent bond R – Y is formed. Depending on the nature of R, the attacking agent Y, the leaving group X and experimental condition, bond cleavage or bond fission can take place in two ways. (i)
Homolysis or homolytic cleavage :
Homolysis of covalent bond may be defined as that type of cleavage where the shared pair of electrons is distributed equally between two bonded atoms resulting in two free radicals. Homolytic fission usually occurs in non-polar bonds on heating at high temperature or in the presence of ultraviolet radiation or in the presence of peroxide compounds. Homolysis can be represented by a fish arrow which shows one-electron displacement. A – A ¾Homolysis ¾¾¾¾ ¾® A× + ×A Free Radical
n ® Cl× + Cl× Cl – Cl ¾h¾ ¾ A – B ¾homolytic ¾¾ ¾¾® A× + B×
(A and B are of approximately equal electronegativity)
(ii)
Heterolysis or heterolytic cleavage :
The type of cleavage of a covalent bond where the shared pair of electrons is donated to one of the bonded atoms resulting with an ion pair is termed as heterolysis or heterolytic cleavage. Heterolytic fission which occurs in a polar bond is represented by curved arrow which shows a two electron desplacement towards more electronagative atom. Case 1 :
A – B
Heterolysis
A + B Cation Anion
where B is more electronegative than A.
H H C H
Cl
Heterolysis
H3C
+
Cl
Carbocation
where chlorine is more electronegative than carbon. Case 2 : where A is more electronagative than B
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687
Heterolysis
H3C – Mg – Br
H3C
+ Mg – Br Carbonation
REACTIVE INTERMEDIATES : Homolytic and heterolytic cleavage of a covalent bond of the substrate result in short lived and highly reactive fragments called reactive intermediates which may be ionic or neutral. (a)
Carbocations :
A carbocation is a positvely charged species where the positive charge resides on a carbon atom with six electrons or three pairs of electron in its valence shell being formed by the heterolysis of a cavalent bond.
CH3 H3C
C
CH3 Heterolysis
Cl
H3C
CH3
t - Butyl cation
Carbocation are classifed as primary (1°), secondary (2°) and tertiary (3°) depending on the nature of the carbon atom bearing positive charge.
H
H
H C
H3C
H Methyl carbocation
CH3
C
H Ethyl carbocation
H3C
CH3
C
H3C
H Isopropyl carbocation
o
o
Primary (1 ) (ii)
+ Cl
CH3
t - Butyl chloride (i)
C
Secondary ( 2 )
C
CH3 t-butyl carbocation o
Tertiary ( 3 )
The observed order stability of the carbocations is : CH3 < CH3 – CH2 < (CH3)2 CH < (CH3)3 C because alkyl group attached directly to the positively charged carbon stabilise the carbocations due to inductive and hyperconjugation effects. The high stability of allyl and benzyl cations can be explained by resonance. Benzyl cation because of having higher number of resonating structures is more stable than allyl cation.
CH2 = CH – CH2 (i) Resonance Allyl cation
CH2 – CH = CH2 (ii) Resonating Structure
... ... CH2 – CH – CH2 Resonance hybrid
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CH2
CH2
(I)
CH2
(II)
(III)
CH2
CH2
(IV)
CH2
(V)
Resonating structures
Resonance hybrid Resonance in Benzyl cation
(iii) The carbocations have trigonal planar structure with bond angle 120°, the carbon atom carrying positive charge being Sp2 hybridized. The unhybridized 2p orbital is empty lying perpendicular to the molecular axis. C
120°
Structure of carbocation (b)
Carbanions :
Carbanions may be defined as the negatively charged species with nagative charge on the carbon atom possessing eight electrons or four pairs of electrons in its valence shell being formed by the heterolytic fission of a covalent bond. (i)
Carbanions can be categorised into three types : primary (1°), secondary (2°) and tertiary (3°) depending on the nature of the carbon atom bearing the negative charge.
CH3
CH3 – CH2
(CH3) CH2
(CH3) C
Methyl anion
Ethyl anion
Isopropyl anion
t-Butyl anion
2° Carbanion
3° Carbanion
1° Carbanions (ii)
The order of stability of carbanions is 1° > 2° > 3° which can be explained by inductive effect. The presence of alkyl group with + I effect intensifies the negative charge on carbon making it unstable. But the stability of allyl and benzyl anions can be explained by the phenomenon of resonance. H2C – CH = CH2 (I) Allyl anion
d–
d–
H2C = CH – CH2 º H2C ... – CH ... – CH2 (II) Resonance Hybrid
ORGANIC CHEMISTRY
CH2
689
CH2
(I)
CH 2
(II)
CH2
(III)
CH 2
(IV)
CH 2
(V)
Benzyl anion
Resonance hybrid
(iii) The carbanion is having a pyramidal i.e. a distorted tetrahedral structure like ammonia where the negatively charged carbon atom is sp 3 hybridized with bond angle 107°. The fourth sp3 hybrid orbital contain a lone pair of electrons.
..
3
sp hybrid orbital containing a lone pair of electron
C Structure of carbanion (c)
Free Radicals :
Free radicals may be defined as the odd electron neutral species which are formed by homolytic fission of a covalent bond. (i)
Free radicals are of three types, such as primary (1°), secondary (2°) and tetiary (3°) depending on the nature of carbon atom carrying the unpaired electron.
.
.
.
.
CH3
CH3 – CH2
(CH3)2 CH
(CH3)3 C
Methyl free radical Primary (1°)
Ethyl free radical Primary (1°)
Isopropyl free radical Secondary (2°)
t-Butyl free radical Teritary (3°)
(ii)
The relative stabiltiy of simple alkyl free radicals has the order .
.
.
.
R3C > R2CH > RCH2 > CH3 3°
2°
1°
1°
which can only be explained on the basis of hyperconjugation. Higher the number of a–H atoms, greater will be the delocalization of the odd electron and hence more stable will the alkyl free radical. (iii) Alkyl free radical is having trigonal planar structure with the central carbon atom being Sp 2 hybridized like carbocation. But in case of free radical the unhybridized 2p orbital contains a single electron.
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2p orbital containing an unpaired electron
.
C sp2 hybridised carbon
120°
Structure of alkyl free radical
(iv) The stability of allyl and benzyl radicals can be explained through resonance. Though these are 10 free radicals, they are more stable than 1 0 alkyl radicals. The stability is attributed to the resonance effect. In both allyl and benzyl radicals the p-orbital on carbon carrying a single electron is in conjugation with p bond and thus delocalisation occurs through p-p overlap.
.
.
]
d
.
d
.
(Resonance stabilisation of allyl radical) The benzyl radical is more stable compared to allyl radical because in benzylic radical, the number of contributing structure is more. d
d
(Resonance stabilisation of benzyl radical) Due to resonance effect, the allyl and benzyl radicals are more stable than 1 0, 20, or 30 alkyl free radicals. Thus we can summarise the stability order of free radicals as Benzylic > Allylic > 3 0 > 20 > 10 > CH3.
16.14 ELECTROPHILES AND NUCLEOPHILES lonic or polar reagents are classified into two subdivisions as (a)
Electrophiles or Electrophilic reagents
(b)
Nucleophiles or Nucleophilic reagents
(a)
Electrophiles or Electrophilic Reagents : The reagents which are positively charged and electron deficient having a tendency to form a bond by accepting an electron pair from another species are called electrophiles (electron loving). An electrophile is thus electron deficient and attacks regions of high electron density (negative centres) in the substrate molecule.
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Electrophilic reagents may be divided into two types : Positive electrophiles and neutral electrophiles. The positive electrophiles refer to those which carry a positive charge and include proton, cations and carbocations. Examples are H+, Cl+, Br+, I+, NO2+, NO+, R3C+, RN2+, H3O+ etc. The neutral electrophiles though deficient in electrons do not carry positive charge. Examples are AlCl3, ZnCl2, BF3, SO3, carbenes. All these reagents have six electrons in their outermost orbit and are thus short of a pair of electrons for acquiring stable configuration and therefore behave as electron seeking reagent. F ·· · B· F ·· F
Cl ·· Al ·· Cl ·· Cl
(Boron trifluoride)
(Aluminium chloride)
·· Cl ·· C ·· Cl (Dichlorocarbene)
It should be noted that an electrophile need not be a positively charged ion, but it must possess an atom or centre which is electron deficient. (b)
Nucleophiles or Nucleophilic Reagents : Anions and molecules with unshared pair of electrons having a tendency to donate a pair of electron to another species, capable of accepting it are nucleophiles (nucleus loving). Nucleophiles are electron rich centres and enter into reactions with electron acceptors. Nucleophiles may be negative nucleophiles or neutral nucleophiles.
Negative nucleophiles are those which carry negative charge. Examples are Cl–, Br– ,I– , R – O , NH 2, OH, CN , CH 3COO , – C etc.
The neutral nucleophiles are rich in electrons due to the presence of unshared electron pair. For example –
··
··
· , R N, R ·S· , R – OH, RS · · ·NH · · H etc. H 2O, 3 3 2
The addition of a negatively charged nucleophile to positively charged substance gives rise to a neutral product while the addition of a neutral nucleophile will give rise to a positively charged product.
CÅ + (Substrate)
CÅ (Substrate)
+
X
C X
(nucleophile) charged
(Product) neutral
X
(nucleophile) neutral
C
XÅ
(Product) charged
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CHAPTER (16) AT A GLANCE 1.
Organic compounds are defined as the compounds of carbon containing hydrogen and one or more additional elements like oxygen, nitrogen, sulphur, phosphorus, halogen etc.
2.
Organic compounds are studied separately due to their characteristic features like vast number, complex nature, homology, isomerism, non-ionic and reversible nature of reaction and vast application in our life.
3.
Carbon is tetracovalent and all the four valencies are equivalent. It possesses the unique property of catenation.
4.
Carbon utilises sp3 hybrid orbitals for bond formation in saturated compounds, sp2 hybrid orbitals in unsaturated compounds containing C = C and sp hybrid orbitals in unsaturated compounds containing C º C.
5.
To elucidate the structure of the organic compounds the following steps are to be followed. (i)
Purification of the organic compounds involving crystallization, sublimation, distillation and differential extraction basing on the difference of one or more physical properties. Chromatography is the modern technology now mostly used for the sepration, identification and purification of organic compounds basing on mainly principles of adsoption and partition, thus leading to different types of chromatography like adsorption chromatography (column chromatography and thin layer chromatography) and partition chromatography (paper chromatography).
(ii)
Qualitaive analysis of the organic compounds to detect the elements present like nitrogen, sulphur, halogens and phosphorus by Lassaigne's test in addition to carbon and hydrogen by estimating the amount of CO2 and H2O formed on oxidation.
(iii) Quantitative analysis of nitrogen is carried out either by Duma's method by Kjeldahl's method. Halogens can be estimated by Carius method, whereas phosphorus and sulphur can be oxidized to form phosphoric acid and sulphuric acid from which the amount present can be calculated. Liebig method is helpful in estimating carbon and hydrogen and percentage of oxygen is usually calculated by taking difference between 100% and the sum of percentages of all other elements present in the organic compound under observation. 6.
Organic compounds are classified basing on carbon skeleton and functional groups.
7.
Compounds containing the same functional group show homology. In a homologous series, two consecutive members differ by a – CH2– group.
8.
Organic compounds are conveniently named according to the IUPAC system which is modified from time to time. According to the recent rule, the prefixes and suffixes are preceded by the locants. A compound may have several names, but a single name should not represent more than one structure.
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693
9.
Organic compounds show structural and stereoisomerism. Structural isomerism arises due to the arrangement of atoms within the molecule resulting in several structural formulae whereas stereoisomerism arises due to the relative arrangement of atoms or groups in space within the molecule.
10.
For a compound to show optical isomerism, the four atoms or groups attached to the carbon atom must be different. Geometrical isomerism arises due to the restricted rotation about a carbon to carbon double bond.
11.
Inductive effect arises due to the electronegativity difference between atoms in a bond. There are +I substituents repelling electrons (alkyl groups, –COO – etc.) and –I substituents attracting electrons (–F, –Cl, –OH, –NO2 etc.) Inductive effect operates through a short distance (i.e. up to about 3A0). The acidity of substituted acids, stability of carbocations etc. can be explained through inductive effect.
12.
Electromeric effect - It is a temporary effect. On the approach of a charged reagent, the pelectron of the multiple bond present in an organic molecule has a tendency to get polarised in the direction of more electronegative atom. The effect is known as electromeric effect.
13.
When several structures can be written for a particular compound by simply redistribution of electron density, no one of these structures satisfactorily account for all the observed properties. The actual structure is represented by a hybrid of all these structures. The molecule is referred to as a resonance hybrid.
14.
Resonance confers stability to the molecule.
15.
Resonance helps to explain certain observed phenomena like acidity and basicity of organic compounds.
16.
Hyperconjugation- The electron releasing tendency of alkyl group gets enhanced if it is linked to a multiple bond present in an organic molecule. The effect is known as hyperconjugation or no bond resonance which causes permanent polarity in the molecule.
17.
Organic reactions may be classified into four fundamental types basing on the nature of products. They are substitution, addition, elimination and rearrangement reactions.
18.
Homolytic cleavage of a covalent bond results in the formation of free radicals whereas heterolytic cleavage results in the formation of ions. Carbon carrying negative charge is called a carbanion and carbon carrying positive charge is called a carbocation.
19.
Reagents having negative charge ( CH 3 , C 2H 5O , COO etc.) or a lone pair of electrons ·· ·· · · etc ) are called nucleophiles. in a complete shell ( NH 3 , H 2 O, H 2 S
20.
Reagents having a positive charge ( Br Å , N O 2 , I Å etc.) or electron deficiency are called electrophiles.
21.
Reactions taking place with nucleophilic reagents are called nucleophilic reactions and the reactions taking place with electrophilic reagents are called electrophilic reactions.
Å
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QUESTIONS A.
SHORT QUESTIONS : (one mark each)
1.
Give the IUPAC names of the following compounds : CH2 CH3
(i)
CH3 – CH – CH – CH – CH 2 – CH3 CH3
CH – CH3 CH3
(ii)
CH3 – CH – CH2 – CH3 Cl CH3
(iii)
CH3 – CH – C – CH3 C2H5 CH3
(iv)
(a)
CH3 CH3
CH – CH – COOH Cl
(b)
(v)
CH3 – CH – CH – COOH | | CH3 Br
(a)
CH2 = CH – CH = CH2
(b)
CH2 = C – CH = CH2 CH3
(vi)
(vii)
CH3 CH3
CH – CH = CH – CH3
CH3 – CH2 – C – CH CH2
(viii)
CH3 – CH2 CO CH3
CH3 CH3
ORGANIC CHEMISTRY
(ix)
695
CH3 – CH – CH – CH3 CH3
(x)
OH
CH3 – CH – CH – CH = CH2 CH3 NO2
(xi)
CH3 – CH – CH – C º C – CH3 CH3 CH3 CH3
(xii)
CH3 – C – CH = CH2 CH3
(xiii)
CH3 – CH – CH2 – CHO CH3
(xiv)
CH3 – CH – CH2 – CH3 CHO
(xv)
CH3 – CH – COOH
(a)
Cl CH3 – CH – CH2 – CHO | Cl
(b)
(xvi)
CH3 – O – CH – CH2 – CH3 CH3
(xvii)
CH3 – CH2 – C º CH
(xviii) CH2(Br) – CH2 – COOH (xix)
C2H5 CH3
CH – CH = CH2
(xx)
CH3 – C º C – C2H5
(xxi)
CH3 – CH2 – COOC2H5
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(xxii)
(a)
CH 2 – CH – CH 2 CN
(b) (xxiii)
CN
CN
CH2 = CH – Cl CH2 = C = CH2 CH3
(xxiv)
CH º C – C – CH = CH2 CH3
COOH (xxv)
CH 3 – CH – CH 2 – CH 2 – COOH CH 3
(xxvi)
CH 3 – N – CH 3 NH2
(xxvii)
CH3 – CH – CH2 – CH – CH2 – CHO C2H5 CH3 O
(xxviii)
Cl – CH = CH – C – C – OCH2CH3 CH3
CH3 NH2
(xxix)
CH3 – CH – CH – CH – CH2 – CH3 CH3
(xxx)
HOOC – CH2 – CH – CH2 – COOH COOH
(xxxi)
CHO COOH
(xxxii)
CH3 – CH – CH2 – COOC2H5 CHO
(xxxiii)
HO – CH 2 – CH – COOH NH 2
ORGANIC CHEMISTRY
697
NH2
(xxxiv)
CH3 – CH2 – CH2 – CH – CH2 – NH2
CH3
(xxxv)
CH3 – C – NH2 CH3
(xxxvi)
(CH3)2 N – CH2 – CH2 – C (CH3)3
(xxxvii)
CH3 – CH = C – CH2 – CH3 CH2 – CH2 – CH3
CH 3 (xxxviii)
CH 3
CH 3 – C – CH – C = CH 2 CH 3 CH (CH 3)2
(xxxix)
(CH3)2 CH – CH2 – C – C º CH CH2
(xxxx)
CH 3 – C – CH2 – C – CH 3 O
O
CH2 – COOH
O
O
(xxxxi)
HO – C – CH2 – CH – CH – CH2 – CH2 – C – OH
(xxxxii)
CH3 CH3 – CH = CH – CH – CH3
CH3
(xxxxiii)
HOOC – CH2 – CH2 – CH2 – COOH
(xxxxiv)
CH3 – CH – CO – CH – Br | | NO2
CH3
(xxxxv) CH3 – CH – CH2 – CH3 | CH2OH (xxxxvi)
CH3 – CH2 – C = CH – CH3 | COOH
698
2.
+2 CHEMISTRY (VOL. - I)
Write down the structural formulae of the following compounds : (i)
1–Bromo – 3 – pentanone
(ii)
2 – Pentanone
(iii) 2 – Methyl – 1 – pentene (iv) 3,3 – Dimethyl – 1 – butene (v)
3 – Methyl – 3 – pentanol
(vi) 1 – Bromo – 2 methylbutane (vii) 4,5 – Dimethyl – 2 – hexyne (viii) 4 – Methyl – 2 – pentene (ix) Ethyl formate (x)
3 – Hydroxybutanoic acid
(xi) 2 – Butanol (xii) 1,3 – Butadiene (xiii) 3 – Pentanol (xiv) 2,2 – Dimethyl – 1 – propanal (xv) 3 – Methylpent – 4 – en – 2 – one. (xvi) 2 – Bromo – 4 – chloropentanal (xvii) Pentane – 2,4 – dione (xviii)
2 – Methyl propane
(xix) Ethyl-4 amino-3-chlorobut-2-enoate (xx) Benzoic acid (xxi) 2-Methyl buta-1, 3-diene (xxii) Propadiene 3.
Which of the following is basic ? (i) CH3CH2OH (ii) CH3COCH3 (iii) CH3CH2NH2 (iv) CH3 – O – CH3
4.
Name the compound which is isomeric with diethyl ether.
5.
Give the IUPAC name of a compound which is isomeric with acetone
6.
(a)
How many sigma and pi-bonds are present in methyl cyanide ?
(b)
How many sigma and pi-bonds are present in vinyl cyanide ?
7.
Name the functional isomer of propanone.
8.
Indicate the hybridised state of carbon atoms in acetylene.
9.
How many isomers are possible for C5H12.
10.
Write formula of one alkynyl group.
11.
How many sigma and pi-bonds are present in acetylene molecule ?
ORGANIC CHEMISTRY
699
B.
Short Questions : (two marks each)
1.
Write the structure of the following compounds : (a) 2 – Methylpentene (b) 2 – butanol
2.
Circle the functional groups in the following : (a) (b)
CH2 = CH – CH2 – CH3; CH3 – COOH; H2N – CH2 – COOH; CH3CONH2 Name the functional groups : O || CH3— ; CH3 – CH2 – NH2 ; CH3 – C – NH2 ; CH3 –
3.
What do you understand by functional group in organic compounds ?
4.
Write a short note on – "Isomerism in organic compounds".
5.
Name the functional groups in the following compounds. CH3CHO; CH3CH2NH2, CH3CH2OH, CH3CONH2
6.
The following names are incorrect. Give correct names. CH3CH 3
(i)
CH3CH2 – CH – CH2 – CH3
(ii)
CH 3 – C – CH – CH2 – CH2 – CH3 CH3
CH2 – CH2 – CH3
(3 – Propylpentane)
(4,5,5 – trimethylhexane)
7. 8.
What is homologous series ? Give an example of it. What is resonance ? Give one example.
9.
Explain with one example the term electrophile.
10.
What is a nucleophile ? Give an example of nucleophilic substitution reaction.
C.
Short questions : (Three marks each)
1.
What is Rf value ? How can it be used for identification of organic compounds in a mixture ?
2.
What is the principle behind the soda lime test for nitrogen? Why is this test not dependable ?
3.
Is Beilstein's test a satisfactory test for the detection of halogens ?
4.
Why is H2SO4 added after addition of ferrous sulphate in Lassaigne's test for nitrogen ? Give reasons.
5.
Why is dry copper oxide used in the detection of hydrogen ?
6.
What is the principle behind Duma's method for the estimation of nitrogen?
7.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
8.
Why is it necessary to used acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test ?
700
D. 1.
2.
3.
4.
5.
6.
+2 CHEMISTRY (VOL. - I)
Multiple Choice Questions The concept of vital force theory was given by (a) Lavoisier
(b) Wohler
(c) Berzelius
(d) Berthelot
Carbon forms a large number of compounds because of (a) catenation
(b) high electron affinity
(c) high ionisation energy
(d) variable valency.
A compound with molecular formula C2H2 must contain (a)
all sigma bonds
(b) all pi-bonds
(c)
three sigma and two pi-bonds
(d) two sigma and three pi-bonds.
Alicyclic compound is (a)
an aromatic compound
(b) a carbocyclic compound
(c)
an acyclic compound
(d) a heterocyclic compound.
The compound R – CO – NH2 is an example of (a) an amide
(b) an amine
(c) a ketone
(d) a cyanide.
The hybridisation of carbons C1 and C3 in the compound is 1
7.
CH3 – 2CH = 3CH – 4CH3
(a) sp2, sp3
(b) sp3, sp2
(c) sp, sp3
(d) sp2, sp2
The IUPAC name of
OHC – CH = CH – CH – CH = CH2
is
CH2 – CH2 – CH2 – CH3
(a) 4 – Butyl – 2, 5 – hexadien – 1 – al.(b) 5 – Vinyloct – 3 – en – 1 – al (c) 5 – Vinyloct – 5 – en – 8 – al 8.
9.
10.
(d) 3 – Butyl – 1,4 – hexadien – 8 – al
Which of the following IUPAC names is correct ? (a) 2 – Methyl – 3 – ethylpentane
(b) 2 – Ethyl – 3 – methylpentane
(c) 3 – Ethyl – 2 – methylpentane
(d) 3 – Methyl – 2 – ethylpentane
The IUPAC name of CH2 = CH – CH (CH3)2 is (a) 1,1 – Dimethyl – 2 – propene
(b) 3 – Methyl – 1 – butene
(c) 2 – Vinylpropane
(d) 1 – Isopropylethylene
The IUPAC name of CH º C – CH = CH2 is (a) 3 – Buten – 1 – yne
(b) 1 – Buten – 3 – yne
(c) But – 1 – yn – 3 – ene
(d) 1 – Butyn – 3 – ene
ORGANIC CHEMISTRY
11.
12.
13.
14.
15.
16.
17.
18.
19.
701
IUPAC name of the compound HOOCCH2CH2COOH is : (a) Buten – 1, 4 – dioic acid
(b) 3 – carboxypropanoic acid
(c) Succinic acid
(d) 1,2 – Dicarboxyethane
Which of the following compounds has isopropyl group ? (a) 2,2,3,3 – Tetramethylpentane
(b) 2,2 – Dimethylpentane
(c) 2,2,3 – Trimethylpentane
(d) 2 – Methylpentane
Isomerism exhibited by acetic acid and methyl formate is (a) functional
(b) geometrical
(c) chain
(d) position.
Electrophiles are (a) electron loving species
(b) electron hating species
(c) nucleus loving reagents
(d) nucleus hating reagents.
Nucleophiles are (a) electron loving species
(b) electron hating species
(c) nucleus loving species
(d) nucleus hating species.
Heterolytic cleavage of a carbon – carbon covalent bond gives (a) carbanion
(b) carbocation
(c) free radical
(d) both carbanion and carbocation
A p – bond is formed by the overlapping of (a) hybridised orbitals
(b) p – orbitals only
(c) any two p-orbitals
(d) orbitals oriented perpendicular to the molecular axis.
Chemically similar compounds differing by – CH 2 group are (a) isomers
(b) allotropes
(c) homologues
(d) tautomers.
Which of the following is the most stable ion ? Å
(b) CH3 - CH2 - CH2 - CH2
(c) CH3 - CÅ - CH3
(d) CH3
I CH 3
20.
Å
(a) CH 3 - CH 2 - C H - CH 3
Å
Which of the following would be optically active ? (a) tert. butanol
(b) sec. butanol
(c) n – butanol
(d) 1 – chloro – 4 – hydroxybutane
702
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
+2 CHEMISTRY (VOL. - I)
Which of the following can exhibit cis – trans isomerism ? (a) H – C º C – Cl
(b) Cl – CH = CH – Cl
(c) CH3 – CHCl – COOH
(d) ClCH2 – CH2Cl
Acetone and propen – 2 – ol are (a) position isomers
(b) keto-enol tautomers
(c) geometrical isomers
(d) chain isomers
A pair of optically active compounds which are not mirror images are called (a) mesomers
(b) tautomers
(c) enantiomers
(d) diastereoisomers
Which are isomers ? (a) Ethanol and Ethoxyethane
(b) Methanol and Methoxymethane
(b) Propionic acid and Ethylacetate
(d) Propionaldehyde and Acetone
The hybridization of carbon atoms in C – C single bond in HC º C – CH = CH2 is (a) sp3 – sp3
(b) sp2 – sp3
(c) sp – sp2
(d) sp3 – sp
Which of the following orders regarding the electronagativity of hybrid orbitals of carbon is correct ? (a) sp > sp2 > sp3
(b) sp < sp2 < sp3
(c) sp > sp2 < sp3
(c) sp < sp2 > sp3
A molecule is said to be chiral if it (a) contains a centre of symmetry
(b) contains a plane of symmetry
(c) can not be superimposed on its mirror image
(d) exists as cis and trans forms
Which of the following belongs to +I group ? (a) – OH
(b) – OCH3
(c) – COOH
(d) – CH3
Which of the following is not a nucleophile ? (a) CN¯
(b) NH3
(c) BF3
(d) OH¯
Which of the following in the most stable radical ? ·
(a) C H 3 ·
(c) R C H 2
·
(b) R C H 2 ·
(d) R 3 C
ORGANIC CHEMISTRY
31.
32.
33.
34.
35.
36.
37.
703
In which of the following molecules, the resonance effect is not present ? (a)
NH2
(b)
NH3
(c)
OH
(d)
Cl
An electrophilic reagent must have (a) a vacant orbital (b) an orbital containing one electron (c) an orbital containing two electrons (d) all completely filled orbitals Methanol and acetone can be separated by (a) fractional distillation (b) distillation (c) steam distillation (d) vacuum distillation Which of the following mixture would steam distillation be the most appropriate method of separation (a) Diethyl ether and water (b) Ethyl alcohol and water (c) Aniline and sodium chloride (d) none of these In quantitiave estimation of hydrogen 0.1g of an organic compound yielded 0.09g of water. The percentage of hydrogen in the compound is (a) 8 (b) 10 (c) 12 (d) 15 In Duma's method for the estimation of nigrogen the sample is heated with copper oxide and the gas evolved is passed over (a) Nickel (b) Copper oxide (c) Copper gauze (d) None of these. In the Carias method for the estimation of halogens, the samle is heated with (a) Conc. HNO3 (b) Conc. H2SO4 and Ag (c) Conc. HNO3 and AgNO3
(d) None of these
E.
SHORT QUESTIONS
1.
Write the IUPAC names of the following compounds
(i)
CH3 CH3
O CH – COOH
(iii) CH3CO.O.COCH3 2.
(ii) (iv)
CH3 – C – CH2 – CH2 – CH3
CH3.CH2CH2COCl
Write the structural formulae of the following : (i) 2,2,3 – Trimethylbutane (ii) 1 – Chloropropan – 2 – one (iii) Benzoyl chloride
704
3.
+2 CHEMISTRY (VOL. - I)
Name each of following compounds on the basis of IUPAC system. (i)
CH3 CO CH2 CH3
(ii)
Cl CH2 CH2COOH
(iii)
CH3 – CH – CH2 – CH – CH2 – CH3 CH2 – CH3
CH3
4.
Arrange the following functional groups in the increasing order of priority. – OH, – COOH,
5.
C = O , – CN, –NO2
Write down all the possible structural isomers of (i) C4H10 (ii) C4H8 (iii) C4H10O
6.
Correct the following IUPAC names of the compounds : (i)
2 – Ethylpentane
(ii)
2,4,4 – Trimethylpentane
(iii) 4 – sec – Butyl – 2, 6 – dimethylheptane (iv) 2 – Methyl – 3 – hepten – 6 – yne. (v)
3 – (2,2 – Dimethylethyl)hexane
(vi) 2 – Chloro – 2 – buten – 4 – ol (vii) Neopentane (viii) 2 – Methyl – 3 – ethyl – 2 – butene (ix) 1,3 – Dimethylpropyne (x)
3 – Ethylbutane
F.
ESSAY TYPE QUESTIONS
1.
What are different types of organic reactions ? Name them with one suitable example in each case.
2.
What are optical isomers ? Mention two conditions for optical activity.
3.
Explain the terms – (a)
Homolytic and heterolytic cleavage
(b)
Nucleophile and electrophile
4.
What is a homologous series ? What are its characteristics ?
5.
Write down the structure of isomeric pentyl radicals. Name them by IUPAC system.
6.
Name the type of isomers the following pair represents (with justification) (i)
CH3CH2CHO and CH3COCH3
ORGANIC CHEMISTRY
(ii)
705
CH3 CH2CH2 CHO and CH3 – CH – CH3 CHO
(iii) CH3 – CH = CH – CH3 and CH2 = CH – CH2 – CH3 (iv) CH3 – CH2 – NH – CH2 – CH3 and CH3NHCH2CH2CH3 (v)
CH3 – CHO and CH2 = CHOH
7.
What is the cause of geometrical isomerism ? Write the structures of the two geometrical isomers of 2 – butene. Name them. Why don't you expect geometrical isomers in case of 2 – butyne ?
8.
Give reasons– (a) (b) (c) (d) (e)
Trichloroacetic acid is more acidic than acetic acid phenol is acidic, but alcohol is neutral Aniline is less basic than methylamine. Acetic acid is acidic. Acid amides are less basic than amines.
9.
Discuss the chemistry for the detection of nitrogen in an organic compound. How is this test modified when both nitrogen and sulphur present in the organic compound are to be detected simultaneously ?
10.
(a)
How are carbon and hydrogen estimated in an organic compound ?
(b)
0.6g of an organic compound gave an combution 0.09g of H2O and 0.55g of CO2. Calculate the percentage composition of the compound if it contains only C, H and oxygen. (Answer C = 25%, H = 1.66%, 0 = 73.34%)
11.
Write a note on chromatography.
12.
Describe Carias method for the estimation of
13.
14.
(i)
Phosphorus
(ii)
Halogens
(iii)
Sulphur
(a)
How is nitrogen estimated in an organic compound by Kjeldahl's method ?
(b)
If NH3 obtained from 0.2g of an organic compound by Kjeldahl's method neutralises 25 ml of 0.1 N H2SO4, what is the percentage of nitrogen in the compound ? (Answer: 17.50)
(a)
How are halogens estimated in organic compound ?
(b)
0.147g of an organic compound containing chlorine was heated with nitric acid and silver nitrate and gave 0.287g of silver chloride. Calculate the percentage of chlorine in the compound (Answer- 48.29)
706
+2 CHEMISTRY (VOL. - I)
ANSWERS A.1. (i)
3, 4 – Diethyl – 2, 5 – dimethylhexane
(ii)
2 – Chlorobutane
(iii)
2, 2, 3 – Trimethylpentane
(iv)
(a) 2 – Chloro – 3 – methylbutanoic acid (b) 2 – Bromo – 3 – methylbutanoic acid
(v)
(a) Buta –1, 3 – diene (b) 2 – Methylbuta – 1, 3 – diene.
(vi)
4 – Methylpent – 2 – ene.
(vii)
2 – Ethyl – 3 – methylbut – 1 – ene
(viii)
Butan – 2 – one
(ix)
3 – Methylbutan – 2 – ol.
(x)
4 – Methyl – 3 – nitropent – 1 – ene.
(xi)
4, 5 – Dimethylhex – 2 – yne.
(xii)
3, 3 – Dimethylbut – 1 – ene.
(xiii)
3 – Methylbutanal. or 3 – Methylbutan – 1 – al.
(xiv)
2 – Methylbuta – 1 – al or 2 – Methylbutanal
(xv)
(a) 2 – Chloropropanoic acid (b) 3 – Chlorobutanal
(xvi)
2 – Methyoxybutane
(xvii)
But – 1 – yne
(xviii)
3 – Bromopropanoic acid
(xix)
3 – Methylpent – 1 – ene.
(xx)
Pent – 2 – yne
(xxi)
Ethylpropanoate
(xxii)
(a) Propane – 1, 2, 3 – tricarbonitrile (b) chloroethene
(xxiii)
Propadiene
(xxiv)
3, 3 – Dimethylpent – 1 – en – 4 – yne
(xxv)
2 – Methylpentane – 1, 5 – dioic acid or 2 – Methylpentanedioic acid.
(xxvi)
N, N – Dimethylmethanamine.
(xxvii) 3 – Amino – 5 – methylheptanal. (xxviii) Ethyl – 4 – chloro – 2, 2 – dimethylbut – 3 – en – 1 – oate. (xxix)
2, 4 – Dimethyl – 3 – hexanamine.
(xxx)
Propane – 1, 2, 3 – tricarboxylic acid.
(xxxi)
2 – Oxoethanoic acid.
ORGANIC CHEMISTRY
707
(xxxii) Ethyl – 3 – formylbutanoate or Ethyl – 3 – methyl – 1 – oxobutanoate. (xxxiii) 2 – Amino – 3 – hydroxypropanoic acid. (xxxiv) 1, 2 – Pentanediamine. (xxxv) 2 – Methyl – 2 – propanamine or tert – Butylamine. (xxxvi) 3, 3, N, N – Tetramethylbutanamine. (xxxvii) 3 – Ethylhex – 2 – ene. (xxxviii) 2, 4, 4 – Trimethyl – 3 – (1 – methylethyl) pent – 1 – ene. (xxxix) 2 – (2 – Methylpropyl) but – 1 – en – 3 – yne. or 2 – Isobutylbut – 1 – en – 3 – yne (xxxx) 2, 4 – Pentanedione. (xxxxi) 4 – (Carboxymethyl) – 3 – methylheptanoic acid (xxxxii) 4 – Methylpent – 2 – ene (xxxxiii)Pentanedioic acid. (xxxxiv)2 – Bromo – 4 – nitropentan – 3 – one (xxxxv) 2 – Methylbutan – 1 – ol (xxxxvi)2 – Ethylbut – 2 – enoic acid 2.
Structural formula : (i)
CH3 – CH2 – CO – CH2 – CH2Br
(ii)
CH3 CO CH2CH2CH3
(iii)
CH3 – CH2 – CH2 – C = CH2 CH3 CH3
(iv)
CH3 – C – CH = CH2 CH3 OH
(v)
CH3 – CH2 – C – CH2 – CH3 CH3
(vi)
CH3 – CH2 – C H – CH2 – Br CH3
708
+2 CHEMISTRY (VOL. - I)
CH3
CH3
(vii)
CH3 – CH – C H – C º C – CH 3
(viii)
CH3 – CH – C H = CH – CH3 CH3 O
(ix)
H – C – OC2H5
(x)
CH3 – CH – CH2 – COOH OH
(xi)
CH3 – CH – CH2 – CH3 OH
(xii)
CH2 = CH – CH = CH3
(xiii)
CH3 – CH2 – CH – CH2 – CH3 OH CH3
(xiv)
CH3 – C – CHO CH3 O
(xv)
CH3 – C – CH – CH = CH2 CH3
(xvi)
CH3 – CH – CH2 – CH – CHO Cl
Br
O (xvii)
O
CH3 – C – CH2 – C – CH3 CH3
(xviii)
CH 3 – CH – CH3 Cl
(xix)
NH2 – CH2 – C = CH – COOC2H5
(xx)
COOH
ORGANIC CHEMISTRY
(xxi)
709
H2C = C – CH = CH2 CH3
(xxii) H2C = C = CH2
3.
(iii) CH3CH2NH2
4.
Butanol
5.
Propanal
6.
(a)
7.
Propanal
8.
sp
9.
Three
10.
CH º C – or CH º C – CH2 – or CH3 – C º C –
11.
Three s and 2p bonds
B.
6.
D.
MULTIPLE CHOICE
5 s and 2p bonds.
(i)
(b)
3 – Ethylhexane
6 s and 3p bonds.
(ii)
2, 2, 3 – Trimethylhexane
1. (c); 2. (a); 3. (c); 4. (b); 5. (a); 6. (b); 7. (a); 8. (c); 9. (b); 10. (b); 11. (a); 12. (d); 13. (a); 14. (a); 15. (c); 16. (d); 17. (d); 18. (c); 19. (c); 20. (b), 21 (b); 22. (b); 23. (d); 24 (d); 25 (c); 26 (a); 27 (c); 28 (d); 29 (c); 30 (d); 31 (b); 32 (a); 33 (a); 34 (c); 35 (b); 36 (c); 37 (c) E.
1.
(i)
2 – Methylpropanoic acid
(ii)
Pentan – 2 – one
(iii)
Ethanoic anhydride
(iv)
Butanoyl chloride
CH3 2.
(i)
CH3 – C –– CH – CH3 (ii) CH3
3.
(i) (iii)
4.
– NO2 , – OH ,
O
O
CH3 – C – CH2 – Cl (iii) C C – C – Cl 6 5
CH3
Butan – 2 – one (ii) 3 – Chloropropanoic acid. 4 – Ethyl – 2 – methylhexane. C = O , – C º N , – COOH H3C
5.
(i)
CH3 – CH2 – CH2 – CH3 ;
(Butane)
H3C
CH – CH3
(2 – Methylpropane)
710
+2 CHEMISTRY (VOL. - I)
(ii)
CH3 – CH2 – CH = CH2
;
(But – 1 – ene) (iii)
CH3 – CH = CH – CH3 (But – 2 – ene)
CH3CH2 – O – CH2 – CH3
;
(Ethoxyethane)
CH3CH2CH2CH2OH (Butan – 1 – ol) H3C
CH3 – O – CH2 – CH2 – CH3
(Methoxy propane)
;
H3C
CH – CH2 – OH
(2 – Methylpropan – 1 – ol)
CH3 CH2 – CH – CH3 OH (Butan – 2 – ol)
H3C H3C H3C
C – OH (2 – Methylpropan – 2 – ol)
6.
(i)
3 – Methylhexane
(ii)
2, 2, 4 – Trimethylpentane
(iii)
4 – Isobutyl – 2, 5 – dimethylheptane
(iv)
6 – Methylhept – 4 – en – 1 – yne.
(v)
4 – Ethyl – 2 – methylheptane
(vi)
3 – Chlorobut – 2 – en – 1 – ol
(vii)
2, 2 – Dimethylpropane
(viii)
2, 2 – Dimethylpent – 2 – ene
(ix)
Pent – 2 – yne
(x)
3 – Methylpentane. qqq
UNIT - XIII
HYDROCARBONS INTRODUCTION Hydrocarbons are the simplest organic compounds composed of only carbon and hydrogen. These are widely distributed in nature in form of petroleum, natural gas and coal. Their use in our daily life is not limited to fuels like LPG, CNG and LNG, but the hydrocarbons are extensively used for the manufacture of polymers like polythene, polystyrene, dyes and drugs. CLASSIFICATION OF HYDROCARBONS Hydrocarbons can be broadly categorised into two types: A. Acyclic or open chain or aliphatic hydrocarbons : These compounds have an open chain of carbon atoms (branched or straight chained) in their molecules which can be further classified into three types depending upon types of carbon-carbon bonds. (i) Alkanes : Alkanes are the saturated hydrocarbons having only carbon - carbon single bonds. For example,
CH3–CH3 Ethane
CH3–CH2–CH3 Propane
CH3 I CH3–CH–CH3 2-Methylpropane
(ii) Alkenes : Alkenes are unsaturated hydrocarbons containing atleast one carboncarbon double bond. CH2 = CH2 Ethene
CH3–CH = CH–CH3 But–2–ene
(iii) Alkynes : Alkynes are unsaturated open chain hydrocarbons with atleast one triple bond between two carbon atoms. CH º CH Ethyne
CH3 – CH – C º CH CH3 3–Methylbut–1–yne
712
+2 CHEMISTRY (VOL. - I)
B. Cyclic or closed chain hydrocarbons : Hydrocarbons containing closed chains or rings of carbon atoms in their molecules are called cyclic hydrocarbons which can be further subdivided into two groups. (i) Alicyclic hydrocarbons : Cyclic hydrocarbons resembling aliphatic hydrocarbons in their properties are termed as alicyclic hydrocarbons. These are of three types. (a) Cycloalkanes : Saturated alicyclic hydrocarbons are called cycloalkanes. For example
Cyclopropane
Cyclobutane
Cyclohexane
Cyclopentane
(b) Cycloalkenes : Unsaturated alicyclic hydrocarbons atleast one carbon-carbon double bond are termed as cycloalkenes.
Cyclohexene
Cyclopropene
(c) Cycloalkyne : Unsaturated alicyclic hydrocarbons containing atleast one carbon-carbon triple bond are callled cycloalkynes. However, due to high strain lower members are unstable. (ii) Aromatic hydrocarbons or Arenes : Hydrocarbons and their alkyl, alkenyl and alkynyl derivatives which contain one or more benzene rings either isolate or fused in their molecules are termed as aromatic hydrocarbons or arenes. For example, CH3
Benzene
Toluene
CH=CH2
Styrene
ALIPHATIC HYDROCARBONS
713
Naphthalene
Diphenyl
The total scheme of classification of hydrocarbons is represented as :
Hydrocarbons
Acyclic or open chain
Saturated (Alkanes)
Cyclic or closed chain
Unsaturated
Alkenes
Alicyclic
Alkynes
Saturated (Cycloalkanes)
Aromatic
Unsaturated
Cycloalkenes
Cycloalkynes
714
+2 CHEMISTRY (VOL. - I)
CHAPTER - 17
ALIPHATIC HYDROCARBONS SATURATED HYDROCARBONS–ALKANES 17.1
SATURATED HYDROCARBONS (PARAFFINS) OR ALKANES
Alkanes are the simplest family of the hydrocarbons. In alkanes, the valencies of carbon atoms are satisfied by single bonds and are almost entirely nonpolar. The chemistry of the alkane is quite simple. A very few chemicals react with them. This is why they are nicknamed as paraffins (Latin : Parum Affinis meaning less affinity). They do not contain functional groups.
17.2
GENERAL FORMULA
Alkanes constitute a homologous series of compounds and are represented by the general formula CnH2n+2, where n is a whole number. In this series one member differs from each adjacent member by one carbon atom and two hydrogen atoms, that is by CH 2. The simplest member of this family contains only one carbon atom (n = 1) and is called methane. It has molecular formula CH4. The next member (n = 2) has the molecular formula C 2H6 and is called ethane. The third member propane is C3H8 and so on. No. of Carbon
CnH2n+2
Name
n = 1
CH4
Methane
n = 2
C2H6
Ethane
n = 3
C3H8
Propane
n = 4
C4H10
Butane
n = 5
C5H12
Pentane
n = 6
C6H14
Hexane and so on
ALIPHATIC HYDROCARBONS
17.3
715
TYPES OF CARBON ATOM IN ALKANES The different carbon atoms forming a carbon chain are distinguished as follows. (i)
A primary (10) carbon atom is one which is linked to one or no carbon atom.
(ii)
A secondary (20) carbon atom is one which is linked to two other carbon atom.
(iii)
A tertiary (30) carbon atom is one which is linked to three other carbon atom.
(iv)
A quaternary (40) carbon atom is one which is linked to four carbon atom.
The hydrogen atoms are designated as primary, secondary or tertiary depending upon the carbon atom to which they are attached. Thus a primary carbon atom should contain atleast three hydrogen atoms. The following compounds will illustrate different types of carbon atoms. 10
10
CH 3
30
CH 10
20
CH 2
CH 3
CH 3 40
C
10
CH 3
0 1
CH 3
The above molecule contains five primary, one secondary, one tertiary and one quaternary carbon atoms.
17.4
IUPAC RULES FOR NAMING ALKANES : The IUPAC rules for the nomenclature of alkanes are as follows. 1.
Use the ending – ane for all alkanes.
2.
Select the longest continuous chain of carbon atoms which forms the parent chain.
3.
Attach a prefix to – ane that specifies the number of carbon atoms in the parent chain.
4.
Number the parent chain from an end such that the carbon atom carrying the substituents gets the lowest numeral. The positions of the substituents are indicated by the locants. (The number of carbon atom, locating the position of the substituents in called LOCANT).
5.
If several substitutents are present, arrange them alphabetically.
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+2 CHEMISTRY (VOL. - I)
6.
Prefix the substituent preceded by the locant to the name of parent hydrocarbon. Two, three and four identical substituents are specified as di–, tri– and tetra– respectively.
7.
Separate the locant from the name of the substituent by a hyphen (–). The name of the alkane is written as one word with the name of the substituent and its numeral serving as prefix.
Example : 1CH 3
(i)
2C
CH3
CH3
2, 2 –Dimethylbutane
3CH 2
Not
2 – Ethyl – 2 – methylpropane
4CH 3
(ii) CH3
CH3
CH3
C
CH
1
2
3
CH2
2, 2, 3 – Trimethylpentane
CH3
4
5
CH3
(iii)
17.5
Not
CH3 — CH2 — CH — Cl | Cl
Not Not
2 – t – Butylbutane
1, 1 – Dichloropropane 1 – Dichloropropane 1, 1 – Chloropropane
STRUCTURE OF ALKANES
In alkanes, the carbon atom is always tetracovalent and the four valencies are directed towards the four corners of a tetrahedron. The tetrahedral carbon atoms are sp3 hybridised and form strong s (sigma) bonds with hydrogens and among themselves. H
´
·
H´· C ·
´ H
H
H
´
·
·
·
C ·
·
´ H
´ H
Electron–dot structure of methane
or
H
C
H
or
CH4
H
Graphic structure of methane
Condensed formula of methane
ALIPHATIC HYDROCARBONS
717
H H ´ ´· ´ H ´· C ´ ´ C · H or ´· · ´ H H
H
H
C
C
H
H
·
H
Electron–dot structure of ethane
H
Graphic structure of ethane
or
CH3 – CH3 or C2H6
Condensed formula of ethane
Higher members are obtained by replacing hydrogen atoms by carbon atoms.
The covalent bonds are denoted by small line (–) and it must be remembered that each carbon atom must be connected with four such covalent bonds. Thus, methane and ethane can be structurally represented as above. The C – H sigma (s) bond is formed by the overlapping of sp3 hybridised orbital of carbon with s – orbital of hydrogen and has the bond length 1.09 A0. The C – C sigma (s) bond is formed by the overlapping of sp3 hybridised orbital of one carbon atom on sp3 hybridised orbital of another carbon atom and has the bond length 1.54A0. Thus alkanes have tetrahedral structures with bond angle 109 O28'. H
109O28' C H
H
H
Tetrahedral structure of methane
17.6
ISOMERISM IN ALKANES Chain isomerism is exhibited by the alkanes.
Chain isomerism is due to the difference in the structure of the carbon chain. Except the first three members – methane, ethane and propane, other alkanes exhibit this chain isomerism. For butane (C4H10) two possible isomers are CH3 – CH2 – CH2 – CH3
CH3 | CH3 – CH – CH3
n – Butane
(isobutane)
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+2 CHEMISTRY (VOL. - I)
The straight chain paraffins are known as normal and usually abbreviated as n. The prefix iso – is used for branched chain hydrocarbon with two methyl groups at the end CH3
of a straight chain
CH
. The prefix neo – is used to denote the hydrocarbon
CH3
which contains three methyl groups on the end carbon (quaternary carbon atom) e.g. CH3 CH3
C
CH2
. For pentane (C5H12), three possible isomers are
CH3
(i)
CH3 – CH2 – CH2 – CH2 – CH3
n – Pentane
CH3
(ii)
CH3
CH
CH2
Isopentane or 2 – Methylbutane
CH3
CH3
(iii)
CH3
C
CH3
Neopentane or 2, 2 – Dimethylpropane
CH3
All higher hydrocarbons show similar chain isomerism and the number of isomers goes on increasing with the increase in the number of carbon atoms. The following table gives the number of isomers for hydrocarbons having different number of carbon atoms. No. of C-atoms in the Hydrocarbon
4
5
6
7
8
9
10
12
No. of Isomers possible 2
3
5
9
18
35
75
355 4347
17.7
15
20 3,66,319
OCCURRENCE
Alkanes are highly widespread in nature. Crude petroleum is a mixture of mostly unlimited quantities of liquid alkanes and higher solid hydrocarbons. Most of the alkanes with lower molecular mass are obtained by the fractional distillation of petroleum and natural gas. Methane, the first member of alkane, is also called marsh gas because it is found in marshy places where it is produced by the bacterial decomposition of dead vegetable matter.
17.8
METHODS OF PREPARATION The alkanes can be prepared by the following general methods.
ALIPHATIC HYDROCARBONS
1.
719
Hydrogenation of alkenes and alkynes Alkanes are formed by passing a mixture of unsaturated hydrocarbon (alkene or alkyne) and hydrogen over finely divided nickel at 200 – 300 0C. This reaction is known as Sabatier Senderens reaction. R
Ni
R C= C
R
R
+ H2
R
R CH
200 3000C
CH
R
Alkene
R Alkane
Ni CH2 = CH2 + H2 Ethene
CH3 CH3 Ethane
200 3000C Ni
R - C º C - R +2H 2 Alkyne
200
R - CH 2 - CH 2 - R Alkane
3000C Ni
+2H2
HC º CH Acetylene
H 3C - CH 3 Alkane
200 3000C
When finely divided Platinum, Palladium or Raney nickel is used as catalyst the above reaction can be carried out at room temperature. Methane can not be prepared by this method. (why ?) 2. Decarboxylation of Carboxylic acids When anhydrous sodium salt of a carboxylic acid is heated strongly with dry sodalime (NaOH + CaO), an alkane is formed. R – COONa (Sodium salt of
+ NaOH
CaO heat
R – H + Na2CO3 (Alkane)
a carboxylic acid)
CH3 – COONa + NaO H (Sodium acetate)
CaO heat
CH4
+ Na2CO3
(Methane)
The process of elimination of carbon dioxide from a carboxylic acid is known as decarboxylation. R – COO
H
NaOH - CaO ¾ ¾¾¾¾¾ ¾®
R – H + CO2 (R is any alkyl group)
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+2 CHEMISTRY (VOL. - I)
The hydrocarbon obtained by this method contains one carbon atom less than the acid from which it is obtained. Laboratory Method of Preparation of Methane : In the laboratory methane is prepared by the decarboxylation of sodium acetate with sodalime. Powdered anhydrous sodium acetate is mixed with two times its mass of dry sodalime. The mixture is heated in a hard glass tube fitted with a delivery tube. The apparatus is arranged as shown in the Fig.17.1 Methane is collected by the downward displacement of water.
Fig. 17.1 Preparation of methane
3.
Wurtz Reaction
When an alkyl halide (mostly) iodide or bromide) is heated with metallic sodium in dry ethereal solution, a higher symmetrical alkane is formed. This reaction is known as Wurtz reaction. Two molecules of alkyl halide react with two atoms of sodium forming a hydrocarbon containing double the number of carbon atoms present in the alkyl halide. R – I + 2Na + I – R Alkylhalide
heat ¾ ¾¾¾ ¾®
Alkylhalide
CH3 – I + 2Na + I – CH3
R – R + 2Nal Alkane
¾ ¾¾¾¾®
CH3 – CH3 + 2Nal
Methyl iodide Methyl iodide (Ethane) Methane cannot be prepared by this method (why ?) Wurtz synthesis is important, because it is used in the preparation of higher hydrocarbon from lower hydrocarbon e.g. methane to ethane. It is not a suitable method for preparing alkanes with odd number of carbon atoms.
ALIPHATIC HYDROCARBONS
721
When a mixture of two different alkyl halides is taken, a mixture of alkanes is obtained. e.g. a mixture of CH3l and C2H5l produces C2H6, C4H10 and C3H8 CH3 I + 2Na + I CH3 C2H5 I + 2Na + I C2H5
dry ¾ ¾¾¾¾¾®
CH3 – CH3 + 2NaI ethane
ether - do ¾ ¾¾¾¾¾¾®
C2H5 – C2H5 + 2NaI butane
CH3
I + 2Na + I C2H5
- do ¾ ¾¾¾¾¾¾®
CH3 – C2H5 + 2NaI Propane
Mechanism : Two mechanisms have been suggested for the Wurtz reaction. Let us start with CH3l in preparing CH3 – CH3 while explaining this mechanism. (i)
Ionic Mechanism : According to this mechanism two atoms of sodium react with methyl iodide molecule to produce methyl sodium and sodium iodide. Methyl sodium then reacts with second molecule of methyl iodide to form ethane and sodium iodide. CH3l + 2Na
¾ ¾¾¾¾®
CH–3 Na+
+ Nal
Methyl sodium CH–3 Na+ + lCH3
¾ ¾¾¾¾®
CH3 – CH3
+ Nal
(Ethane)
(ii)
Free radical Mechanism : According to this mechanism sodium atom attacks methyl iodide to produce methyl free radical and sodium iodide. Two methyl free radicals then combine to give ethane. CH3l + Na
¾ ¾¾¾¾®
Methyle iodide
.
.
CH3 + CH3
.
CH3 + Nal Methyl free radical
¾ ¾¾¾¾®
CH3 – CH3 (Ethane)
4.
Kolbe's synthesis :
This is also known as Kolbe's electrolytic method. When concentrated solution of sodium or potassium salt of a fatty acid is electrolysed, alkane is obtained at the anode and hydrogen gas is liberated at the cathode.
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R – COO– + K+
R – COOK
At the Anode : 2 R – COO –
R – R + 2CO2
At the Cathode : 2 K+
+2e
2K 2KOH + H21
2 K + 2H2O
For example, a solution of potassium acetate on electrolysis gives ethane at anode and hydrogen at cathode. CH3COO – + K +
CH3COOK
At the Cathode : 2 K+
+2e
2 K + 2H2O
2K 2KOH + H2-
Methane can not be prepared by this method. Mechanism : Several mechanisms have been proposed for the Kolbe's reaction. Out of them free radical mechanism in the most favoured one. i.e.
CH3COOK
CH3COO– + K+
At the Anode : The acetate ion is discharged at the anode to form a free radical. –
CH3COO
CH 3COO ·
Acetate ion
Acetate free radical
+e
The acetate free radical then breaks up into the methyl free radical and carbon dioxide. CH 3COO ·
·
C H 3 + CO 2
Then two such methyl radicals combine to form a molecule of ethane. ·
·
C H3 + C H3
CH3 – CH3 (Ethane)
ALIPHATIC HYDROCARBONS
5.
723
Clemmensen's reduction : Aldehydes and ketones are reduced to alkanes by zinc amalgam and concentrated hydrochloric acid. (Clemmensen's reagent) Zn/Hg
RCHO + 4[H] (Aldehyde)
Conc.HCl Zn/Hg
RCOR + 4[H] (Ketone) 6.
Conc.HCl
RCH3 + H2O (Alkane)
R – CH2 – R + H2O (Alkane)
Reduction of Alkyl halides : Reduction of alkyl halides by Zn - Cu couple and ethyl alcohol results in the formation of alkanes. Zn – Cu R – X + 2[H]
R – H + HX Ethyl alcohol Zn – Cu
CH3 – I + 2[H] 7.
Ethyl alcohol
CH3 – H + HI
Reduction of alcohols, ketones and fatty acids : Alcohols, ketones and fatty acids when heated with concentrated hydriodic acid in presence of red phosphorus at 1500C under pressure yield alkanes. Red P (i)
R – OH + 2HI (Alcohol)
150 C, press. 0
R – H + H2O + I2 (Alkane)
Red P CH3 – OH + 2HI (Methanol)
150 C, press. 0
CH4 + H2O + I2 (Methane)
Red P (ii)
R – CO – R + 2HI (Ketone)
150 C, press. 0
R – CH2 – R + H2O + 2I2 (Alkane)
Red P CH3 – CO – CH3 + 4HI (Acetone)
CH3CH2CH3 + H2O + 2I2 150 C, press. (Propane) 0
724
+2 CHEMISTRY (VOL. - I)
Red P
(ii)
R – COOH + 6HI (fatty acid)
R – CH3 + 2H2O + 3I2 (Alkane)
150 C, press. 0
Red P CH3COOH (acetic acid)
+ 6HI
CH3 – CH3 + 2H2O + 3I2 (Ethane)
150 C, press. 0
8. Corey - House synthesis - This reaction is very useful for the preparation of unsymmetrical alkanes and higher members of the alkane family. This reaction occurs in two steps. In the first step, the alkyl halide R – X is treated with lithium metal and solvated in anhydrous ether to form alkyl lithium compound, R-Li. The starting R – X can be primary, secondary or tert. alkyl halide. R – X + 2Li
Anh. ether
R - Li
+ LiX
Alkyl lithium The second step requires the alkyl lithium compound to be treated with cuprous iodide forming lithium dialkyl cuprate compound (known as Gilman reagent). 2R-Li + Cu I
R2Cu Li +
LiI
Lithium dialkyl cuprate This in turn reacts with the second alkyl halide which couples to the compound. R – R¢ + R Cu + Li X
R2Cu Li + R¢ X
(R and R¢ groups may be same or different) Thus, (CH 3 )2 CHCl
(i) Li / Ether (ii) CuI
[(CH ) CH] CuLi 3 2
2
CH3CH2Cl
(Isopropyl chloride) (CH3)2CHCH2CH3 + (CH3)2CH – Cu + LiCl (2-methylbutane) 9. Hydrolysis of Grignard Reagent Alkyl magnesium halides are known as Grignard reagents. When an alkyl magnesium halide in ethereal solution is treated with compounds containing active hydrogen such as water, alcohol, amine etc, an alkane is formed. R – Mg X + H2O
R – H + Mg (OH)X (Alkane)
R – Mg X + R¢ OH CH3 Mg I + H2O CH3 CH2Mg Br + C2H5OH
R – H + Mg (OR¢) X CH4 + Mg (OH) X CH3 – CH3 + Mg (OC2H5) Br (Ethane)
(Ethoxy magnesium bromide)
ALIPHATIC HYDROCARBONS
10.
725
Other methods of preparation of methane : (a)
Methane can be prepared by the action of water on aluminium carbide.
Al4C3 Aluminium carbide
+ 12H2O
3CH4 (Methane)
+ 4Al(OH)3
(b) Lower alkanes, i.e. methane and ethane can be prepared by the direct combination of carbon and hydrogen in an electric arc. 12000C C + 2H2
CH4 (Methane) 12000C
2C + 3H2
C2H6 (Ethane)
17.9. GENERAL PROPERTIES PHYSICAL PROPERTIES : (1)
State – The first four alkanes (i.e. methane, ethane, propane and butane) are colourless gases with petrol like smell, the next thirteen (C 5 – C17) are colourless liquids and higher alkanes (C18 onward) are colourless solids.
(2)
Solubility – In hydrocarbon both the carbon-carbon and carbon-hydrogen bonds are non-polar, so the hydrocarbon molecules are non-polar in nature and are insoluble in water, whereas they dissolve in non-polar solvents like benzene, carbon tetrachloride etc. Their solubility decreases with the increase in their molecular mass.
(3)
Density – The density of normal alkanes steadily rises for lower members with the increasing chain length, but it reaches a limiting value of about 0.79 with n - hexadecane. Thus alkanes are always lighter than water.
(4)
Boiling points – The boiling point rises regularly as the number of carbon atom in the alkane increases. But branched chain isomers have lower boiling points than the straight chain isomers, which is due to decrease in their surface area.
(5)
Melting points – Unlike boiling point, the melting point of alkanes does not increase regularly with the increase in chain length. The increase in melting point is relatively more in moving from an alkane having odd number of carbon atoms to the next higher alkane, than in moving from an alkane having even number of carbon atoms to the next higher alkanes.
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This is because the intermolecular forces in a crystal depend not only on the size of the molecule, but also on how they are packed into crystal. Alkanes having even number carbon atoms have their end carbon atoms on opposite sides, and alkanes having odd number carbon atoms have their end carbon atoms on the same side of the molecule. C C
C C
C
C
C
C
C
C
(Even numbered carbon chain)
C C
C
(Odd numbered carbon chain)
Probably, alkanes with even number of C – atoms fit-well into the crystal lattice, so as to permit greater intermolecular attraction. Therefore, the increase in the melting point with increase in chain length is not uniform.
M.P
1
2 No. of C
3 atom
4
5
6
7
Physical properties of the alkanes are given in the following table 17.1 and 17.2.
ALIPHATIC HYDROCARBONS
727
TABLE – 17.1 Physical Properties of the alkanes No. IUPAC Name
Structure
Boiliing Point 0 C
Melting Point 0 C
Density g/ml at 200C
Heat of Combustion KCal/mole
1.
Methane
CH4
– 162
– 183
0.424
210.8
2.
Ethane
CH3CH3
– 88
– 172
0.546
368.4
3.
Propane
CH3CH2CH3
– 42
– 188
0.501
526.3
4.
Butane
CH3(CH2)2CH3
0
– 135
0.579
684.0
5.
Pentane
CH3(CH2)3CH3
36
– 130
0.626
838.3
6.
Hexane
CH3(CH2)4CH3
69
– 95
0.659
989.3
7.
Heptane
CH3(CH2)5CH3
98
– 91
0.684
1149.9
8.
Octane
CH3(CH2)6CH3
126
– 57
0.703
1302.7
9.
Nonane
CH3(CH2)7CH3
151
– 54
0.718
1463.8
10. Decane
CH3(CH2)8CH3
174
– 30
0.730
1610.2
11. Undecane
CH3(CH2)9CH3
196
– 26
0.740
1776.3
12. Dodecane
CH3(CH2)10CH3
216
– 10
0.749
1932.6
s
TABLE – 17.2 Physical Properties of hexane isomers. Isomer
Structure
BP 0 C
M.P 0 C
Density at 200C
n – Hexane
CH3(CH2)4CH3
68.7
– 94
0.659
63.3
– 118
0.664
60.3
– 154
0.653
58.0
– 129
0.661
49.7
– 98
0.649
CH3 3 – Methylpentane
CH3 CH2 CHCH2
CH3
CH3 2 – Methylpentane
CH3 CH2 CH2 CHCH3
2.3 – Dimethylbutane CH 3
CH3
CH3
CH
CH
CH3
CH3 2,2 – Dimethylbutane CH3
CH2
C CH3
CH3
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+2 CHEMISTRY (VOL. - I)
CHEMICAL PROPERTIES Alkanes are relatively stable to most of the common reagents at room temperature. They do not react with acids like hydrochloric acid and sulphuric acid, bases like sodium hydroxide or potassium hydroxide, oxidising agents like potassium permanganate or sodium dichromate, reducing agents like stannous chloride and lithium-aluminium hydride and active metals like sodium or potassium. The relative stability or inactivity of alkanes may be explained by considering the nature of C – C and C – H bonds presents in their molecules. (sp s) s bond. sp3 S Almost non-polar +
C
H
C
H
(sp3 +
C
C
C
sp3) s bond. (non-polar) C
Since the electronegativities of carbon (2.60) and of hydrogen (2.1) do not differ appreciably, the bond electrons in C – H are equally shared between them. Thus C – H bonds in alkanes are almost non-polar and the same is true for C – C bonds. Hence polar and ionic reagents (acids, alkalis etc.) find no reaction sites on alkane molecules to which they can be attracted. Alkanes mainly undergo two types of reactions : (A)
Substitution reactions
(B)
Thermal and catalytic reactions.
(A)
Substitution Reactions : In substitution reaction, an atom or a group present in a compound is replaced by another without the compound undergoing any change in its structure. The products thus obtained are called substitution products. In the substitution reactions of alkanes the direct replacement of one or more hydrogen atom by halogen atoms or other monovalent groups takes place.
1.
Halogenation : In general, when an alkane, RH reacts with a halogen molecule X 2, a hydrogen atom of the alkane is replaced by one of the halogen atoms, fluorine (fluorination)
ALIPHATIC HYDROCARBONS
729
or chlorine (chlorination) or bromine (bromination) or iodine (iodination) The
products of this reaction are alkyl halide and hydrogen halide and the reaction by which they are produced is called halogenation. RH Alkane
+
X2 Halogen
RX + HX Alkyl halide Hydrogen halide
The order of reactivity of halogens is : F2
>
Cl2
>
Br2
>
I2
The ease of substitution in an alkane is in the following order : C C
C
>
H
C i.e.
(a)
>
Tertiary
C CH2 C Secondary
>
C – CH3
>
Primary
Chlorination : Chlorination may be brought about by light, heat or catalyst and the extent of chlorination depends largely on the amount of chlorine used.
For example, methane reacts with chlorine in the presence of ultraviolet light or at high temperature (3000C) to yield methyl chloride or chloromethane and hydrogen chloride. uv light CH4 (Methane)
+ Cl2
or heat
CH3Cl + HCl Methyl chloride or Chloromethane
The reaction does not stop at this stage and the remaining H – atoms of CH 3 – Cl are successively replaced by chlorine atoms. CH3Cl + Cl2
CH2Cl2 Methylene chloride or Dichloromethane
+ HCl
CH2Cl2 + Cl2
CHCl3 Chloroform or Trichloromethane
+ HCl
CHCl3 + Cl2
CCl4 Carbon tetrachloride or Trichloromethane
+ HCl
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In the presence of direct sunlight, methane reacts with chlorine with explosion producing carbon. Direct CH4 + 2Cl2
C + 4HCl Sunlight
Mechanism of chlorination : The chlorination of alkanes takes place through the formation of free radicals as intermediates. An important feature of the mechanism is that the chlorine atom consumed in the first step is replaced by another chlorine atom in the second step. This type of process is called a chain reaction. Thus, for the chlorination of methane the following steps have been proposed. (1)
Cl
:
Cl. + Cl.
Cl
Chain Initiation
H (2)
Cl.
+ H : C
H
+ HCl CH3 . (Methyl free radical)
H (3)
CH3.
Chain propagation
CH3Cl + Cl. (Methyl chloride)
+ Cl2
The propagation steps 2 and 3 occur in competition with chain termination steps 4, 5 and 6 (4) (5) (6)
CH3. Cl. CH3.
+ Cl. + Cl. + CH3
.
CH3Cl Cl2 CH3 – CH3
Chain termination
When the concentration of CH3Cl produced as a result of chain propagation steps increases sufficiently, then it can combine with a chlorine free radical to produce chloromethyl radical and HCl. This radical participates further in chain reaction to form dichloromethane and chlorine free radical which can continue the chain by reacting with another molecule of CH3Cl
ALIPHATIC HYDROCARBONS
731
H Cl. + H : C
HCl +
Cl
. CH Cl 2
(Chloromethyl radical) H
. CH
2Cl
+
CH2Cl2 + Cl. (Dichloromethane)
Cl : Cl
Similarly, trichloromethane and tetrachloromethane are obtained by further chain reactions. (b)
Bromination : Bromination is similar to chlorination, but it is slower. The substitution products obtained are exactly similar. The reaction with methane is given below.
CH4
Br2
CH3Br
Br2
Bromo– methane methane (c)
CH2Br2
Br2
Dibromo– methane
Br2
CHBr3 Tribromo– methane
CBr4 Tetrabromo–
Iodination : Iodine reacts ith alkanes reversibly and slowly. The HI formed is a powerful reducing agent and is capable of reducing alkyl iodides to alkanes.
CH4 + I2 CH3I + HI (Methane) (Methyl iodide) However, alkanes can be iodinated in presence of an oxidising agent such as iodic acid (HIO3) or nitric acid which destroys HI as it is formed. 5HI + HNO3 3H2O + 3I2 (d) Fluorination : Fluorine is the most reactive of the halogens towards alkanes. Pure fluorine reacts with alkanes explosively under most conditions. Fluorination in alkanes can take place by the action of fluorine diluted with nitrogen. 2.
Nitration : Alkanes do not react with nitric acid at ordinary temperature. Under certain
conditions, alkanes react with nitric acid forming nitroalkanes. This process is known as nitration. Nitration of lower alkanes is carried out in the vapour phase at elevated temperature (150 – 475 0C) Vapour phase RH Alkane
+
HONO2 Nitric acid
RNO2 Nitroalkane
+
H 2O
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+2 CHEMISTRY (VOL. - I)
Vapour phase CH4 Methane
+
HONO2
nitration
CH3NO2 + Nitromethane
H2O
Higher hydrocarbons can be nitrated in liquid phase and form a mixture of mononitroalkanes, which consists of every possibility of chain fission of alkanes. For example, propane gives a mixture of 1–nitropropane, 2–nitropropane, nitroethane and nitromethane.
CH3CH2CH3 (Propane)
NO2 HNO3 ` CH3CH2CH2NO2 + CH3 – CH – CH3 + CH3CH2NO2 + CH3NO2 4000C (I – nitropropane) (2 – nitropropane) (nitroethane) (nitromethane) 25% 40% 25% 10%
The amount of 1–nitro and 2–nitropropanes indicates that 2 0H > I0H in reactivity. 3.
Sulphonation : It is the process of replacing a hydrogen atom by a sulphonic acid group – SO3H.
Sulphonation of normal alkanes from hexane onwards may be carried out by treating the alkanes with fuming sulphuric acid H2S2O7 (i.e. H2SO4 + SO3). Lower hydrocarbons are not sulphonated. RH (Alkane)
+
HOSO3H (Sulphuric acid)
RSO3H + (Alkyl sulphonic acid)
H2O
Replacement of tertiary hydrogen atom easily takes place by sulphonic acid group. For example, SO3H CH3
CH
CH3 + HOSO3H
C
CH3 + H O 2
CH3
CH3
(Isobutane) B.
CH3
(t – Butylsulphonic acid)
Thermal and catalytic reaction : (1)
Thermal decomposition, pyrolysis or cracking : Thermal decompositon of alkanes to produce lower hydrocarbons (saturated or unsaturated) is known as pyrolysis or cracking.. When alkanes are heated as high temperature in the absence of air, cracking occurs.
Alkane (large molecule) (i)
Crack
Alkane (small molecule) + Alkene + Hydrogen
Ethane when passed through a hot metal tube (5000C) in the absence of air, it breaks to yield a mixture of methane, ethylene and hydrogen.
ALIPHATIC HYDROCARBONS
733
5000C
3CH3 – CH3 (Ethane) (ii)
2 CH2 = CH2 + H2 + 2 CH4 (Ethylene) (Methane)
Cracking of propane at 6000 C, gives a mixture of ethylene. methane and hydrogen. 5000C
3CH3 – CH3 (Ethane) 6000C
3CH3 – CH2 – CH3 (Propane)
2 CH2 = CH2 + H2 + 2 CH4 (Ethylene) (Methane) CH3 – CH = CH2 + H2 + 2CH2 = CH2 + 2CH4 (Propylene) (Ethylene) (Methane)
When cracking is carried out in the presence of a catalyst (finely divided silica-alumina), the process is called catalytic cracking. (2) Oxidation : (a) Combustion : The process in which alkanes readily burn on heating in the presence of air or oxygen producing CO2 and H2O is called combustion. CH4 + 2O2 ® CO2 + 2H2O Since this is an exothermic reaction, alkanes which are the constituents of LPG, Diesel, Kerosine oil are widely used as fuels. General combustion reaction can be expressed as æ 3n + 1 ö C n H 2n + 2 + ç ÷ O 2 ® nCO 2 + (n + 1)H 2 O è 2 ø During incomplete combustion of alkanes with limited supply of air or oxygen, CO is produced along with unburnt carbon in the form of carbon black or soot which is used in the manufacture of ink, paints and polishes.
2CH4(g) + 3O2(g) ® 2CO(g) + 4H2O (a)
Catalytic oxidation : Controlled oxidation of alkanes gives different products depending upon the conditions. The reaction takes place with oxygen at high pressure in presence of a catalyst.
RCH3 Alkane (i)
O
RCH2OH Alcohol
O
RCHO Aldehyde
O
RCOOH Carboxylic acid
When a mixture of methane and oxygen in the ration 9 : 1 is passed through a copper tube at 2000C and 100 atmospheric pressure, methyl alcohol is formed. CH4 + O
(ii)
CH4(g) + O2(g) ® C(s) + 2H2O
Cu
CH3OH 200 C, 100 atm 0
When methane mixed with oxygen is passed over molybdenum, it is oxidised to formaldehyde. CH4 + O2
Mo
HCHO + H2O
ALIPHATIC HYDROCARBONS
735
CONFORMATION OF ALKANES Alkanes contain carbon-carbon sigma (s) bonds. The electron distribution of the sigma molecular orbial is symmetrical around the internuclear axis of the C – C bond. Thus, ethane which contains such a C - C sigma bond maintain a full degree of overlap while its two ends rotate. Hence the energetic barrier to rotation about sigma bond is generally very low. The different spatial arrangements formed by rotation about a single bond are called conformations or conformers or rotamers. H
H H
H
C C
H rotation by 0 180
C
H
H
C
H
H
H H
H
Fig. : 17.2 Free rotation about C – C bond in ethane Visualising Conformation : Several methods are available for visualising conformation. One of these methods uses wedges to denote bonds that are extending out from the plane of the page towards the reader and dashes to indicate bonds that are going into the plane of the page away from the reader. This notation is generally used to represent the tetrahedral geometry of sp 3 hybridised carbon. Conformations of Ethane Ethane molecule (C2H6) contains C – C single bond and each carbon atom is attached to three hydrogen atoms. Keeping one carbon atom stationary, one can rotate the other carbon atom around the C – C axis. This rotation results infinite number of spatial arrangements of hydrogen atoms attached to one C-atom with respect to hydrogen atoms attached to the other C-atom. These are called conformational isomers or conformers or rotamers. Out of several conformers there are two extreme cases. One in which the H-atoms attached to two C-atoms are as close as possible is called eclipsed conformation and the other in which such H-atoms are as far apart as possible is called staggered conformation. Any other intermediate conformation is called a skew conformation. Eclipsed and staggered conformations can be represented by Sawhorse and Newman projections. 1. Sawhorse Projection Here, the molecule is viewed along the molecular axis. On the paper, it is projected by drawing the central C – C bond as a long straight line diagonally from upper right side to lower left side. The upper end is the rear C and the lower end is the front C-atom. The lines are inclined at an angle of 1200 to to each other. Eclipsed and staggered conformation are shown below (fig17.3) H H H
H
H
H
H
H
(i) Eclipsed
H
H H
H
(ii) Staggered Fig : 17.3 Sawhorse projections of ethane
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+2 CHEMISTRY (VOL. - I)
2. Newman Projection A Newman projection can be used to specify the conformation of a particular bond with clarity and detail. Newman projection represents the head-on look down the bond of interest. The circle in the projection represents the rear C-atom with three hydrogen atoms attached to it by the shorter lines at an angle of 120O to each other. The front carbon atom is represented by a point at the centre of the circle from which three lines are drawn at an angle of 120 O referring to three hydrogen atoms. rear hydrogen
H
H
H
H t Fron
H C
n Carbo
Rear Carbon
C H
H
H
H look down through
Front hydrogen
H
Fig : 17.4 How to draw Newman projection Newman projection can be characterised by the angle formed between the bonds on the front atom and bonds on the rear atom. Such angles are termed as dihedral angle. The full 3D shape of any molecule can be described by its bond length, bond angle and dihederal angle. In the eclipsed conformation, the C – H bonds in the front and back carbons are aligned with each other with dihedral angle of 0 degree. In the staggered conformation, the C – H bonds on the rear C lie between those on the front C with dihedral angle of 60 degrees.
Dihedralangle is actually 00 but it is tilted slightly to make near bonds visible H
00 H
H
60 0 H
H H
H
(Eclipsed) (eclipsed)
H
H
H
H H
(Staggered)
Fig. 17.5 Eclipsed and staggered conformations of Ethane
ALIPHATIC HYDROCARBONS
737
Energetically, not all conformations are equally formed. The eclipsed conformation of ethane is less stable than staggered one by 3 kcals / mol. The staggered conformation is most stable of all possible conformations of ethane, since the angle between C – H bonds of the front and rear carbon one maximised at 600. In the eclipsed form, the electron densities on the C – H bonds are closer together than they are in the staggered form. When the C – H bonds are brought into a dihedral angle of 0 degree, the electron clouds experience repulsion, which raises the energy of the molecule. The eclipsed conformation of ethane has 3 such C – H eclipsing interaction, so we can infer that each eclipsed C – H costs roughly 1 kcal/mol. H
H
H
H H
H
Eclipsing interaction, steric hindrance between C – H bonds at O0 dihedral angle
Fig. 17.6 Eclipsing interaction in ethane Eclipsing interactions are the examples of a general phenomenen called steric hindrance, which occurs whenever bulky portions of a molecule repel other molecule or other part of the same molecule. Because such hindrance causes resistance to rotation, it is also called torsional strain. The 3 kcal / mol needed to overcome the resistance is the torsional energy. This energy is very small and at room temperature, the collisions of ethane molecules supply sufficient kinetic energy to overcome this energy barrier. Due to this rapid rotation it is impossible to isolate any particular conformer. Therefore, conformers are not true isomers because of rapid interconversion. The variation of energy of the conformations of ethane with rotation about C–C single bond is shown in Fig. 17.7. H H Eclipsed (less stable) H H
H H
Eclipsed
Eclipsed
Eclipsed
Potential energy
3 Kcal/mole
Staggered H
H
O
0
H
H
H
H
H
H
H
H
H
H
Staggered (Stable)
Staggered (Stable)
60
O
120
O
O
180
O
240
O
300
360
O
Angle of rotation
Fig. 17.7 Variation of energy during rotation about C–C single bond in ethane molecule.
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CHAPTER (17) AT A GLANCE Preparation of alkanes : Reduction, Ni / 3000C
CnH2n (Alkene)
Reduction, Ni / 3000C
CnH2n – 2 (Alkyne)
R – H (Alkane)
Distil with sodalime
RCOONa (Sodium alkanoate)
Na / Dry ether
RX. (Alkylhalide)
R – R (Alkane)
Kolbe’s electrolysis
R COONa (Alkylhalide)
Zn – Hg Couple R – CHO (Aldehyde)
R – CH3 (Alkane)
Conc.HCl Zn – Cu Couple
R – X (Alkyl halide)
R – H (Alkane)
C2H5OH HI / Red P
R – OH (Alcohol)
R–X
Reduction
(i) Li / ether (ii) Cu I
R2Cu Li
(Alkyl halide R – X + Mg (Alkyl halide)
R'X
R – R¢ [Corey House synthesis] (Alkane)
ether
R – Mg – X
H2O R – H
(Grignard reagent)
(Alkane)
ALIPHATIC HYDROCARBONS
739
Chemical properties of alkanes :
Chlorination / Bromination / RCl, RBr, RI, RF Iodination / Fluorination HNO3 (Nitration) R – H (Alkane)
R – NO2
H2SO4 (Sulphonation)
R – SO3H + H2O
Catalytic oxidation Alcohol / Aldehyde / Acid Chemical oxidation Alcohol Combustion CO2 + H2O
Cracking / Pyrolysis
Lower hydrocarbon Isomerisation Isomeric alkanes. Aromatisation (For n-alkanes having six or more C-atoms
Aromatic hydrocarbons
QUESTIONS
A.
SHORT QUESTIONS : (One mark each) 1.
How many isomers are possible for a compound of molecular formula C 4H10?
2.
Marsh gas mainly contains
3.
What is the angle between any two bonds in methane ?
4.
Give the IUPAC name of
—————————
(H2S, CO, C2H2, CH4)
CH3 — CH2 — CH — CH — CH3 CH3 B.
CH3
SHORT QUESTIONS : (Two marks each) 1.
How do you get ethane from acetylene ?
2.
How can you convert methane to ethane ?
740
C.
+2 CHEMISTRY (VOL. - I)
3.
Write a short note on 'Wurtz reaction'.
4.
What happens when sodium propionate is heated it sodalime ?
5.
What happens when ethane reacts with nitric acid at 400 0C ?
6.
What is Wurtz reaction ? Give an example of it.
7.
State with equation what happens when aqueous solution of sodium acetate is electrolysed.
8.
What happens when ethyl iodide is heated with sodium in dry ethereal solution? Give equation.
9.
Two moles of sodium acetate will produce how many moles of methane and how many grams of methane ?
OTHER SHORT QUESSTIONS : 1.
Complete the reaction (a)
CH4 + 2F2
(b)
Al4C3 + H2O
2.
A hydrocarbon having the formula CnH2n +2 diffuses twice as fast as SO2 at the same temperature. Calculate the value of n. [Ans. n = 1]
3.
Name the alkanes which are obtained when a mixture of methyl iodide and ethyl iodide is treated with metallic sodium and dry ether. Explain the formation of alkanes with equations.
4.
There are five isomers of C6H14. Write their structures and IUPAC names.
5.
Identify A and B. (i)
A + H2O
(ii)
A + Na
CH4 + B Ether
C2H6 + B. Heat
(iii) CH3COONa + A
CH4 + B.
KOH
(iv) CH3COOH
Kolbe’s reaction
A
Zn Cu Couple
(v) 6.
CH3Br
and alcohol
A
B. Cl2
Balance and complete the following equation : (i)
Hexane + O2
D
B.
ALIPHATIC HYDROCARBONS
(ii)
741
2 – Bromopropane + Na 5000C
(iii) Methane + HNO3 (iv) Isobutyl bromide + Zn + HCl Pt (v)
Neopentyl chloride + H2
(vi) n – Butane
AlCl3 / HCl D
ANSWERS
(i) (ii)
D 2C6H14 + 19O2 2(CH3)2 CHBr + 2Na
12CO2 + 14H2O (CH3)2CH – CH(CH3)2 + 2NaBr
5000C
(iii) CH4 + HNO3
CH3NO2 + H2O 2H
(iv) (CH3)2CH CH2 Br
Zn HCl
(CH3)3CH
2H
(v)
(CH3)3CCH2 Cl
(vi) CH3CH2CH2CH3 7.
Pt
(CH3)4C
AlCl3 / HCl
CH3 CH3
CH
CH3
Fill in the blanks : (i) Ethane can be prepared by the hydrolysis of ——————— . (ii) Ethane can be prepared by the hydrogenation of ——————— . (iii) Hydrogen iodide reduces
————————
to alkane in the presence of
————
(iv) n – Hexane has ———————— boiling point than neohexane. (v) Neopentane has ————————. number of quaternary hydrogen atoms. (vi) A saturated alkyl group has the general formula (vii) The common name of tribromomethane is
————————
————————
.
.
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+2 CHEMISTRY (VOL. - I)
Answers :
8.
D.
E.
(i) (iv) How (i) (ii)
C2H5MgX (ii) Ethene (iii) alkyl halide, red phosphorous. higher (v) nil (vi) CnH2n+1 (vii) Bromoform. would you carry out the following conversions ? Methane to Ethane and vice versa. Ethane to Propane and vice versa.
LONG QUESTIONS : 1. Write notes on : (a) Wurtz reaction. (b)
Kolbe's reaction.
(c)
Substitution reaction in alkanes.
(d)
Pyrolysis.
(e)
Sabatier and Senderens reaction.
(f)
Combustion
2.
Describe any three methods for the preparation of aliphatic saturated hydrocarbons and their general chemical reactions.
3.
How methane can be prepared ? What are the chief properties of these compounds ? Explain the statement "Methane is a saturated hydrocarbon."
4.
How does methane react with halogens ? Give the order of reactivity of halogens with alkanes and the order of reactivity of alkanes with halogens.
5.
Describe a method for the preparation of ethane. From ethane how would you obtain methane and vice versa ?
6.
Explain the following with suitable examples. (a)
Isomerisation of alkanes.
(b)
Chain isomerism in alkanes
(c)
Cracking of alkanes.
(d)
Substitution reactions of alkanes.
MULTIPLE CHOICE QUESTION : 1.
Which of the following compound in likely to have the lowest boiling point ? (a)
2.
n – Butane
(c) n – Pentane Marsh gas mainly contains. (a) C2H2 (c) H2S
(b)
Ethane
(d)
Propane
(b) (d)
CH4 CO
ALIPHATIC HYDROCARBONS
3.
4.
The (a) (c) The
743
highest boiling point in expected for isooctane (b) 2, 2, 3, 3 – tetramethylbutane n – Octane (d) n – Butane reactions conditions leading to the best yield of C 2H5Cl are UV
(a)
C2H6 (excess) + Cl2
light dark
(b)
C2H6 + Cl2
room temp. UV
(c)
C2H6 + Cl2 (excess)
light UV
(c) 5.
6.
7.
8.
9.
10.
C2H6 + Cl2
light
Which of the smallest alkane is obtained by Wurtz reaction : (a) CH4 (b) C2H6 (c) C6H10 (d) C6H14 The compound having one isopropyl group is (a) 2, 2, 3, 3 – tetramethylpentane (b) 2, 2 – dimethylpentane (c) 2 – methylpentane (d) 2, 2, 4, 4 – tetramethylpentane The maximum number of the isomer of C5H12 are (a) 3 (b) 2 (c) 4 (d) Unlimited When sodium is treated with an equimolar mixture of CH 3Br and C2H5Br which of the following is not formed. (a) Ethane (b) Butane (c) Propane (d) Methane An alkane is more likely to react with (a) Free radical (b) An alkali (c) An electrophile (d) A nucleophile On electrolysis of concentrated solution of sodium acetate ——————————— gas is liberated. (a) Methane (b) Ethane (c) Ethylene (d) Butane
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+2 CHEMISTRY (VOL. - I)
11.
12.
13.
14.
15.
16.
When sodium acetate is heated with sodalime ——————————— is formed. (a) Methane (b) Ethane (c) Ethylene (d) Acetaldehyde Which of the following is most stable. (b) (CH3)2CH Å (a) CH3 Å (c) (CH3)3C Å (b) CH3CH2 Å [Hint : The stability of carboinium ion is in the order tert > sec > prim] The reactivity of hydrogen atom in an alkane towards substitution by bromine atom is (a) 1OH > 2OH > 3OH (b) 1OH < 2OH < 3OH (c) 1OH > 2OH < 3OH (b) 1OH < 2OH > 3OH Isomerization in alkane may be brought about by using (a) Al2O3 (b) Fe2O3 (c) AlCl3 and HCl (b) Conc. H2SO4 The major product obtained in the photobromination of 2-methyl butane is (a) 1-bromo-2-methylbutane (b) 1-bromo-3-methylbutane (c) 2-bromo-3-methylbutane (d) 2-bromo-2-methylbutane Electrolysis of concentrated solution of sodium salt of propanoic acid produces the hydrocarbon (a) Methane (b) Ethane (c)
Propane
(d)
Butane
ANSWERS 1. (b)
2. (b)
3. (b)
4. (a)
5. (b)
6. (c)
7. (a)
8. (d)
9. (a)
10. (b)
11. (a)
12. (c)
13. (b)
14. (c)
15. (d)
16. (d)
qqq
CHAPTER - 18
UNSATURATED HYDROCARBONS - ALKENES 18.1
ALKENES (OLEFINS)
Alkenes are hydrocarbons containing carbon to carbon double bond(s) (C = C) in their molecules. They have the general formula CnH2n, where n is the number of carbon atoms. Alkenes contain two hydrogen atoms less then the corresponding alkanes and are thus designated as unsaturated hydrocarbons. 2H
CnH2n + 2
CnH2n
(Alkane)
(Alkene) 2H
CH3 – CH3
CH2 = CH2
(Ethane)
(Ethane or ethylene)
These hydrocarbons are called olefins. The name 'olefin' arose from the fact that the ethylene was called 'olefiant gas' (oil forming gas), since it forms oily liquid when treated with chlorine. The carbon to carbon double bond (C = C) in an alkene consititutes the functional group and largely determines its chemical behaviour. The double bond is commonly referred to as the olefinic or ethylenic bond. Since there can be no alkene with one carbon, the first member of the series has the molecular formula C2H4 (n = 2) and is commonly known as ethylene. The second member of this family has the molecular formula C 3H6 (n = 3) and is commonly known as propylene, These may be represented by the following structures. H
H
H
C= C H
H
Ethylene (C2H4)
CH3
H
C=C
H
Propylene (C3H6)
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+2 CHEMISTRY (VOL. - I)
In general alkene may be represented by the formula R1
R3 C= C
R2
18.2
where R1,2,.3,4. = H or alkyl group
R4
NOMENCLATURE OF ALKENES
According to the IUPAC system of nomenclature, the suffix for the class of olefins is –ene. While naming the alkene the following rules are followed. 1.
Use the ending ene for alkenes.
2.
Choose the parent chain containing the double bond, as the longest chain of carbon.
3.
Number the parent chain from the end that gives the lowest possible number to the carbon involved in the double bond.
4.
Attach the numerical prefix to – ene that specifies the position of the double bond, preceded by the parent hydrocarbon part.
5.
Use the suffix diene, triene if the alkene contain two or three double bonds respectively.
6.
Use the number of first carbon involved in each double bond as its locant.
7.
Indicate the side chains or substituents as in the case of alkanes.
Examples :
1.
CH2 = CH2
Ethene or Ethylene.
2.
CH3 – CH = CH2
Propene or Propylene
CH3
3.
CH3
C
CH2
2 – Methylpropene Not
4.
CH3 – CH2 – CH = CH2
2 – Methyl prop-1-ene.
But – 1 – ene Not 1 – Butene or 3 – Butene
5.
CH3 – CH2 – CH2 – CH = CH – CH3
Hex – 2 – ene Not
2 – Hexene
Not
4 – Hexene.
UNSATURATED HYDROCARBONS - ALKENES
747
CH2 – CH3
18.3
6. CH3 – C – CH2 – CH2 – CH = CH2 5 – Chloro – 5 – methylhept – 1 – ene. Not 5 – Chloro – 5 – methyl – 1 – heptene Cl Not 5 – Chloro – 5 – ethylhex-1 – ene Not 2 – Chloro – 2 – ethylhex-5 – ene STRUCTURE OF ALKENE Alkenes can be represented by the general structural formula R1
R3 C
C
R2
R4
where R = Alkyl group (same or different) or H atoms. The simplest member ethylene can be represented as H C H
H
H
H C
or
C H
All the atoms lie in one plane.
C
H
H
(Lewis electron-dot structure)
The simplest alkene, ethylene (C2H4) contains one carbon-carbon double bond, which consists of a sigma ( s ) bond and a pi (p ) bond. The strong sigma ( s ) bond is formed by the end-on overlap of the sp2 – orbitals of two carbon atoms and the weak pi (p ) bond, by the side-wise overlap of their 2pz orbital. The s-orbital of hydrogen atoms overlaps on sp2 orbitals of carbon atoms forming sigma ( s ) bonds. H 0
C
1200
1.34A0
C
1200
s
H
H
(a) Fig. 18.1 : (a)
H
A 09 . 1
1200
(b)
End-on overlap of two sp 2 orbitals forming a sigma bond, lateral overlap of two 2pZ orbitals forming a pi (p ) bond. End-on overlap of s-orbitals of hydrogen on sp2 orbitals of carbon forming sigma ( s ) bond. (b) Planar representation of ethylene molecule.
748
+2 CHEMISTRY (VOL. - I)
The H – C – C and H – C – H bond angles are 1200 and the C – H bond length is 1.09A0, whereas C – C bond length is 1.34A0 as compared to C – C bond length 1.54A0 in ethane. The bond strength of a carbon double bond is found to be 147 kcal / mole ( s bond = 83kcal, p bond = 64 kcal) which is more than carbon carbon single bond. Because of the higher bond strength the two nuclei are pulled closer and the bond length is shorter by about 0.2A0 than the carbon-carbon single bond.
C
1.34A0
C
C
1.54A0
C
The alkenes are more reactive than alkanes because of the high electron density of p – electron cloud, which makes the double bond a nucleophile. Due to lateral overlap, the pi-bond is weak and pi–electron pair is more mobile and available easily for electron seeking reagents (electrophiles). The energy required to break a p –bond is only 70% of that required for breaking a s – bond. Hence the double bond, instead of being a source of strength is really a vulnerable point of attack by outside reagents. Symmetrical and Unsymmetrical alkenes : Alkenes are said to be symmetrical if the atoms and groups attached on all the carbon atoms of the carbon to carbon double bond are same, such as in but – 2 – ene. CH3 – CH = CH – CH3 (But – 2 – ene) Alkenes are said to be unsymmetrical, if one carbon of C = C is substituted with atoms and groups different from those in other carbon, such as in propylene. CH3 – CH = CH2 (Propylene) 18.4
ISOMERISM The presence of double bonds in alkenes increases the possibility of isomerism. Alkenes
exhibit three types of isomerism. 1.
Chain isomerism : Alkenes having the same molecular formula but differing in the structure of the parent
chain due to branching are chain isomers.
UNSATURATED HYDROCARBONS - ALKENES
749
Example : CH3 CH3 – CH2 – CH = CH2 (But – 1 – ene) 2.
CH3 – C = CH2 (2 – Methylpropene)
Position isomerism :
Alkenes having same molecular formula, but differing in the position of double bond in the same chain are position isomers. CH3 – CH2 – CH = CH2 (But – 1 – ene) 3.
CH3 – CH = CH – CH3 (But – 2 – ene)
Geometrical isomerism :
In alkenes there is no free rotation about the carbon carbon double bond. If two different atoms or groups are attached to the double bonded carbon atoms, two isomers are possible, which differ only in the spatial arrangement of these groups or atoms about the double bond. These are referred to as geometrical isomers and designated as cis-and trans–isomers. When the two similar groups lie on the same side of the double bond, the isomer is called cis– and when they lie on the opposite side of the double bond, the isomer is called trans-isomer. Thus two geometrical isomers are possible for but – 2 – ene. The molecule in which two methyl groups are on the same side of C = C is the cis but – 2 – ene and the molecule in which two methyl groups are on opposite sides of double bond, is the trans isomer. CH3
CH3 C
H
CH3
C
H C
H
cis But – 2 – ene (m=0.33D)
H
C CH3
trans But – 2 – ene (m=0)
The two isomers differ in their physical and chemical properties and can be distinguished from each other. For example, trans but-2 ene is non-polar whereas the cisbut-2-ene is polar having dipole moment 0.33 Debye. 18.5
STABILITY OF ALKENES
The stability of alkenes may be explained on the basis of heat of hydrogenation of alkenes. Heat of hydrogenation is the energy difference between the starting alkene and the product alkane. The table No. shows the heat of hydrogenation of a few alkenes. Table 18.1 Heat of hydrogenation of some alkenes Name Structure – D H, Kcal/mole Ethene CH2 = CH2 32.8 Propene CH3 – CH = CH2 30.1 1-Butene CH3CH2CH = CH2 30.3 cis-2-butene cis – CH3 – CH = CH. – CH3 28.6 trans-2-butene trans CH3 – CH = CH – CH3 27.6 1, 3 butadiene CH2 = CH – CH = CH2 57.1
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+2 CHEMISTRY (VOL. - I)
Consider 3 alkenes which can be reduced to butane. H2 , Pt
CH3 – CH2 – CH = CH2
H2 , Pt H 2 , Pt
cis CH3 – CH = CH – CH3
> CH
3
– CH2 – CH2 – CH3
trans CH3CH = CH – CH3 The greater the value of heat of hydrogenation, the higher is the energy of the starting alkene and less is the stability. So, from the table it can be concluded that out of 3 butenes trans 2-butene is the most stable and 1-butene is the least stable. Conclusion 1.
Alkenes with more alkyl groups on the pi bond carbons are more stable. This is due to inductive effect of alkyl groups which release electron density towards sp 2 hybridised carbons. CH2 = CH2, RCH = CH2, RCH = CHR, R2C = CH2, R2C = CHR, R2C = CR2
Increase in stability This can also be explained on the basis of hyperconjugation. Conjugated dienes are more stable than dienes with same number of isolated double bonds. This is due to delocalisation of pi-electron density. Trans-alkenes are more stable than cis alkenes This is because of fewer steric repulsions in trans - isomer.
2. 3.
H
repulsion
H
C H C H
H H C
C
H H
H
H
H H C
C
C H
less stable (cis)
18.6
H
C
H H
more stable (trans)
METHODS OF PREPARATION
The preparation of alkene involves thes elimination of an atom or group of atoms from adjacent carbons and subsequent formation of a carbon – carbon double bond. C
C
— C = C — + X – Y, where X = H or halogen p
Y = OH or halogen
X
Y
Two atoms or groups attach to vicinal carbon atoms are removed with the simultaneous formation of a p – bond. There is also a redistribution of hybridisation of C – atoms from sp3 to sp2. It is brought about by different routes.
UNSATURATED HYDROCARBONS - ALKENES
751
These are (1)
Dehydration of alcohols : (a)
When a primary alcohol is heated with concentrated sulphuric acid at 160 – 1700C, a molecule of water is eliminated and an alkene is formed. In place of concentrated sulphuric acid, other dehydrating agents like glacical phosphoric acid, phosphorus pentoxide, anhydrous zinc chloride or alumina can be used. When vapours of alcohol are passed over heated alumina (Al 2O3) at 3500C, dehydration takes place and alkene is formed. H
R
C
C OH H
C
H
Conc. H2SO4, 160
Conc. H2SO4, 160
CH3CH2OH
(b)
C = C H
H
+ H2O
H (Alkene)
1700C CH2 = CH2 + H2O
Al2O3, 3500C
or
Ethanol
R
Al2O3, 3500C
or
H
1700C
(Ethene or Ethylene)
Dehydration of secondary and tertiary alcohols is best carried out using dilute sulphuric acid at lower temperature. OH
H 60% H2SO4
CH3
CH
CH2
1000C
Propan – 2 – ol (a secondary alcohol) CH3 CH3
C CH2
OH
CH3 – CH = CH2 + H2O
20% H2SO4
H
2– Methylpropan – 2 – ol (a tertiary alcohol)
80 900C
(Propene)
CH3 C
CH2
CH3
Isobutene or 2 – Methylpropene
The ease of dehydration of alcohols is in the following order. Tertiary alcohol > Secondary alcohol > Primary alcohol.
+ H2O
752
+2 CHEMISTRY (VOL. - I)
(c)
With unsymmetrical secondary and tertiary alcohol, dehydration may occur in two ways. e.g.
CH3
CH
CH
CH2
H
OH
H
CH3 CH2 CH = CH2 But 1 ene (20 35%)
H2O ( 1 : 1) H2SO4
CH3 CH = CH CH3 But 2 ene (65 80%)
Experiments show that hydrogen attached to the adjacent carbon atoms having least number of hydrogen atoms is eliminated most easily. Thus in the above reaction, the major product is but–2–ene (65 – 80%) This elimination occurs in accordance with Saytzeff's rule for the dehydration of alcohols. According to this rule, "always the most alkyl substituted olefin is formed as the major product during the dehydration of alcohols" as more substituted the alkene, greater is its stability. (2)
Dehydrohalogenation of alkyl halides :
When an alkyl halide is refluxed with alcoholic KOH (solution of KOH in ethyl alcohol) solution, a halogen atom (X) and a hydrogen atom (H) from two adjacent carbon atoms are eliminated in the form of halogen hydracid (HX) and result in the formation of an alkene. H R
H
C
C
H
X
H H
H
Alcoholic KOH R
D
C C (Alkene)
H
+ KX + H2O
Alkyl halide Alcoholic KOH CH3 – CH2 – Br
D
CH2 = CH2 (Ethylene)
The reactivity of alkyl halides is in the following order. Alkyl iodide > Alkyl bromide > Alkyl chloride
+ KBr + H2O
UNSATURATED HYDROCARBONS - ALKENES
753
The case of dehydrohalogenation of alkyl halides is in the following order. Tertiary > Secondary > Primary With unsymmetrical secondary and tertiary alkyl halides, dehydrohalogenation may occur in two ways. e.g. CH3
CH
CH
CH2
H
X
H
CH3 CH2 CH = CH2 But 1 ene (20%)
HX
(2 Halobutane)
CH3 CH = CH CH3 But 2 ene (80%)
According to Saytzeff's rule the major product is the most alkyl substituted alkene. S o but – 2 – ene is the major product here. (3)
Dehalogenation of vicinal dihalides :
Dehalogenation involves the removal of a halogen molecule from the reactant molecule. Compound having two halogen atoms on adjacent carbon atoms is called a vicinal dihalide or vic – dihalide. Compound having two halogen atoms on the same carbon atom is called a geminal dihalide or gem-dihalide.
R
H
H
C
C
H R
H + Zn methanolic solution
X
H
Heat
C C (Alkene)
H
+ ZnX 2
X
(vic-dihalide)
CH3
H
H
C
C
Br
Br
Heat
H + Zn
methanolic solution
CH3 CH = CH2 + ZnBr2 (propene)
(1, 2 – Dibromopropane)
(4)
Electrolysis of dibasic acids :
When sodium or potassium salt of a dibasic acid (for example, succinic acid) is electrolysed, alkene is liberated at the anode.
754
+2 CHEMISTRY (VOL. - I)
CH2
COONa
CH2 COONa (Sodium succinate)
CH2
COO
CH2
COO
+ 2Na+
At the Anode : CH2
COO
CH2
COO
2e
CH2 CH2
+ 2CO2
At the Cathode :
(5)
2Na+
+
2e
2Na
2Na
+
2H2O
2NaOH +
H 2-
Partial Reduction of Alkynes :
Alkynes on partial reduction either by Lindlar's catalyst or by Birch reduction result in cis - or trans - alkenes depending on the nature of the catalyst used. (i) Catalytic reduction of alkynes in the presence of palladium supported over CaCO 3 or BaSO4 and partially poisoned by addition of PbCO3, S or quinoline (Lindlar's catalyst) gives cis-alkenes. º C – CH3 CH3 – C = But–2–yne
H2 – Pd/CaCO3 + S Lindlar’s catalyst
H3C H
CH3 C=C
H
Cis–but–2–ene (ii) If alkynes are reduced with sodium in liquid ammonia (Birch reduction), transalkenes are the major products. º C – CH3 H3C – C =
Na / liq. NH3
H3C
Birch reduction
H
H3 C=C
CH 3
Trans-but-2-ene
18.7
PROPERTIES Physical Properties : The physical properties of alkenes are similar to those of corresponding alkanes. (1) State : Lower members ethene, propene and butene are gases. Members containing upto 18 carbon atoms are colourless liquids and beyond this, higher members are colourless solids. (2) Solubility : They are insoluble in water but are soluble in organic solvents such as ether, alcohol, chloroform, benzene etc. Alkenes with lower molecular mass are slightly more soluble in water because of polarizability of the p – electrons.
UNSATURATED HYDROCARBONS - ALKENES
(3)
755
Density : They are lighter than water and just like alkanes, have limiting density of rather less than 0.8.
(4)
Melting and Boiling point : The melting and boiling point of alkenes show a regular gradation with the rise in molecular mass. Further, branched-chain alkenes have lower boiling points than straight-chain alkenes.
(5)
Odour : Alkenes possess characteristic odour. The lower members act as general anaesthetics.
Table – 18.2 Properties of some alkenes IUPAC Name
Structure
Boiling Point 0 C
Melting Point 0 C
Density (g / ml) at 100C
– 103.8
– 169
0.566
– 47.7
– 185
0.609
– 6.5
– 185
0.625
Ethene
CH2 = CH2
Propene
CH3 = CH – CH3
But – 1 – ene
CH3 = CH – CH2– CH3
Pent – 1 – ene
CH2 = CH – (CH2)2– CH3
30
– 165
0.641
Hex – 1 – ene
CH2 = CH – (CH2)3– CH3
64
– 140
0.676
Hept – 1 – ene
CH2 = CH – (CH2)4– CH3
94
– 119
0.697
Oct – 1 – ene
CH2 = CH – (CH2)5– CH3
121
– 102
0.715
Non – 1 – ene
CH2 = CH – (CH2)6– CH3
147
– 81
0.729
Dec – 1 – ene
CH2 = CH – (CH2)7– CH3
171
– 66
0.741
Chemical Properties : The carbon carbon double bond in an alkene is an unsaturated group and can undergo a wide variety of addition reactions. Double bond contains a s bond and a p – bond. The p– bond is weaker than s – bond. During addition reaction the p– bond breaks and the sigma bond remains intact. In such a process the p– bond is replaced by two new sigma bonds. Also the p– bond in alkene acts as a source of electrons. Hence electrophiles, which are positively charged species and can accept electrons, would be attracted by the p– electron cloud while nucleophiles with completed shell structure will be repelled. Therefore, the characteristic reaction of alkenes are the electrophilic addition reactions.
756
+2 CHEMISTRY (VOL. - I)
The mechanism of addition of X – Y to an alkene is given below. If Y is more d+ delectronegative than X, the addendum can be represented as X - Y. . Thus,
Alkene
p – Complex
Addendum
Rearranges X C
C
Y
Y (Three centered s-complex)
Addition product (Trans)
Here the positive end of addendum attached to the p– electron cloud gives positive charge to the p– complex, leaving the nucleophile Y . Then the p– complex rearranges to a carbocation which reacts with the nucleophile Y to form the addition product. In this process, the hybridisation of carbon changes from the planar arrangement of alkene (sp2 hybridisation) to a tetrahedral structure (sp3 hybridisation). p C == C s
+
Alkene sp – hybridisation at carbon (Planar structure) 2
X
s
Y
Addendum
X s C s
C s Y
Addition product sp3 – hybridisation at carbon Tetrahedral structure
The carbon – carbon double bond adds H2, Cl2, Br2, HX, HOX, H2SO4 and H2O and is attacked by strong oxidising agents including ozone. (1)
Addition of hydrogen :
In the presence of powdered metal catalyst and under pressure and heat, hydrogen adds to double bond of alkene forming the corresponding alkane. The process is known as catalytic hydrogenation.
UNSATURATED HYDROCARBONS - ALKENES
757
Catalyst C == C
+ H2
heat Pressure
C
C
H
H
(Alkene)
(Alkane)
Hydrogenation of alkenes can be effected by passing the vapour of alkene and hydrogen over finely divided nickel at 200 – 3000C and pressure of 50 – 100 atmosphere. With the more active catalyst like Pt, Pd or Raney Ni addition occurs at room temperature and atmospheric pressure. Examples : Ni CH2 = CH2
+ H2
CH3 – CH3
0
(Ethene)
200 – 300 C, Pressure
(Ethane)
Ni CH3 – CH = CH2 + H2 (Propene)
(2)
CH3 – CH2– CH3
0
200 – 300 C, Pressure
(Propane)
Addition of chlorine or bromine :
When an alkene in treated with chlorine or bromine in an inert solvent like carbon tetrachloride, halogens add rapidly to the carbon carbon double bond to form dihaloalkanes. Iodine reacts very slowly with alkenes, the di-iodide being unstable regenerate alkene. X C == C
+ X2
CCl4
C
Room temp.
(Alkene)
C
X (1, 2 – Dihaloalkane)
X = Cl, Br
I C == C
C
+ 12
C
I Example :
Br CH2 = CH2
(Ethene)
+
Br2
CCl4 Room temp
CH2
CH2
Br (1, 2 – Dibromoethane)
[Note : The reaction with bromine is used as a test for unsaturation. When an alkene is treated with a 5% solution of bromine in carbon tetrachloride, the reddish brown colour of bromine is immediately discharged]
758
+2 CHEMISTRY (VOL. - I)
Mechanism : The addition of halogens (say Br2) to alkene forming 1, 2 – dihaloalkanes takes place through the following steps. (i)
When bromine molecule comes in the proximity of an alkene, the negative p– electron cloud of the alkene causes polarisation of the bromine molecule. dd+ Br — Br
Br – Br (p– cloud) (ii)
The positive end of the polarised molecule is then attracted near the p – electron cloud to form a low stability p – complex. Br d -
C == C
+
dd+ Br —— Br
Br d + C == C (p – complex)
(iii) The positively charged bromine atom is then attracted by both carbons of the double bond and then the Br – Br bond breaks. The negatively charged bromine leaves as Br– and bromonium ion is formed. Br d Å
Br d + C == C
Br C
+
(p– complex) (iv)
C
+ Br
(Bromonium ion)
Nucleophilic attack of bromide ion (Br–) yields dibromo compounds. Å
Br
Br C
C
C
+ Br
C Br
(1, 2–Dibromo compound)
(3)
Addition of hydrogen halide :
When alkenes are treated with halogen acids HX (HCl, HBr, HI) alkyl halides or haloalkanes are formed and the reaction is known as hydrohalogenation. The order of reactivity of hydrogen halides is HI > HBr > HCl
UNSATURATED HYDROCARBONS - ALKENES
759
H
C == C
C
+ HX
C X
(Alkene)
(Alkyl halide) X = Cl, Br, I
Addition to symmetrical alkenes : The addition of hydrogen halide to a symmetrical alkene gives only one product because it does not matter as to which carbon of the double bond the halogen is attached. Example :
CH2 = CH2
+ HBr
H
Br
Br
H
CH2
CH2 or
CH2
CH2
(Ethylene) H
(Ethylbromide) H
H + HBr
C == C CH 3
CH3
H
Br
C
C
(But – 2 – ene)
or
CH 3
CH3
H
H
Br
H
C
C
H
CH3
CH 3
(2 – Bromobutane)
Mechanism : Hydrohalogenation of symmetrical alkene follows carbocation mechanism, as given below. Step – I carbocation.
Hydrogen halide forms a p – complex which rearranges to give
Br d Hd+ Rearranges CH2 = CH2 + HBr
CH2 == CH2
Å
CH3 — CH 2 + Br (Carbocation)
Step – II Nucleophilic attack of Br to the carbocation results in the formation of alkyl halide. Å
CH3 — CH 2 + Br
(Carbocation)
CH3 – CH2 – Br
(Ethyl bromide)
760
+2 CHEMISTRY (VOL. - I)
Since only one type of carbocation is possible from a symmetrical alkene, only one product is formed. Addition to unsymmetrical alkene : When hydrogen halide is treated with an unsymmetrical alkene, two addition products are possible as the halogen atom is placed on one or the other carbon of the double bond. Thus, propylene reacting with HBr, can form n - propyl bromide or isopropyl bromide. Br CH3 CH CH3 (isopropyl bromide)
CH3 CH = CH2 + HBr (Propylene)
CH3 CH2 CH2Br (n-propyl bromide)
But experimentally, if has been found that, isopropyl bromide is obtained predominantly. Such a direction of addition of an unsymmetrical reagent to an unsymmetrical alkene was first observed by Vladimer Markownikoff (1870), a Russian Chemist and his work is honoured till today by a rule named after him – Markownikoff's rule. Markownikoff's rule : The rule states the negative part of the addendum (molecule to be added) goes to that carbon atom which contains least number of hydrogen atoms during the addition across a carbon-carbon double bond of an unsymmetrical olefin. In otherwords, when an unsymmetrical reagent (X +Y–) adds to an unsymmetrical alkene, the positive portion of the reagent (X +) adds to that carbon of the double bond, which has greater number of hydrogen atoms.
R
Å
C == C H
H
X
R
H + X— Y
C H
Y
H
C H
Thus hydrogen bromide will add to propene to give isopropyl bromide and to 2–Methylpropene to give t–butyl bromide as the major product.
UNSATURATED HYDROCARBONS - ALKENES
CH3 — CH = CH2
761
+ HBr
CH3
(Propene)
CH3
C
H
CH
CH2
Isopropyl bromide or 2 – Bromopropane
CH2
CH3
Br
CH3
+ HBr
Br
H
C
CH2
CH3
(2–Methylpropene)
t–Butyl bromide or 2 – Bromo – 2 – methylpropane
Anti-Markownikoff's Rule : (Kharasch Peroxide effect)
Kharasch & Mayo (1933) observed that the addition of HBr to unsymmetrical alkene, in the presence of organic peroxide (R – O – O – R) takes place opposite to that predicted by Markownikoff. This phenomenon of anti-Markownikoff addition caused in presence of peroxide, is called Kharasch peroxide effect. Thus, when propylene reacts with HBr in the presence of a peroxide, the product is mainly n-propyl bromide, whereas in the absence of peroxide, the main product is isopropyl bromide. Presence of Peroxide
CH3 CH2 CH2 Br (n-propyl bromide) (Anti-Markownikoff product)
CH3 CH = CH2 + HBr Br Absence of Peroxide
CH3 CH CH3 (isopropyl bromide) (Markownikoff product)
The addition of HBr, in the presence of peroxide is a free radical chain reaction. (a)
R – O – O – R decomposes to give radicals R –O –O – R
(b)
R – O· + R – O·
R – O · radical reacts with HBr to form free bromine radical. R – O · + HBr
(c)
Homolysis
R – OH + Br ·
The bromine radical then attacks the alkene giving two possible bromoalkyl free radicals.
762
+2 CHEMISTRY (VOL. - I)
.
CH3 CH2 CH2Br 20 Free radical (more stable)
.
CH 3 - CH .. - CH 2 + Br Br
.
CH3 CH CH2 (10 Free radical) (less stable)
Since the order of stability of free radical is 3 0 > 20 > 10, the 20 free radical is formed prodominantly. (d)
The more stable radical (20 free radical) reacts with HBr forming anti-Markownikoff product and another bromine free radical which propagates the chain reaction. H
.
CH 3 - CH - CH 2 Br + H : Br
CH3
(20 free radical)
CH
CH2Br + Br .
(n - propyl bromide)
HCl and HI do not give anti-Markownikoff products in the presence of peroxide, because
(4)
(i)
the H – Cl bond is stronger than H – Br bond. It is not broken by alkoxy free radicals obtained from peroxide.
(ii)
The H – I bond is weaker than H – Br. It is broken by alkoxy free radical obtained from peroxide, but the iodine atoms so formed readily combine with each other giving iodine molecule rather than attacking the double bond of alkene.
Addition of Hypohalous acids (HOX) : Alkenes when treated with hypohalous acid form halohydrins.
C = C
+
HOX
(Alkene)
OH
X
C
C
(Halohydrin) X = Cl or Br
UNSATURATED HYDROCARBONS - ALKENES
763
In hypohalous acid, the OH group behaves as the negative part of the adding species (HO– X+). The addition follows Markownikoff's rule Examples :
CH2 = CH2 + HOBr
OH
Br
CH2
CH2
Ethelene bromohydrin or 2–Bromoethanol OH CH3 – CH = CH2 + HOCl
(5)
Cl
CH3 CH CH2 1 – Chloropropan – 2 – ol.
Addition of sulphuric acid :
Alkenes when treated with cold conc. H2SO4 at room temperature, form alkyl hydrogen sulphates.
C = C
+
(Alkene)
H – OSO3H
H
OSO3H
C
C
(Sulphuric acid)
With unsymmetrical alkenes, the additioon of sulphuric acid follows Markownikoff's rule. For example :
CH3 – CH = CH2
d +d + HOS O 3 H
CH3
(Propene)
OSO3 H
H
CH
CH2
(isopropyl hydrogen sulphate)
The resulting alkyl hydrogen sulphate upon hydrolysis give respective alcohols. OSO3H CH3
CH
CH3
(isopropyl hydrogen sulphate)
OH + H2O
CH3
CH
CH3 + H2SO4
(isopropyl alcohol)
764
+2 CHEMISTRY (VOL. - I)
(6) Addition of Water (Hydration) Alkenes undergo acid- catalysed addition of water to give alcohols. Ethanol is industrially prepared by the method. Sulphuric acid is commonly used as catalyst. The addition obeys Markownikoff’s rule. CH2 = CH2 + H – OH
H2SO4
CH3 – CH2 OH
(Ethene)
(Ethanol)
CH3 – CH = CH2 + H – OH (Propene)
(7)
H2SO4
CH3 – CH – CH3 | OH (2-Propanol)
Oxidation :
Alkenes are easily oxidised than alkanes. The products of oxidation depend upon the nature of oxidising agent. (i)
(ii)
Combustion : Alkenes burn in air with luminous flame to form carbon dioxide and water. 2CnH2n + 3nO2
2nCO2 + 2nH2O
CH2 = CH2 + 3O2 (Ethylene)
2CO2 + 2H2O
Hydroxylation : Alkenes when treated with cold dilute aqueous or alkaline solution of potassium permanganate (Baeyer's reagent) add two hydroxyl groups across the double bond to form 1, 2 – dihydroxy compounds, known as glycol. The pink colour of KMnO4 is discharged and a brown precipitate of MnO2 is formed. 2KMnO4 + H2O ® 2KOH + 2MnO2 + 3 [O] brown ppt.
C = C
+ (H2O + O)
Cold, dilute alkaline KMnO4 soln.
OH
OH
C
C
(glycol)
+ MnO2 (brown ppt.)
The reaction is known as hydroxylation of alkenes. For example :
CH2 = CH2 (Ethylene)
+ (H2O + O)
Cold, dilute alkaline KMnO4 soln.
OH OH H2C
CH2 Ethylene glycol or, 1, 2 – Dihydroxyethane
UNSATURATED HYDROCARBONS - ALKENES
765
Cold, dilute alkaline KMnO4 soln. CH3 – CH = CH2
+ (H2O + O)
CH3
(Propylene)
OH
OH
CH
CH2
Propylene glycol or, 1, 2 – Dihydroxypropane
(iii) Chemical oxidation : With strong oxidising agents like warm acidified potassium permanganate or aqueous chromic acid, the double bond is reptured and the produts are aldehyde, ketone and acid depending upon the nature of alkene. Example :
CH2 = CH2
Acidic KMnO4
+ 4O
(Ethylene)
(Formic acid)
CH3– CH = CH2
Acidic KMnO4
+ 4O
hot
(Propylene)
CH3 CH3
CH3COOH
+ HCOOH
(Acetic acid) (Formic acid)
C = CH2
3O + 30
Acidic KMnO4
CH3
hot
CH3
(Isobutylene) (iv)
HCOOH + HCOOH
hot
C = O + HCOOH
(Acetone)
(Formic acid)
Epoxidation : Lower alkenes react with oxygen at 200 – 4000C in the presence of silver as catalyst forming epoxides. O
Ag
C = C
+
½ O2
C
200 4000C
(Alkene)
(Epoxide)
The oxidation reaction of alkene giving epoxides is epoxidation. Thus, O Ag CH2 = CH2 + ½ O2
C
200 4000C
CH2
CH2
(Epoxyethane)
766
+2 CHEMISTRY (VOL. - I)
The better method of epoxidation is by the treatment of alkene by peracids such as perbenzoic acid, trifluoroperacetic acid. O
O CH2 = CH2 + C6H5 (Ethylene)
C
O
OH
(Perbenzoic acid)
Mechanism :
O
CH2
(Epoxyethane)
O
C
C
O O
+
C
C
O
H
H (Alkene)
OH
C6 H 5
O
C
+
C
(Benzoic acid)
C6 H 5
C
(v)
C6H5
CH2 +
(Peracid)
(Epoxide)
O (Acid)
Ozonolysis : When ozone gas is bubbled into a solution of an alkene in an inert solvent (CHCl3 or CCl4) at low temperature, an addition compound known as ozonide, is formed.
C = C
+ O3
O
O
C
C O
(Alkene)
(Ozonide)
The reaction is known as ozonisation of alkenes. Ozonisation breaks both p and s bonds of alkene. Examples :
CH2 = CH2 (Ethylene)
+ O3
O
O
CH2
CH2 O
(Ethylene ozonide)
CH3 – CH = CH – CH3 (2 – Butene)
+ O3
CH3
O
O
CH
CH
CH3
O (2 – Butene ozonide)
UNSATURATED HYDROCARBONS - ALKENES
767
Ozonides are unstable and when treated with reducing agents like zinc and water are cleaved at the position of double bond to give two carbonyl fragments. The cleavage of alkenes by the action of ozone and subsequent hydrolysis is known as ozonolysis. The products of ozonolysis are aldehydes, ketones or an aldehyde and a ketone depending on the nature of the alkenes. This reaction is highly helpful in detecting the position of the double bond in alkenes or other unsaturated compounds.
C = C (Alkene)
+O3
O
O
C
C
O (Zn+H2O)
C + C
Carbonyl compounds
O
(Alkene)
O
(Ozonide)
+ ZnO + H2O
Carbonyl compounds
Examples : O CH2 = CH2
O3
O
H2C
CH2
(Zn+H2O)
(formaldehyde)
O CH3
CH3 C = C
CH3
H
CH3
CH3
2–Methylbut–2–ene
H2C = O + O = CH2 + ZnO + H2O
O
O
C
C
CH3 CH3 (Zn+H2O)
C=O+O=C
(acetone)
CH3
H
O
CH3
(Ozonide)
(acetone)
H
(acetaldehyde) + ZnO+H2O
(8)
Polymerisation :
A polymer is a long chain molecule with repeating structural units. The repeating structural units are generally known as monomer and the process of formation of polymers is known as polymerisation. Examples are polyethylene, polypropylene etc. n(Monomer) n(CH2 = CH2) Ethylene (monomer)
(Monomer)n or Polymer (
CH2
CH2
)n
Polyethylene (polymer)
This type of polymerisation shown by alkenes without any loss of atoms is known as addition polymerisation. The alkene serves as monomer. Polyethylene is manufactured by heating ethylene to about 200 0C under high pressure (1500 – 2000 atm) in the presence of very small amount of oxygen.
768
+2 CHEMISTRY (VOL. - I)
n(CH2 = CH2)
Ethylene
0.01%; O2 2000C
(
CH2
High pressure
CH2
)n
(Polyethylene)
It is widely used for making wire, bottles, tubing, toys and packing materials. Substituted ethylenes CH2 = CHZ (Z = Cl, CN, C6H5 etc) also undergo polymerisation to form a wide variety of commercially important synthetic polymers. Z Peroxide n CH2 = CHZ
(
catalyst
(substituted ethylene)
CH2
CH
)n
vinyl polymer
When Z = Cl, CN and C6H5 the polymers are known as polyvinyl chloride (PVC), polyvinyl cyanide (orlon) and polystyrene respectively. Cl Polymerisation n CH2 = CHCl
(Vinyl chloride)
(
CH2
CH
)n
Polyvinyl chloride (PVC) CN
Polymerisation n CH2 = CHCl
(
(Vinyl cyanide)
CH2
CH
)n
Polyvinyl cyanide (orlon) Polymerisation
n C6H5 – CH = CH2
(
CH
CH2
)n
C6H5
(Styrene)
Polystyrene Polymerisation
n F2C = CF2
(Tetrafluoroethylene)
(
CF2 CF2
)n
(Polytetrafluoroethylene) Teflon Teflon is used as electrical insulator and for making gaskets, valve seals and other lining on nonstick kitchenware. (9) Substitution reaction in Alkenes : All alkenes contain carbon carbon double bond and all except ethylene, also contain a saturated alkyl group as a part of the molecule. The alkyl groups of alkenes undergo substitution reaction at high temperature. Thus propylene reacts with chlorine at a temperature of 5000 – 6000C to yield allyl chloride. Bromine behaves similarly.
UNSATURATED HYDROCARBONS - ALKENES
769
500 – 6000C CH3 – CH = CH2 + Cl2
gas phase
Cl – CH2 – CH = CH2 3 – Chloroprop – 1 – ene or (allylchloride)
When the alkyl group has more than one carbon atom, substitution occurs at the a carbon atom (allylic carbon atom). Thus, 6000C CH3– CH2– CH = CH2+ Cl2 But – 1 – ene
gas phase
CH3 – CH – CH = CH2 Cl 3 – Chlorobut – 1 – ene
CHAPTER (18) AT A GLANCE ALKENE General molecular formula – CnH2n Preparation of Ethylene :
(i)
CH2Br
Distil with Zn
CH2Br (1, 2 – Dibromoethane) (ii)
CH3 – CH2 – Br
(iii) CH3 – CH2 – OH
(iv)
CH2 – COOK
Heat with alc. KOH D with conc H2SO4 at 1700C or Al2O3, 3500C
Kolbe's Electrolysis
CH2 – COOK
(v)
CH º CH
Reduction by Lindlar's Catalyst H2 – Pd / CaCO3 + S
CH2 = CH2 (Ethylene)
770
+2 CHEMISTRY (VOL. - I)
Properties : Adddition of H2, Ni at 200 – 300 0C
Addition of halogen, X 2 (X2 = Cl2, Br2, I2)
CH3 – CH3 (Ethane)
CH2 – X CH2 – X (Ethylene dihalide)
Addition of halogen and HX (HCl, HBr, HI) CH3 – CH2 – X (Ethyl halide)
CH2 = CH2 (Ethylene)
OH Addition of HOCl Addition of H2O alkaline KMnO4 [H2O + O]
Cl
CH2 – CH2 (Ethylene chlorohydrin) CH3CH2OH (Ethanol) CH2 – OH CH2 – OH (Ethylene glycol) O
O2, Ag, 200 – 400 0C
Ozone
H 2C CH2 Epoxyethane O CH2
CH2
Zn/H2O O O H–C–H+H–C–H
O O (Ethylene ozonide) Polymerisation (
CH2 CH2 (Polyethylene)
)n
UNSATURATED HYDROCARBONS - ALKENES
771
QUESTIONS (A)
Short questions (One mark each) 1.
What type of compounds undergo addition reaction ?
2.
Ethylene reacts with dilute KMnO4 to give
3.
Write down the structural formula of but – 2 – ene.
4.
A double bond is composed of
5.
Shape of ethylene molecule is
6.
Write the IUPAC name of the compound.
———————— ————————
————————
.
sigma & one
————————
bone.
.
C 2H 5 CH
CH = CH 2
CH 3
(B)
7.
Which metal is used as catalyst in the hydrogenation of unsaturated hydrocarbons ?
8.
What is the polymerised product of ethylene.
9.
How many olefins are obtained when a mixture of 1–chloropropane and 2–chloropropane is heated with alcoholic KOH solution ? Name it/them.
Short questions (Two marks each) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
12.
What happens when propene is treated with hydrogen gas in the presence of finely divided platinum as catalyst. Give equation. State Markownikoff's rule. What happens when ethylene reacts with ozone ? What happens when sodium succinate solution is subjected to electrolysis ? Write with equation what happens when propene and HBr react. Which alkene is formed when an aqueous solution of sodium ethanoate is electrolysed ? How is ethylene obtained from ethyl alcohol ? Give equation. What is Kharasch Effect ? Discuss with example. How can you detect unsaturation in an organic compound ? Give equation for the test. How many isomers can be obtained from the molecular formula of dichloroethene ? Write their structures. Which of the isomer will have zero dipole moment ? An unsaturated hydrocarbon having the molecular formula C 5H8 gives two molecules one of formaldehyde and one molecule of 2-oxopropanal on ozonolysis. Write the structure of the hydrocarbon. What happens when 2-bromobutane is treated with alc. KOH ?
772
+2 CHEMISTRY (VOL. - I)
(C) Other short and objective Question : 1. Complete the following reactions. Acidic
(i)
CH2 = CH2
KMnO4 Dil.alkaline
(ii)
CH2 = CH2
KMnO4
(iii) Propene + Br2 Peroxide
(iv)* Propene + HCl (v)
Peroxide
Propene + HBr
Peroxide
(vi)* Propene + HI (vii) But – I – ene + HOCI (viii) Ethylene + H2SO4 (ix)
CH2
COOK
CH2
COOK
Electrolysis
Alcoholic
(x)
CH3 – CH(Br) – CH2 – CH3
KOH
[* Hint : In (IV) and (VI) normal addition product as per Markownikoff's rule will be obtained]. 2.
Identify the compounds A, B and C. (i) (ii)
O3
A A
Zn/H2O
B
Electrolysis
CH3COCH3 + CH3CHO + H2O
Baeyer’s
B
HO – CH2 – CH2 – OH
reagent
(iii) A + Br2
CCl4
Zn
B
CH3 – CH = CH2 Mg
HI
(iv) CH2 = CH2
(v)
CH3 – CH2 – CH2 I
A
dry ether
Alc. KOH
B
H2O
HBr
A
B
HBr
C.
C. Aq. KOH
C.
UNSATURATED HYDROCARBONS - ALKENES
3.
4.
5.
773
How would you carry out the following conversions ? (i) Ethyl alcohol to ethylene and vice versa. (ii) Ethane to ethylene and vice versa. (iii) Ethyl bromide to ethylene and vice versa. Explain the following. (i) p – bond is weaker than sigma bond. (ii) Ethane is insoluble in conc.H2SO4, but ethene dissolves readily. (iii) In the series HI, HBr, HCl, HI is easily added to the carbon carbon double bond. (iv) But-1-ene reacts with hydrogen bromide in the presence of peroxide to give 1 – bromobutane but not 2 – bromobutane. Fill in the blanks : (i) Ethylene is prepared by the electrolysis of —————— . (ii) An alkene has the general formula —————— . (iii) Propylene can be obtained from methyl acetylene by —————— . (iv) An alkene decolourises the —————— colour of —————— solution in CCl4. (v) Hydration of —————— will give butan-2-ol. (vi) But-2-ene exhibit both —————— and —————— isomerism. (vii) Baeyer's reagent reacts with alkene to form —————— . (viii) The reduction of RC º CR to RCH = CHR with Na / liquid NH3 produces _______ isomer. (ix) The dehydration of alcohols producing alkenes is catalysed by _______.
Answers :
6. 7.
[(i) Potassium succinate, (ii) C nH2n, (iii) partial hydrogenation, (iv) red, bromine, (v) but-2-ene, (vi) position and geometrical, (vii) glycol, (viii) trans (ix) acid Write the structural formula and IUPAC names for all isomeric pentenes C 5H10. Include cis and trans isomers. Explain what is meant by sp2 - hybridisation at carbon.
(D) Long Questions : 1. How ethylene is prepared from ethyl alcohol ? Explain it with equation. How does ethylene react with (a) HOCl (b) alkaline KMnO4 (c) Ozone and treatment with water ? 2. Give a method for ascertaining the position of a double bond in a molecule. 3. Write notes on (a) Markownikoff's rule. (b) Ozonolysis (c) Peroxide effect. (d) Epoxidation (e) Saytzeff rule (f) Polymerisation 4. Discuss the structure of ethylene molecule. Describe two methods of preparation of alkenes.
774
+2 CHEMISTRY (VOL. - I)
5. 6. 7.
8.
(E)
What are alkenes ? Give the general methods of preparation of alkenes. How ethylene is prepared from ethyl chloride ? How can you convert ethylene into (a) Ethane (b) Acetylene (c) Ethyl alcohol (d) Ethyl bromide ? How are alkenes prepared from (a) alkyl halides (b) alcohols What happens when propene reacts with the following substances (a) HOCl (b) alkaline KMnO4 (c) Ozone How is ethene prepared from (a) ethyl alcohol (b) Sodium Succinate ? What happens when ethene reacts with the following substances ? (a) HOBr (b) Ozone (c) Alkaline KMnO4.
Multiple choice questions : 1. Propene belongs to (a) Paraffins (b) Alkynes (c) Olefins (d) Alkanes 2. Alkene is formed by the dehydration of an (a) Alcohol (b) Acid (c) Aldehyde (d) Amide 3. Which one of the following decolourises dilute alkaline solution of KMnO 4. (a) ethane (b) ethylene (c) methane (d) CCl4 4. Which one of the following gases is produced when potassium succinate is electrolysed ? (a) ethane (b) ethylene (c) acetylene (d) propane 5. The product obtained after the hydrolysis of ethylene ozonide is (a) CH3CHO (b) CH º CH (c) HCHO (d) C2H6. 6. The reaction of HCl with ethylene is an example of (a) Substitution (b) Polymerisation (c) Condensation (d) Addition 7. The middle carbon atom in CH2 = C = CH2 has the hybridisation (a) sp (b) sp2 (c) sp3 (d) dsp2 8. In the addition of haloacids to alkenes, Kharasch peroxide effect was shown only by (a) HF (b) HCl (c) HBr (d) HI 9. Addition of a molecule of HBr to but – 1 – ene in the presence of peroxide gives. (a) n – Butane (b) 1 – Bromobutane (c) 2 – Bromobutane (d) 1,2 – Dibromobutane 10. 2 – Chlorobutane when treated with alcoholic KOH gives (a) 2 – Butanol (b) 2 – Butene (c) 4 – Butene (d) none of above 11. Out of the following alkenes, the one which shows geometrical isomerism is (a) 1 – Butene (b) 2 – Butene (c) Ethene (d) Propene
UNSATURATED HYDROCARBONS - ALKENES
12.
13.
14.
15.
16.
17.
18.
775
Olefins are converted to paraffins by (a) Hydrolysis (b) Halogenation (c) Dehydrogenation (d) Hydrogenation Anti-Markovnikov addition of HBr is not observed in (a) Propene (b) 1–butene (c) 2–butene (d) 2–pentene The ease of dehydration of an alcohol with conc. H2SO4 is (a) 3° > 2° > 1° (b) 1° > 2° > 3° (c) 3° > 2° < 1° (d) 3° < 2° > 1° The addition of halogen to an alkene involves the formation of an intermediate (a) Carbocation (b) Carbanion (c) Free radial (d) Halonium ion The addition of Br2 to trans-2-butene produces (a) (+) 2, 3 - dibromobutane (b) (–) 2, 3-dibromobutane (c) (+) 2, 3 - dibromobutane (d) meso-2, 3-dibromobutane The ozonolysis of an olefin gives only propanone. The olefin is (a) Propene (b) but–1–ene (c) but–2-ene (d) 2, 3 dimethyl but–2–ene Dipole moment is shown by (a) 1, 4–dichloro benzene (b) cis–1, 2 dichloro ethene (c) trans–1, 2–dichloroethene (d) Trans–but–2–ene
ANSWERS 1. 7. 13.
(c) (a) (c)
2. 8. 14.
(a) (c) (a)
3. 9. 15.
(b) (b) (d)
4. 10. 16.
(b) (b) (d)
qqq
5. 11. 17.
(c) (b) (d)
6. 12. 18.
(d) (d) (b)
776
+2 CHEMISTRY (VOL. - I)
CHAPTER - 19
UNSATURATED HYDROCARBONS-ALKYNES Alkynes are unsaturated hydrocarbons with carbon-carbon triple bond (C º C). They have the general formula CnH2n–2 . The simplest member of this family, acetylene has the molecular formula C2H2. Alkynes have two hydrogen atoms less than the corresponding alkenes. They are also known as 'acetylenes' after the name of the first member, acetylene. The triple bond, –CºC, in an alkyne constitutes the functional group and largely determines the chemical behaviour.
19.1
NOMENCLATURE
In the IUPAC system, this class have the suffix-yne. They are named after the corresponding alkane by replacing the suffix –ane by –yne. The IUPAC system of naming the alkyne follows the following rules. (i)
the rules for numbering the carbon atoms are the same as for alkenes.
(ii)
hydrocarbons with two triple bonds are called alkadiynes.
(iii)
hydrocarbons with both double and triple bond are called alkenynes.
In numbering alkenynes, the double bond is given preference over the triple bond, provided it does not violate the lowest set of locants rule H–CºC–H CH3–CºC–CH3 (Ethyne) (But-2-yne)
19.2
CH3–CºC–H (Propyne)
HCºC–CºCH (Butadiyne)
CH3–CH2–CºCH (But-1-yne)
CH3–CH=CH–CºCH (Pent-3-en-1-yne). (Not Pent - 2 - en - 4 - yne)
STRUCTURE OF ALKYNE Alkynes can be represented by the general formula R–CºC–R. where R may be alkyl group or H. The simplest member, acetylene can be represented as H–CºC–H.
UNSATURATED HYDROCARBONS - ALKYNES
777
In the formation of acetylene molecule sp- hybridised carbon atoms are used. 2pz
p 2py
s s sp
sp s
2pz 2py
p
sps s
sp
Fig. 19.1 Orbital structure of acetylene Acetylene is a linear molecule. The carbon-carbon triple bond consists of one sigma( s ) and two pi (p) bonds. (Fig.18.1) The C–C–H bond angle is 1800. The C–C bond length is 1.20 A0. The C–H bond length is 1.06 A0. The bond length and bond angle in acetylene molecule are shown below. 1800
H—C
C—H
1.06A0 1.20A0
19.3
ISOMERISM
Alkyne being a linear molecule, do not exhibit cis-trans isomerism. Acetylene and methylacetylene do not exhibit isomerism. However, the following types of isomerism are shown by higher alkynes. (i)
Chain Isomerism : Alkynes differ in the structure of parent chain. CH3 CH3––CH2––CH2––CºCH Pent–1–yne.
(ii)
CH3––CH––CºCH (3–Methylbut –1–yne)
Position Isomerism : Alkynes differ in the position of the triple bond CH3––CH2––CºCH
CH3––CºC—CH3
(But–1–yne)
(But –2–yne)
778
(iii)
19.4
+2 CHEMISTRY (VOL. - I)
Functional isomerism Alkynes are isomeric with alkadienes. CH3––CºC—CH3 (But–2–yne)
CH2=CH—CH=CH2 (Buta–1, 3–diene)
METHODS OF PREPARATION
1. Dehydrohalogenation of vicinal dihalides :
When vicinal or 1,2- dihalo compounds are treated with alcoholic KOH, two molecules of hydrogen halides are eliminated from adjacent carbon atoms to give alkynes. H
X
alc . KOH alc . KOH R—C—C—R ¾¾¾¾¾¾® R—C=C—R ¾ ¾¾¾¾® R—CºC—R - HX - HX (Alkyne) X H X H (1, 2- dihalide) (Haloalkene) H
Br
alc . KOH alc . KOH R—C—C—H ¾¾¾¾¾¾® H—C=C—H ¾¾¾¾¾¾® H—CºC—H - HBr - HBr (Acetylene) Br H Br H (1, 2- Dibromoethane) (Bromoethene) 2. Dehydrohalogenation of gem– dihalides : Gem– or 1,1- dihalides when treated with alcoholic KOH or sodamide double dehydrohalogenation takes place forming alkynes. H
Br
alc . KOH alc . KOH or sodamide or sodamide ¾® R—C=C—R ¾ ¾ ¾ ¾ ¾¾® R—CºC—R R—C—C—H ¾ ¾¾¾¾¾ HBr
HBr
H Br
H Br (Alkyne)
When R=H, we get H–CºC–H (acetylene). 3. Dehalogenation of tetrahalides : Tetrahalides in which the four halogen atoms are attached to two adjacent carbon atoms, when treated with zinc dust in ethanol yield alkynes. X X R—C—C—H +2Zn ¾Alcohol ¾¾¾¾® R—CºC—H +2ZnX2 X X (1,1,2,2- Tetrahaloalkane)
(Alkyne)
UNSATURATED HYDROCARBONS - ALKYNES
779
Br Br H—C— C—H +2Zn ¾Alcohol ¾¾¾¾® HC º CH + (Acetylene) Br Br
2ZnBr2
4. Kolbe's electrolytic method : When a concentrated solution of sodium or potassium salt of maleic or fumaric acid is electrolysed, acetylene is obtained at the anode. H––C––COOH
H––C––COOH
H––C––COOH
HOOC––C––H
(maleic acid)
(fumaric acid)
CH––COONa
CH––COO– ¾ ¾®
CH––COONa
–
+ 2Na+
CH––COO
(sodium maleate) At the Anode CH––COO–
-2e
¾ ¾¾®
CH––COO– (maleate ion)
CH
CH (acetylene)
+ 2CO2
At the Cathode 2Na+ + 2e
¾ ¾® 2Na
2Na + 2H2O ¾¾® 2NaOH + H 2 5. Laboratory method of preparation of acetylene : In the laboratory acetylene is prepared by the action of water on calcium carbide (CaC 2) Calcium carbide required for this purpose is manufactured by heating lime stone with coke in an electric furnace.
D CaO + CO CaCO3 ¾¾® 2 CaO + 3C ¾2275K ¾ ¾¾® CaC 2 + CO C
CH + 2H2O ¾ ¾®
Ca C
(Calcium carbide)
CH
+ Ca(OH)2
(Acetylene)
Acetylene thus obtained is contaminated with impurities like NH 3, H2S, PH3, AsH3 etc. The gas is purified by bubbling it through acidified copper sulphate solution, which absorbs NH3, PH3, H2S, and AsH3 except acetylene. Acetylene is collected by the downward displacement of water.
780
+2 CHEMISTRY (VOL. - I)
WATER ACETYLENE
SAND
ACIDIFIED CuSO4
CALCIUM CARBIDE
Fig. 19.2 Preparation of Acetylene The apparatus consists of conical flask fitted with a dropping funnel and a delivery tube. The end of this delivery tube is dipped in acidified CuSO 4 solution. A mixture of CaC2 and sand is taken in the conical flask and water is added from the dropping funnel. The gas after bubbling through CuSO4 solution, is collected by the downward displacement of water. 6. From Iodoform : Iodoform when heated with silver powder, pure acetylene is formed . CH I3 + 6 Ag + I3 CH
¾ ¾¾®
(Iodoform)
CHºCH
+ 6 AgI
(Acetylene)
7. Synthesis : Berthelot (1860) synthesised acetylene by striking an electric arc between two carbon electrodes in an atmosphere of hydrogen inside a silica tube. The yield is only 7%. 2C + H2 19.5
Electric spark 25000 C
¾¾¾¾¾¾¾ ¾®
CHºCH
PROPERTIES
(A) Physical Properties : 1. State : The first three members (C2 C4) are gases, next fourteen (C5 to C18) are liquids and the higher alkynes (C19 ) are solids. 2. Colour and odour : All alkynes are colourless and odourless except acetylene which has characteristic garlic odour. 3. Solubility : Alkynes are compounds of low polarity and hence they are slightly soluble in water. Alkynes are soluble in organic solvents like ether, benzene, carbon tetrachloride etc. 4. Melting point, boiling point and density : The melting point, boiling point and density of alkynes are slightly higher than the corresponding alkanes. These properties regularly increase with the increase in molecular mass. Again, branched chain alkynes have lower melting points than their corresponding straight chain isomers.
UNSATURATED HYDROCARBONS - ALKYNES
781
(B) Chemical properties : The chemical properties of alkynes are mainly due to the (i) Presence of electrons : Alkynes are characterised by the presence of triple bond (i.e. one sigma and two pibonds). Due to the presence of pi bonds, which contain loosely held pi electrons, alkynes undergo electrophilic addition reactions like alkenes. However, the addition reactions of alkynes differ from those of alkenes in the aspect that, CºC is less reactive than C=C . (ii) Presence of acidic hydrogen atoms : Hydrogen atom attached to carbon – carbon triple bond can be easily removed by a strong base and hence acetylenes are considered as weak acids. ¾® R — C º C R— C º C : H + Base ¾
+ H — Base.
As we move from ethane to ethylene and then to acetylene, there is a gradual increase in s-character. H
H
H—C—C—H
Hydrocarbon
H
Type of hybridisation s-character in C
H
C=C
H
H
sp3 25%
sp2 33.3%
H H
H—CºC—H
sp 50%
Due to more s-character of the sp- hybrid carbon in alkyne, the electron pair constituting the C—H bond is closer to the carbon nucleus than in an alkene and alkane. We know that higher the electronegativity of an atom greater in the ability to hold bonding electrons to it. The electronegativity of the three types of carbon atom is sp > sp2 > sp3
Hence, hydrogen atom in an alkyne will tend to be more positive and can be easily removed as a proton. The general reactions of alkynes are given below. (I)
ADDITION RECTIONS :
Alkynes can add two molecules of a reagent, while an alkene can add one molecule only.
X +X
2 ¾® — C = C — —C º C —¾¾¾
Alkyne
Alkyne (a)
X
X +X2
¾¾¾ ¾®
X
—C—C— X
X
saturated product
Addition of Hydrogen (Catalytic hydrogenation) : In the presence of catalyst like finely divided Pt, Pd or Raney nickel, alkynes add up two molecules of hydrogen forming the corresponding alkenes first and finally the alkanes.
782
+2 CHEMISTRY (VOL. - I)
H
+H Pd
2® ¾ H—CºC—H ¾ ¾¾
C=C
H
(Acetylene)
H
+H
2® ¾ ¾¾ ¾ CH3 — CH3 Pd H
(Ethylene)
(Ethane)
Catalytic hydrogenation of disubstituted alkyne with Lindlar's catalyst (PdCaCO3) gives the cis-alkene, whereas reaction with Na or Li in liquid ammonia gives trans–alkene (Birch reduction) CH3 — CºC—CH3 + H2
CH3
Lindlar's Catalyst
¾ ¾¾¾¾®
CH3
C=C
H
H
(cis But-2-ene) CH3 — CºC—CH3 + H2
Na liq.NH 3
¾ ¾¾¾ ¾®
CH3
H C=C
H
CH3
(trans But-2-ene)
(b)
Addition of halogens : Alkynes add two molecules of halogens (chlorine or bromine) in the dark at room temperature forming first dihaloalkenes and then a tetrahaloalkanes . X X
X
X
X¾ X¾ 2 ® R — C = C —R ¾+ 2® R — C — C — R ¾¾ R—C º C —R ¾+¾¾
(Alkyne)
(Dihaloalkene)
(Alkyne)
(Dihaloalkene) (X = Cl or Br)
X
X
(Tetrahaloalkane)
For example,
Cl Cl
Cl2 ® R—CºC—R ¾ ¾¾
Cl
Cl
Cl2 H—C—C—H H— C = C —H (Catalyst) (1,2-Dichloroethylene) Cl Cl Kieselguhr (diluent) (1,2-Dichloroethyle (1,1,2,2-Tetrachloroethane)
SbCl5
(Acetylene) (Acetylene)
Acetylene reacts with bromine water to give the dibromide, whereas with bromine (liquid) in the absence of any solvent gives tetrabromide. CHBr2 CHBr2
Br
¬¾¾2¾
(1,1,2,2-Tetrabromoethane)
CH CH
Br2 /H 2¾ O® ¾ ¾¾¾¾
(Acetylene)
CHBr CHBr
(1,2-Dibromoethylene)
UNSATURATED HYDROCARBONS - ALKYNES
783
Acetylene di-iodide only is obtained with iodine, when the reaction is carried out in alcoholic sulution. HCºCH + I2
Alcohol
I—CH=CH—I
¾ ¾¾¾¾®
(Acetylene)
(1,2-Di-iodoethene)
The order of reactivity of halogen is
Cl2 > Br2 > I2 . (c)
Addition of hydrogen halides : Alkynes add two molecules of halogen halides forming haloalkenes in the first step, then gemdihalides in the second step. The addition of hydrogen halide to unsymmetrical alkynes follows Markownikoff's rule in both the steps. The order of reactivity is HI > HBr > HCl. X
X
R—CºC—H
HX
¾ ¾¾®
HX
¾ ¾¾®
R — C = CH2
(Alkyne)
(Haloalkene)
(Alkyne)
(Haloalkene) Br
R—CºC—H
(Acetylene)
HBr
¾ ¾¾ ¾®
(Acetylene)
R — C — CH3 X
(Dihaloalkane) Br
H
H
HBr ¾® H — C — C — H H — C = C — H ¾ ¾¾ Br
1-Bromoethene or Vinyl bromide
H
1,1-Dibromoethane or Ethylidine bromide
Dilute HCl reacts with acetylene at 650C in the presence of mercuric ion when only one molecule of HCl is added giving vinyl chloride. Hg2+ ¾® CH =CH—Cl CHºCH+HCl ¾ ¾¾ 2 650 C
Peroxide have the same effect on the addition of HBr to alkyne as on alkene. H
R—C º C — H + HBr ¾Peroxide ¾¾¾¾ ¾®
(d)
Br
R—C = C—H .
Addition of Hypohalous acid : Alkynes add two molecules of hypohalous acid (HOX) in two steps forming a dihaloaldehyde in case of acetylene or a dihaloketone in case of an alkyl acetylene. Addition takes place according to Markownikoff's rule. OH Cl
OH Cl
O
Cl
- H2O HOCl R—CºC—H ¾HOCl ¾® R — C — C — H ¾¾¾® R — C = C — H ¾¾¾¾® R — C — C — H ¾¾¾ OH Cl
(unstable)
Cl
(Dihaloketone)
784
+2 CHEMISTRY (VOL. - I)
When R = H, H—CºC—H (Acetylene)
HOCl
¾¾¾¾®
O
OH Cl
OH Cl H—C = C—H
Cl
HOClH — C — C — H ¾¾® H — C — C — H
¾¾¾¾®
Cl
OH Cl
(Acetylene)
(unstable)
(Dichloroacetaldehyde)
When R = CH3 , OH Cl
OH Cl CH3—CºC—H ¾HOCl ¾¾¾® (Propyne)
CH3— C = C —H ¾HOCl ¾¾¾® CH3 — C — C — H OH Cl (Unstable)
O
Cl
CH3 — C — C — H Cl (1,1 Dichloroacetone) (e) Addition of Sulphuric acid : Two molecules of sulphuric acid add to an alkyne molecule at room temperature to give dihydrogen sulphate derivative of alkyne in two steps, which is in accordance with Markownikoff's rule. OSO3H
CH3—CºCH + H2SO4
CH3 — C = CH2
H2SO4
OSO3 H CH3 — C — CH3 OSO3H
(Markownikoff's product)
(Isopropylidine hydrogen sulphate)
(f) Addition of Water : Water adds to alkynes in presence of dilute H 2SO4 and mercuric sulphate catalyst. The addition takes place according to Markownikoff's rule to give initially an adduct called enol (–ene + ol) . Enols are unstable and are immediately converted to stable keto compounds.
–CºC– + H–OH
H2SO4 HgSO4
60°C
H
H
C=C
CH — C
OH
(enol)
O
(keto)
UNSATURATED HYDROCARBONS - ALKYNES
785
Thus when acctylene is passed thriugh 42% H2SO4 in presence of 1% HgSO4 at 600 C , acetaldehyde is formed. 42% H 2SO¾ 4® H—C º H—C + H—OH ¾ ¾¾¾¾¾ [CH2=CH—OH] 1% HgSO 4 60°C Vinyl alcohol
CH3—CHO
(Acetaldehyde)
(Unstable) Propyne gives acetone under this condition. H 2SO4 ¾® CH3—Cº CH + H2O ¾ ¾¾ HgSO4 ,600 C
CH3 — C = CH2
CH3 — C — CH3
OH
O
(Propyne)
(Acetone)
(g) Addition of Hydrogen cyanide : Alkynes add hydrogen cyanide in presence of barium cyanide to form alkenyl cyanides. Thus, acctylene with HCN gives vinyl cyanide (acrylonitrile), used for the preparation of synthetic fibre "orlon". H — C º C—H (Acetylene)
+ HCN
Ba(CN)
2® ¾ ¾¾¾¾ ¾
CH2 = CH — CN Vinyl cyanide (acrylonitrile)
II. SUBSTITUTION REACTION : (a) Formation of metallic derivatives : A hydrogen atom attached to triply bonded carbon atom is slightly acidic in nature and hence can be easily replaced by only strong base to form the corresponding salt called acetylide or alkynide. Alkynides are generally unstable and explosive. These are easily converted into alkynes when heated with dilute acids. (i) Sodium alkynides : Acetylene and terminal alkynes react with sodium metal at 475K or sodamide in liquid ammonia as solvent to form sodium acetylide and sodium alkynide respectively with evolution of H2 gas. liquid NH
2R — C º CNa + H2(g) (Sodium alkynide)
liquid NH
¾¾¾ ¾® HCºC Na+ ¾liquid NH 3 Monosodium acetylide
2R—CºCH + 2Na
3® ¾ ¾¾¾¾¾ ¾
CHºCH + NaNH2
3® ¾ ¾¾¾¾¾ ¾
–NH3
Na NH2
NaCºCNa + NH3 Disodium acetylide
786
+2 CHEMISTRY (VOL. - I)
(ii)
Acetylene reacts with ammoniacal solution of cuprous chloride and silver nitrate to form the corresponding copper acetylide (red) and silver acetylide (white) precipitate. H—CºC—H + 2[Cu(NH3)2] OH
¾ ¾®
CuC º C Cu + 2H2O + 4 NH3 . copper acetylide (red)
H—CºC—H + 2[Ag (NH3)2] OH– Tollen's reagent
¾ ¾®
AgC º C Ag + 2H2O + 4 NH3 . silver acetylide (white)
R—CºC—H + [Ag (NH3)2] OH
¾ ¾®
R—C º C Ag + H2O + 2 NH3 . silver alkynide (white)
These alkynides are unstable and explode in dry state. Hence, they should be decomposed while still wet by warming with dilute mineral acid which liberate back the alkynes. AgC º C Ag + 2HNO3 ¾¾® HCºCH + 2 AgNO3 (b) Alkylation of Alkynes : Monosodium acetylide on treatment with alkyl halides preferably a bromide, forms a higher homologue of acetyene. HCºCNa + CH3Br ¾¾® HCºC —CH3 + NaBr (monosodium (Propyne) acetylide) The above reaction may be carried out with 1-.alkynes. l
Na R Br ¾¾¾ ¾® RCºCNa ¾ ¾¾¾ RCºCH ¾ ¾® RCºCR/ + NaBr.. liquid NH NH2
3
(c) Substitution by halogens : Acetylene and 1-alkynes react with NaCl or NaOBr to form the corresponding di- and monohalo derivatives. HCºCH + 2 NaOCl ¾¾® Cl – C º C – Cl (Dichloroacetylene)
+ 2NaOH
HCºCH + 2 NaOBr ¾¾® Br – C º C – Br (Dibromoacetylene)
+ 2NaOH
RCºCH + NaOX
¾ ¾® RC º CX + NaOH
(X=Cl or Br)
(1-haloalkynes) (III) OXIDATION REACTIONS : (i) Combustion : All alkynes burn in oxygen producting carbon dioxide and water as the end products. At the same time tremendous amount of heat is liberated.
UNSATURATED HYDROCARBONS - ALKYNES
787
CnH2n–2 + 3n - 1 O2 ¾¾® nCO2 + 2n - 2 H2O + Heat 2
2
2 HCºCH + 5O2 ¾¾® 4CO2 + 2H2O + 312 kcal. Therefore, oxyacetylene flame is used for cutting and welding of metals. (ii) Chemical oxidation : (a)
Alkaline potassium permanganate oxidises acetylene to salt of oxalic acid. CH
COOH
KMnO4 ® + 4(O) ¾alkaline ¾¾¾¾¾¾¾
COOH
CH (Acetylene) (b)
COO—
alkali ¾ ¾¾ ¾®
(oxalic acid)
COO— (oxalate)
Acidified potassium dichromate oxidises acetylene to acetic acid, K2Cr2O7 ® HCºCH + (O) +H2O ¾ ¾¾¾¾ CH3COOH H 2SO4
(c)
Acidified KMnO4 oxidises acetylene to formic acid with cleavage of the triple bond. . KMnO4 ® 2 HCOOH HCºCH + 3 (O) + H2O ¾acid ¾¾¾¾¾¾ . KMnO4 ® RCOOH + R/ COOH R—CºC—R/ + 3(O) + H2O ¾acid ¾¾¾¾¾¾
(iii) Ozonolysis : Alkynes react with ozone. The rate of formation of ozonide is slower than that of alkenes. These ozonides on decomposition with 2n/water yield carboxylic acids. Zn/H2 O
/ ¾® R—C — C—R R—CºC—R/ + O3 ¾
O Examples :
O
(ozonide)
O
O
¾® H—C — C—H + O3 ¾
Zn/H2O
O
(ozonide)
H—C — C—H ¾¾® 2H—C — OH
¾® CH3—C — C—H CH3—CºC—H + O3 ¾
(Propyne)
(Formic acid)
O
O
O
O
(Glyoxal) + H2O2
O—O
(Acetylene)
R—C — C—R/ ¾¾® R—C — OH + (Diketone) O + H2O2 R/—C—OH
Examples : H—CºC—H
O
O
O
O
Zn/H2 O
O
CH3—C — C—H
+ H2O2
O
(Ozonide)
O
O
CH3—C—OH + H—C — OH (Acetic acid) (Formic acid)
788
(IV)
+2 CHEMISTRY (VOL. - I)
POLYMERISATION REACTION :
(i) Alkynes when passed through a red hot quartz or iron tube in the presence of chromic acid polymerise to yield aromatic hydrocarbons. Thus, acetylene molecule trimerizes to produce benzene H C HC
H C CH
Red hot Iron tube
HC
CH
HC
CH
¾ ¾¾¾¾®
HC
CH C H
C H
(Acetylene) (Benzene) (3 molecules) (ii) In the presence of nickel cyanide at high pressure, acetylene tetramerizes to produce cyclooctatetraene HC
CH
H C
H C
2 , 15-20 atm ¾Ni(CN) ¾¾¾¾¾¾¾¾ ¾®
60 - 700 C.
C H
C H HC
THF (Solvent)
(Cyclooctatetraene)
CH
(iii) When passed through CuCl solution containing NH4Cl, acetylene forms a mixture of mono-and divinyl acetylene. Here linear polymerisaton takes place. CH
CH
CuCl CH2=CH—CºC—CH=CH2 CHºCH + CHºCH ¾ ¾¾¾® CH2=CH–CºCH NH 4 Cl CuCl/NH4Cl monovinylacetylene
Divinylacetylene
(iv) Linear polymerisation of ethyne takes place under suitable conditions to produce polyacetylene which conducts electricity. Thin films of polyacetylene can be used as electrodes in batteries. n HC º CH Polymerization – (CH = CH) n
Ethyne
Polyethyne
(V) ISOMERISATION : 1-Alkynes when treated with KOH in ethanol readily isomerize to the more stable 2alkynes.
UNSATURATED HYDROCARBONS - ALKYNES
789
KOH ¾¾¾ ¾® R—CH2—CºCH ¬¾¾¾¾ R—CH=C=CH2 Ethanol
1-Alkyne
¾¾¾ ¾¾¾ ® ¬
R—CºC—CH3
Allene
2-Alkyne
The above rearrangement which take place through the formation of allene are called acetylene- allene rearrangement, which is reversible. 19.6
19.7
USES OF ACETYLENE : 1.
Acetylene is used for the artificial ripening of fruits.
2.
Acetylene is used for illumination purposes in hawker's lamp and light houses.
3.
Oxyacetylene flame is used in welding, cutting and cleaning iron and steel.
4.
It is used for the manufacture of acetaldehyde, acetic acid, alcohol etc.
5.
Industrial solvents like westron and westrosol are prepared from acetylene.
6.
It is used in the commercial production of polymers like synthetic rubber, plastics, fibres etc.
DISTINCTION OF ETHANE, ETHENE AND ETHYNE Test
Ethane
Ethene
Ethyne
1.
Burning
Burns with a nonluminous flame
Burns with a luminous flame
Burns with a luminous flame
2.
Bromine water
No change in colour of the solution
Colour of bromine water discharged
Colour of bromine water discharged.
3.
Baeyer's reagent
No reaction
Decolourises
Decolourises
4.
Ammoniacal silver nitrate solution.
No reaction
No reaction
White ppt.
5.
Ammonialcal cuprous chloride solution
No reaction
No reaction
Red ppt.
790
+2 CHEMISTRY (VOL. - I)
CHAPTER (19) AT A GLANCE ALKYNES :
General Formula : CnH2n—2
1. Preparation of Acetylene :
X
H
H—C—C—H
warm with
KOH¾® ¾alc. ¾¾¾¾
H X (vic-Dihalide) H
X
H—C—C—H
warm with
KOH¾® ¾alc. ¾¾¾¾
H X (gem-Dihalide) X
X
H—C—C—H
Zn / alcohol ¾ ¾¾¾ ¾®
HC—COONa HC—COONa (sodium maleate)
Kolbe's
¾Electrolysis ¾¾¾¾®
CaC2
Water ¾¾¾¾¾¾¾ ®
CHI3
with ® ¾Distil ¾¾¾¾¾
(Iodoform)
2C + H2
C2H2 (Acetylene)
¾¾¾¾¾®
X X (Tetrahalide)
at room. temp
Ag powder.
spark ¾Electric ¾¾¾¾¾¾ ¾® 25000 C
UNSATURATED HYDROCARBONS - ALKYNES
791
2. Properties of Acetylene :
H
H 2 / Pd
H
C=C
H 2 /Pd
H
H
(Ethane)
X
X X2
H—C—C—H (Acetylene dihalide) CHBr2
Br2
H
H—C—C—H
H
Ethene X
X2
H
H
X
H— C — C — H X
X
(Acetylenetetrahalide)
CHBr2
(Tetrabromoethane )
CH— Br
Br2 / H 2 O
CH— —Br (Dibromo ethylene) CH—I I 2 / Alcohol CH—I HX(X = F, (Diodoethene) CH2 Cl,Br,I)
HCl / Hg 2+
CHX Vinyl halide
CH3 CHX 2 (Ethylidene dihalide
CH2 =CH —Cl (vinyl chloride)
OH
OH Cl HOCl
H 2SO 4
CH= CH
H—C = C—H
HOCl
H — C — C —H OH Cl (Unstable)
OSO 3H CH3 — C — CH 3
OSO 3H (Isopropylidine hydrogensulphate )
(Acetylene)
H 2O 42%H 2SO 4 , 1%HgSO 4 HCN
Ba(CN) 2 Na in liq.NH 3
H H
C=C (Unstable)
Cl
O
H—C—C—H Cl (Dichloroacetaldehyde )
H
H
OH
H O (Acetaldehyde)
CH —C
H
CH2 =CH —CN (vinylcyanide)
CH = CNa
(Monosodium acetylide)
Na in liq. NH 3
Cl
NaC = CNa (Disodium acetylide )
792
+2 CHEMISTRY (VOL. - I)
combustion [O], Alkaline KMnO4
¾® CH º CH ¾ (Acetylene)
®
CO2 +H2O + heat COOH
®
COOH (oxalic acid)
[O],K2Cr2O7 ® H2SO4
CH3COOH (acetic acid)
[O], acid KMnO4
®
2 HCOOH (Formic acid) O
ozone ®
CH
CH
O O (Acetylene ozonide) Chromic acid polymerisation ® Pass through red hot tube
C6H6 (Benzene)
QUESTIONS
(A) Short Questions (one mark each) 1.
In acetylene, the triple bond between two carbon atoms consists of One sigma and two pi bond, One sigma and one pi bond Two sigma and the pi bond.
2.
How will you convert acetylene to acetic acid ?
3.
Indicate the hybridisaton state of carbon atom in acetylene.
4.(a) Which compound is formed when acetylene is passed into 42% H 2SO4 at 600C in the presence of HgSO4 ? (b) How is acetylene converted to acetaldehyde ? 5.
How many sigma & pi bonds are present in acetylene molecule ?
6.
What is the IUPAC name of the compound. CH3—C º C—C2H5
UNSATURATED HYDROCARBONS - ALKYNES
793
7.
How do you get acetylene from ethylenedibromide ?
8.
What is Baeyer's reagent ?
9.
Name a reagent which can differentiate 1- Butyne from 2- Butyne through a precipitation reaction
(B) Short Questions (two marks each) 1.
Complete the following reaction : dilH2SO¾ 4® CH º CH ¾ ¾¾¾¾ [A] ¾KCN ¾¾ ¾® [B]. Hg ++
2.
What happens when acetylene is treated with hydrogen bromide ?
3.
How is acetylene prepared from calcium carbide ?
4.
How can you obtain benzene from acetylene ?
5.
How to get ethane from acetylene ?
6.
State a reaction with equation to establish the acidic character of ethyne molecule.
7. 8.
How would you distinguish ethylene and acetylene ? Give one specific test for each. Identify A, B and C. Explain the reaction. 2SO¾ 4® 2O® A ¾dilH 2 /Ni ® C. ¾¾¾¾ B ¾H CaC2 ¾H ¾¾ ¾¾¾ HgSO 4
9.
What is the type of hybridisation of orbitals of carbon involved in a acetylene molecule?
10.
Explain why acetylene exhibits slightly acidic character whereas ethane does not have such a property.
11.
What happens when acetylene is passed through ammonical silver nitrate solution. Give equation.
12.
Suggest a test to distinguish but-1-yne and but-2-ene. Give equation.
(C) Other short and objective questions : (Three marks each) 1.
Write the structural formulae and IUPAC names of all isomeric butynes C 4H6.
2.
Describe what is meant by sp-hybridisation at carbon.
3.
Explain why acetylene is linear.
4.
Write the reaction of ozonolysis of 2-butyne.
5.
Write the structural formulae of alkyne which on ozonolysis gives a mixture of acetic acid and propionic acid.
6.
Explain why the hydrogen atoms of acetylene are acidic.
7.
What happens when acetylene is passed through ammoniacal CuCl solution ?
794
8.
+2 CHEMISTRY (VOL. - I)
Complete the following equations : (i)
Alcoholic ¾® CH3 — CHBr — CH2Br ¾ ¾¾¾¾ KOH
(ii)
Ammoniacal R—CºCH ¾¾¾¾¾¾¾®
(iii)
9.
10.
11.
AgNO3 liquid CH º CH + 2 Na ¾ ¾¾¾® NH 3
(iv)
R—CºCH+HOCl ¾ ¾®
(v)
42%. H2SO¾ 4® CH3—CºCH ¾¾¾¾¾¾ HgSO4
Identify the compound A, B and C. (i)
3 ® B ¾water A ¾O ¾¾ ¾® CH3COOH + CH3COOH ¾ ¾
(ii)
2Br2 ® Br CH — CHBr A ¾Electrolysis ¾¾¾¾¾® B ¾ ¾¾ 2 2
(iii)
Amm.¾® HCl C ¾dil A + H2O ¾¾® B ¾¾¾¾ ¾¾¾ ¾® CHºCH CuCl
(iv)
CH¾ ,H 2¾ O® C. 3I ® B ¾H CHºCH ¾Na ¾ ¾® A ¾ ¾¾ ¾¾¾ 2+
(v)
42% H 2SO¾ 4® CH2 = CH2 + Br2 ¾¾® A ¾Alc.KOH C. ¾¾¾ ¾® B ¾ ¾¾¾¾¾ 0
+
Hg
How the following conversions are made :
1%HgSO 4 60 C
(i)
Ethane to Ethyne and vice versa.
(ii) Fill in (i) (ii) (iii) (iv) (v) (vi) (vii)
1, 1, 2, 2- tetrabromoethane to acetylene and vice versa. the blanks : Acetylene is prepared by the electrolysis of ———. An alkyne has general formula ———. 1- Alkyne forms white precipitate with ——— solution. Acetylene is converted to ethane by ———. An alkyne decolourises ——— reagent. Acetylene forms a red precipitate when passed through ———. ——— is an excellent solvent and is known as westron.
Answer : (i) Potassium fumarate or maleate
(ii) CnH2n–2
(iii)
ammoniacal silver nitrate
(iv) hydrogenation
(v)
Baeyer's
(vi) ammonical cuprous chloride solution
(vii)
acetylene tetrachloride.
UNSATURATED HYDROCARBONS - ALKYNES
(D)
795
Long Questions :
1. How is acetylene prepared from (a) calcium carbide (b) by electrolysis ? How will you prove the acidity of hydrogen in acetylene ? What are its uses ? 2. How acetylene is prepared from calcium carbide ? What happens when acetylene reacts with HI ? How would you get benzene from acetylene ? 3. How acetylene is prepared in the laboratory ? How can you get ethane and benzene from acetylene ? Give equations : How will you prove that the acetylenic hydrogen atoms are acidic ? 4. How acetylene is prepared from calcium carbide ? What happens when acetylene reacts with HOBr ? Give equation. 5. Discuss the structure of acetylene molecule. Discuss two methods of preparation of alkynes. 6. How will you show that acetylene is an unsaturated compound ? How will you use it to prepare (a)
Benzene
(b)
Acetaldehyde
(c)
Ethane
(d)
copper acetylide
(e)
Ethene
(f)
Propyne
7. Compare and contrast the properties of saturated hydrocarbon with unsaturated hydrocarbons. Explain why the hydrogen atom in acetylene is acidic. 8. How acetylene is prepaned from the following compounds ? (i) maleic acid (ii) calcium carbide Explain the acidic character of acetylenic hydrogen. How does acetylene react with (i) Alkaline KMnO4 solution (ii) HOCl 9. How acetylene is prepared in the laboratory ? Describe with diagram and chemical equation. How acetylene reacts with (i) Br2 (ii) Na in liq NH3, Give two important uses of acetylene. 10.
How acetylene is prepared from the following compounds. (i) Calcium Carbide, (ii) Maleic acid H ow doesacetylene reactw ith (a)H Br(b)N a/liq.N H 3 Give two important uses of acetylene.
(E)
Multiple Choice Question : 1. Propyne belongs to (a) Paraffins (b) Alkenes (c) Olefins (d) Alkynes. 2. The general formula of alkyne is (a) CnH2n (b) CnH2n–2 (c) CnH2n+2 (d) CnHn. 3. The C—C bond is shorter in (a) Ethane (b) Ethylene (c) Acetylene (d) Propane. 4. When potassium maleate is electrolysed, the compound produced at the anode is (a) Ethyne (b) Ethane (c) Ethene (d) Propyne.
796
+2 CHEMISTRY (VOL. - I)
5.
6.
7. 8.
9.
10. 11. 12.
13.
14.
15.
16.
The product of the reaction CH3—CºCH+2HCl is (a) CH3.C.Cl2 — CH3 (b) ClCH2 — CHCl—CH3 (c) ClCH2 — CH2—CH2— Cl (d) CH3 — CH2—CHCl2 After ozonolysis of but-2-yne the product on hydrolysis is : (a) Formic acid (b) Acetic acid (c) Formic acid and acetic acid (d) Propionic acid. The alkyne R—CºC—R does not react with (a) HOCl (b) Br2 (c) O3 (d) Ammoniacal cuprous chloride Ethene and ethyne may be distinguished by using (a) Baeyer's reagent (b) Tollen's reagent (c) Bromine water (d) Schiff's reagent A yellowish white precipitate was obtained on addition of ammoniacal silver nitrate solution to a hydrocarbon. The hydrocarbon could be (a) Ethane (b) Ethene (c) Ethyne (d) Toluene Which of the following belongs to acetylene series : (a) C7H14 (b) C9H16 (c) C12H20 (d) C15H30 What is the number of possible alkynes with formula C 5H8 ? (a) 12 (b) 3 (c) 4 (d) 5 Acetylene reacts with sodamide in liquid ammonia and then reacts with CH 3I producing (a) But-2-yne (b) 3-Methylbut-1-yne (c) Propyne (d) Acetylene tetraiodide Acidic hydrogen is present in (a) Ethyne (b) Ethene (c) Benzene (d) Ethane Lindlar's catalyst is (a) Na in liquid NH3 (b) Pt in ethanol (c) Ni in ether (d) Pd with BaSO4 and quinoline When propyne is treated with aqueous H2SO4 in the presence of HgSO4, the major product is (a) Propanol (b) Propyl hydrogen sulphate (c) Acetone (d) Propanol The compounds 1–butyne and 2–butyne can be distinguished by using (a) Bromine water (b) KMnO4 solution (c) Tollens' reagent (d) Chlorine gas. ANSWERS 1. (d)
2. (b)
3. (c)
4. (a)
5. (a)
6. (b)
7. (d)
8. (b)
9. (c)
10. (b)
11. (b)
12. (a)
13. (a)
14. (d)
15. (c)
16. (c)
qqq
CHAPTER - 20
AROMATIC COMPOUNDS: AROMATIC HYDROCARBONS
20.1
INTRODUCTION
The term ‘aromatic’ was derived from the Greek word ‘aroma’ meaning pleasant smell. The aromatic compounds were so named because of their sweet odour. The sweet smelling compounds often contained groups like —OCH3, —CHO, —CH = CH — COOH attached to a phynyl (C6H5—) group. Coal tar was the main source of aromatic compounds. Compounds like phenol (C6H5OH) and benzene (C6H6) were the important aromatic compounds obtained as a result of distillation of coal tar. But subsequently the original meaning of term aromatic was abandoned when some odourless and vile-smelling substances containing C 6H5 group were discovered. Thus aromatic compounds were better known as benzenoid compounds i.e it included benzene and its derivatives. But now a days the scope of the term aromatic is not limited to benzenoid compounds only. It also includes some nonbenzenoid compounds e.g the heterocyclic compounds. Though they do not contain C6H5 group yet are found to be aromatic in nature. The aromatic compounds contain a higher percentage of carbon than the aliphatic ones and they produce sooty flames when heated over a copper foil. 20.2
AROMATIC HYDROCARBONS OR ARENES Aromatic hydrocarbons are known as arenes. Benzene is the simplest member of the
class. The general formula for benzene and its homologues is C nH2n—6. Benzene (n=6) has the molecular formula C6H6. Other members include toluene (C6H5—CH3)C7H8 (n = 7), xylene (C6H4 (CH3)2) C8H10 (n=8) etc. Two benzene rings fused together constitute naphthalene. So naphthalne and its derivatives constitute another series of arenes. The general formula is C nH2n—12, the first member being naphthalene C10H8 ( n = 10).
798
+2 CHEMISTRY (VOL. - I)
The general formula of arenes can be represented by C nH2n—6y, where y is the number of rings in the molecule. In benzene series y = 1 and in naphthalene series y = 2. CH3 Benzene 20.3
Toluene
Naphthalene
MEANING OF AROMATICITY The unique properties exhibited by the aromatic compounds is termed as aromaticity. So
the characteristics which attribute to the aromaticity of the compounds are : (i)
High degree of unsaturation : They are highly unsaturated compounds but very much resistant to addition reactions. It is to be noted that the aliphatic unsaturated compounds can readily undergo addition reactions. CH2 = CH2 + Br2 CCl 4 (Yellow)
(ii)
CH2 – CH2
+ Br2(aq) ® No decolourisation
Br Br (colourless)
Substitution reaction : They undergo electrophilic substitution reactions though they are highly unsaturated. Examples of such reactions include nitration, sulphonation, halogenation of benzene. Cl AlCl3 + Cl2 ¾Anh ¾¾¾¾ ¾®
(iii)
(iv)
(v)
(vi)
+ HCl
Benzene Cholorobenzene Low heat of hydrogenation and combustion : The aromatic compounds have low values of heat of hydrogenation and combustion. This is in favour of their stability. Higher percentage of carbon (sooty flame) : Unlike aliphatic compounds they contain a higher percentage of carbon and hence produce sooty flame having no residue when heated over a copper foil. Size of the ring & their structure : These are cyclic compounds having five, six or seven membered rings. They are flat or planar molecules. Hückel’s rule :- All aromatic compounds must satisfy Hückel’s (4n + 2) p rule. Hückel’s Rule or (4n + 2) p rule : According to this rule, if in a planar cyclic system of overlapping p-orbitals the number of p electrons is (4n + 2), then the system will have aromatic character. Here ‘n’ is either ‘0’ or a positive integer (n=0, 1, 2, 3 ....)
AROMATIC HYDROCARBONS
799
Benzene 6 p electrons n = 1, (4n + 2) p =(4x1 + 2) = 6 p electrons (4n+2)p rule satisfied AROMATIC
Naphthalene 10 p electrons n = 2, (4n + 2) p = (4x2 + 2) p = 10 p electrons (4n+2)p rule satisfied AROMATIC
e.g.(i) Benzene and naphthalene contain 6 and 10 p electrons respectively. They satisfy Hückel’s (4n+2) p rule and hence aromatic. (ii)
(iii)
Some cyclic ions obey Huckel’s rule and exhibit aromatic character.
+ Cyclopropenylcation
– Cyclopentdilnylanion
2p electrons
6p electron
Cycloheptatrienylcation 6p electrons
1, 3 Cyclobutadiene and 1, 3, 5, 7 Cyclo-octatetrene contain 4p and 8p electrons respectively. They do not satisfy Huckel’s (4n+2)p rule and hence are not aromatic.
1, 3–Cyclobutadiene 4 p electrons (4n + 2) p rule not satisfied ANTI AROMATIC (iv)
+
1, 3, 5, 7–Cyclooctatetrene 8 p electrons (4n + 2) p rule not satisfied NON - AROMATIC
Hückel extended his rule to five and six membered nonbenzenoid heterocyclic systems.
N .. Pyridine 6 p electrons (4n + 2) p rule is satisfied AROMATIC
.. O ..
.. .S.
Furan
Thiophene
.. N | H Pyrrole
There are four p electrons & 2 p-electrons on the heteroatom inside the ring. Thus, (4n+2) rule is satisfied. AROMATIC
800
+2 CHEMISTRY (VOL. - I)
Table 20.1 Examples of aromatic compounds obeying Hückel’s Rule n
(4n+2) p electrons
0
2p
Structures & Names of Aromatic Compounds Cyclopropenylcation
+ 1
2
.. N .. H Benzene Pyridine Pyrrole
6
N
10
Quinoline
Furan Thiophene
NH Indole
14 Anthracene
20.4
.. .S.
N Naphthalene
3
.. O ..
Phenanthrene
MOLECULAR ORBITAL STRUCTURE OF BENZENE X-ray diffraction measurements reveal that benzene consists of a planar hexagon of six
carbon atoms. C—C bond length is 1.39 A0 and C—C—C bond angle is 1200. Therefore, it stands to reason that each carbon atom in benzene is sp2 hybridised. C*6 = 1s2 2s1 2p1x 2p1y 2p1z — unhybridised (excited) hybridise form three sp2 hybrid orbitals Each carbon utilises two of its hybrid orbitals in forming sigma bonds with two other carbon atoms and one hybrid orbital in forming a sigma bond with hydrogen. Thus each carbon atom is left with an unhybridised p-orbital. The following figure depicts the sigma bond frame work of benzene resulting in planar hexagonal structure.
AROMATIC HYDROCARBONS
801
H H
H
sp2 HYBRID ORBITALS
C
C
C
C
|
H
H | C
H
or H
H H
C | H
|
H
H
OVERLAPPING
Fig 20.1 : s bond structure of benzene There are six unhybridised p-orbitals one for each carbon atom with two equal lobes. One of the lobes lies above and the other below at right angles to the plane containing the sigma frame work. These p-orbitals overlap laterally with each other in two different ways that lead to the formation of two Kekule structures (I & II) as shown below is Fig.20.2. These two Kekule structures thus correspond to localised p bond formation.
OVERLAP
OVERLAP
s ORBITAL FRAMEWORK BEFORE P-ORBITALS OVERLAP
I CORRESPONDING KEKULE STRUCTURE
II CORRESPONDING KEKULE STRUCTURE
Fig.20.2 Overlapping of p-orbitals that leads to formation of two Kekule structures (I & II)
Again since the internuclear distances between the carbon atoms of the hexagon are equal, there appears no reason why a p-orbital will overlap in one direction only and not in both the directions. Therefore overlapping of p-orbital with adjacent p-orbitals in both the directions is considered.
802
+2 CHEMISTRY (VOL. - I)
The six unhybridised p-orbitals each containing an electron laterally overlap with each other forming a delocalised p molecular orbital, half of which is above the plane and half, below the plane just sandwitching the six carbon atoms in between, (fig. 20.3).
DELOCALISATION OF
SIDEWISE OVERLAP
Separate p- orbitals on benzene ring may overlap on either side.
ELECTRONS
The six p- orbitals are delocalised, the lobes above and below the ring separately.
Or
p molecular orbitals of benzene having a continuous annular cloud, one above and one below the carbon sextet.
Fig : 20.3 Formation of continuous p election annular clouds in benzene molecule. Delocalisation of electrons stabilises the benzene ring. The actual energy of benzene in found to be less than that of the arrangement corresponding to any one of the two Kekule structures. The difference in energy between the energy of actual structure of benzene and that of the Kekule structure is known as resonance stabilisation energy or resonance energy. Thus, the actual structure of benzene is not represented by any one of the two Kekule structures. It is a hybrid of the two structures. Benzene is represented by a regular hexagon with an inscribed circle that symbolises the six delocalised p-electrons. This structure is known as the graphic formula of benzene.
Fig : 20.4 Graphic formula of Benzene Resonance description of the structure of Benzene : The concept of resonance plays a vital role in describing the actual structure of benzene when the proposed Kekule structures were found to have certain drawbacks. According to this concept benzene was considered to be a resonance hybrid of the two Kekule canonical structures as shown below. I
II Two Kekule forms
Resonance hybrid
AROMATIC HYDROCARBONS
803
The resonance hybrid of the two Kekule structures is in direct confirmity with the X-ray diffraction studies, according to which (a)
All the carbon - carbon bond lengths are equal i.e. 1.39 A 0 and are intermediate between C — C single bond distance (1.54A0) and C = C double bond distance (1.34A0).
(b)
The six carbon atoms are linked forming a hexagonal flat structure with bond angle 120°.
Stability of Benzene as explained by Resonance Theory: The unusual stability of benzene was explained by the resonance hybrid structure of benzene. The resonance stabilisation energy or simply the resonance energy was really responsible in explaining the unusual stability of benzene. It was found that the hybrid structure of benzene was more stable than any of the Kekule structures by about 36 kcal/mole of enegy. This energy, commonly known as the resonance energy could be calculated indirectly from the measurements of heat of hydrogenation as follows. The addition of hydrogen to a carbon- carbon double bond is an exothermic process. Hence some amount of heat energy is given out during the process of hydrogenation. For example: + H2 Cyclohexene
+ 28.6 kcal/mole Cyclohexane
+ 2 H2 Cyclohexadiene
+ 55 kcal/mole Cyclohexane
+ 3 H2 Benzene
+ 49.8 kcal/mole Cyclohexane
The heat of hydrogenation of one double bond in cyclohexene is 28.6 kcal/mole whereas that of cyclohexadiene is 55 kcal/mole (which is nearly twice the value for cyclohexene) as in this case there are two double bonds. If Kekule structure of benzene is considered the value of heat of hydrogenation would be expected to be 3x 28.6 = 85.8 kcal/mole. But when benzene is hydrogenated 49.8 kcal/mole of energy is liberated. The difference (85.8- 49.8) = 36 kcal/mole between the calculated and observed value is known as the resonance energy of benzene. This is described by the following figure. (Fig.20.5).
804
+2 CHEMISTRY (VOL. - I)
Resonance energy (36 kcal)
CYCLOHEXATRIENE 86 kcal
CYCLOHEXADIENE CYCLOHEXENE
BENZENE
55 kcal 50 kcal
28.6 kcal
CYCLOHEXANE
CYCLOHEXANE
CYCLOHEXANE
Fig : 20.5 Calculation of Resonance energy by the value of heat of hydrogenation. Form the above discussion we can conclude that the hybrid structure of benzene is thermodynamically more stable than any of the imaginary Kekule structures. 20.5
NOMENCLATURE OF AROMATIC COMPOUNDS :
1.
Hydrocarbons :- The trivial name of parent monocyclic arene is Benzene. Other members
particularly lower arenes are also named by their trivial names approved by IUPAC. For example: CH3 |
6
CH3 | CH3 2
CH3
6
3
5
1|
2 3
|
5
1|
4
Methylbenzene (TOLUENE)
4
1,2 - Dimethyl benzene (ORTHOXYLENE) CH2CH3 |
Ethylbenzene CH3 | |
CH3 H3C 1,3,5 - Trimethylbenzene (MESITYLENE)
CH3
6
CH3
1,3 - Dimethyl benzene (METAXYLENE) CH |
CH3 CH3
Isopropylbenzene (CUMENE)
5
1| 4|
2 3
CH3
1,4 - Dimethyl benzene (PARAXYLENE) CH2CH |
Isobutylbenzene
CH3 CH3
AROMATIC HYDROCARBONS
2.
805
Aryl group (Ar):- Just as ‘R’ is used to represent alkyl group, the symbol ‘Ar’ is used to represent an aryl group. The hydrocarbon group left after removal of a hydrogen atom of benzene, the parent hydrocarbon is called a phenyl group which is represented by the symbol Æ , Ph or C6H5. CH3 |
CH3 |
|
CH3 |
|
|
e.g. |
— —
p-tolyl
—CH
—CH2— benzyl
3.
m-tolyl
— —
o-tolyl
(Phenyl or Æ or Ph)
—C—
benzal
benzo
Halogen derivatives (a)
Nuclear substitution product (obtained by replacement of H atom of benzene ring by halogen atom) Cl |
CH3 |
Cl |
|
Chlorobenzene (b)
o-chlorotoluene
Cl
|
CH3 | Cl
| Cl p-chlorotoluene
o-dichlorotoluene
Side chain substitution product (obtained by replacement of H atom of side chain by halogen atom) CH2Cl |
Benzylchloride
CHCl2 |
Benzalchloride
C Cl3 |
Benzotrichloride
806
Hydroxy derivatives (a)
Nuclear substitution products (Phenols) OH |
OH |
OH |
|
OH | CH3
|
CH3
(Phenol)
2-Methylphenol (o-cresol)
3-Methylphenol (m-cresol) OH |
| CH3 4-Methylphenol (p-cresol) OH |
|
OH | OH
|
OH
1,2 - Dihydroxybenzene (Catechol)
| OH
1,3 - Dihydroxybenzene (Resorcinol)
1,4 - Dihydroxybenzene (Quinol)
OH |
|
OH | OH
1,2,3 - Trihydroxybenzene (Pyrogallol) (b)
OH
OH
|
OH
|
4.
+2 CHEMISTRY (VOL. - I)
1,3,5 - Trihydroxybenzene (Phloroglucinol)
Side chain substitution products (Aromatic alcohols):
CH2OH |
CH2CH2OH |
Phenyl methanol
b-Phenyl ethyl alcohol
(Benzyl alcohol)
(Phenyl ethanol)
AROMATIC HYDROCARBONS
5.
807
Ethers OC2H5 |
OCH3 |
6.
Methyl phenyl ether
Ethyl phenyl ether
(Anisole)
(Phenetole)
Aldehydes & Ketones CHO |
Benzaldehyde
Orthohydroxy benzaldehyde (Salicylaldehyde)
o- Tolualdehyde
|
CHO | CH
|
CHO | OH
3
Besides there are some aryl substituted aliphatic aldehydes. Here, —CHO group is attached to the side chain.
e.g.
CH2CHO |
Phenylacetaldehyde
A phenones.
CH=CH-CHO |
Cinnamaldehyde
C = O group when attached to a benzene ring, the compound is known as
COCH3 |
COC6H5 |
e.g. Methyl phenyl ketone
Diphenyl ketone
(Acetophenone)
(Benzophenone)
808
7.
+2 CHEMISTRY (VOL. - I)
Carboxylic Acids.
|
Benzoic acid
o- Methyl benzoic acid (o-Toluic acid)
COOH | COOH
COOH |
COOH | OH |
COOH | CH3
COOH |
o- Hydroxy benzoic acid (Salicylic acid) COOH |
|
|
COOH
| COOH
Phthalic acid Isophthalic acid Terephthalic acid (Benzene -1,2 dicarboxylic acid) (Benzene -1,3 dicarboxylic acid) (Benzene -1,4 dicarboxylic acid) CH=CH-COOH |
Cinnamic acid 8.
Acid derivatives (i)
Acid chlorides COCl |
O || C – Cl C – Cl || O
Benzoyl chloride (ii)
Phthaloyl chloride
Acid amides CONH2 |
CO NH
Benzamide
CO Phthalimide
AROMATIC HYDROCARBONS
(iii)
809
Acid anhydrides
O || C O C || O
Phthalic anhydride (iv)
Esters COOCH3 |
Methyl benzoate 9.
COOC2H5 |
Ethyl benzoate
Amines NH2 |
Aminobenzene (Aniline) NH2 | NH2 m- Phenylene diamine
o- Phenylene diamine, (1,2 - Diaminobenzene)
NH2 |
CH3 |
| NH2
| NH2
p- Phenylene diamine
p-Aminotoluene or 4 - Amino toluene or p - toluidine
Nitro compounds NO2 |
NO2 |
|
NO2 | CH3
|
10.
o- Aminotoluene 2- Amino toluene (o-toluidine)
NH2 | NH2 |
|
NH2 | CH3
Nitrobenzene
o-Nitrotoluene
NO2
1,3 - Dinitrobenzene (m- Dinitrobenzene)
810
+2 CHEMISTRY (VOL. - I)
NO2 | Cl
CH2NO2 |
|
CH3 | NO2 |
O2N
| NO2 2,4,6-Trinitrotoluene (TNT) 11.
o-Chloronitrobenzene
Phenyl nitromethane (side chain substitution product)
Sulphonic acids SO3H |
SO3H |
|
CH3 | SO3H SO3H
Benzene sulphonic acid
Benzene-1, 3-disulphonic acid
o- Toluene sulphonic acid
NH2 | | SO3H p- Amino benzene sulphonic acid 12.
Cyanides & isocyanides CºN |
Phenylcyanide (Benzonitrile) 13.
N® C |
Phenyl isocyanide
Grignard reagent MgBr |
|
CH3 | MgBr
Phenyl magnesium bromide
o- Tolyl magnesium bromide
AROMATIC HYDROCARBONS
14.
811
Diazonium salts – + N º NCl |
|
CH3 | N+2Cl–
Benzene diazonium chloride 20.6
o- Toluene diazonium chloride
METHODS OF PREPARATION OF BENZENE Carbonisation of bituminous coal is done to get coal chemicals and coke.
Carbonisation of coal Carbonisation of coal is undertaken to manufacture coke which is used in iron and steel industry as a reducing agent. When coal is heated above 673K in closed retorts (in absence of air) coal becomes soft and plastic. The volatile materials are obtained as distillate. Low temperatrue carbonisation (723-973K) : In this method, coal is heated in absence of air up to 973K. A large amount of coal tar is produced along with ammonia, light oil, coke oven gas and coke. (1)
From coal tar : It is a thick, black viscous liquid obtained as a result of low temperature carbonisation of coal. It is considered to be a rich source of aromatic compounds. Coal tar is subjected to fractional distillation. COAL TAR
HOT VAPOURS
PREHEATER
IRON STILL BENZOL + WATER PITCH
FURNACE
To fractionating columns for getting (1) Light oil (2) Middle oil (3) Heavy oil (4) Green oil
Fig.20.6 Fractional distillation of coal tar Four main fractions are usually collected leaving behind a residue of ‘pitch’ which is used mostly for surfacing roads.
812
+2 CHEMISTRY (VOL. - I)
Table 20.2 Fractions obtained in coaltar distillation. Fraction
Temp. range
Chief components
I
Light oil
primary > secondary > tertiary. CH3 CH – > (CH3)3 C – CH3 – > CH3 – CH2 – > CH3 This is because the alkyl group must possess at least one H atom on the C atom joined to the ring carbon. tert-butyl benzene on nitration gives only para isomer. This is due to steric hindrance caused by bulky methyl groups not allowing the electrophile to enter at the ortho position. C(CH3)3 C (CH3)3 | | + NO2
¾ ¾®
tert-butyl benzene
2)
| NO2 p-nitro tert - butyl benzene
Stability of s - complex (Carbocation) In electrophilic substitution reaction of mono substituted benzene there is formation of complex which immediately rearranges to form s complex. The stability of the s complex formed in case of ortho, meta and para attack of the electrophile enables us to predict the directive influence of the group already present in the benzene nucleus.
AROMATIC HYDROCARBONS
i.
833
When the group already present is electron donating. ..
S | E+
+
Electrophile
S |
H
+
E
S | +
,
ortho
S | H, E
+ H E para
meta - complexes
Wheland structures for the carbocation ( s - complexes) ..
H E
+
S |+
+
H E
S+ || H E
..
S |
..
S |
..
S |
H E
+
H E
ortho
+ H E
+
H E
..
(+)
S | +
H E
H E
+S ||
+
H E
..
.. S |
H E
..
..
S |
+
..
meta
H E
S |
S |
..
S |
..
S | +
S || +
H E
H E
para
Thus there are more number of resonating forms for Wheland carbocation is case of ortho and para substituents. The ortho and para attack of the electrophile results in more stabilized carbocation. It can therefore be concluded that the electron donating groups allow the electrophile to enter at the ortho and para positions only or in otherwords it can be said that the electron donating groups are o –p directing.
834
ii)
+2 CHEMISTRY (VOL. - I)
when the group already present is electron withdrawing x=y | +
E
x=y | H
+
+
Electrophile
E
x=y |
x=y | +
,
meta
ortho
where (-x = y) is an electron withdrawing group
H, E
+ H E para
complexes
Wheland structures for the carbocation +d - d x=y | H E + 1
x=y | H + E ortho
+d - d x=y |
x=y | +
H E
+ 4
meta x=y | + H E Para
H E
+d - d x=y | +
H E 7
+d - d x=y | H E
+
+
3 Less stable
2
+d - d x=y | 5
+d
+d - d x=y | H + E
H E
-d x=y | +
H E 8 Less stable
+d - d x=y | + 6
H E
+d - d x=y | + H E 9
In structures 3 and 8 the two +ve charges are adjacent to each other. The repulsive interaction leads to high energy and therefore the contributing structures are less stable. They do not contribute to the resonance hybrid. Consequently there are 3 resonating structures for meta attack and two for each ortho and para attack. Hence the carbocation ion intermediate formed in case of meta attack is comparatively more stable than that for ortho and para attack. Thus the electron withdrawing groups allow the electrophile to enter at the meta position or in otherwords it can be concluded that the electron withdrawing groups are meta directing.
AROMATIC HYDROCARBONS
20.9
835
CARCINOGENICITY AND TOXICITY
Benezene and polynuclear hydrocarbons containing more than two benzene rings fused together are toxic and carcinogenic by nature. Their carcinogenic activity is further influenced by the presence of certain groups like –CH3, –OH, –CN, –OCH3, etc. Examples are :
1
1 2 2
1, 2 - Benzanthracene
8
9
CH3
4
1
7
2
6
3 5
10
CH3
1, 2 - Benzpyrene
4
9, 10 - Dimethyl1, 2 -benzanthracene
5
3
6
2 1
1, 2, 5, 6 Dibenzanthracene
Polynuclear hydrocarbons are formed by the incomplete combustion of organic matter such as coal, petroleum, tobacco etc and hence widely present in the environment causing human cancer. Naturally the root of the fatal disease cancer can be substantially evaded by the elimination of carcinogens from the environment. When these polynuclear hydrocarbons enter into human body, undergo biochemical reactions and are converted into their oxides called epoxides and then into dihydroxyepoxides. On reaction with the purine bases such as guanine, present in DNA and RNA of the human cells, these hydroxides affect both structure and characteristics. Even at this stage purine because of bigger size can not fit into the double helix of DNA. This damage causes mutation and consequently leads to cancer. Schematic representation of carcinogenic effect of polynuclear hydrocarbons (PNH) can be given as: PNH Cancer.
O2
PNH epoxide ® PNH dihydroxyepoxide DNA or RNA Mutation ®
836
+2 CHEMISTRY (VOL. - I)
CHAPTER (20) AT A GLANCE 1.
Aromatic compounds contain a higher precentage of carbon atom and produce a sooty flame when heated over a copper foil.
2.
The general formula of arenes is CnH2n–6y where y refers to no. of rings.
3.
Hückel’s Rule :- All aromatic compounds must satisfy Hückel’s (4n + 2) p rule.The rule states that in a cyclic system of overlapping p-orbitals the no. of p electrons should be (4n+2).
4.
The six carbon atoms in benzene are sp2 hybridised, C—C bond length is 1.39 Ao and C – C – C bond angle is 1200.
5. 6. 7. 8. 9. 10.
Resonance stabilisation energy for benzene is 36 kcal/mole. Coal tar is a thick, black, viscous liquid obtained as a result of low temperature carbonisation of coal. It is a rich source of aromatic compounds. Benzene can be isolated from the light oil fraction of coal tar. Benzene undergoes electrophilic substitution reactions like nitration, sulphonation, halogenation and Friedel Craft’s reaction. 2,4,6 Trinitrotoluene (TNT) is a well known explosive. Friedel Craft’s reaction : Benzene reacts with alkyl halide or acyl halide in presence of anh. AlCl3 forming the corresponding alkyl benzene and acyl derivative of benzene
11.
respectively. The reactions are known as F—C alkylation and F—C acylation. Nitrobenzene can not exhibit Friedel Craft reaction.
12.
g — isomer of benzene hexachloride (BHC) is known as Gammexane.
13.
Benzene upon oxidation in presence of V2O5. forms maleic anhydride.
14. 15.
The arrangement of substituents on the benzene ring is called orientation. The electron donating groups activate the benzene nucleus and they are ortho and para directors. The electron withdrawing groups deactivate the benzene nucleus and they are meta directors. The halogen atom is o-p directing but is a ring deactivator. The alkyl groups do not possess unshared pair of electrons but they are o-p directing and act as ring activators. Hyperconjugation is no-bond resonance. Benzene and polynuclear hydrocarbons containing more than two benzene rings fused together are toxic and carcinogenic activity is influenced by substituents like methyl, hydroxy, cyano, methoxy etc. which disrupt cellular metabolic processes.
16. 17. 18. 19. 20.
AROMATIC HYDROCARBONS
837
QUESTIONS A.
Very Short answer type (1 mark) (i)
Give the structure of Gammexane.
(ii)
Give the structural of formula of p-xylene
(iii)
Give the graphic formula of benzene
(iv)
State Huckel’s rule
(v)
Complete C6H5 COONa + NaOH
CaO
———— + —————
(vi)
What is the electrophile used in case of nitration of benzene ?
(vii)
Name the fraction from which benzene can be isolated in coal tar distillation.
(viii)
What is the main source of aromatic compounds ?
(ix)
A cyclic hydrocarbon molecule has all the carbon and hydrogenn atoms in a single plane. All C – C bonds have same length which is less than 1.54A0, but more than 1.34A0. What is the C – C – C bond angle in the molecule ?
(x) (B)
Name the reaction where Lewis acid is used as catalyst for aromatic substitution.
Short answer type (2 marks) (i)
Write a note on ‘Aromaticity’
(ii)
Starting from benzene how can you prepare cyclohexane ?
(iii)
What happens when toluene in oxidised by acidified KMnO4 solution ?
(iv)
Aromatic compounds undergo substitution reactions, explain with one reaction.
(v)
Although benzene is highly unsaturated normally it does not undergo addition reactions Give reasons.
(vi)
What happens when acetylene is passed through heated iron tube ?
(vii)
What happens when benzene reacts with methyl chloride in presence of anh. AlCl 3?
(viii)
What happens when [each 2 marks] (a)
Benzene is oxidised in presence of V2O5.
(b)
Benzene is subjected to ozonolysis.
(c)
Benzene is treated with Cl2 in presence of sunlight.
(ix)
Write the position isomers of xylene.
(x)
What happens when excess of Cl2 is passed through boiling toluene in presence of sunlight.
838
+2 CHEMISTRY (VOL. - I)
_
(xi)
Nitration of toluene is easier than benzene why ?
(xii)
What is TNT ? How is it formed from toluene ?
(xiii)
Nitrobenzene does not respond to Friedel Craft reaction — why ?
(xiv)
What happens when benzene reacts with Cl2 in presence of UV light ?
(xv)
Metanitrophenol is less acidic than orthonitro phenol — why ?
(xvi)
What is Friedel Craft’s acylation ?
(xvii) State and explain Huckel’s rule (xviii) What are the numbers of p-electrons in benzene and nephthalene ? (xix)
Name four o- and p- directing groups.
(C)
Long answer type : (10 marks)
1.
What are different reactions exhibited by aromatic compounds ? Give an example of each reaction and explain.
2.
(a)
How is benzene obtained from light oil coal tar distillation.
(b)
How the following groups are introduced in the benzene ring ? —CH3, —NO2, —SO3H
3.
Write notes on Friedel Craft’s reaction
4.
Discuss the molecular orbital structure of benzene.
5.
Predict the product & suggest the mechanism. (a)
+ (CH3)2CHCH2Cl
(b)
+ CH3CH2COCl
Anh.AlCl3
Anh.AlCl3
6.
Discuss any one electrophilic substitution reaction in benzene.
(D)
Multiple Choice type question with answers.
1.
When phenol is distilled with zinc dust the product is
2.
(a)
Toluene
(b)
Benzene
(c)
Xylene
(d)
None of the above
In Benzene the ‘C’ atoms are (a)
sp3 hybridised
(b)
sp2 hybridised
(c)
sp hybridised
(d)
Unhybridised
AROMATIC HYDROCARBONS
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
839
Coaltar is a main source of (a) Aromatic compounds (b) Aliphatic compounds (c) Cycloalkanes (d) Heterocylic compounds The following products are formed by fractional distillation of Coal tar. Which one is basic ? (a) Phenol (b) Toluene (c) Anthracene (d) Pyridine All bonds is benzene are equal due to (a) tautomerism (b) I-effect (c) Resonance (d) Isomerism Phenol is less acidic than (a) o–Cresol (b) p-methoxy phenol (c) p-nitrophenol (d) Ethanol Nitro group in nitrobenzene is (a) ortho director (b) meta director (c) para director (d) ortho & para director Nitration of benzene is (a) Nucleophilic substitution (b) Electrophilic substitution (c) Homolytic substitution (d) Electrophilic addition The compound that is most reactive towards electrophilic nitration is (a) Toluene (b) Benzene (c) Benzoic acid (d) Nitrobenzene Anhydrous AlCl3 is used in Friedel Craft reaction because it is (a) electron rich (b) soluble in ether (c) insoluble (d) electron deficient In which can the C—C bond length in same (a) But -2-ene (b) But-l-ene (c) Benzene (d) Prop-l-yne Which of the following pairs will give the same bond angle ? (a) Ethane & ethyne (b) ethane & ethene (c) Ethene & benzene (d) Ethyne & benzene The reaction of Toluene with Cl2 in presence of FeCl3 gives predominantly (a) m-chlorotoluene (b) benzoyl chloride (c) o - & p chlorotoluene (d) benzal chloride Nitration is easy is case of (a) Toluene (b) Nitrobenzene (c) Chlorobenzene (d) Sulphobenzene Benzene reacts with chlorine in presence of sunlight to form (a) 666 (b) BHC (c) Gammexane (d) All
840
16.
17.
18.
19.
20.
21.
22.
23.
24.
(E)
+2 CHEMISTRY (VOL. - I)
C — C bond lenght in benzene is (a)
1.54 A0
(b)
1.20 A0
(c)
1.39 A0
(d)
1.34 A0
Benzoic acid when heated with sodalime gives (a) benzene (b) sodium benzoate (c) benzaldehyde (d) benzyl alcohol p-Nitrophenol is a stronger acid than phenol because nitro group is (a) Electron donating (b) Electron withdrawing (c) Basic (d) Acidic Which is the correct statement ? (a) Benzyl alcohol is more acidic than phenol (b) Ethanol is a powerful oxidising agent (c) Phenol is more acidic than propanol. (d) Ethane has higher boiling point than ethanol The compound that is not a Lewis acid is (a) BF3 (b) AlCl3 (c) BeCl2 (d) SnCl4 The direct iodination of benzene is not possible because (a) Iodine is an oxidizing agent (b) Resulting C6H5 I is reduced to C6H6 by HI (c) HI is unstable (d) The ring gets deactivated. In aniline, the –NH2 group (a) activates the benzene ring by both inductive and resonance effect. (b) deactivates the ring by both inductive and resonance effect. (c) activates the ring by resonance effect and deactivates by inductive effect. (d) activates the ring by inductive effect and deactivates it by resonance effect. The electrophile in aromatic sulphonation is + SO H (a) (b) SO3 3 – (d) H2SO4 (c) HSO3 Chlorination of toluene in the presence of FeCl 3 gives predominantly (a) Benzoyl chloride (b) m–chlorotoluene (c) benzyl chloride (d) o– and p–chlorotoluenes Fill in the blanks : 1. Aromatic hydrocarbons burn with ——— flame. 2. ————— is an important source of aromatic compounds. 3. ————— fraction of coaltar mainly contains benzene.. 4. Polymerisation of acetylene gives ————. 5. g-Isomer of benzene hexachloride is called ————— 6. Aromatic compounds are stabilised by ————.
AROMATIC HYDROCARBONS
841
7.
An electron withdrawing substituent in benzene orients the incoming electrophile in the ———— position.
8.
Toluene on oxidation with CrO2Cl2 gives ———
9.
The number of p electrons is the aromatic ring in equal to (4n+2) This is in accordance with ———— rule.
10.
Benzene on oxidation with V2O5 gives ————
11. 12.
Benzene on ozonolysis gives ————. Anthracene is a ———— compound.
13.
Benzene when treated with Conc. H2SO4 at 800C produces ————
14.
No. of p electrons in naphthalene is ————
ANSWERS D. MULTIPLE CHOICE QUESTION
E.
1. b
6. c
11. c
16. c
21. b
2. b
7. b
12. c
17. a
22. c
3. a
8. b
13. c
18. b
23. b
4. d
9. a
14. a
19. c
24. d
5. c
10. d
15. b
20. c
1. Luminous
2. Coaltar
5. Gammaxene
6. Resonance 7. Meta
8. Benzoic acid
9. Huckel’s
10. Maleic anhydride
11. Glyoxal
12. Benzenoid
13. Benzene sulphoric acid
14. 10
..
3. Light oil
qqq
4. poy ethyne or poly acetylene
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UNIT – XIV CHAPTER - 21
ENVIRONMENTAL CHEMISTRY 21.1
INTRODUCTION
The word ‘Environment’ literary means surroundings. So the ‘Environment’ is considered as a composite term for the conditions in which an organism lives and consists of air, water, soil, sunlight, food etc. which are the basic needs of all organisms to survive. The basic components of the environment are : (a)
Abiotic or non-living components which are further sub-divided into (i)
Atmosphere or air
(ii)
Hydrosphere or water
(iii)
Lithosphere or the rocks and soil.
(b)
Biotic or living components comprising of flora and fauna.
(c)
Energy components consisting of solar energy geochemical energy, geothermal energy, nuclear energy etc.
Thus the environmental studies deal with the sum of all social, economical, biological, physical and chemical interrelations with our surroundings. Focus on environmental studies was concentrated in the year 1972 when United Nations Conference on the Human Environment was held at Stockholm showing alarming concern over environmental pollution. Its main object is to enlighten the people about the importance of protection and conservation of our environment and the need to restrain human activities leading to indiscriminate release of pollutants into the environment. ENVIRONMENTAL CHEMISTRY is a part of enviromental education. It may be defined as the study of the sources, reactions, transport, effect and fate of chemical species in the air, water and soil and the effect of human activity upon these. Precisely environmental chemistry is the science of chemical phenomena in the environment.
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ENVIRONMENTAL POLLUTION The undesirable change in the physical, chemical or biological characteristics of air, water
and soil is known as pollution. It may harmfully affect life and therefore becomes a potential health hazard to a living organisim. In otherwords, the direct or indirect change in any component of the biosphere that is harmful to living organisms is known as pollution. The substance that causes pollution is known as a pollutant. It may include chemical or geochemical substances like dust, sediment, grit etc, biotic component and its products or physical factor like heat which badly affect the environment. Environmental Pollutants : The pollutants that affect air, water and soil are environmental pollutants. They are : 1)
Gases -
Carbon monoxide (CO), Sulphur dioxide (SO2), Oxides of nitrogen (NO and NO2), Halogens (Cl2, Br2, I2)
2)
Metals -
Mercury, Lead, Zinc, Iron, Nickel, Tin, Cadmium, Chromium etc.
3)
Deposited matter -
Smoke, Soot, Dust, Dirt, Tar etc.
4)
Fluorides
5)
Acid droplets -
6)
Organic substanses - Ether, Benzene, Acetic acid etc.
7)
Agrochemicals -
Sulphuric acid, Nitric acid.
Fertilisers and Biocides (Pesticides, Fungicides, Nematicides, Bactericides etc)
8)
Radioactive waste
9)
Solid waste
10)
Photochemical
Ozone, Photochemical smog, PAN (Peroxy Acetyl Nitrate), Aldehyde
oxidants 11)
Ethylene etc.
Noise The order of priority of pollutants and the medium as stated in UNEP (United Nation
Environment Programme) document is as follows.
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Order of priority
Medium
SO2 + suspended particles
Air
Strontium, Caesium
Food
Ozone
Air
DDT and organochlorine compounds
Biotic, Man
NO3, NO2
Air, Drinking water
Nitrogen oxides. 4.
Mercury compounds
Food, water
Lead and cobalt. 5.
Petroleum hydrocarbons
Sea
Carbon monoxide
Air
6.
Fluorides
Water (Fresh water)
7.
Asbestos
Air
Arsenic
Drinking water
Mycotoxins and
Food.
8.
microbial contaminants
21.3
AIR POLLUTION
The word pollution has been derived from the Latin word “pollutioneim” meaning the art of making dirt. The earth is surrounded by an insulating blanket known as atmosphere. We live in an ocean of air. The air envelope, called the atmosphere protects the earth from extremes of heat and cold and from the harmful rays of sun. Without air life is impossible, since air contains substances that all living things require. Pollution of air is imbalance in the quality of air caused as a result of introduction of some foreign elements from natural and man made sources to the air so that it becomes harmful to living organisms. Composition of air The air envelope that surrounds the earth is composed of mixture of gases. Normal composition of clean air is as follows:
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Gases
Percent(by volume)
Nitrogen
78.084
Oxygen
20.997
Argon
0.934
Carbon dioxide
0.0314
Methane
0.0002
Hydrogen
0.00005
Other gases
minute
21.3.1 Sources of air pollution : Pollution of air results from emission of gaseous substances from industry,automobiles,thermal power stations, domestic combustion etc.The important sources of air pollution are discussed below. 1:
Industrial chimney wastes : Chemical industries like paper and pulp mills, fertiliser plants, petroleum refineries, steel plants etc. produce certain solid, liquid or gaseous substances by which the land, water and air get polluted. The most common gases responsible for causing pollution are carbon dioxide, carbon monoxide, sulphur dioxide, oxides of nitrogen, hydrogen sulphide etc.
2.
Automobiles: The toxic vehicular discharges now-a-days have become a serious threat to ambient air quality. The vehicles discharge a lot of poisonous gases like carbon monoxide (about 77%), oxides of nitrogen (about 8%) and unburnt hydrocarbons (about 15%). Besides these gases there are particulates of lead and traces of aldehydes, esters, ethers, peroxides and ketones which are chemically active and these combine in presence of sun light forming highly toxic photochemical smog in the atmosphere. Noise is a serious type of pollution arising from road traffic, air craft and machineries. Loud speaker and record players also add noise to air. Noise can cause psychological disorders due to constant strain on nervous system.
3.
Burning of fossil fuels and fires Poisonous gases like carbon dioxide, carbon monoxide, sulphur dioxide, oxides of nitrogen, methane etc. are released into the atmosphere by increasing use of fossil fuels, coal etc. The concentration and quality of these pollutants depend upon the type of fuel being used. For example SO2, coming out through the chimneys of most of the
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factories gets oxidised to sulphur trioxide which when react with rain water gets converted to sulphuric acid. This is known as acid rain. Acid rains cause global ecological problem affecting land flora and fauna. 4.
Thermal Power stations For augmentation of energy generation a number of thermal power stations are set up in our country by NTPC (National Thermal Power Corporation). The consumption of coal by thermal plants is several million tonnes. The main pollutants are sulphur dioxide, oxides of nitrogen, fly ash and hydrocarbon.
5.
Radioactive pollution : Radioactive pollution has become a potential health hazard now-a-days . The pollution is due to production of nuclear fuel, testing of nuclear weapons, use of nuclear energy in power plants and more over the use of radio isotopes in medicine, industries and scientific research.
The air pollutants from various sources are as follows: 1.
Carbon compounds : These include both CO and CO2. CO is produced as vehicular exhaust whereas CO2 is released by complete combustion of fossil fuels.
2.
Sulphur compounds : These are mainly SO2, H2S and H2SO4. The pollutants are released by combustion of fossil fuels.
3.
Oxides of nitrogen : These mainly include NO, NO2 and HNO3.The pollutants are released by power plants, automobiles and industries.
4.
Hydrocarbons : These include benzene, benzpyrenes etc. which are released into the atmosphere by industries and automobiles.
5.
Ozone : The level of O3 rise in the atmosphere due to different human activities.
6.
Metals : These include lead, nickel, beryllium, tin, vanadium, titanium, cadmium etc which are realeased during metallurgical process and also as automobile exhaust in the form of solid particles, liquid droplets or gases.
7.
Fluorocarbons: These are released by the industries and insecticide spray.
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Particulate matter: These include dust, grit, fly ash and other suspended particulate matter (SPM)
9.
Photochemical products : These mainly include photochemical smog, PAN etc which are released mostly by automobiles.
10.
Other Toxicants : These include complex chemical substances released during manufacturing processes.
21.3.2 Harmful effects of air pollutants : 1.
Carbon dioxide : As a result of burning of fossil fuel for domestic cooking, heating and the fuel used in power plants, industries, carbon dioxide is released into the atmosphere. An increase in concentration of CO2 in the atmosphere contributes significantly to global warming, green house effect. Green-house effect CO2 is present exclusively in the troposphere. The temperature on the surface of the earth is maintained by the energy balance between the sun rays striking the earth and the heat radiated back into the outer space. This is, ofcourse under normal condition i.e. with normal concentration of CO2. But when there is an increase in CO2 concentration, the thick layer of CO2 functions like the glass panels of a green house or the glass windows of a motor car. This layer only allows the sunlight to filter through it reaching the earth but prevents the heat being reradiated by earth back to the outerspace. Thus, most of the heat is absorbed by CO2 layer and water vapours in the atmosphere which ultimately results in heating up of the earth’s atmosphere. The effect is known as Green house effect.
2.
Carbon monoxide : The chief source of CO is automobile exhaust. Besides this, natural processes like forest fires, natural gas emission, volcanic eruption, marsh gas production etc. serve as other sources of carbon monoxide. CO causes difficulty in breathing, causes headaches and also irritation of mucous membrane. On inhalation, carbon monoxide passes into the blood stream through lungs. Inhaled CO combines with blood haemoglobin forming carboxy haemoglobin. This does not permit the transport of oxygen from lungs to the cells. Thus, oxygen carrying capacity of the blood gets diminished. This leads to oxygen deficiency known as hypoxia. The gas is fatal over 1000 ppm causing unconciousness in an hour and ultimately death in a few hours.
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Sulphur dioxide The major sources of SO2 are burning of fossil fuels in thermal power plants and smelting industries. Petroleum refineries and automobiles also release SO 2 into the atmosphere. SO2 causes intense irritation in eyes and respiratory tract. In upper respiratory tract it is absorbed and cause swelling and stimulate mucous secretion. Exposure to 1ppm level of SO2 leads to constriction of air passage causing significant breathing problem. It reacts with moist air, fogs, H2O2, O3 etc. forming H2SO4 which acts as a stronger irritant than SO2. It corrodes lime stone and metals.
4.
Oxides of nitrogen Vehicular exhaust is the major source of oxides of nitrogen which include both nitric oxide and nitrogen dioxide. NO is produced by the combustion of N 2 and O2 during lightening discharge and by bacterial oxidation of NH 3 in the soil. Besides, these oxides of nitrogen are also released into the air by electric power plants and other chemical industries. NO, on combination with atmospheric oxygen produces NO 2 gas. NO2 gas is the chief constituent of photochemical smog. At high concentration NO 2 causes variety of diseases like lungs cancer, pneumonia and emphysema (inflammation).
5.
Hydrogen sulphide The oceans release about 30 million tons of H2S gas every year whereas land releases about 60 to 80 million tons. About 3 million tons of H 2S are released by the industries per year. But the chief industrial sources of H 2S are users of sulphur containing fuels. Other sources include decaying vegetation and organic matter, sulphur springs, coal pits, volcanic eruption etc. At low concentration H2S causes headache, nausea, coma and even death. At 5 ppm its unpleasant odour may affect digestive system destroying appetite. Higher concentration may cause conjectivities and irritation of mucous membrane.
6.
Hydrocarbons The chief source of hydrocarbons are the motor vehicles. About 40% of the vehicular exhaust hydrocarbons are unburnt fuel components, the rest are the products of combustion. Plants and bacteria are the natural sources of hydrocarbons. Human activities may contribute nearly 20% of hydrocarbons emitted to the atmosphere. Hydrocarbons have carcinogenic effect on lungs. Aromatic hydrocarbon benzene, emitted from gasoline causes lungs cancer. The most potent cancer inducing hydrocarbon pollutant is benzpyrene. It is present in small amount in smoke, charcoal boiled stakes, tobacco and gasoline exhaust. Decay of garbage and aquatic vegetation may produce
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marsh gas (methane) which is a gaseous pollutant. Higher concentration of it may cause explosion. Hydrocarbons combine with oxides of nitrogen under UV-component of light forming pollutants like PAN (Peroxy acetyl nitrate) and O 3. These pollutants are known as photochemical smog which cause irritation of eye, nose and throat and allied respiratory distress. Acid rains The important gaseous pollutants are oxides of nitrogen and sulphur. These oxides are produced by combustion of fossil fuels, automobile exhaust, power plants and domestic fires. These oxides are swept up into the atmosphere and react with water in the atmosphere to form sulphuric and nitric acids. These acids formed in the air soon come down to the earth with the rain and so the name acid rain. The acid rain is infact a cocktail of H 2SO4 and HNO3. The ratio of the two acids depends on the relative quantities of the oxides of S and N emitted. 60-70% of acidity is attributed to H 2SO4 whereas 30-40% to HNO3. Acid rains have assumed global ecological problems and their impacts are far reaching. They increase acidity of soil, cause acidification of lakes and streams, thereby affecting aquatic life, affect production of crops and moreover pose an invisible threat to human health. They also corrode buildings, statues, monuments, bridges, railings etc. Levels of heavy metals like Al,Mn, Zn, Cd, Pb, Cu etc. in water increases beyond safe limit. Many useful bacteria and blue green algae are destroyed due to acidification, thus causing ecological imbalance. 7.
Ozone Ozone is formed in the atmosphere through chemical reactions involving certain pollutants like SO2, NO2, aldehydes etc. upon absorption of UV radiation. The ozone layer of the atmosphere has two interesting and interrelated effects. It absorbs UV light, thereby protecting all the life on earth from harmful effects of radiation. Again, by absorbing UV radiation it heats up the stratosphere causing temperature inversion. As a result of inversion of temperature, the vertical mixing of the pollutants becomes limited and dispersal of pollutants is caused over larger areas near the earth’s surface. Thus, a dense cloud of pollutants hangs over the atmosphere in highly industrialised areas causing several unpleasant effects. So, the layer of ozone acts as a destroyer as well as a protector. Increase in ozone concentration near earth’s surface reduces yielding of crops to a marked extent. It has also adverse effect on human health. Thus, while a higher concentration of ozone in the atmosphere protects us, at the same time it becomes harmful when it comes in direct contact with us.
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Ozone is intimately connected with life sustaining processes. Any depletion of ozone layer has catastrophic effect on life systems on the earth. Some pollutants may enter stratosphere, react with ozone and are converted to other products thereby depleting ozone layer. Major pollutants responsible for depletion of ozone include chlorofluoro carbons (CFC), nitrogen oxides (coming from fertilisers) and hydrocarbons. Depletion of ozone leads to change in temperature and rainfall failure on earth. Harmful U.V radiations due to thinner ozone layer or hole in it directly affect us causing skin cancer known as melanoma. The other disorders are cataracts, destruction of aquatic life, vegetation and loss of immunity. 8.
Fluorocarbons Industrial processes of phosphate fertilisers, ceramics, aluminium, fluorinated plastic, uranium and other metal involve release of fluorides into the atmosphere. The pollutant is in gaseous and particulate state. Fluoride level of air is 0.05mg/m 3. Fluoride in air mainly comes from smoke of industries, volcanic eruptions and insecticidal sprays. Though flurocarbons are helpful in protecting tooth decay in man, yet higher percentage of these are highly toxic. Fluoride pollution in man and animals is mainly through water. In plants, the tip burn disease is due to higher concentration of fluorides.
9.
Metals The most common metals present in the air are Zn, Cd, Pb and Hg. The main sources of the metals are industries and human activities. i)
Mercury: It is a liquid volatile metal. The compounds of Hg are used during the manufacture of fungicide, preparation of paints, cosmetics etc. Inhalation of 1mg/ m3 Hg containing air for 3 months may cause death. Nervous system is badly affected. Liver, the vital organ of the body may be damaged. Other symptoms include headache, anxiety, loss of appetite, eye sight problem etc.
ii)
Lead : Lead compounds are used as antiknock substances, therefore are released in to the air with the vehicle exhaust as lead halides. Inhalation of lead causes reduced haemoglobin formation. Kidney and liver get affected by lead poisoning.
iii)
Cadmium: It is released to air by the industries and other human activities. It is poisonous even at very low level. It is emitted as vapour and quickly forms the oxide, sulphate and chloride. It affects liver and kidney and causes hypertension, emphysema. It is also carcinogenic in mammals.
iv)
Zinc : It is released into air by industries in the form of white zinc oxide fumes. It is highly toxic to man.
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Particulate matter Small solid particles like dust, smoke particles and liquid droplets suspended in air are known as suspended particulate matter. The natural sources are soil and rock debris, volcanic eruption and forest fire. Besides they may be released into air by fuel combustion, industrial operation, vehicular exhaust etc. The particulate matter is injurious to health. Inhalation leads to respiratory diseases like tuberculosis and also cancer. Cotton dust inhalation leads to occupational disease ‘Byssinosis’.
11.
Radioactive emission Disintegration of radioactive substance takes place with emission of radiation like alpha, beta and gamma rays. These radiations upon reaction with living tissues damage them. The harmful effects of radiation are cancer, damage of central nervous system. Besides these, the skin, eye and the living tissues are badly affected.
12.
Toxicants There are some toxic substances which are responsible for causing serious health hazards. They are: i)
Asbestos : It is a mineral fibre used in cement pipes, roofing product, cement sheets etc. The fibre is nondegradable and it causes cancer.
ii)
Arsenic : It is a byproduct in metal refining process. It is carcinogenic.
iii)
CCl4 and CHCl3 : CHCl3 degrades slowly forming highly toxic phosgene. Both CCl4 and CHCl3 are carcinogenic.
iv)
Chromium : It is used in stainless and alloy steel. It has carcinogenic properties.
v)
Nickel : It is used in petroleum and metal products, electrical goods. It is strongly carcinogenic.
vi)
Vinyl chloride : It is used for preparation of polyvinyl chlorides (PVC). It causes brain and lungs cancer.
21.3.3 Units of measurement of air pollutants The concentration of air pollutants is expressed in forms of 1)
Parts per million (ppm) : It is the volume of the pollutant present in one million volume of air. The measurement of volume of pollutant and that of air are measured at standard temperature (250C) and pressure (76cm of Hg).
2)
Microgram per cubic meter (mgm/m3) It is the amount of pollutant in mgm (1mgm=10–6gm) present in one cubic metre of air. In highly polluted air, the concentration may be expressed as mg/m 3.
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21.3.4 Control of Air pollution Although air pollution can not be totally eliminated, yet its harmful effects may be minimised. The people should be aware of environmental health problems arising out of rapid, unplanned industrial growth and road traffic. The two main sources of pollution are industries and motor vehicles. The control of air quality in both the cases can be made to minimise air pollution. 1.
Control of Industrial Pollution The air quality control implies limiting the emission of pollutants. To check the air pollution by industrial plant chimney wastes several measures are adopted for removal of particulate matter and gaseous pollutants from the wastes. The most common equipment used for removal are cyclone collectors, electrostatic precipitators, bag filter and scrubber. i)
Cyclone collectors The waste gas is subjected to centrifugation. The suspended particles move towards the wall of cyclone body and then to its bottom. The dust particles get thrown on the periphery and the clean gas escapes out from the centre of cyclone.
ii)
Electrostatic precipitators These are working on the principle of charging the dust by the application of high voltage electricity. The particles settle down finally. These are safe and simple to operate.
iii)
Bag filters It consists of filter bags made up of bag materials like cotton, glasswool, teflon, ceramic fibre or polyester etc. The dust laden gas is passed through it. The dust gets filtered and clean air escapes out. The dust collected are removed by shaking the bag periodically.
iv)
Scrubbers The harmful gases released by different factories can be removed by spraying cool water in a device called scrubber. The gaseous pollutants are absorbed in appropriate liquid to bring the pollutant from gaseous to liquid or soid state. The different types of scrubbers used are spray type, ventury scrubbers, impingment scrubbers etc.
Pollution due to industries may be eliminated by the use of electricity in place of coal fuel. Solar engery may also be taken up for the purpose. The extent of air pollution can be minimised by self cleaning process of air. This can be achieved by increasing vegetation in the nearby locality and providing green belt between residential and industrial areas.
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The control of air pollution may also be made by implementation of law. All Industries should follow the absolute emission standard fixed by Pollution Control Board. 2.
Control of vehicular pollution i)
21.4
Checking the emission of pollutants from vehicular exhaust: a)
use of good quality fuel.
b)
use of additives to improve combustion
c)
injecting air into the exhaust for converting exhaust compounds to less toxic materials. Recently thermoreactor, a simple attachment, has been devised which when fitted to exhaust tail pipe can convert CO into pure oxygen.
d)
updating the design of engine.
e)
more exact timing of fuel feeding.
f)
using new proportion of gasoline and air.
g)
perfect tuning of caburettor and maintenance of the engine.
h)
installation of catalytic converters to convert NO x to nitrogen.
ii)
Use of filters : The emission of hydrocarbons can be controlled by the use of filters. The filters capture and recycle the escaped hydrocarbons in this engine.
iii)
Control through law : Through motor vehicle act some laws are to be enforced for proper design of engines.
WATER POLLUTION
Water is very essential for the existence of every living being. We use water for drinking, cooking our food, growing our crops, irrigating our lands, generating electricity, running several industries and moreover for keeping ourselves clean. Water is very familiar to every one of us. More than three fourths of earth’s surface is covered with water. It is present in the earth crust and in the form of rain, cloud, snow and hail in the atmosphere. Pollution of water is defined as “the addition of any substance to water that is harmful to living beings”. In chemical sense water is never pure. Dissolved as well as suspended impurities are present in it. These include suspended matter like clay, slit and sand, dissolved minerals like Ca, Mg and Na salts and dissolved gases like N2, NH3, CO2 and H2S. These natural impurities are present in very low amounts and therefore do not pollute much. Polluted water is bad smelling, turbid, unpleasant and is unfit for drinking, bathing and washing purposes.
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21.4.1 Sources of Pollution The chief sources of water pollution are discussed below: a)
Sewage and other waste: The water borne waste derived from domestic waste, animal and food processing plants is known as sewage. It includes human excreta, cloth, paper, soap, detergent etc. This is due to uncontrolled dumping of wastes into ponds, lakes, streams and rivers. Decomposition of sewage and other wastes is largely aerobic process, therefore leads to oxygen depletion which adversely affect fish and aquatic life. Public sewage contains microorganisms like bacteria, moulds, yeasts, viruses, eggs of helminthes algae etc. So, discharge of sewage into rivers and other water bodies create health hazards to mankind. These pallutants chiefly consist of carbonaceous organic materials which are converted to CO2 and water by oxidation. The extent of pollution is measured in terms of biological oxygen demand (BOD) or chemical oxygen demand (COD). BOD is the amount of oxygen required for biological oxidation by microbes in any unit volume of water.
2)
Industrial wastes The sources of industrial effluents (pollutants) include wastes from breweries, tanneries, paper, pulp, textile, sugar and steel industries. The pollutants are metallic wastes, suspended solids, phenols, toxins, acids, salts, dyes, cyanides etc. However, these pollutants can be categorised as follows:
3.
i)
Inorganic substances like chlorides, carbonates, nitrogen etc. pollute water making it unfit for drinking and also encourage the growth of microorganism.
ii)
Organic substances like phenol, alcohol etc. increase the BOD depleting the oxygen content.
iii)
Heavy metals like lead, mercury, arsenic and poisonous substanes like cyanides, acetylene etc. cause harmful and irreversible damages to plant and animal life.
iv)
Colour producing dyes change the colour of water, decrease the oxygen level thereby affecting aquatic life.
v)
Acids and alkalies rapidly change pH of water affecting fish and other aquatic life.
Agricultural discharges Modern agriculture has become a major source of pollution, since it relies heavily on a wide range of synthetic chemicals which include various types of fertilisers and biocides (pesticides, herbicides and weedicides). Excess use of fertilisers and pesticides cause maximum damage to natural ecosystem.
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i)
Fertilisers : Artificial fertilisers applied to modern agriculture ultimately reach the rivers, lakes and streams etc. through irrigation, rainfall, drainage etc. affecting the ecosystem. Due to excessive use of nitrogeneous fertilisers many crops lack potassium. Again, excessive potash treatment is bad. It decreases the amount of valuable nutrients in foods such as ascorbic acid (vitamin C) and carotene. Excess use of fertilisers leads to accumulation of nitrates and phosphates in water. It normally makes water unfit for drinking, but also create harmful diseases. The water if taken by us are converted to nitrites by microbial flora of intestine. The nitrites then combine with haemoglobin of the blood forming methaemoglobin, so that oxygen carrying capacity of the blood is inhibited. Severe damage to vascular and respiratory system is caused. Similarly, accumulation of phosphates in water leads to depletion of dissolved oxygen due to excessive algal growth. It results in death of fish and other aquatic life.
ii)
Pesticides: These are the chemicals used for killing plant and animal pests. The chemcials donot degrade or degrade very slowly in nature and therefore, are toxic to the ecosystem. The most toxic biocides include DDT (dichlorodiphenyl trichloroethane), BHC (Benzene Hexachloride), Chlordane, Methoxychlor, Aldrin, Endrin and PCB (Polychlorinated biphenyl). These pesticides are quite persistent and accumulate in the tissues of aquatic and other animals. DDT, gammaxane have been banned by some countries due to their long acting toxicity.
Radioactive wastes Most of the nuclear arm tests are carried out in oceans. As a result radioactive materials like strontium - 90 and caesium -137 are released. The amount of C-14 and tritium also increase on the water surface of oceans. These radioactive pollutants kill the marine life and enter bodies of all living organisms directly or indriectly in the food chain causing pathological and genetic damage to us. A lot of heat is generated by nuclear tests destroying the marine life in the vicinity.
5.
Thermal pollution Heat is generated as a result of burning of fuel in industrial plants. Nuclear plants use water as coolant. The water, after absorbing heat is let out into seas and rivers. Due to rise in temperature of water some marine species may die and some others may not reproduce at all. Due to rise in temperature water lacks the ability to dissolve gases in it. It has been found that for 100C rise in temperature the oxygen content in water is reduced by 17 percent. This oxygen depletion is more pronounced in sea water than in fresh water since salt water has low specific heat. The oxygen depletion is detrimental to marine life.
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Shipping water pollution or oil pollution The major shipping water pollution is due to oil. Oil is spilled into the seas during the process of transport, loading and unloading from ships. Besides production and refining of oil has become another source for oil pollution. Off shore oil drilling also adds quite a huge quantity of oil into the sea. Oil can remain in sea for at least more than a year. Hydrocarbons are chief components of oil. The microorganisms from the sea decompose hydrocarbons and during the process the oxygen content of water is greatly reduced. Spreading of oil over water isolate water from the contact with atmospheric oxygen. A continuous film of oil is formed which inhibits photosynthesis and formation of oxygen. Thus, aquatic life (both plant and animals) is adversely affected. Oxygen content reduction results in death of sea fish and sea birds. More than one million sea birds are reported to have been killed by oil pollution.
21.4.2 Effects of water pollution Major effects of water pollution are studied under two heads namely (1) Human and animal health (2) Loss of recreational area. 1)
Human and animal health Pollution of water poses a great danger to both human and animal health. Oil and thermal pollution ultimately result in destroying aquatic plants and animals. Pollution due to use of artificial fertilisers and radioactive waste have become potential health hazards to both human and animals. Besides, contamination of water with some heavy metals like mercury, lead, arsenic, cadmium, copper, barium, zinc etc. has toxic effect on the body which are discussed below: i)
Mercury : The concentration of mercury in aquatic environment is increased day by day due to industrial activities such as manufacture of electrical equipments, paper making, mining etc. Mercury can not be easily excreted. Once it enters the food chain its concentration gradually goes on increasing. The disease, resulting from mercury poisoning is usually known as “Minimata disease” since there was outbreak of mercury poisoning in Japan among the residents who ate food from Minimata bay contaminated with methyl mercury. Mercury poisoning causes diseases like abdominal pain, headache, diarrhoea, blurring of vision, apathy and mental derangement.
ii)
Arsenic : It is a long acting slow poison in small quantity, but at the same time causes infant death if taken in large doses. Arsenic poisoning causes liver cirrhosis, lung cancer, hyperkeratosis, ulcer in GI tract and mental disorder.
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iii)
Lead : The aquatic environment around industrial belt is rich in lead concentration. Poisoning of lead causes loss of apetite, damage of liver, kidney and brain, convulsion, vomiting and anaemia.
iv)
Copper : High concentration of copper in water causes hypertension, uremia, sporadic fever.
v)
Cadmium : High concentration of cadmium in drinking water can cause bone deformation, growth retardation, diarrhoea, kidney damage, hypertension and anaemia.
vi)
Barium: Poisoning due to barium results in paralysis, colic pain, vomiting, excessive salivation, paralysis.
vii)
Selenium : It is not very common. Its high concentration can cause fever, nervousness, vomiting, blindness, low blood pressure.
viii)
Cobalt : High concentration of cobalt in water causes bone deformation, low blood pressure, paralysis etc.
Loss of recreational areas Recreational activities in sea beaches suffer serious economic damages due to pollution of water. When water contains some toxic substances or there is high bacterial count in water it gets polluted. Water becomes depleted in oxygen as microorganisms die and decay. The fishes present in sea requiring a high level of oxygen either die or move out. In heavily polluted areas the death of algae, rotting mats of floating debris, solid wastes and some oil spills have caused considerable damage to valuable beach resources. Pollution due to mercury and other chemicals also hamper the recreational activities.
21.4.3 Prevention and Control of Water pollution : Pollution of water due to biodegradable pollutants may be controlled at source by their treatment for reuse and recycling. The non-degradable pollutants such as heavy metals, mineral oils, biocides, plastic materials etc. are removed by suitable methods. The various ways and techniques used for control of water pollution are discussed below: i)
Stabilisation of ecosystem The most scientific and reliable way to control water pollution is made by stabilisation of ecosystem. The principles underlying the method include reduction of waste input, harvesting and removal of biomass, trapping of nutrients, aeration and fish management. Species diversity and ecological balance in water bodies may be restored by various biological and physical methods to prevent water pollution.
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Reutilisation and recycling of waste There are various kinds of wastes such as industrial effuents, sewage of municipal and other systems, thermal pollutants and radioactive wastes which may be recycled to beneficial use. Recycling is the process by which the municipal and industrial waste water can be treated for reuse. For example, the municipal wastes like sewage and sullage may be recycled to generate cheaper fuel gas and electricity. A new technology has been developed by the NEERI, Nagpur for management of radioactive wastes and chemical wastes of atomic power plants, reclamation of waste water and to supply cheap piped gas and generate electricity by recycle of urban wastes.
iiii)
Removal of pollutants Appropriate methods such as adsorption, electrodialysis, ion exchange, reverse osmosis etc. are used for removal of various pollutants like radioactive, chemical and biological pollutants present in water body. The most widely and commonly used technique is reverse osmosis. This is based on the removal of salts and other substances by forcing the polluted water through a semi-permeable membrane under a pressure exceeding osmotic pressure. As a result, water flows in the reverse direction to the normal osmotic flow. To desalinate the brackish water and for purifying the water from the sewage, reverse osmosis is commonly used. Council for Scientific and Industrial Research (CSIR) has developed techniques for removal of different pollutants from polluted water. These are as follows: 1.
Ammonia : This can be removed by ion exchange technique. A weak acidic Cation exchange takes place which removes NH3 in the form of ammonium sulphate. Ammonium sulphate is used as fertiliser.
2.
Mercury : Mercury -selective ion exchange resin is used for the removal of mercury from chlor-alkali effluent plants.
3.
Phenolics : Waste water of pulp and paper mills, petroleum refineries, tanneries and resin plants contain phenol and its derivatives. These can be removed by using polymeric adsorbents.
4.
Colouring matter : The waste water from printing and saree dyeing industries could be decolourised by an electrolytic decomposition technique.
5.
Sodium salts: Reverse osmosis method can be used for the removal of sodium salts. Sodium sulphate from a rayon mill effluents could be easily removed by this technique.
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21.4.4 International standards for drinking water : Drinking water must be pure otherwise it may cause serious diseases. International standards have been laid down for the water to be used for drinking as given below with the chemicals allowed to be present and maximum limits upto which they are permitted. TABLE - 21.1 Chemical/ Condition
Source
Tolerable Limit
Use/Harms of higher conc.
(i) Fluoride
Added externally
1 ppm or 1 mg dm–3
Protects tooth deccay High conc. (> 10 ppm) are harmful to bones and teeth.
(ii) Lead
Lead pipes used for transport of water
50 ppb or mgdm–3
Lead poisoning (damages kidneys, liver and brain)
(iii) Other metals
-
Zn= 5 ppm Cu = 3 ppm Fe = 0.2 ppm Al = 0.2 ppm Mn = 0.05 ppm Cd = 0.005 ppm
-
(iv) Sulphates
-
< 500 ppm
Higher conc. has laxative effect
(v) Nitrates
-
50 ppm
Excess causes methemoglobinemia (blue baby syndrome) which may be linked to stomach cancer.
(vi) pH
-
5.5 - 9.5
21.5
SOIL POLLUTION
The top layer of the earth’s crust is known as soil, which is the most vital part of the lithosphere. The term ‘soil’ is derived from the Latin word ‘solum’ which means the earthly material on which the plants grow. It provides support for all terrestrial organisms. The process of formation of soil invloves the weathering of rocks and subsequent changes due to the action of micro organisms like bacteria, fungi by secretion of organic acids, enzymes, CO 2, production and addition of organic matter upon their death. In an average soil, inorganic matter, organic matter, soil water and air present by 40, 10, 25 and 25% respectively. Soil is the soul of biosphere which provides nutrients, water and minerals for the growth of the plants on which all the animals
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depend. But unfortunately population explosion, rapid urbanisation, industrialization, increase in the use of automobiles etc. are damaging the quality of soil and resulting into soil pollution. 21.5.1 Sources of soil pollution : Soil pollution mainly results from the following sources : (i)
Indudstrial water : The rapid growth of industries has resulted in the release of a lot of industrial wastes containing toxic acid disastrous chemicals which are usually non-biodegradable. These are mainly discharged from pulp and paper mills, oil refineries, sugar factories, textiles, drugs, glass industries etc. The industrial wastes affect and alter the chemical and biological characteristics of soil which finally enter into food chain, disturb the biochemical processes and finally induce serious hazards to living organisations.
(ii)
Urban wastes : Urban wastes commonly known as refuse contains garbage and rubbish materials like plastics, glasses, fibres, polyethene bags, food wastes, fuel residues, abandoned vehichles etc. These wastes emit poisonous gases, toxic hydrocarbons and pathogenic bacteria causing various diseases.
(iii)
Agricultural wastes : Agricultural wastes include farm wastes which include roots and stems of crops, straw, hay etc and also the excess fertilizers, pesticides, herbicides etc. These enter the food chain, thus affecting the health of human beings causing serious metabolic and physiological disorders. Agricultural pollutants are : (a)
Fertilizers : Fertilizers act as nutrients, but excess nitrates and phosphates have hazardous effects.
(b)
Pesticides : Common pesticides used are 1. Insecticides like DDT, Gammaxene, Aldrin etc. 2. Herbicides - Triazines, Sodium chlorate and Sodium arsenite. 3. Fungicides - Organo mercury compounds.
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(c)
Soil conditioners : Though used to protect the soil fertility, these contain several toxic metal like, Pb, As, Hg, Cd, Co etc.
(d)
Farm wastes : Farm wastes from dairies poultries are usually collected as a wet slurry on the land which may seep through the soil and pollute the ground water.
Radioactive wastes : Radioactive substances produced from nuclear explosion, nuclear reaction and from nuclear testing laboratories enter into the soil and accumulate there causing land pollution. These wastes are highly dangerous and cause cancer in living organisms and also degrade genetic materials like DNA and RNA inducing mutation.
(v)
Pollution by Biological agents : Soil pollution is also due to biological agents present in the excreta of human beings, animals and birds. The sewage sludges contain a number of viruses and viable intestinal worms which take shelter in soil and cause a number of diseases by attacking the living organisms.
21.5.2 Effect of soil pollution : Detrimental effects of soil pollution are : (i)
Industrial wastes contianing toxic metals like Hg, Pb, Cd, As, Cu etc. kill bacteria and beneficial micro-organisms in the soil.
(ii)
Direct discharge of industrial effluents through sewage system causes several soil borne and water borne diseases.
(iii)
The dumped urban wastes spread several chronic diseases becoming a serious threat to mankind.
(iv)
The fertilizers though increases the yield, inhibit the synthesis of vitamin-C and carotene in vegetables and fruits.
(v)
Entry of polychlorinated biphenyl (PCB) into human body causes deformities in foetus, nervous disorders, liver and stomach cancer.
(vi)
Pesticides being non-biodegradable by nature seep into ground water through soil and contaminate the entire public water supply system.
(vii)
Pathogenic bacteria in soil act as carrier of a number of chronic diseases like cholera, typhoid, dysentry etc from soil to man and vice versa.
(viii)
The radioactive wastes enter into the food chain and causes the disruption of physiological process and metabolic change.
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21.5.3 Control of soil pollution : Since soil is vital for the living organisms to exist, protective measures must be adopted to keep soil pollution free, some of which are: (i)
Organic measure should be used instead of agro-chemicals.
(ii)
The effluents released should be subjected to proper treatment before their release into soil.
(iii)
The garbage produced should be burnt in a closed chamber.
(iv)
Integrated pest control method should be applied.
(v)
Proper care must be taken for the planned disposal of radioactive wastes.
21.6
CHEMICAL REACTIONS IN ATMOSPHERE
Atmosphere is the thick multilayered gaseous envelope that surrounds the earth, protecting the living organisms from the unfriendly environment of the outer space. It can be divided mainly into four zones or spheres of varying temperature, density and chemical composition. These are troposphere, stratosphere, mesosphere and thermosphere containing gases which undergo photochemical reactions. (a)
Reactions in troposphere : Troposphere is the lowest zone of the atmosphere lying just above the earth surface where all biological activity takes place. The must important reaction in troposphere are those involving CO2 and H2O. (i)
CO2 molecules present absorb sunlight and get excited. CO2 + hy ® CO2* (Excited molecules) Due to collision of the excited CO2* molecules excess energy present in them is converted into heat, hence increasing the temperature of the atmosphere.
(ii)
Similarly, water vapour present in the troposphere absorb the solar energy reaching the earth and when the earth cools, the energy is re-emitted as infrared radiation. Some of this re-emitted radiations is obsorbed by H 2O vapour and CO2 and again radiated back to the surface of the earth. Such warming of the earth by absorption and re-emission of solar radiation is called “Green House Effect” or Global Warming.
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Reactions occurring in the stratosphere : (i)
The upper stratosphere contains a considerable amount ozone which is a product of UV radiations acting on oxygen molecule splitting into two free oxygen atoms. These combine with molecular oxygen to form ozone. O2 hv O + O O + O2 ® O3 Ozone is unstable and breaks down aborbing UV radiation to form oxygen molecule and an oxygen atom. Heat is released warming up the stratosphere. O3 hv O2 + O + Heat Thus, ozone cycle is completed in the stratosphere.
(ii)
Ozone layer in the stratosphere protects us from the harmful UV radiation acting as a protective layer for life on the earth, otherwise humans will have to suffer from skin cancer. But ozone layer is being depleted due to human activities releasing oxides of nitrogen and chlorofluoro carbons (CFCs) or freons into the atomosphere. NO can react with O3 thus decreasing its amount. NO + O3 ® NO2 + O2 NO2 + O3 ® NO + 2O2
(iii)
Freons are nonreactive, nonflammable, nontoxic organic molecules being used in refrigerators, air conditioners and in the production of plastic foam. Once they are released in the atmosphere, they alongwith the other atmospheric gases reach the stratosphere where they are decomposed by UV radiation to form chlorine free radical. . . CF2Cl2 hv CF2Cl + Cl Chlorine free radical then react with O3 to form chlorine monoxide radicals and molecular oxygen. . . Cl + O3 ® ClO + O2 Reaction of chlorine monoxide radical with atomic oxygen produces more chlorine radicals. . . ClO + O ® Cl + O2 In this way the reaction is continued causing the depletion of O 3 layer.
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Reaction occurring in the mesosphere and thermosphere : Due to photochemical reactions free ions and electrons are generated in these regions. NO hv NO+ + e– O2 hv O2+ + e– N2 hv N2+ + e– N2+ + O hv NO+ + N O hv O+ + e– He hv He+ + e– O2 hv O + O N2 hv N + N Since free ions and electrons are unstable they will collide with each other to form neutral species. However they may stay free in the thermosphere i.e. ionosphere where pressure and density are very low.
21.7
SMOG
The word smog is a combination of two words smoke and fog. It was named so because for the first time it was observed to be formed by the condensation of some kind of fog on the carbon particles present in the smoke produced by the combustion of fuels like coal and petroleum. Smog, the most common example of pollution, is mainly of two types. These are : A.
Classical smog or London smog or sulphurous smog : It is a mixture of fog, smoke and sulphur dioxide (SO 2) gas, which was first observed in London in 1952 killing about 4000 people and for this reason it is called ‘London Smog’. Its formation is initiated by a mixture of SO 2, particulates and high humidity in the atmosphere. SO2 being oxidized to SO3 combines with water to form sulphuric acid (H2SO4) droplets, which causes bronchitis and respiratory problems, pneumonia and eye diseases. Since it is a mixture of reducing pollutants, it is also known as reducing smog. 2SO2 + O2 ® 2SO3 SO3 + H2O ® H2SO4
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Photochemical smog or Los Angeles smog : This type of smog was so named because it was first observed in Los Angeles in 1950. This type of smog occurs in warm, dry and sunny climate. The main components of photochemical smog result from the action of sunlight on nitrogen dioxide and hydrocarbons released by automobiles and factories. Since it is having high concentration of oxiding agents and is therefore called as oxidising smog. (a)
Formation of photochemical smog : NO2 and hydrocabons are produced in large amount due to heavy vehicular traffic during the day time. In the presence of sunlight NO2 undergoes photolysis to form NO and atomic oxygen, which in turn combines with oxygen molecule to form ozone (O3). The ozone thus formed reacts with NO to regenerate NO2 and O2. Thus, NO2 cycle is completed. NO2 hv NO + O O + O ® O3 O3 + NO ® NO2 + O2 Ozone is a toxic gas and both NO2 and O3 are strong oxidizing agents and can react with unburnt hydrocarbons in the polluted air producing chemicals like formaldehyde, acrolein and peroxyacetylnitrate (PAN). 3 CH4 + 2O3 ®
3 CH2 = O + 3 H2O Formaldehyde
O
CH2 + CH = C – H Acrolein (b)
H3CCO3 NO2 Peroxyacetylnitrate (PAN)
Harmful effects of photochemical smog : High concentration of ozone, PAN, aldehydes and ketones in the atmosphere is quite destructive influencing both the physiological and metabolic activities of the living beings.
(i)
Both ozone and PAN act as powerful eye irritants.
(ii)
Presence of NO2 in smog causes nose and throat irritation and chronic diseases in lungs and hearts.
(iii)
PANs cause dizziness and headache.
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(iv)
Photochemical smog leads to cracking of rubber and causes extensive agricultural and forestry damages.
(v)
It also damages metals, stones, building materials and painted surfaces.
Control of photochemical smog : It can be controlled by the following measures :
21.8
(i)
Emission of nitrogen oxides and hydrocarbons by the automobiles into the atmosphere can be prevented by using catalytic converters in the automobiles.
(ii)
Plantation of certain plants like Pinus, Juniparus, Pyrus and Vitis can metabolise nitrogen oxide and check the formation of photochemical smog.
ACID RAIN
Acid rain may be defined as the rain water containing sulphuric acid and nitric acid being formed from the oxides of sulphur and nitrogen present in the air as pollutants having a lower pH than that of normal rain water. Normal rain water has a pH of 5.6 i.e. slightly acidic due to the presence of H + ions formed by the reaction of CO2 gas present in the atmosphere with rain water giving carbonic acid (H2CO3) which is a weak acid. CO2(g) + H2O(l) ® H2CO3(aq) H2CO3(aq)
H+(aq) + HCO3– (aq)
When the pH of the rain water drops below 5.6 it is called acid rain. A.
Sources of acid rain : Acid rain is due to the presence of various pollutants like oxides of nitrogen, oxides of sulphur and halogen radicals or molecules in the atmosphere which are discharged by the following two factors. (i)
Natural sources : Natural phenomena like volcanic eruptions, forest fires, lightening, burning of fossil fuel, decomposition of organic matters etc. release large quantities of these pollutants into the atmosphere.
(ii)
Human activities : Due to human activities mainly involving combustion of fuels like coal, wood, petroleum products or from chemical industries release large quantities of oxides of nitrogen and sulphur into the atmosphere.
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Formation of acid rain : (i)
After the release of the pollutants, oxides of nitrogen undergo oxidation reaction followed by reaction with water vapour present in the atmosphere to form HNO 3. NO + O3 ® NO2 + O2 2 NO2 + O3 ® N2O3 + 2 O2 2 NO2 + O3 ® N2O5 + O2 N2O5 + H2O ® 2 HNO3 (Nitric acid) N2O3 + H2O ® 2 HNO2 (Nitrous acid)
(ii)
The oxidation of SO2 to SO3 is catalysed by aerosol containing metal ions like Cu(ll), Fe (ll) etc which subsequently reacts with water vapour to form H 2SO4. 2 SO2 + O2 ® 2 SO3 SO3 + H2O ® H2SO4 (Sulphuric acid) SO2 + H2O ® H2SO3 (Sulphurous acid)
(iii)
A little percentage of HCl is also formed. Cl2 ¾hu ¾® ¾ Cl× + ×Cl H2 ¾hu ¾® ¾ H× + ×H H× + ×Cl ¾hu ¾® ¾ HCl (Hydrochloric acid) All these acids mixing with rain water form acid rain and pH falls below 5.6, at times pH can also fall to 2.0. The concentration of these acids as well as the quantity of water in which the acids are dissolved determines the pH of rain water.
C.
Harmful effects of acid rain : (i)
Acid rain causes extensive damages to buildings, statues and sculptural material especially those made up of limestone, marble, slate etc. Acid rain attacks CaCO 3 (marble) as follows : CaCO3 + H2SO4 ® CaSO4 + CO2 + H2O This is the main reason for which the wonderful monuments Taj Mahal at Agra is being slowly disfigured, discoloured and lustreless. In order to save Taj Mahal, Govt. of India has announced an action plan aiming at clearing the air in Taj Trapezium i.e. area which includes the towns of Agra, Firozabad, Mathura and Bharatpur. Measures taken are :
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(a)
All industries in these areas would switch over to the use of natural gas or LPG instead of coal or oil.
(b)
People will be encouraged to use LPG instead of coal, kerosene or firewood.
(c)
Vehicles in the nearby area would be advised to use low sulphur content fuel.
(ii)
Acid rain increase the soil acidity affecting land flora and fauna.
(iii)
It damages iron and steel structures.
(iv)
It corrodes water pipes . As a result heavy metal like iron, lead and copper are leached into drinking water which have toxic effects.
(v)
Diseases caused by bacteria and pathogens can be spread by acid rain.
(vi)
Acid rain may cause respiratory and skin diseases.
(vii)
It causes acidification of aquatic bodies which lead to the killing of quatic plants and animals.
(viii)
It retards the growth of plants and reduces the productivity of crops like potato, peas, beans etc.
Control measures : Acid rain can be controlled by implementing certain measures. These are:
21.9
(i)
Fuels, devoid of sulphur or having low sulphur content should be used.
(ii)
The leakage chlorine or its discharge should be stopped.
(iii)
The vehicular exhaust should be minimised by using control valves in the outlet of the exhaust pipeline of the automobiles.
(iv)
SO2 gas released can be removed as calcium sulphate (CaSO 4) by spraying with lime water.
OZONE LAYER
Ozone layer serves as the earth’s protective umbrella as it protects us from the harmful ultraviolet (UV) radiations coming from the sun to reach the surface of the earth which is responsible for skin cancer (melanoma) in humans. In the upper stratosphere at an altitude of 25– 30 km, the concentration of ozone is about 10 ppm forming ozone layer. A.
Formation of ozone layer : Ozone, a light bluish gas, is formed in the stratosphere by the action of highly energetic ultraviolet radiation coming from the sun on dioxygen (O 2) molecules.
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In the first step, UV radiations split apart molecular oxygen into free oxygen (O) atoms. O2(g)
hu O(g) + O(g) uv
In the second step, the oxygen atoms combine with molecular oxygen to form ozone. O(g) + O2(g) ® O3(g) Ozone thus formed again absorbing UV radiation breaking into dioxygen and an oxygen atom. O3(g)
¾hu ¾® ¾ O2(g) + O(g) + Heat
Heat is given out warming up the stratosphere. In this way ‘ozone cycle’ goes on in the stratosphere. Thus a dynamic equilibrium exists between the production and decomposition of ozone molecules. B.
Depletion of ozone layer : Due to continuous and non-interrupted release of chlorofluorocarbons and oxides of nitrogen the ozone layer is constantly depleted. These two ozone depleting agents are: (i)
Chlorofluorocarbons (CFC) : CFC, commonly known as freons are the compounds containing chlorine, fluorine and carbon. These compounds are chemically unreactive, nonflammable, nontoxic and adourless, for which these are used in refrigerators, air-conditioners, in the production of plastic foams as blowing agents and by the electronic industry for cleaning computer parts etc. Once CFCs are released in the atmosphere, they mix with the normal atmospheric gases and ultimately reach the stratosphere where they start decomposing in the presence of UV radiation coming from the sun releasing chlorine free radical. CF2Cl2(g) CFCl2(g)
¾hu ¾® ¾ Cl×(g) + ×CF2Cl(g) ¾hu ¾® ¾ Cl×(g) + ×CFCl2(g)
The chlorine radical then react with stratospheric ozone to form chlorine monoxide radicals and molecular oxygen. Cl×(g) + O3(g) ® ClO×(g) + O2(g) Then chlorine monoxide radical react with atomic oxygen producing chlorine radical again repeating the decomposition of ozone and thus depleting the ozone layer. ClO×(g) + O(g) ® Cl×(g) + O2(g)
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(ii)
Nitric oxide (NO) : When fossil fuel in burnt in an automobile engine nitrogen molecule and oxygen molecule combine at high temperature to yield nitric oxide (NO) and nitrogen dioxide (NO2) N2(g) + O2(g) ¾1483K ¾ ¾¾® 2NO(g) NO reacts immediately with oxygen to form NO2. 2NO(g) + O2(g) ® 2NO2(g) Production of NO2 is accelerated when NO reacts with ozone in the stratosphere, thereby decreasing the amount of ozone. NO(g) + O3(g) ® NO2(g) + O2(g) NO2 in turn reacts with oxygen atoms available in the stratosphere due to decomposition of ozone and oxygen producing back NO. Thus, there is no loss of NO but ultimately O3 gets depleted. NO2(g) + O(g) ® NO(g) + O2(g)
C.
The Ozone Hole : Due to depletion of ozone especially by CFCs taking place in all parts of stratosphere, a large ozone hole has been reported mainly in the stratosphere over Antartica in 1980s where ozone level dropped by 30 percent. This is because in other parts of the stratosphere, chlorine monoxide radicals combine with the oxides of nitrogen and chlorine free radicals combine with the methane present in the stratosphere thus ceasing the chain reaction and preventing much ozone depletion. ClO×(g) + NO2(g) ® ClO NO2(g)
.
Cl×(g) + CH4(g) ® CH3(g) + HCl(g). But in Antarctica, the climatic conditions are quite different. In winter special type of clouds called polar stratospheric clouds (PSC) are formed over Antarctica which are of two types : (i)
Type I Clouds are those which are formed at – 77°C containing solidified nitric acid trihydrate (HNO3 × 3H2O)
(ii)
Type II Clouds are those which are formed at –85°C containing some ice. These clouds provide surface on which chlorine nitrate formed gets hydrolysed to from hypoclorous acid and reacts with hydrogen chloride to form moleculer chlorine.
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ClO NO2(g) + H2O(g) ® HOCl(g) + HNO3(g) ClO NO2(g) + HCl(g) ® Cl(g) + HNO3(g) HOCl and Cl2 thus formed are easily converted back into reactive chlorine atoms even under mild conditions thus starting the chain reaction leading to depletion of ozone. Depletion of ozone over Antarctica takes place during spring when sunlight returns and the sun’s warmth breaks up the clouds and HOCl and Cl2 are photolysed by sunlight to form reactive cholorine atoms which destroy the ozone layer. . HOCl(g) ¾hu ¾® ¾ OH(g) + Cl×(g) Cl2(g) ¾hu ¾® ¾ 2Cl×(g) A tight whirlpool of wind surrounds Antarctica due to the presence of PSC, which is called Polar Vertex. It is so rigid that it keeps Antartica separated from the surrounding ozone rich air of non-polar regions, Ultimately ozone hole remains unfilled. But when spring passes away, the intensity of sunlight increases and the Polar Vertex breaks down, thus helping the ozone hole to be filled up. D.
Effect of depletion of ozone layer : Due to the depletion of ozone layer in the stratosphere, the harmful UV radiation will reach the surface of the earth causing the following destructive effects.
E.
(i)
It can cause skin burns and skin cancer (melonoma).
(ii)
Exposure of eye to UV radiation damages the cornea and lens of eye and may cause cataract or even blindness.
(iii)
It injures plant proteins and causes depletion of chlorophylls and harmful mutations of cells.
(iv)
It causes killing of many phytoplanktons and damage to fish productivity.
(v)
It increases evaporation of surface water through the stomata of the leaves and decreases the moisture content of the soil.
(vi)
Ozone depletion brings about significant changes in the climate.
(vii)
UV radiation damage paints and fibres, causing them to fade faster.
Protection of ozone layer : Ozone layer can be protected by taking some controlling measures as follows : (i)
Use of CFC must be either restricted or completely banned by finding out some alternatives.
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(ii)
Use of plastic foam has to be boycotted.
(iii)
Suitable methods should be adopted to recapture CFC released from the airconditioners and refrigerators.
21.10 GREEN HOUSE EFFECT AND GLOBAL WARMING The Green house is that body which allows short wavelength solar radiation to pass through it but does not allow the long wavelength infrared radiation to escape. Due to rapid industrialization, gases like CO 2, chlorofluorocarbons (CFCs), methane, nitrogen oxides, ozone etc are accumulated in the atomosphere which behave like wall of a green house and transmit short wave solar radiations but does not allow the longer wavelength heat radiations to be reflected back into outer space. This means green house it transparent to solar radiation but not to heat radiation. Thus, the green house effect may be defined as the progressive warming up of the atmosphere at the surface of the earth due to blanketing of infrared radiation from the earth’s surface by the green house gases, which is otherwise termed as Global Warming. The green house effect was initially essential for older climates to grow few plants in winter, which require higher temperature for their growth and survival. But now-a-days our atmosphere is enriched with green house inducing gases causing global warming. The fundamental principles underlying green house effect are :
A.
(i)
Absorption of infrared radiation by the green house inducing gases.
(ii)
Re-emmission back towards the earth’s surface which results in heat trap and increases the mean global temperature.
Consequences of Green House effect : (i)
Average global level of CO2 has been reported to have increased by 26 percent between 1960 and 1986 leading to an average increase of global temparature by 1.5°C to 4.5°C.
(ii)
Increase in temperature is more prominent in polar region which will lead to melting of glaciers and polar ice caps raising the levle of sea water by about 1.5 meter and flooding of low lying areas all over the earth.
(iii)
Global warming causes evaporation of terrestrial water content leading to shortage of drinking water.
(iv)
It may decrease the crop productivity due to reduction of soil moisture.
(v)
Increase in the global temperature increases the incidence of infectious diseases like malaria, dangue, yellow fever and sleeping sickness.
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Control of Green House effect : Following measures are to be adopted to control the Green house effect. (i)
To minimise the use of automobiles by using bicycle, public transport system or go for carpool, so that consumption of fossil fuels can be drastically reduced.
(ii)
Banning of deforestation and developing greeneries by massive plantations.
(iii)
Banning of the use of chlorofluorocarbons (CFCs)
(iv)
To avoid burning of dry leaves and wood.
(v)
To avoid smoking in public places.
21.11 POLLUTION DUE TO INDUSTRIAL WASTES The industries use raw materials, process them and produce finished products along with some by-products which are thrown into the environment as industrial wastes in the form of gas, liquid or solid, thus polluting either the air or water or soil. Industrial wastes may be classified as : 1.
2.
Biodegradable and non-biodegradable wastes : (i)
Biodegradable wastes are generated by cotton mills, food processing units, paper mills and textile factories.
(ii)
Non-biodegradable wastes are generated by iron and steel plants, thermal power plants, fertilizer industries which pose a serious threat to humankind.
Process waste and chemical wastes : (i)
Process waste is produced during washing and processing of raw materials which may be organic or inorganic in nature, but both are toxic to living oranisms. Organic wastes are liberated from food processing units, distilleries, breweries, sugar mills etc. Inorganic process wastes may be generated by caustic soda industry, paint industry, petroleum industry, iron and steel plants, thermal power plants etc. Fly ash from thermal power plants contaminate atmospheric tract causing respiratory tract disorder. Fertilizer industries produce gypsum. Iron and steel plants produce slag. The industries may also cause thermal pollution and noise pollution.
(ii)
Chemical wastes includes heavy metals and their ions, detergents, acids, alkalis and various other toxic substances, produced by fertilizer industries, paper and pulp industries, sugar mills etc. These are usually liberated into nearby water bodies like rivers, lakes etc. which may alter the pH, BOD (Biological Oxygen Demand) and COD (Chemical Oxygen Demand).
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The aquatic animals and plants absorb, accumulate the chemical wastes destroying the trophic levels and food chain of the ecosystem. The disposal of non-biodegradable industrial solid wastes may be done by the following ways in order to control environmental pollution. (a)
Fly ash and slag from the steel industry are to be utilized by the cement industry.
(b)
Toxic water can be destroyed by controlled incineration.
(c)
Wastes should be subjected to proper treatment before their discharge.
21.12 GREEN CHEMISTRY AS AN ALTERNATIVE TOOL FOR REDUCING POLLUTION A.
Introduction : Green chemistry or Sustainable chemistry may be defined as the sustainable, safe and non-polluting chemical science which enables man to manufacture products with minimum consumption of materials and energy and also production of minimum waste. Important principles of Green Chemistry are :
B.
(i)
Minimizing or eliminating the need for waste clean-up by emphasizing waste preventions.
(ii)
Minimizing energy consumption.
(iii)
Planning and using chemical products with minimum toxicity.
(iv)
As far as possible, use and production of hazardous substances which cause harm to man and environment should be avoided.
(v)
Materials and the involved process which may lead to explosions, high temperatures or pressure should avoided.
Use of environmental friendly reagents : In ideal case green chemistry helps in avoiding feedstock and catalysts, toxic reagents or solvents and drastic conditions eliminating generation of hazardous intermediates and byproducts. It involves the use of mild and environmental friendly reagents such as sunlight, microwaves, sound waves and enzymes. (i)
Use of sunlight and microwaves : Recently many photochemical reactions are being carried out using microwave ovens where no toxic solvents are used, but only the reactants in appropriate ratios are mixed on a solid support such as alumina. The use of microwaves have not only reduced the time of reaction, but have also increased the yields.
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Use of sound waves : The branch of chemistry where sound waves have been used to carry out certain known chemical reactions instead of microwaves is called sonochemistry.
(iii)
Use of enzymes : Now many biochemical methods have been utilized to prepare precursors and intermediates of certain medicines and antitiotics using enzymes which work in aqueous solution at mild canditions. For example, antibiotics like ampicillin and amoxycillin have been prepared using this method.
C.
Achievements of Green chemistry : Some of the recent achievements of green chemistry are : (i)
Catalytic dehydrogenation of diethanolamine : Herbicide can be synthesized by the catalytic dehydrogenation where the reactants and the products are eco-friendly unlike the old classical reactions in which cyanide and formaldehyde were used.
(ii)
Synthesis of ibuprofen : Ibuprofen is now synthesized by a new technique giving 99% yield using smaller quantities of solvents and being associated with no waste products.
(iii)
Replacement of chlorofluorocarbons (CFCs) by CO2 as blowing agent : CFCs are responssible for ozone depletion, global warming and formation of smog. Now CFCs have been replaced by CO2 as blowing agent.
(iv)
Replacement of organotins by ‘Sea-nine’ as antifouling compound in sea marines : Previously organotins were used as antifouling agents in sea marines, which are highly stable persisting for a longer time causing pollution. Now these have been replaced by safer compounds called ‘sea-nine’.
21.13 STRATEGY FOR CONTROL OF ENVIRONMENTAL POLLUTION The human activities such as the rapid growth of population, industrialization, urbanization etc. play a vital role for environmental pollution, which has become a global problem. In order to check environmental pollution and to save our earth from the future destructive situation the following measures should be adopted and implemented by each individual in addition to Government and NGO’s co-operation.
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(i)
Public awareness must be created through meetings and trainings how to control pollution.
(ii)
Individuals should learn three R’s i.e. Recycle, Recovery and Reuse about the waste materials.
(iii)
Waste materials should not be thrown into the water bodies or sewer.
(iv)
Instead of using chemical fertilizers and pesticides in the agriculture, traditional manures e.g. cow dung and compost manures are to be used.
(v)
Every one must take care of waste management.
(vi)
Waste materials must be put in waste containers or at selected places.
(vii)
Deforestation must be checked and massive plantation programmes are to be encouraged at the road sides, on the river bank, on the sea shore and unused lands.
(viii)
All of us should accept the famous slogan ‘Think Globally, Act Locally’.
21.14 WASTE MANAGEMENT : Waste management is the process by means of which the wastes are converted into valuable resources by means of the improved techniques without producing any sort of environmental problems. Since the waste materials create many environmental problems including health hazards to humanbeings if not properly disposed of, it is now highly essential to properly implement the waste management. Main sources of waste materials are : (i)
Domestic waste includes sewage and municipal garbage.
(ii)
Industrial waste which includes toxic materials
(iii)
Construction sites, medical etc.
Environmental pollution due to these wastes can be substantially reduced by applying the following methods. A.
Recycling : Recycling is based on the principle that waste is considered as a resource, where the waste materials can be used as raw material to manufacture some useful products again e.g. (i)
Used glass bottles as well as broken pieces of glass.
(ii)
Iron scrap for manufacture of steel.
(iii)
Plastic wastes and polyethene bags.
(iv)
Used newspapers and magazines for making papers again.
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B.
Composting : It is a biological degradation or breaking of organic materials such as cow dung, garbage, residues of plants and animals through biochemcial process. This process takes place under aerobic conditions giving rise to manure, which increases the soil nutrients and improve crop yeilds.
C.
Land fills : In this method, solid wastes are dumped into low lying areas in layers with insecticides like DDT being sprayed on the top to prevent breeding of mosquitoes , flies etc.
D.
Incineration : Incineration involves the buring of combustible wastes including household wastes, chemical wastes and biological wastes at high temperature leaving the ash which can be used as a land-filling material. But the gases produced must be suitably treated before they escape into the atmosphere and cause pollution. Disposal of polychlorinated biphenyls (PCBs) is possible by this method as high temperature breaks the C–Cl bonds. However, the ash produced might contain toxic heavy metal causing serious health hazards.
E.
Pyrolysis : This is the combution of waste in the absence of oxygen, where the end products obtained are the combustible gases, tar, charcoal etc. and are used as resource materials in industries.
F.
Sewage treatment : Treatment of sewage can be carried out through the following steps:
G.
(i)
Bigger materials are filtered through screens, which can be used to fill low lying land.
(ii)
It is allowed to stand in tanks, so that heavy solids settle down (called sludge) while oils and grease float on the surface which can be easily removed.
(iii)
The organic materials present in it are allowed to undergo microbial oxidation.
(iv)
Finally the waste water is treated to remove phosphate followed by coagulation, filtration and disinfecting it by adding chlorine.
Digesting : By this method the waste is disposed by microorganisms in the absence of oxygen releasing mainly CO2 and CH4. 2 [CH2O] ® CO2(g) + CH4(g) Methane can be used as a fuel.
H.
Disposal into ocean : The oceans are the vast, natural sink for waste materials. The types of waste disposed into the oceans are sewage sludge, dredgs spoils, solid waste, construction and demolition materials. Although it is a simple and cheap method, it has certain disadvantages, e.g. (i)
It causes water pollution leading to marine pollution and causing health hazards.
(ii)
The lighter solid waste float and return to the beach creating enviromental problems. _____
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CHAPTER (21) AT A GLANCE 1.
Environmental Chemistry deals with the chemical reaction in the environment where the reacting species are generated either naturally or by human activities.
2.
Environmental Pollution may be defind as the process involving the addition of any undesirable material to air, water and soil naturally or due to human activities to such a level of concentration adversely affecting the quality of environment and having harmful effects on plants, animals and humanbeings.
3.
Troposphere and Stratosphere are the two important zones of atmosphere mainly affected by pollution. Ozone layer present in stratosphere protects our earth from harmful UV ray coming from the sun. But now ozone layer is being depleted by mainly CFCs, increasing the chance of occurrence of skin cancer.
4.
The lowest region of atomsphere i.e. troposphere where usually the living organisms live is polluted by oxides of sulphur, nitrogen and carbon mainly responsible for acid rain.
5.
Increased concentration of green house gases, mainly CO 2 is responsible for Global Warming due to re-emission of sun’s energy absorbed by the earth followed by in absorption by CO2 molecules and H2O vapour present near the earth surface and then its radiation back to the earth. If not timely controlled it may submerge the coastal land due to melting of polar ice caps.
6.
Rapid urbanisation, excessive industrialisation and use of excesive pesticides, fertilizers in agriculture are the main source of air, water and soil pollution.
7.
It is now high time to think seriously to adopt and implement very strong controlling measures to save our earth from environmental pollution. Most important measures include (i) Waste management (ii) Green chemistry as an alternative tool for reducing pollution. (iii) Individual awareness.
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QUESTIONS A.
B.
VERY SHORT ANSWER QUESTIONS (1 mark) 1.
Ozone is present in which region of the atmosphere ?
2.
Give an example of non-biodegradable waste.
3.
Which gas is mianly responsible for ozone layer depletion ?
4.
What is the colour of ozone gas ?
5.
Unpolluted rain water contains which acid ?
6.
In the absence of ozone layer which rays will enter into atmosphere ?
7.
Which gas can remove oxygen from oxyhaemoglobin ?
8.
What is the other name of Los Angeles smog ?
9.
What is CFC ?
10.
What is PAN ?
11.
What is BOD ?
12.
What COD ?
13.
Which disease is caused due to hole in the ozone layer ?
14.
Why photochemical smog is so called ?
15.
Which acids are present in the acid rain ?
16.
Name two natural sources of air pollution.
17.
What is the nature of photochemical smog ?
18.
What are polar stratospheric clouds ?
19.
Name any two methods for waste management.
20.
Which national monument is now being corroded by acid rain ?
21.
Which is more poisonous CO or CO2 ?
22.
Which gas is used during photosynthesis by the plants ?
SHORT ANSWER QUESTIONS (Carrying 2 marks) 1.
What do you mean by environmental chemistry ?
2.
Define environmental pollution.
3.
What are pollutants ?
4.
Define green house effects.
5.
What causes ozone to break down ?
6.
Explain how CO is harmful to a person exposed to it.
7.
What are the effects of soil pollution ?
8.
How sulphuric acid and sulphurous acids are formed in the atmosphere ?
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9.
What are biodegradable and non-biodegradable pollutants ?
10.
What is Los Angeles smog ? How is it produced ?
11.
How ozone layer is formed and acting as a protecting umbrella ?
12.
Write down the difference between London smog and photochemical smog.
13.
How plant nutrients and pesticides act as water pollutants ?
14.
Why ozone hole has been observed only in Antarctica ?
SHORT ANSWER QUESTIONS (Carrying 3 marks) 1.
Give two examples from each of solid, liquid and gaseous pollutants.
2.
What is CFC ? How does it destroy ozone ?
3.
Define acid rain ? Name any one acid and its formation which causes acid rain ?
4.
What are the pollutants attacking Taj Mahal and how ?
5.
What is smog ? How Los Angeles smog is produced ?
6.
What do you know by depletion of ozone layer ? How it is caused ?
7.
What are air pollutants ? Give two examples.
8.
What do you know by incineration ? How is it helpful for waste management ?
9.
What do you mean by global warming ? Suggest one method to control it.
10.
Which one is more dangerous, carbon monoxide or carbon dioxide and why ?
LONG ANSWER QUESTIONS : 1.
What is smog ? How is classical smog different from photochemical smog ? How can they be controlled ?
2.
What do you mean by green chemistry ? How will it be helpful to decrease environmental pollution ?
3.
What are the major causes of water pollution ? What measures can you suggest to control it ?
4.
Explain the reactious occurring in the atmosphere.
5.
What is soil ? What are the sources of soil pollution ? Can it be controlled ?
6.
What do you mean by acid rain ? How it is formed and what are its consequences ?
7.
Define green house effect. Discuss its causes and effects. Suggest some remedial measures to control it.
8.
Write short notes on. (a)
Global warming
(b)
Smog
(c)
Acid rain
(d)
Pollutants
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(e)
Waste management
(f)
Industrial wastes
(g)
Ozone hole
MULTIPLE CHOICE QUESTIONS : 1.
Which is true for photochemical smog ? (a) is reducing in nature
(b) It is formed in winter
(c) It is a mixture of smoke and fog (d) It causes irritation in eyes 2.
3.
4.
5.
6.
7.
8.
Ozone layer is present in (a) Troposphere
(b) Stratosphere
(c) Mesosphere
(d) Exosphere
Which of the following is not a green house gas ? (a) CO2
(b) CH4
(c) Chlorofluorocarbons
(d) O2
Which of the following is present in maximum amount in acid rain ? (a) HNO3
(b) H2SO4
(c) HCl
(d) H2CO3
Which is not considered as a pollutant ? (a) NO2
(b) O3
(c) CO2
(d) CH4
Depletion of ozone layer causes (a) Breast cancer
(b) Lung cancer
(c) Skin cancer
(d) Blood cancer
Which is not involved in formation of photochemical smog ? (a) NO
(b) Hydrocanbon
(c) SO2
(d) O3
Which has the greatest affinity for haemoglobin ? (a) CO
(b) NO
(c) O2
(d) CO2
ANSWERS 1. (a) 2.(b)
3.(d)
4. (b) 5. (c) 6. (c) 7. (c) 8. (a) qqq