Data Loading...
+2 CHEMISTRY,PART-II (PAGE 499-792) Flipbook PDF
+2 CHEMISTRY,PART-II (PAGE 499-792)
2,529 Views
77 Downloads
FLIP PDF 4.2MB
ALCOHOLS
4.
5.
6.
7.
8.
9.
10.
483
Butan - 1 - ol and Butan - 2 - ol are (a)
Chain isomers
(b) Functional isomers
(c)
Position isomers
(d) Optical isomers
Which of the following compound is called carbinol ? (a)
CH3OH
(b) C2H5OH
(c)
C3H7OH
(d)
12.
13.
CH — OH
How many structural isomers of alcohol with molecular formula C 4H9OH are possible ? (a)
5
(b) 7
(c)
3
(d) 6
Which one is primary alcohol ? (a)
Butan – 2 – ol
(b) Propan – 2 – ol
(c)
Butan – 1 – ol
(d) 2,3 dimethylhexan – 4 – ol
Action of nitrous acid on ethyl amine gives (a)
C2H6
(b) C2H5OH
(c)
C2H5OH and C2H4
(d) C2H5OH and NH3
Ethyl alcohol is heated with conc. H2SO4 . The product formed is, (a)
CH3CO . OC2H5
(b) C2H6
(c)
C2H4
(d) C2H2
The compound 'B' formed in the following sequence of reaction . CH3CH2CH2OH
11.
CH 3 CH 3
¾PCl ¾5 ®
A ¾alc.KOH ¾¾¾¾® B, is
(a)
Propyne
(b) Propanal
(c)
Propane
(d) Propene
The compound which reacts fastest with Lucas reagent at room temperature is, (a)
Butan – 1 – ol
(b) Butan – 2 – ol
(c)
2–methylpropan –2–ol
(d) 2 – methylpropan – 1 – ol
Primary and Secondary alcohols on action of reduced copper gives (a)
aldehydes and ketones respectively
(b) ketones and aldehydes respectively
(c)
only ketones
(d) only aldehydes.
When vapours of an alcohol are passed over hot reduced copper, alcohol is converted into alkene, the alcohol is (a)
Primary
(b) Secondary
(c)
Tertiary
(d) None of these
484
14.
15.
16.
17.
18.
19.
20.
21.
22.
+2 CHEMISTRY (VOL. - II)
Dehydrogenation of 2 – butanol gives (a)
2 – butene
(b) butanone
(c)
Butanal
(d) None of these
Which of the following compounds is oxidised to prepare methyl ethyl ketone ? (a)
Propan – 2 – ol
(b) Butan – 1 – ol
(c)
Butan – 2 – ol
(d) t – butyl alcohol
Which of the following gives secondary alcohol (a)
CH3CO CH3
(b) CH3CHO
(c)
CH3COOH
(d) CH3—O—CH3
Fermentation of starch solution to ethyl alcohol does not require (a)
diastase
(b) invertase
(c)
maltase
(d) zymase
The boiling point of ethyl alcohol should be less than (a)
Propane
(b) dimethyl ether
(c)
Formic acid
(d) None of the above
A low boiling alcohol failed to give the Lucas test, but gives a positive iodoform test. The alcohol is : (a)
Methanol
(b) Propan – 1 – ol
(c)
Butan – 1 – ol
(d) Ethanol
A compound 'X' having molecular formula C4H10O when oxidised produces an acid having formula C4H8O2. The compound 'X' is (a)
secondary alcohol
(b) primary alcohol
(c)
tertiary alcohol
(d) an ether
The enzyme which can catalyse the conversion of glucose to ethanol is (a)
Zymase
(b) Invertase
(c)
Maltase
(d) Diastase
An organic compound 'X' on treatment with acidified K2Cr2O7 gives a compound 'Y' which reacts with I2 and sodium carbonate to form tri-iodomethane. The compound 'X' is, (a) (c)
23.
CH3OH CH3CHO
(b) CH3COCH3 (d) CH3CH(OH) CH3
The compound which does not respond to iodoform test is, (a) (c)
CH3OH CH3COCH3
(b) CH3CHO (d) C2H5OH
ALCOHOLS
24.
The reaction between alcohol and carboxylic acid is called. (a) (c)
25.
29.
30.
31.
32.
33.
34.
C–C C–H
(b) C – O (d) O – H
(CH3)3 C – OH C3H7 OH
(b) (CH3)2 CH OH (d) (C2H5)2 CH OH
1 – Chlorobutane on reaction with alcoholic potash gives, (a) (c)
28.
(b) Saponification (d) Hydrogenation
Which one of the following is an isomer of diethyl ether ? (a) (c)
27.
Hydrolysis Esterification
In CH3CH2OH, the bond that undergoes heterolytic cleavage most readily is, (a) (c)
26.
485
But – 1 – ene Butan – 1 – ol
(b) Bnt – 2 – ene (d) Butan – 2 – ol
Ethyl alcohol is miscible with water in all proportions. It is because, (a)
It is acidic in nature
(b) It dissociates in water
(c)
It is basic in nature
(d) It forms hydrogen bonding with water
Which of the following compounds has the highest boiling point ? (a)
CH3CH2CH3
(b) CH3CH2OH
(c)
CH3Cl
(d) CH3—O — CH3
Which of the following compounds has the lowest boiling point ? (a)
n – butyl alcohol
(b) sec – butyl alcohol
(c)
tert – butyl alcohol
(d) 2 – methyl propan – 1 – ol
Hydrogen bonding is maximum in, (a)
Ethyl chloride
(b) Ethanol
(c)
Diethyl ether
(d) Triethylamine
Which one of the following is soluble in water ? (a)
CCl4
(b) C6H6
(c)
CH3OH
(d) CH4
Hydrochloric acid reacts fastest with, (a)
Propan – 1 – ol
(b) Propan – 2 – ol
(c)
2 – methyl propan – 1 – ol
(d) 2 – methylpropan – 2 – ol
A mixture of water and alcohol can be separated by (a)
evaporation
(b) decantation
(c)
distillation
(d) filtration
486
35.
+2 CHEMISTRY (VOL. - II)
Secondary alcohol on oxidation gives, (a)
Ketone
(b) Aldehyde
(c)
Ether
(d) Hydrocarbon
ANSWERS
1. c
2. b
3. d
4. c
5. a
6. b
7. c
8. b
9. c
10. d
11. c
12. a
13. c
14. b
15. c
16. a
17. b
18. c
19. d
20. b
21. a
22. d
23. a
24. c
25. d
26. a
27. a
28. d
29. b
30. c
31. b
32. c
33. d
34. c
35. a
qqq
PHENOLS
487
CHAPTER - 16
PHENOLS Phenols are compounds with the general formula Ar – OH, where Ar is phenyl, substituted phenyl or other aryl groups. Phenols differ from alcohol in having –OH group directly attached to an aromatic ring. Phenols are often considered as derivatives of benzene, the simplest member of the family being phenol. Methyl phenols are called cresols. Phenols are classified as mono–, di – or trihydric phenols according as they contain one, two or three –OH groups attached directly to aromatic nucleus. Monohydroxy benzene is called phenol. OH
OH
Phenol
Cl
o-Chlorophenol
OH
CH3
o- cresol
OH
CH3 m-cresol
OH
CH3 p-cresol
The functional group in phenol is phenolic –OH group. Phenol is also known as carbolic acid. 16.1 METHODS OF PREPARATION : The following general methods of synthesis of phenols are used for their preparation. (i) Hydrolysis of Aryl halides : Aryl halides yield phenol when hydrolysed with aqueous sodium hydroxide. For example chlorobenzene is hydrolysed to phenol at high temperture and pressure by NaOH. The is known as Dow’s Process. Cl +2 NaOH
370,o 4000 psi NaCl, H2O
+
ONa
Chlorobenzene
OH
HCl
+ NaCl
The above hydrolysis can also be effected by water in presence of copper as catalyst at 4000C temperature and 4000 psi pressure.
488
+2 CHEMISTRY (VOL. - II)
Cl
Cu-catalyst + H2O
(ii)
OH
o
+
4000 Psi
HCl
Alkali fusion of sodium arenesulphonates Simple phenols are prepared conveniently by fusing sodium arenesulphonates with so-
dium hydroxide at about 300OC. A mixture of sodium phenoxide and sodium sulphite is produced from which phenol is liberated by acidification. Ar – SO3– Na+
+ 2 NaOH
Sodium arenesulphonate Ar O Na + HCl –
ArO– Na+
+
Na2SO3 + H2O
Sodium phenoxide ArOH + NaCl
+
For example, phenol is synthesised from benzene as follows. Benzene sulphonic acid can be obtained from benzene upon sulphonation with fuming sulphuric acid. Benzene sulphonic acid is fused with NaOH at 300 0 C. A mixture of sodium phenoxide and sodium sulphite is produced which upon acidification yields phenol. SO3H NaOH
H2SO4, SO3
OH
ONa
HCl
3000C Benzene sulphonic acid
Sod. phenoxide
Phenol
(iii) From diazonium salts : When an aqueous solution of aryldiazonium chloride is heated, the diazonium group is replaced by –OH group and phenol is formed. +
Cl
OH
2
benzene diazonium chloride
+ HCl
Phenol
This reaction is usually conducted in an acid medium to prevent the coupling of phenol with unreacted diazonium salt. Application : The method is useful for the preparation of m-chlorophenol which cannot be prepared by electrophilic chlorination of phenol.
Cl
Cl
+
Cl+ H2O
OH m-chlorophenol
+ HCl
PHENOLS
489
Mechanism : In this process, water H2O is the nucleophile.
+
N2
..
+
N2 +
+
+ -H
OH2
HO2
OH Phenol
(iv) From Cumene : The most recent commercial synthesis of phenol is from cumene, isopropyl benzene. The process involves two steps. (a)
Preparation of cumene (isopropylbenzne)
Cumene is synthsised from benzene and propylene, both derived from petroleum by FriedelCrafts Reaction. CH3 C
Anhyd.AlCl3
H
CH3
Cumene (b) Cumene is then oxidised by aerial oxygen at 130 0C in the presence of metal catalyst (or initiator) to cumene hydroperoxide, which is converted by aqueous acid (dilute sulphuric acid) into phenol and acetone. This process is of great importance as it provides the principal source for the commercial preparation of phenol and acetone. CH3
CH3 C
H
O2 130 C, Cat
C
0
CH3
CH3
O OH
Cumene hydroperoxide + H2O, H
CH3
OH +
C=O CH3 Acetone
490
+2 CHEMISTRY (VOL. - II)
16.2 PROPERTIES : (A) Physical properties : The simplest phenols are either liquids or low melting solids having quite high boiling points. Phenol is a colourless crystalline solid with low melting point (43 0C), fairly soluble in water (9g/100gH2O). The solubility is due to the hydrogen bonding with water molecules. Most of the other phenols are insoluble in water, but soluble in organic solvents. Phenol has a characteristic odour and is very poisonous. While comparing the physical properties of nitrophenols, it has been observed that onitrophenol has low m.p. and is volatile. It has low solubility (0.2g/100g H 2O at 250C). This is because of the proximity of –OH and – NO2 group in the benzene ring for which they form intramolecular hydrogen bonding. O
H +O N O
O
Intramolecular hydrogen bonding in o-nitrophenol
+ N O
O
O H
+ N
O H
O
Intermolecular hydrogenbonding in p-nitrophenol.
This gives rise to a stable six membered ring (chelation) and can not form associated molecules. Therefore, it exists as a single unit and it has low m.p, low b,p, low solubility and high volatility compared to its other isomers. On the other hand, due to the greater distance between – OH and –NO2 group in m-nitrophenol and p-nitrophenol, such intramolecular hydrogen bonding is not possible. They form associated molecules by intermolecular hydrogen bonding and have higher m.p., b.p than o-isomer. They are much less volatile than o-isomer. Their solubility is more in water because they form hydrogen bonding with water. This is the reason why o-nitrophenol is steamvolatile and it can be separated from m-and p-isomers by steam distillation. (B) Chemical properties ACIDITY OF PHENOLS The acidity of a compound is defined as its ability to release a proton in presence of water. Phenols when dissolves in water form phenoxide ions, hence acidic in nature. .. —O .. — H Phenol
—O
–
+ H
+
Phenoxide ion
Phenols are weaker acids than carboxylic acid. This is evident from their K values. (K -10 -5 for phenol is 10 and that of CH COOH is 10 ) 3
PHENOLS
491
Both phenol and phenoxide ion exhibit resonance. A number of resonating structures can be written for both phenol and phenoxide ion. .. OH |
+
.. OH || –
+
.. OH ||
+d
+
OH – ||
OH || d-
|||
..
..
.. OH |
–
Phenol
Resonance hybrid
I
II
III
IV
V
O– |
O– |
O ||
O ||
O || –
–
|||
–
O ||
–
Phenoxide ion I
II
III
IV
Resonance hybrid
V
In case of phenol forms I and II are the two Kekule resonating structures. In forms III, IV and V, there is charge separation. Since separation of charge costs energy, these forms are known as high energy forms and they do not contribute to the resonance hybrid. One the otherhand, in case of phenoxide ion structures I to V contribute to the resonance hybrid. In this case there is no charge separation, rather there is charge delocalisation. Thus resonance hybrid of phenoxide ion is more stable than that of phenol. The equilibrium between the two is represented as .... — O—
H
Phenol (resonance hybrid structure)
.... – — O
+
H+
Phenoxide ion (resonance hybrid structure)
Since the phenoxide ion is more resonance stablised thean phenol itself, the equilibrium + is shifted towards right releasing H ions. This explains why phenol behaves as a weak acid. Effect of substituents on acidity of phenol Presence of electron releasing group decreases the acidity of phenol whereas presence of electron withdrawing group enhances the acidity. When an electron releasing group (say CH 3 – , C H – etc) present the –ve charge on the phenoxide ion gets intensified. As a result the anion 2 5 becomes less sable and acidity of phenol decreases.
492
+2 CHEMISTRY (VOL. - II)
O
– O + H3O+
H + H2O
R R = Electron releasing
R Destabilised
On the other hand, when an electron withdrawing group ( -X, -CHO, - NO , - COOH 2 etc.) is present, electrons from the ring system are withdrawn and that results in greater dispersal of the –ve charge of the phenoxide anion. The phenoxide ion thus gets stabilised and acidity of phenol is enhanced. O
– O + H3O+
H + H2O
R R = Electron withdrawing
R Stabilised
Thus, methyl phenols are less acidic than phenol whereas chlorophenols are more acidic than phenol. Again, since inductive effect of a substituent operates only through the covalent bonds, its intensity gets reduced with increase in distance from the reaction centre. In case of isomeric chlorophenols, the K ortho > K meta >K para. a a a However in all the cases the acidity of phenol can not be explained by taking only inductive effect into account. For example, among nitrophenols meta isomer is the weakest acid whereas among alky1 phenols, meta isomer is the strongest. This can be explained by taking resonance effect into account along wit inductive effect. The phenoxide anion gets stabilized by the dispersal of –ve charge. – O O O O | || || || – –
O
– O
–O
+
– O
| N
– O
|| |
– O
|
O
|| N
O
|
|
+N
|| |
| N
|| |
+
+
|
|
||
||
The lower acidity of metanitro isomer is due to the fact that m-nitro phenoxide anion is stabilised by inductive effect only and no resonance effect is operating due to absence of conjugate system. – O O | || – O + + O N N | | O– O– (no further resonance with NO2 group)
PHENOLS
493
The values of K for three nitrophenols are metioned below. a OH |
OH |
|
OH | NO2
NO2 o-nitrophenol Ka = 600 x 10–10
| NO2
m-nitrophenol Ka = 50 x 10–10
p-nitrophenol Ka = 690 x 10–10
It is important to mention here that presence of electron withdrawing group at the ortho position in some cases makes the phenol weaker than the para isomer. For example, o –fluorophenol and o-nitrophenol are weaker than their corresponding p-isomers. The reason behind this is that the acidic hydrogen is involved in intramolecular hydrogen bonding with fluorine and oxygen. O—H | —F
||
O
o - Fluorophenol
– + O N— H | | —O
o - Nitrophenol
(a) Esterification Reaction : Phenol reacts with acid chlorides or acid anhydrides in presence of aqueous alkali solution to give phenyl esters. The alkali first forms the phenoxide ion which then reacts with the acid chloride to form the ester.
O Na+ + H2O
OH + Na2CO3 (Phenol)
(Sodium phenoxide)
O +
O Na
O
+ CH3 - C - Cl
(Sodium phenoxide)
(Phenyl acetate)
O +
O Na + (Sodium phenoxide)
O - C- CH3 +NaCl
C - Cl
O O-C (Phenyl benzoate)
+ NaCl
494
+2 CHEMISTRY (VOL. - II)
Introduction of acetyl group (CH3CO–) in phenol forming an ester is called acetylation. Acetylation of salicylic acid gives acetyl salicylic acid (aspirin). COOH OH
COOH +
(CH3CO)2O
Salicylic acid
OCOCH3
H+
+
CH3COOH
Acetylsalicylic acid (aspirin)
When a mixture of phenol and benzoylchloride is shaken with excess of aqueous NaOH, the ester phenyl benzoate is formed and this is called benzoylation and the reaction is known as Schotten Baumann reaction. (b) Electrophilic Substitution reactions As already discussed the –OH group activates the benzene ring to electrophlic attack and is ortho paradirecting. (i)
HALOGENATION :
When treated with excess of aqueous bromine solution phenol produces 2, 4, 6– + tribromophenol. Br is the electrophile. Br2 + Br2
Br+ + Br3–
OH |
OH Br
+ 3Br+
Br | Br
2, 4, 6 tribromophenol
But when phenol is treated with limited amount (1 mole) of Br 2 dissolved in less polar solvents like CHCl3, CS2 etc monobromophenols are produced OH |
Br2
OH |
OH |
Br +
CS2 o-bromo phenol
| Br p- bromophenol
+ HBr
PHENOLS
(ii)
495
SULPHONATION When treated with conc H2SO4 phenol is readily sulphonated to produce o-phenol (+)
sulphonic acid and p–phenol sulphonic acid. SO3H is the electrophile. + – SO3H + HSO4 + H2O
2H2SO4 OH |
OH |
ConcH2SO4
OH |
SO3H +
| SO3H
o- phenol sulphonicacid
p- phenol sulphonicacid At room temperature, ortho isomer is the main product while at 100 0c, para isomer is the main product. (iii)
NITRATION Phenols undergo nitration when treated with dilute HNO3. The products are o-nitro
phenol and p-nitro phenol. NO +2 is the electrophile.
– + NO2 + NO3+ H2O
2HNO3
OH |
dil HNO3
OH |
NO2
o-nitro phenol
(iv)
OH | + | NO2 p- nitrophenol
Reimer – Tiemann Reaction
When phenol is treated with chloroform (CHCl ) is presence of aq. NaOH at 330 K, 3 to –CHO group is introduced mainly in the ortho position to –OH group. The process is known formylation. Though ortho isomer is the major product, yet para isomer is obtained to a smaller extent. Bot the isomers can be separated by steam distillation after acidification. This reaction is known as Reimer - Tiemann Reaction.
496
+2 CHEMISTRY (VOL. - II)
Reimer Tiemann reaction : OH |
–
O |
CHO
–
|
+ CHCl3
aq. NaOH 330K
O |
| CHO H3O+ OH |
|
H3O+ OH | CHO
o - Hydroxy benzaldehyde or Salicylaldehyde (Major product)
| CHO p - Hydroxy benzaldehyde (Minor product)
Mechanism:
Formation of very reactive intermediate dichlorocarbene by the action of base on chloroform.
– OH +
H – CCl3
– H2O
–
– Cl
: CCl3
:CCl2 Dichlorocarbene. (electrophile)
The carbon atom in :CCl2 contains only six valence electrons, thus acts as an electrophile in aromatic substitution. Electrophilic substitution in benzene ring – O O || |
|
: CCl2
–H+
CHCl2
|
H – CCl2
– O |
o - (dichloromethyl) phenoxide ion.
PHENOLS
497
Hydrolysis and acidification : –
OH |
– OH |
Cl
C
Cl
O |
– OH
+
O — H + OH
–
– CHO
–
+ Cl
|
O |
Cl
C
+
H2O
– CHO
OH |
CHO
|
|
H3O+
Salicylaldehyde (Major product) Ortho-isomer predominates due to its greater stability resulting from intramolecular hydrogen bonding which is shown below.
H
|
H — O O || | C
Intramolecular hydrogen bonding in salicylaldehyde If CCl4 is taken in place of CHCl3, salicylic acid is obtained instead of salicylaldehyde OH | + CCl4
NaOH(aq)
OH |
— COOH
—
OH
OH | + | COOH
o - Hydroxy benzoic acid or Salicylic acid (Major product)
p - Hydroxy benzoic acid (Minor product)
498
+2 CHEMISTRY (VOL. - II)
(c) Oxidation : Phenol is easily oxidised without disruption of its carbon skeleton to form p-benzoquinone.
OH + 2[O]
CrO2Cl2
O
Phenol
O + H2O p-Benzoquinone
Similarly, hydroquinone on oxidation with silver salts yield p-benzoquinone
HO
O
OH + 2Ag+ OH– Hydroquinone
O + 2Ag + 2H2O p- benzoquinone
The oxidation of phenols is explained by the fact that the presence of – OH group in the ring furnishes electrons to the ring which thus renders susceptible to oxidation especially in alkaline medium. (d) Reaction with zinc dust : When a phenol is distilled with zinc dust, the –OH group is replaced by a hydrogen atom. Zinc removes the oxygen of the phenolic group as zinc oxide, to yield the parent hydrocarbon.
OH + Zn Phenol
+ ZnO Benzene
CHAPTER (16) AT A GLANCE 1.
Dow’s Process : Chlorobenzene reacts with aqueous sodium hydroxide under high temperature and pressure to form phenol.
2.
Acidity of phenol is due to resonance stabilisation of phenoxide ion.
3.
Presence of electron releasing group decreases the acidity of phenol whereas presence of electron withdrawing group increases its acidity.
4.
Reimer-Tiemann reaction : Phenol is treated with chloroform in presence of aq.NaOH at 330 K forming salicylaldehyde upon acidification.
5.
Phenolic –OH group can be acetylated or benzoylated forming esters. Benzoylation process is known as Schotten Baumann reaction.
PHENOLS
499
QUESTIONS A.
Very Short Answer Type (1 Mark) (i) (ii) (iii) (iv) (v)
B.
2. 3. 4. 5.
Complete the following reaction giving the names of the products : (i) C6H5OH + Zn dust (ii) C6H5OH + Br2 water conc H2SO4 (iii) C6H5OH + HNO3 (iv) C6H5OH + HNO2 (v) C6H5OH + CHCl3 + KOH (vi) C6H5OH + CCl4 + KOH (vii) C6H5OH + CO2 Give the resonating structures for phenol and phenolate ion. How will you distinguish between ethyl alcohol and phenol. Give the reaction of phenol with benzene diazonium chloride. How will you obtain the following form phenol. (i) Salicyladehyde (ii) Benzene (iii) Picric acid (iv) Nitrophenol (v) Salicylic acid (vi) p-cresol
Short answer type (3 Marks) (i) (ii) (iii) (iv) (v)
D.
.
Short Answer Type (2 Marks) 1.
C.
Phenol is acidic because is more stable than reaction Formylation of phenol is Benzene can be obtained by heating phenol with Phenol reacts with to give o-and p- Nitrophenols. Phenol gives with CrO2Cl2
What is Dow’s process ? What is Reimer-Tiemann reaction ? Why phenol is acidic, but alcohol is neutral. Explain. How can you carry nitration is phenol ? Explain. What is the directive influence of phenolic group ? Explain with reasons.
Long answer type (7 marks each) (i)
How is phenol prepared from benzene ? How can phenol be converted to (i) benzene (ii) salicylaldehyde. (ii) How can phenol be prepared from cumene ? Convert phenol to aspirin and pbenzoquinone. (iii) Starting from benzene diazonium chloride, how can phenol be prepared ? How can phenol be sulphonated ? What is the electrophile for this reaction ?
500
E.
+2 CHEMISTRY (VOL. - II)
Multiple choice type questions (i)
When phenol is distilled with zinc dust, the product is (a) toluene (b) benzene (c) xylene (d) none of these
(ii)
Phenol is less acidic than (a) acetic acid (b) p-methoxy phenol (c) o-nitrophenol (d) ethanol
(iii) Electrophilic substitution in phenol takes place at (a) o-position only (b) p-position only (c) o- and p-position (d) m-position only. (iv) Sodium salt of benzene sulphonic acid on fusion with caustic soda gives (a) C6H5OH (b) C6H6 (c) C6H5COOH (d) none of these (v)
Phenol gives salicylaldehyde on heating with CHCl3 & NaOH. The reaction is called (a) Cannizzaro’s reaction (b) Claisen condensation (c) Reimer-Tiemann reaction (d) Perkin reaction
(vi) Chlorobenzene on heating with NaOH at high temperature and pressure gives (a) Phenol (b) Benzaldehyde (c) Benzene (d) Chlorophenol (vii) Phenol reacts with Br2 in CCl4 at low temperature to give (a) m-bromophenol (b) o- and p-bromophenol (c) p-bromophenol (d) 2,4,6tribromophenol (viii) Intramolecular hydrogen bonding is found in (a) phenol (b) o-nitrophenol (c) p-nitrophenol (d) m-nitrophenol (ix) p-nitrophenol is stronger acid than phenol because nitrogroup is (a) Electron donating (b) Electron withdrawing (c) Acidic (d) Basic (x)
The most convenient method of removing a phenolic group from a compound is by (a) reduction with Sn + HCl (b) heating strongly (c) heating strongly with copper (d) distilling with zinc dust. ANSWER
A. (i) phenoxide, phenol (ii) Riemer-Tiemann (iii) Zinc (iv) dil HNO 3 (v) p-benzoquinone D. (i) b (ii) a (iii) c (iv) a (v) c (vi) a (vii) b (viii) b (ix) b (x) d.
qqq
ETHERS
501
CHAPTER - 17
ETHERS The general formula of ethers is R–O–R'. Both the alkyl or aryl groups (R/Ar or R'/Ar') may be same or different. The functional group of ether is 'OR / OAr' (alkoxy/aryloxy). 17.1
CLASSIFICATION OF ETHERS :
Ethers are classified as simple or symmetrical ethers if the two alkyl or aryl groups attached to the oxygen atom are the same, and mixed or unsymmetrical ethers if the two groups are different. (a)
Simple or Symmetrical ethers : CH3—O—CH3 Dimethyl ether
(b)
C6H5—O—CH3 Methyl phenyl ether
C6H5—O—CH2C6H5 Benzyl phenyl ether
Aliphatic ethers in which R and R' are both alkyl groups. For example, CH3—O—CH3 Demethyl ether
2.
C6H5—O—C6H5 Diphenyl ether
Mixed or Unsymmetrical ethers : CH3—O—CH2CH3 Ethyl methyl ether
1.
CH3CH2—O—CH2CH3 Diethyl ether
CH3—O—CH2CH3 Ethyl methyl ether
CH3CH2—O—CH2CH3 Diethyl ether
Aromatic ethers in which either one or both R and R' are aryl groups.
Aromatic ethers are further subdivided into phenolic ethers and diaryl ethers. Ethers in which one of the groups is aryl while the other is alkyl are called phenolic ethers or alky aryl ethers. On the other hand, ethers in which both the groups are aryl are called diaryl ethers. For example, C6H5—O—CH3 Methyl phenyl ether (Phenolic ether)
C6H5—O—C6H5 Diphenyl ether (Diaryl ether)
502
+2 CHEMISTRY (VOL. - II)
17.2
NOMENCLATURE OF ETHERS :
1. Common system : The common names of ethers are derived by naming the two akyl or aryl groups linked to the oxygen atom as separate words in alphabetical orders and adding the word ether. In case of symmentrical ethers, the prefix di is used before the name of the alkyl or aryl group. 2. IUPAC system : The IUPAC system, ethers are named as alkoxyalkanes. The ethereal oxygen is taken with the smaller group and forms a part of the alkoxy group while the larger group is considered to be a part of the alkane or arene. The common and IUPAC names of some ithers are given below: Structural formula
Common name
IUPAC name
CH3—O—CH3
Dimethyl ether
Methoxymethane
CH3—O—CH2CH3
Ethyl methyl ether
Methoxyethane
CH3CH2—O—CH2CH3
Diethyl ether
Ethoxyethane
CH3—O—CH2CH3CH
Methl n-proply ether
1-Methoxypropane
Isopropyl methyl ether
2-Methoxypropane
Di-isopropyl ether
2-(2-Propoxy) propane
tert-Butyl methyl ether
2-Methoxy-2-methylpropane
Di-tert-Butyl ether
2-(2-Methyl-2 propoxy) methylphropane
C6H5—O—CH3
Methyl phenyl ether (Anisole)
Methoxy benzene
C6H5—O—CH2CH3
Ethyl phenyl ether (Phenetole) Ethoxybenzene
C6H5—O—C6H5
Diphenyl ether or phenyl ether Phenoxy benzene
C6H5—O—(CH2)6CH3
n-Heptyl phenyl ether
1-Phenoxyheptane
Isopentyl phenyl ether
(3-Methyl-1-butoxy) benzene
—
CH3
2
3
1
1
—
—
CH3—O—CH—CH3 CH3
3
CH3
2
2
CH3—CH—O—CH—CH3 CH3
— —
1
2
CH3—O—C—CH3 3
CH3 CH3
— —
CH3
— —
1
2
CH3—C—O—C— CH3 3
3
CH3
1
CH3
2
3
4
—
C6 H5 —O—CH2CH2—CH—CH3 CH3
CH3O—CH2—CH2—OCH3
1, 2 Dimethoxyethane
ETHERS
17.3
503
STRUCTURE OF ETHERS :
Since ether is regarded as dialkyl or diaryl dervatives of water, its structure is similar to that of water. The structure is V-shaped with bond angle »110°. Bond angle in ether is greater than water (105°) due to the repulsion between two bulky alkyl groups. As size of R increases, bond angle also increases. 141 pm H H
C
H
.. O .. 111.7°
C
H H H
Methoxymethane 17.4
PREPARATION OF ETHERS : Ethers are prepared by the following general methods.
1. From alcohols : (a) Simple or symmetrical ethers are prepared by heating a primary alcohol with concentrated sulphuric acid at lower temperature (413K). Two molecules of the alcohol eliminate a molecule of water (dehydration) to form ether. R— O—H + H —O—R 2CH3CH2OH Ethyl alcohol
H 2SO4 413K
H2SO4 R—O—R + H2O 413k CH3CH2—O—CH2CH3 Diethyl ether
Diethyle ether boils at 35°C (308K) and is removed from the reaction mixture as it is produced. Note : (a) The success of this method depends upon the careful control of experimental conditions. The temperature is kept at 413K and alcohol is used in excess. If the temperature is increased to 443K, the product from ethyl alcohol is mainly ethylene which is produced by intramolecular dehydration of alcohol. CH3CH2OH Ethyl alcohol
H 2SO4 CH2 = CH2 + H2O 413K Ethylene
Remember, lower temperature favours the formation of ether whereas higher temperature favours the formation of alkene.
504
+2 CHEMISTRY (VOL. - II)
H2SO4 Excess alcohol 413K
CH3 CH2 —O—CH2 CH3 Diethyl ether
CH3CH2OH H2SO4 443K
CH2= CH2 Ethylene
Mechanism : The formation of ether is a bimolecular reaction (S N2) involving the attack of the alcohol on a protonated alcohol molecule. It consists of the following three steps: .. + CH3 – CH2 – O .. – H + H
(i)
+ CH3 – CH2 – O
+ .. CH3 – CH2 – O + CH3 – CH2 – O | H + (iii) CH3 – CH2 – O + CH2 – CH3 | H ..
(ii)
H H
+ CH3 – CH2 – O – CH2 – CH3 + H2O H | H + CH3 – CH2 – O – CH2 – CH3 + H H
(b) Mostly primary alcohols and to some extent secondary alcohols, react with sulphuric acid at 413K to form ethers, tertiary alcohols almost exclusively yield alkenes.
CH3—CH—CH + CH3—CH—O—CH—CH3 = 2 Propylene Diissopropylether (Major product) (Minor product)
Conc.H 2SO4 413K
CH3 —C—OH CH3 tert-Butyl alcohol (3° Alcohol)
CH3
—
Conc.H 2SO4 413K
CH3
— —
—
CH3—CH—OH Isopropyl alcohol (2° Alcohol)
CH3
CH3
—
—
CH3
CH3—C = CH2 + H2O Isobutylene
The order of dehydration of alcohols leading to the formation of ethers follows the sequence: primary > secondary > tertiary. Limitations : (i)This methods is generally not suitable for the preparation of unsymmetrical ethers since complex mixtures are obtained. For example. ROH + R'OH
Conc.H2SO4
ROR + ROR' + R'OR'
ETHERS
505
(ii) This method of preparation of ethers is suitable only for industries since reaction conditions must be carefully controlled. In place of H2SO4 other protonic acid like H3PO4 can also be used for preparation of ethers. (c) Catalytic dehydration of alcohols to ether can also be achieved by passing the vapours of an alcohol over heated alumina at 513–523K. For example, CH3CH2— OH + H —OCH2CH3 Ethyl alcohol (two molecules)
Al2 O3 CH3CH2—O—CH2CH3 + H2O 513–523K Diethyl ether
(d) By the action of diazomethane on alcohols. Methyl ethers can also be prepared by the action of diazomethane (CH2N2) on alcohols in presence of tetrafluoroboric acid (HBF 4) as catalyst. HBF4 CH3CH2—O—CH3 + N2 CH3CH2OH + CH2N2 Ethyl alcohol Diazomethane Ethyl methyl ether 2. From Sodium Alkoxide, Williamson Synthesis : It is an important laboratory method for the preparation symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide. The sodium alkoxide is prepared by the action of sodium on a suitable alcohol. – + 2R' – OH + 2Na 2R' – ONa + H2O Alcohol Sod. alkoxide R – X + R' – ONa Alkyl halide
R – O – R' + NaX Ether
Ethers containing substituted alkyl group (secondary or tertiary) may also be prepared by this method. Mechanism : The reaction involves SN2 attack of an alkoxide ion on primary alkyl halide. d+ d– R¢ – ONa+ + R – X R¢ – O – R + NaX Sod. Alkoxide Alkyl halide Ether
Limitations : Williamson synthesis is very sucessful with primary alkyl halides. The alkoxide ions are both powerful nucleophiles and bases and bring about dehydrohalogenation of the tertiary and secondary halides to form alkenes preferentially. Phenols are also converted to ethers by this method. In this, phenol is used as the phenoxide moiety.
506
+2 CHEMISTRY (VOL. - II)
–+ ONa
OH
O–R R–X
+ NaOH Phenol
+ NaX
Sod. Phenoxide
Ether
This method can also be used for the preparation of both symmetrical and unsymmentrical ethers. For example, d+ d– + CH3CH2 —O Na + CH3CH2—Br 330K
Ethyl bromide
— —
CH3 –
CH3 —C—O Na CH3
+
d+
–
Diethyl ether CH3
— —
Sod. ethoxide
+
CH 3CH 2—O—CH 2CH 3 + Na Br
d-
+ CH3—Br Methyl bromide
+
–
CH3—C—OCH3 + Na Br CH3
Sod. tert-butoxide
tert-Butyl methyl ether
Similarly, alkyl aryl ethers (phenolic ethers) can be easily prepared by treating sodium phenoxide with suitable alkyl halides. For example,
–
+
O Na
–
OCH3
+ CH3—Br Methyl bromide
Sod. phenoxide
–
D
+
O Na ;
D
+ CH3CH2 — I
(–NaBr) Anisole
OCH2CH3
Sod. phenoxide
+
O Na
Ethyl iodide
(–Nal) Phenetole
O–CH2–CH=CH2
+ CH2 = CH – CH2 – Br
D
+ NaBr
Allyl bromide Sod. phenoxide
o-Allyphenol or o-Allyl phenyl ether
However, these ethers cannot be prepared by treating bromobenzene or iodobenzene with sodium salts of the corresponding alcohols, i.e. sodium methoxide, sodium ethoxide or sodium allyl alkoxide
ETHERS
507 Br
CH3O–Na+ or CH3CH2O–Na+ or CH2 = CHCH2O–Na+ Sod. methoxide Sod. ethoxide Sod. allyl alkoxide
Bromobenzene
No reaction
This is due to the reason that the halogen is attached to the sp 2 carbon atom of the aromatic ring and the halides are much less reactive than alkyl halides towards nucleophilic substitution reaction. Further, this method cannot be used for preparing diaryl ethers since aryl halides do not undergo nucleophilic substitutions easily. –
+
O Na + Sod. phenoxide
Br Heat
O
Bromobenzene
+ NaBr
Diphenyl ether
3. From alkyl halides : Simple ethers can be prepared by boiling alkyl halides with dry silver oxide. R–I R–I
+ Ag2 O
R – O – R + 2AgI Ether
4. From Grigrand reagent : This is a good method for preparing higher ethers from lower members. Thus, the action of a lower halogenated ether react with Grignard reagent to give a higher member. Thus, Br CH3O CH2Cl + BrMgC2H5 CH3O CH2CH2CH3 + Mg Cl MonocloroEthyl magnesiumMethyl propyl dimethyl ether bromide ether 5. From alkenes : Alkenes can be converted into ethers by the following reaction. R – CH = CH2 Alkene
17.5
(i) Hg (OAc)2 / R ¢OH (ii) NaBH 4 / OH
–
OR¢ R – CH – CH3 Ether
PHYSICAL PROPERTIES OF ETHERS :
1. Colour, state and odour : Lower members like dimethyl ether and ethylmethyl ether are gases at ordinary temperature while the other lower homologues are colourless liquids with characteristic pleasant odour 'ether smell'. 2. Dipolar nature : Ether molecules are slightly polar and thus have dipole moment. The dipole moment of ethers is due to the polar character of the C–O bonds. The dipole moment of dimethyl ether is 1.3D and that of diethyl ether is 1.18D.
508
+2 CHEMISTRY (VOL. - II)
3. Boiling point : The weak polarity of ethers do not appreciably affect their boiling points which are comparable to those of the alkanes of comparable molecular masses but are much lower than the boiling points of isomeric alcohols. The large difference in boiling points of alcohols and ethers of same molecular masses (isomeric) is due to the presence of hydrogen bonding in alcohols. 4. Solubility : The solubility of lower ethers in water is due to the formation of hydrogen bonds between water and ether molecules. R
H .. O ..
R
H
.. O .. .. O ..
H
H
Diethyl ether is 7.5 percent soluble in water. As the molecular mass increases, the solubility of ethers in water decreases due to the corresponding increase in hydrocarbon portion of the molecule. Ethers are, however, fairly soluble in common organic solvents such as alcohol, benzene, chloroform, acetone etc. Diethyl ether is itself a good solvent for organic solutes. Apart from the polar character of C–O bond in ether, ether has the ability to solvate cations (electrophile) by donating an electron pair from their oxygen atom. These properties make diethyl ether a good solvent. 5. Density : All ethers are lighter than water. Their density and boiling points show a gradual increase with increase in molecular masses. 17.6
CHEMICAL PROPERTIES : In ethers two alkyl (or aryl) groups are linked to an oxygen atom. O
R Alkyl or aryl group
Ethereal oxygen
R¢ Alkyl or aryl group
The aryl – oxygen bond is more stable than the alkyl–oxygen bond. Owing to the absence of active groups and multiple bonds, ethers are comparatively inert substances. The reagents like ammonia, alkalies, dilute acids and metallic sodium have no action upon them in cold. They are not readily oxidised or reduced. However, under specific conditions, ethers undergo the following reactions in accordance with their structural features. I.
Reactions of ethereal oxygen : Ethers behave as Lewis bases on account of the presence of two lone pairs on the oxygen
atom. 1. Formation of oxonium salts : Ethers like alcohols are weakly basic and thus dissolve in strong inorganic acids (e.g. H2SO4, HClO4, HBr etc.) to form stable oxonium salts. thus,
ETHERS
509
R
R
+
R
Br–
..
O .. – H
..
.. O + HBr
R
Ether
Dialkyloxonium bromide
Oxonium salts are stable in high concentration of acids. When their acid solutions are diluted with water, the oxonium salts are dissociated into original ether and acid because water is a stronger base than ether. 2. Formation of coordination complexes : Ethers form relatively stable coordinate complexes with Lewis acids (e.g. BF3, AlCl3 and RMgX etc.) The complexes are known as etherates. CH3CH2
.. O + BF3
.. O
..
CH3CH2 CH3CH2
CH3CH2
Diethyl Ether
BF3
Boron trifuoride etherate
..
.. 2 (CH3CH2)2O
+
Diethyl ether
R
R
X
(CH3CH2)2O
Mg
Mg
Grignard reagent
O(CH2 CH3 )2 X
Grignard reagent etherate
Due to the formation of these complexes, Grignard reagents dissolve in ether. That is why Grignard reagents are usually prepared in ethers. II.
Reactions involving the cleavage of C–O bonds in ethers :
1. Action of Halogen acids : The cleavage of C–O bond in ethers takes place under drastic conditions with excess of hydrogen halides. First, the ethers are cleaved to form alkyl halide and alcohol. The alcohol thus formed reacts with excess of the halogen acid to form the corresponding halide. R–X
R – O – R + HX Ether
R – O – H + HX
+
Alkyl halide
R – OH Alcohol
R – X + H2O
Thus, dialkyl ether gives two alkyl halide molecules. Mechanism : The cleavage of ethers by halogen acids occurs by the following mechanism: Step 1 : Ethers being Lewis bases, undergo protonation to form oxonium ion.
Ethyl methyl ether
—
H .. + – CH3—O—CH .. 2CH3 + H I
+
–
CH3—O .. —CH2CH3 + I Oxonium ether
510
+2 CHEMISTRY (VOL. - II)
Step 2 : Iodide ion is a good nucleophile. Due to steric hindrance, it attacks the smaller alkyl group of the oxonium ion formed in step 1 and displaces the alcohol molecule by SN2 mechanism as shown below : –
H –
+
I + CH3—O —CH2CH3
S N2 Slow
I
+
CH3
Oxonium ion
O .. —CH2CH3
CH3—I + CH3CH2—OH
Transition state
Methyl iodide
Ethanol
Step 3 : When axcess of HI is used, ethanol thus formed reacts with another molecule of HI of form ethyl iodide. —
H .. + – CH3CH2—O—H + H I ..
+
–
CH3CH2—O .. —H + I
—
H –
+
I + CH3CH2—O —H
CH3CH2— I + H2O Ethyl iodide
Reactivity of halogen acids : The order of reactivity of halogen acids follows the sequence: HI > HBr > HCI. Greater the nucleophilicity of the halide ion, more reactive is the halogen acid. Since the nucleophilicity of the halide ions follows the sequence I – > Br– > Cl–, the reactivity of of halogen acids follows the same sequence i.e. HI > HBr > HCl. Site of cleavage : (a) However, when one of the alkyl groups is a tertiary group the halide first formed is a teritiary halide.
CH3—C—O—CH3 + HI
— —
CH3
— —
CH3
CH3 OH + CH3 —C—I
CH3
CH3
It is because in step 2 of the reaction, the departure of leaving group (HO–CH 3) creates a more stable carbocation [(CH3)3C+] and the reaction follows SN1 meachanism. CH3
—
+ CH3—C — O—CH3 slow
CH3 H
— —
— —
CH3
+
CH3—C + CH3 OH CH3
ETHERS
511
+
CH3
–
CH3—C + I
fast
CH3
— —
— —
CH3
CH3—C— I CH3
Ethers with two different alkyl groups are also cleaved in same manner and the mechanism of the cleavage is ascertained by the nature of the leaving group and the stability of intermediate carbocation ion. R – O – R¢ + HX
R – X + R¢ – OH
(b) Cleavage of alkyl aryl ethers : Alkyl aryl ethers (or phenyl) ethers are cleaved at the alkyl-oxygen bond due to the more stable aryl-oxygen bond. In case of phenolic ethers, the products are always phenol and an alkyl halide. O–R
OH + HX
Aryl ether
+R–X Phenol
Alkyl halide
Thus, in case of anisole, when it reacts with HI, the products are phenol and methyl iodide. O–CH3 + HI
OH + CH3 I +
In the first step, as per the mechanism, the methylphenyl oxonium ion,
C6 H5 —O—CH3 H
is formed by protonation of anisole. The bond between O–CH3 is weaker than the bond between O–C6H5 because the carbon of phenyl group is sp2 hybridised and there is a partial double bond character. Therefore attack by I– ion exclusively breaks the weaker O–CH3 bond forming methyl iodide and phenol. Phenols thus formed, do not react further to give aryl halides because sp2 hybridised carbon of phenol does not undergo nucleophilic substitution reaction by the halide ion to form the corresponding aryl halides. Cleavage of benzyl alkyl ethers : In case of benzyl alkyl ethers containing a primary alkyl group such as benzyl ether, the reaction proceeds by S NI mechanism. Since the benzyl carbocation is more stable than the methyl carbocation, therefore, cleavage of C–O bond gives methyl alcohol and benzyl carbocation. This carbocation then reacts with I – ion to form benzyl iodide as shown below :
512
+2 CHEMISTRY (VOL. - II)
–I
–
–I
–
+.. CH2 —O—CH3
C–O cleavage Slow
—
.. + – CH2 —O—CH 3+ H I .. Benzyl methyl ether
H
Oxonium ion +
CH2
+
Benzyl carbocation
I– Nucleophilic attack, Fast
CH 3OH Methyl alcohol
CH2 I + CH 3OH Benzyl iodide
Methyl alcohol
Cleavage of diaryl ethers : Diaryl ethers such as diphenyl ether are, however, not cleaved by HI. The reason being that the C–O bond has some double bond character due to resonance between the lone pairs of electrons on the O atom and the C atoms of the aryl groups directly linked to the O atom. 2. Hydrolysis : On heating with dilute sulphuric acid under presence, ethers are hydrolysed to alcohols. Dil H 2SO4 , D under pressue
R–O–R + H2O
2R – OH Alcohol
3. Action of Sulphuric Acid : Cold concentrated sulphuric acid has no action on ethers except that it dissolves them forming oxonium salts. However, if the solution is heated, cleavage of carbon-oxygen bond takes place leading to the formation of alcohol and alkyl hydrogen sulphate. C2H5–O–C2H5 Diethyl ether
Conc. H2SO4 D
C2 H5 –OH + C2 H5 HSO4 ethyl alcohol
ethyl hydrogen sulphate
4. Action of Phosphorus Pentachloride : Phosphorus pentachloride brings about the cleavage of both the carbon oxygen bonds of ethers forming alkyl chlorides. D CH3CH2 – O – CH2CH3 + PCl5 ¾¾® 2CH3CH2Cl + POCl3 Diethyl ether
Ethyl Chloride
5. Reaction with acid chlorides and anhydrides : Acid chlorides react with ethers when heated in the presence of anhyd. ZnCl2 or AlCl3 to form alkyl halides and esters. C2H5 – O – C2H5 + CH3COCl Diethyl ether
Acetyl chloride
Anhyd.ZnCl2 D
C2H5Cl Ethyl chloride
+ CH3COOC2H5 Ethyl acetate
ETHERS
513
However, with anhydrides only esters are formed C2H5 – O – C2H5 + (CH3CO)2O Diethyl ether
III.
Anhyd.ZnCl2 D
2 CH3COOC2H5
Acetic anhydride
Ethyl acetate
Reactions involving the alkyl group :
1. Action of air and light : Formation of peroxides. When exposed to air and light for a long time ethers are oxidised to form hydroperoxides or simple peroxides. CH3CH2 – O – CH2CH3 + O2
Light
Diethyl ether
—
OOH
CH 3 – CH – O – CH2CH3 1-Ethoxyethyl hydroperoxide
It is a free radical reaction and oxidation occurs at the carbon atom next to the ethereal oxygen to form hydroperoxides. These peroxides are very dangerous compunds since they decompose violenly at high temperatures. Therefore, serious explosions may occur during distillation of old samples of ethers if peroxides are not removed. A simple method to remove peroxide is to shake the old sample with ferrous salt solution when the ether peroxides are reduced to ethers and Fe 2+ salts are oxidised to Fe3+ salts. The Fe3+ ions thus formed are removed by washing with water. 2. Halogenation : Ethers react with Cl2 or Br2 to give substitution products. The extent of halogenation, however, depends upon the reaction conditions. For example, diethyl ether reacts with chloride in the dark to give a, a¢ -dichlorodiethyl ether.
CH3CH2 – O – CH2CH3
–HCl
Cl2 dark
a
a¢
CH3 – CH – O – CH–CH3
–HCl
a Chlorodiethyl ether
Diethyl ether
Cl
—
a
CH3CH2 – O – CH–CH3
Cl
—
Cl2 dark
—
Cl
a, a¢ -Dichlorodiethyl ether
However, in the presence of light and excess of chlorine, all the hydrogen atoms are substituted to give perchlorodiethyl ether. hv CH3CH2 – O – CH2CH3 + 10 Cl2 ¾¾® CCl3CCl2 – O – CCl2CCl3 + 10 HCl Diethyl ether
(excess)
Perchlorodiethyl ether
..
.. OR
I
+ OR
..
+ OR
..
+ ..
II
III
IV
OR
..
3. Electrophilic substitution : In aryl alkyl ethers the alkoxy group (–OR) is ortho, para directing and activates the aromatic ring towards electrophilic substitution in the same way as in phenol. However, aromatic ethers are less reactive than phenols. OR
V
514
+2 CHEMISTRY (VOL. - II)
(i) Halogenation : Phenylalkyl ethers undergo usual halogenation in the benzene ring, e.g., anisole undergoes bromination with bromine in ethanoic acid even in the absence of iron (III) bromide catalyst. It is due to the activation of benzene ring by the methoxy group. Para isomer is obtained in 90% yield. OCH3
OCH3
OCH3 Br
Br 2 in Ethanoic acid
+
Anisole
Br
po-Bromoanisole (major)
o-Bromoanisole (minor)
(ii) Friedel-Crafts reaction : Anisole undergoes Friedel-Crafts reaction, i.e. the alkyl and acyl groups are introduced at ortho and para positions by reaction with alkyl halide and acyl halide in the presence of anhydrous aluminium chloride (a Lewis acid) as catalyst. OCH3
OCH3
OCH3 CH3
Anhyd, AlCl 3 + CH3Cl CS2
+ CH3
2-Methoxytoluene (Minor)
OCH3
4-Methoxytoluene (Major)
OCH3 + CH3COCl
OCH3 COCH3
Anhyd. AlCl 3
+
Ethanoyl chloride
COCH3 2-Methoxy acetophenone (Minor)
4-Methoxy acetophenone (Major)
(iii) Nitration : Anisole reacts with a mixture of concentrated sulphuric and nitric acids to yield a mixture of ortho and para nitroanisole. OCH3
OCH3
OCH3 NO2
H2SO4 HNO 3
+ NO2 2-Nitroanisole (Minor)
4-Nitroanisole (Major)
ETHERS
17.7
515
USES OF ETHERS : Important uses of diethyl ethers : 1.
As a solvent both in laboratory and industry. Ethers are almost inert in nature and have good dissolving power. In industry it is used as solvent for oils, resins, gums etc.
2.
As an extracting solvent in laboratory and industry.
3.
As a reaction medium for the preparation of Grignard and organometallic reagents, also for carrying out LiAlH4 reductions.
4.
Diethyl ether finds its use as an anaesthetic agent in surgery.
5.
A number of naturally occurring ethers are used as perfumes and flavouring agents because of their pleasant odour.
6.
Diphenyl ether is used as a heat transfer medium because of its high boiling point, 531K.
7.
Lower ethers are volatile liquids which on evaporation produce low temperatures. They are therefore, used as refrigerants.
Diethyl Ether, Ether, C2H5OC2H5 : This is the most important representative of this class and often referred to as simply ether. It can be prepared in the laboratory by: (i)
Williamsons synthesis
(ii)
On a large scale by dehydration reaction of ethyl alcohol by Conc. H 2SO4
Equal volumes of alcohol and conc. H2SO4 are heated in a flask fitted with a dropping funnel and cold water running condenser. When the temperature of the content in the flask reaches 413K, ether begins to distil over. Alcohol is now run in from the dropping funnel at the same rate at which ether distills over. The process is continuous and hence the name Continous Etherfication Process. Estimation of Methoxy (or Ethoxy) Groups : The methoxy (or ethoxy) groups are estimated by Zeisel's method. A known mass of the sample is heated with conc. HI to form methyl iodide (or ethyl iodide). The iodide is then treated with alcoholic silver nitrate to give a precipitate of AgI ROCH3 + HI ® ROH + CH3I CH3I + AgNO3 ® AgI + CH3NO3 In this reaction : – OCH3 is equivalent to AgI, mass of methoxy group is 31 and that of 31 × mass of AgI × 100 AgI 234.88. Thus, % of methoxy group = 234.88 × mass of ether
516
+2 CHEMISTRY (VOL. - II)
CHAPTER (17) AT A GLANCE 1.
Two important methods of preparation of ethers are : (i)
Dehydration of alcohol and
(ii)
Williamson synthesis
2.
Weakly polar ethers have much lower boiling points compared to their isomeric alcohols.
3.
The lower members are fairly soluble in water due to the formation of hydrogen bonding.
4.
Diethyl ether is a good solvent and all the ethers are lighter than water.
5.
Owing to the absence of of active groups and multiple bonds, ethers are comparatively inert. However, under specific conditions ethers undergo the following reactions in accordance with their structural features. (i)
Reaction at ethereal oxygen
(ii)
Reactions involving cleavage of C–O bonds particularly with HI, HBr and PCl 5
(iii) Reactions involving the alkyl group. Ethers undergo halogenation at the alkyl groups. In aryl alkyl ethers the alkoxy group activates the aromatic ring and directs the incoming group to ortho and para positions in electrophilic substitution reactions.
QUESTIONS I. Multi choice type questions. (one mark) : 1.
2.
3.
Which of the following will exhibit highest boiling point ? (a)
CH3CH2O CH2CH3
(c)
CH3O CH2CH2CH3
(b)
CH3CH2CH2CH2OH
(d)
CH3CH2CH2CH2CH2OH
Which of the following cannot be prepared by Williamson synthesis ? (a)
Methoxybenzene
(c)
Di-tert-butyl ether
(b)
Methoxy ethane
(d)
tert-butyl methyl ether
In the reaction, (a) (b)
Br
OCH3
Br2 Acetic acid the products are
OCH3 + H2 Br + CH3Br
(c)
Br + CH3OH
(d)
OH + CH3Br
ETHERS
4.
5.
6.
7.
8.
9.
517
The ether C6H5 – O – CH2C6H5 when treated with HI produces (a)
C6H5CH2I
(c)
C6H5CH2OH
(b)
C6H5I
(d)
C6H5 – O – C6H5
Which of the following ethers is not cleaved by HI ? (a)
Methyl phenyl ether
(c)
Ethyl methyl ether
(b)
Diphenyl ether
(d)
Ethyl phenyl ether
The molecular formula of alkyl ethers is (a)
CnH2nO
(c)
CnH2n + 2O
(b)
CnH2n + I
(d)
CnH2nO CnH2n
Ethers react with Conc. H2SO4 to form (a)
Alkyl free radicals
(c)
Oxyanion
(b)
Zwitter ion
(d)
Oxonium ion
Ether gives chemical reactions due to (a)
C – O bond cleavage
(c)
Lone pair present on oxygen
(b)
C – H bond cleavage
(d)
All of the above
Which of the following pairs will give ether ? (a)
C2H5ONa and C2H5I
(b)
H C2H5OH 373K
Dry Ag2 O
(c)
C2H5I
(d)
All of the above
+
10.
11.
12.
Which of the following solvents is used for the preparation of Grignard reagent. (a)
Ethyl alcohol
(c)
Cyclohexonol
(d)
Diethyl ether
(d)
Benzene
When (CH3)3 COCH3 is treated with hydriodic acid, the fragments after the reaction obtained are : (a)
(CH3)3C – I + HOCH3
(c)
(CH3)3 CH + CH3OCH3 + I2
(b)
(CH3)3C – OH + I – CH3
(c)
(CH3)3 C – OH + CH4 + I2
C6H5COCH3, when treated with HI at 373K the following are the products. (a)
CH3OH and C6H5I
(c)
C6H5I and CH3I
(b)
CH3I and C6H5OH
(d)
C6H5OH and CH3OH
ANSWERS 1. (d), 2. (c), 3. (a), 4. (a), 5. (b), 6. (c), 7. (d), 8. (d), 9. (d), 10. (b), 11. (a), 12. (b)
518
+2 CHEMISTRY (VOL. - II)
II. Fill in the blanks 1.
Ethers exhibit functional isomerism with ______ .
2.
C4H10O has ______ metamers, one of them is diethyl ether while the others are ______ and ______.
3.
Williamson's synthesis involves the reaction of an ______ with an ______.
4.
Ethers behave as weakly ______ substances due to the presence of two lone pairs of electrons on the oxygen atom.
5.
______ is widely used as a solvent for the preparation of Grignard reagent.
6.
Aliphatic ethers are purified by shaking with a solution of ferrous salt to remove ______ which are formed on prolonged standing in contact with air.
7.
Cleavage of phenolic ether, anisole, by HI, gives ______ and ______.
8.
Alkyl aryl ethers are best prepared by treating sodium salt of ______ with ______ halide. ANSWERS
1. alcohols, 2. three, methyl n-propyl ether, isopropyl methyl ether, 3. alkoxide, alkyl halide, 4. basic, 5. Diethyl ether, 6. peroxides, 7. Phenol, methyl iodide, 8. phenol, alkyl. III. Choose the True (T) and Flase (F) statement : (one mark) 1.
Cleavage of ethyl methyl ether with HI at 373K gives methyl iodide and ethanol.
2.
Phenetole reacts with HI at 373K to give iodobenzene and ethanol.
3.
Dimethyl ether and ethyl alcohol have the some boiling point as both have same molecular masses.
4.
tert-Butyl ether can be prepared by heating sodium ethoxide with tert-butyl bromide.
5.
Alkyl aryl ethers are less reactive than phenols towards electrophilic substitution reactions.
6.
Both symmetrical and unsymmetrical ethers can be prepared by Williamson's synthesis. ANSWERS
1. (T), 2. (F), 3. (F), 4. (F), 5. (T), 6. (T). IV. Very short answer type questions (one mark) : 1.
Give the IUPAC names the following : (i)
CH3CH2CH2OCH3
(ii)
CH3COCH2CH2Cl
(iii) C6H5OC2H5 (iv) CH3 – O – CH2 – CH – CH3 | CH3
ETHERS
2.
519
Write the structure of the compounds whose names as follows : (i)
2-Ethoxy-3-methyl pentane
(ii)
1-Ethoxypropane
(iii) 1-Phenoxyheptane (iv) Anisole and phenetole 3.
Name one metamer of diethyl ether.
4.
Which method is used for the estimation of a methoxy group in organic compounds.
5.
Which class of ethers are not cleaved by HI.
6.
Which product is formed by the Friedel crafts reaction of anisole with CH 3Cl.
7.
Between anisole and phenol, which is more reactive towards electrophilic substitution reaction.
8.
Give the order of reactivity of halogen acids towards the cleavage of carbon-oxygen bond of ethers.
9.
Which reagent brings about the cleavage of carbon-oxygen bond of ethers leading to the formation of only alkyl halides.
10.
Give the order of polarity of alcohol, phenol and ether.
V. Short answer type questions (two marks) : 1.
Give the reason of the higher boiling point of ethanol in comparision to methoxy methane.
2.
Write the names of reagents and equation for the preparation of the following ethers by Williamson's synthesis:
3.
4.
(i)
Ethoxy ethane
(iii) 1-Methoxyethane
(ii)
Ethoxy benzene
(iv) 1-Propoxypropane
Write the equations for the following reaction (i)
Nitration of anisole
(ii)
Bromination of anisole in ethanoic acid medium.
Give tthe major products that are formed by heating each of the following with HI CH3 (i)
CH3 – CH2 – CH – CH2 – O – CH2– CH3 CH3
(ii)
CH3 – CH2 – CH2 – O – C – CH2 – CH3 CH3
(iii)
– CH2 – O –
520
5.
+2 CHEMISTRY (VOL. - II)
Predict the product of the following reactions : (i)
CH3 – CH2 – CH2 – O – CH3 + HI
(ii)
(CH3)3 C – O – C2H5 + HI OC 2 H5
(iii)
+ HBr OC 2 H5
(iv)
Conc.H2 SO4 Conc.HNO3
6.
Explain why diphenyl ether is not cleaved by HI.
7.
(CH3)3 C – O – CH3 + HI
(CH3)3 CI + CH3OH
Justify the preferential formation of the products. 8.
Predict the products A and B OCH3
9.
HI 373K
A+ B
Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why ? ONa
Br
(i)
+ CH3ONa NO 2
(ii)
+ CH3Br NO 2
10.
How you will distinguish C2H5– O – C2H5 and C6H5 – O – C6H5 by treating with HI ?
11.
Why can ethers be cleaved preferentially by hot conc.HI and HBr but not by conc.HCl.
VI. Short answer type questions (3 marks) : 1.
Illustrate with examples the limitations of Williamson sysnthesis for the preparation of certain type of ethers.
2.
How is 1-propoxypropane synthesised from propan-1-ol ? Write the mechanism of this reaction.
3.
Preparation of ethers by acid dehydration of secondary and tertiary alcohol is not suitable method. Give reason.
4.
Write the mechanism of the reactions of HI with methoxymethane.
5.
Write the equations of the the reactions of hydrogen iodide with : (i)
methoxybenzene
(ii)
benzyl ethyl ether
(iii) 1-propoxypropane
ETHERS
521
6.
Explain how the alkoxy group (–OR) is ortho and para directing and activates the aromatic ring towards electrophilic substitution reactions.
7.
Give an example for the synthesis of unsymmetrical ether by Williamson synthesis.
8.
Write the reaction of Williamson synthesis of 2-ethoxy-3-methyl-pentane starting from ethanol and 3-methylpetan-2-ol.
9.
Explain why is bimolecular dehydration not appropriate for the preparation of ethyl methyl ether.
10.
Compound (A), C4H10O, is found to be soluble in sulphuric acid. (A) does not react with metallic sodium or potassium permanganate solution. When (A) is heated with excess of HI, it is converted to a single alkyl halide. What is the structural formula of (A).
11.
Write the equation of the reaction of hydrogen iodide with (i)
1-propoxypropane
(ii)
methoxy benzene
(iii) benzyl ethyl ether 12.
Write a suitable reaction for the preparation of t-butyl ethyl ether.
13.
Give the majo products that are formed by reacting each of the following ether with HI. (i)
methyl propyl ether
(ii)
phenyl methyl ether
(iii) benzyl phenyl ether 14.
Preparation of ethers by acid dehydration of secondary and tertiary alcohol is not suitable method. Give reason.
VII. Long answer type question (7 marks) : 1.
Outline two important methods of prepration and three properties of dialkyl ethers.
2.
How does diethyl ether react with (i) PCl5,
(ii) Conc.H2SO4
(iii) Conc.HI
(iv) O2
3.
Describe Williamson synthesis of ethers. What are its limitations? Discuss the mechanism.
4.
How is diethyl ether prepared in the laboratory? How will you distinguish it from ethanol and diphenyl ehter. What happens when it is treated with hot HI?
5.
Explain the following with an example. (i)
Unsymmetrical ether
(ii)
Cleavage of benzyl alkyl ethers by HI
(iii) Williamson ether synthesis 6.
Discuss the electrophilic substitution reaction like halogenation, nitration and FriedelCrafts reactions of aryl alkyl ethers.
7.
Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming group to ortho-and para-positions in the benzene ring.
qqq
522
+2 CHEMISTRY (VOL. - II)
UNIT - XII
CHAPTER - 18
ALDEHYDES AND KETONES 18.1
INTRODUCTION :
The aldehydes and ketones belong to the class of oxygen containing organic compounds that have a C = O group, for which they are termed as carbonyl compounds. (i)
In aldehydes, the carbonyl group is attached to one hydrogen atom and one alkyl or aryl group or another hydrogen atom, whereas in ketones, the carbonyl group is attached to either one alkyl and one aryl group or to two alkyl (or aryl) groups. O
O (a)
R—C—H Aldehyde
Examples :
H—C—H Formaldehyde O
(Where R = H or any alkyl or aryl group)
H 3C — C — H Acetaldehyde O O —C—H Benzaldehyde O
O (b)
R — C — RI Ketone (Where R and R¢ may be same or different alkyl or aryl groups)
Examples :
H3C — C — CH3 Acetone O O — C — CH3 Acetophenone
(ii)
When two alkyl (or aryl) groups i.e. R and R¢ are same, the ketone is termed as a simple ketone, where as when R and R¢ are different, the ketone is named as a mixed ketone.
ALDEHYDES AND KETONES
523
O
O H3C — C — CH3 Acetone R = RI = — CH3 Simple ketone
O — C — CH3 Acetophenone R=— O RI = — CH3 Mixed ketone H
(iii) Aldehydes are characterised by the functional group — C = O (aldehydic group) and the functional group of ketones is C = O (ketonic group). (iv) Aldehydes and ketones have some similar characteristics because of the presence of same carbonyl group, but differ in some properties due to the presence of a hydrogen atom on the carbonyl group of aldehydes. (v)
In case of aldehydes the —CHO group always occurs at the end of the carbon O chain, where as the — C — group in case of keones occurs within the carbon chain.
(vi) Carbonyl compounds may be aliphatic or aromatic in nature depending on the presence of only alkyl group or an aryl group (may be in addition to alkyl or aryl group in case of ketones). (a)
Aliphatic aldehydes and ketones O H—C—H Formaldehyde
(b)
O CH3 — C — H Acetaldehyde
CH3 — C — CH3 Acetone
Aromatic aldehydes and ketones O O —C—H Benzaldehyde
O
O
O — C — CH3
O —C— O
Acetophenone
Benzophenone
18.2
NOMENCLATURE OF CARBONYL COMPOUNDS :
A.
Nomenclature of Aldehydes :
(i)
Common system : (a)
O
In the common system, the names of aliphatic and aromatic aldehydes are derived from the common names of the acids which they form on oxidation by replacing the terminal ‘ic acid’ by the suffix aldehyde.
524
+2 CHEMISTRY (VOL. - II)
O
O
H—C—H+O Formaldehyde
H — C — OH Formic acid
Formic acid – ic acid + aldehyde = formaldehyde (b)
In case of substituted aldehydes, the positions of the substituents are indicated by Greek letters a, b, g, d etc with the carbon atom next to the —CHO group being designated as a. g
b
a
CH3 — CH — CH2 — CHO CH3 b – Methylbutyraldehyde (ii)
IUPAC system : (a)
In IUPAC system the aliphatic aldehydes are named as alkanals which are derived from the names of the corresponding alkanes by replacing the ending – e with – al. O
O CH3 — C — H Ethanal
H—C—H Methanal (b)
O CH3 — CH2 — CH2 — C — H Butanal
In case of substituted aldehydes, the longest carbon chain is choosen having —CHO group and the substituents are prefixed in the alphabetical order along with numerals indicating their positions in the carbon chain with the carbon atom of the —CHO group given number 1. CH3
Br
CH3
O
CH3 — CH — CH2 — CHO
CH3 — CH2 — CH — CH — C — H
3-Methylbutanal
3-Bromo-2-methylpentanal
4
3
2
1
5
5
4
3
4
2
3
2
1
1
OHC — CH2 — CH2 — CH2 — CHO Pentan - 1, 5 - dial
(c)
When aldehyde group is attached to an alicyclic ring, the suffix carbaldehyde is added after the full name of the cycloalkane, with the numbering starting from the carbon atom containing –CHO group.
ALDEHYDES AND KETONES
525
O C
O
CH3
H
2
C
H
1
Cyclohexanecarbaldehyde (d)
2-Methylcyclohexanecarbaldehyde
The suffix carbaldehyde is used when an unbranched carbon chain is attached to more than two aldehyde groups, where the compound is considered as a derivative of the parent alkane which does not include the carbon atom of the functional groups. CHO OHC — CH2 — CH2 — CH — CH2 — CH2 — CHO 5
4
3
2
1
Pentane - 1, 3, 5 - tricarbaldehyde But : CH2—CHO OHC — CH2 — CH — CH2 — CH2 — CHO 1
2
3
4
5
6
3 - (Formylmethyl) hexan - 1, 6- dial (e)
The name benzenecarbaldehyde or simply benzaldehyde is given to the simple aromatic aldehyde where aldehyde group is attached directly to the benzene ring. In case of substituted benzaldehydes, the positions of substituents in the benzene ring with respect to the –CHO group are indicated by the prefixes ortho(o) meta(m) and para(p) or by the numbers 1, 2, 3 etc. with the carbon having –CHO group being considered as number 1. O C—H
CHO OH
Benzaldehyde
2- Hydroxybenzaldehyde (o-Hydroxybenzaldehyde or Salicylaldehyde)
The common and IUPAC names of some aldehydes are given in the Table 18.1
526
+2 CHEMISTRY (VOL. - II)
Table 18.1: Common and IUPAC names of some Aldehydes. Formula
Common Name
IUPAC name
Formaldehyde
Methanal
Acetaldehyde
Ethanal
CH3CH2CHO
Propionaldehyde
Propanal
CH3CH2CH2CHO
Butyraldehyde
Butanal
Isobutyraldehyde
2-Methylpropanal
b-Methyl butyraldehyde
3-Methylbutanal
Acraldehyde.
Prop-2 - enal
O || H–C–H O || CH3 – C – H
3
CH3 CH3
2
CH –1CHO
CH3 | CH3 – CH – CH2 – CHO b a O || 3 CH2 = 2CH – 1C – H
(Acrolein) CH3 – CH = CH – CHO
Crotonaldehyde
But-2 - enal
g-Methylcyclohexane
3-Methylcyclohexanecarb-
-carbaldehyde
aldehyde
Phthaldehyde
Benzen-1, 2-dicarbaldehyde
P-Bromobenzaldehyde
4-Bromobenzaldehyde
b 3 CH
g3 2
a CHO 1
CHO CHO CHO
Br A.
Nomenclature of Ketones :
(i)
Common system : (a) In the common system, the names of aliphatic and aromatic ketones are derived by naming the two alkyl or aryl groups bonded to the carbonyl group in alphabetical
ALDEHYDES AND KETONES
527
order as two separate words and adding the suffix ketone. In case of symmetrical ketones the prefix di is attached to the name of the alkyl group. COCH3
O
O
CH3 — C — C2H5
CH3 — C — CH3
Ethyl methyl ketone
Dimethyl ketone (Acetone)
Methyl phenyl ketone
The positions of the substituents are indicated by Greek letters, aa¢, bb¢ etc starting from the carbon atoms next to the carbonyl group taken as aa¢.
(b)
CH3
O
CH3 — CH — C — CHI 2 — CHI 3 b
a
a
b
a -Methyl diethyl ketone (c)
In case of aromatic ketones generally the namings are done by adding the acyl group as prefix to phenone. O
O
C–CH3
C–CH2–CH 3
C
Acetophenone
(ii)
O
Propiophenone
Benzophenone
IUPAC System : (a)
In the IUPAC system, ketones are named as alkanones, which are derived by replacing the terminal -e of the corresponding alkanes by the suffix - one.
(b)
During the naming the longest chain carrying the carbonyl group is choosen and the numbering begins from the end giving lowest number to the carbonyl group. If the chain carries two keto groups, prefix di is attached to -one. O
O
O
O
CH3 — C — CH3
CH3 — CH2 — C — CH3
CH3 — C — CH2 — C — CH3
Propan-2-one
Butan-2-one
Pentan-2, 4-dione
1
2
(c)
3
4
3
2
1
1
2
3
4
5
In case of substitued ketones, the parent chain containing the keto group and the substituents are prefixed in alphabetical order along with numerals indicating their positions in the carbon chain. CH3
O
O
CH3 — CH — C — CH2 — CH3
1
2
3
4
2-Methylpentan-3-one
5
CH3 — CH = CH — CH2 — C — CH3
6
5
4
3
4-Hexen-2-one
2
1
528
+2 CHEMISTRY (VOL. - II)
O
The same principle is applied to cyclic ketones, the number 1 being given to the carbonyl group. (d)
1 2
3 CH When both aldehyde and keto groups are present in 3 the same compound, aldehyde group is given 3-Methylcyclopentanone importance because of its higher reactivity and will be assigned number 1. Keto group will be given the name oxo prefixed by its position. For example.
O CH3 — C — CH2 — CH2 — CHO 5
4
3 2 4-Oxopentanal
1
Common and IUPAC names of some ketones are given in Table 18.2. Table 18.2 : Common and IUPAC names of Ketones Formula O || CH3 – C – CH3 O || CH3 – C – CH2 – CH3 O || CH3 – CH2 – C – CH2CH3 O || CH3 – C – CH2CH2CH3 O CH3 || | CH3 – C – CH – CH3 CH3 O 3 2 || 1 4 | CH3 – C = C H – C– C H3 5
Common name
IUPAC name
Acetone
Propanone
Ethyl methyl ketone
Butan - 2 - one.
Diethyl ketone
Pentan - 3 - one.
Methyl n-propyl ketone
Pentan-2-one
Isopropyl methyl ketone
3-Methylbutan-2-one
Mesityl oxide
4-Methylpent-3-en-2-one.
a-Methylcyclohexanone
2-Methylcyclohexanone
Biacetyl
Butan-2, 4-dione
m-Bromoacetophenone
3-Bromoacetophenone
Propionylacetaldehyde
3-Oxopentanal
O CH3 O
O
CH3 — C — C — CH3 COCH3
Br
O CH3 — CH2 — C — CH2 — CHO
ALDEHYDES AND KETONES
18.3
529
ISOMERISM IN ALDEHYDES AND KETONES : 1.
Aldehydes: Aldehydes show three types of structural isomerism i.e. chain, functional isomerism and position isomerism.
(a)
(b)
Chain isomerism: The molecular formula C4H8O represents compounds like (i)
CH3 CH2 CH2CHO Butanal
(Straight chain)
(ii)
CH3 – CH – CHO | CH3
(Branched chain)
2– Methylpropanal.
Functional isomerism: Aldehydes exhibit functional isomerism with ketones. The molecular formula C3H6O represents two compounds.
(c)
(i)
CH3CH2CHO
(ii)
Propanal CH3COCH3 Propanone.
Position isomerism : Aromatic aldehydes exhibit position isomerism. CHO
CHO
CHO
CH3 CH3 CH3 o-Methylbenzaldehyde 2.
m-Methylbenzaldehyde
p-Methylbenzaldehyde
Ketones: Ketones exhibit chain isomerism, functional isomerism and metamerism. (i) Chain isomerism: Molecular foumula C5H10O represents the following compounds.
(ii)
O || CH3 – CH2 – CH2 – C – CH3 Pentan -2-one O 4 3 2|| 1 CH3 – CH – C – C H3 | CH3 3 –Methylbutan - 2- one. Functional isomerism: Ketones exhibit functional isomerism with aldehydes. O || CH3 – C – CH3 CH3 – CH2 – CHO Propanone Propanal
530
+2 CHEMISTRY (VOL. - II)
(iii)
18.4
Metamerism: The same molecular formula represents two ketones which differ in the nature of alkyl groups linked to the divalent keto group. This is also termed as position isomerism. O O || || CH3 – CH2 – CH2 – C – CH3 CH3 – CH2 – C – CH2 – CH3 Pentan -2-one Pentan-3-one
STRUCTURE OF CARBONYL GROUP :
Electron diffraction and p bond p bond spectroscopic studies confirm that 1200 º carbonyl group is planar in nature. The s bond O120 s bond 0 C C=O 2 carbon atom of carbonyl group is sp hybridised, thus it has three sp 2 hybrid 1200 p bond orbitals. It forms three sigma bonds, one p bond Fig.18.1 Structure of Carbonyl group with oxygen and other two with other two atoms or groups. All these sigma bonds lie in one plane at an angle of 120 0 to each other. The unhybridised p–orbital of carbon laterally overlaps with half filled p–orbital of oxygen, thereby forming a p bond. The electron cloud of p bond. lies both above and below the C – O bond. Since oxygen atom is more electronegative the electron cloud is displaced more towards oxygen. This results in polarisation of C – O bond. The carbon end becomes slightly +ve whereas the oxygen end, slightly –ve. Hence the carbonyl carbon is an electrophilic (Lewis acid) and carbonyl oxygen, a nucleophilic (Lewis base) centre. +d -d C=O Due to polarity of C – O bond, the aldehydes and ketones are associated with large values of dipole moment ( m =2.3 – 2.8 D) and are more polar than ethers. Due to its polar nature, the carbonyl group can be represented as a resonance hybrid of the following two structures.
C=O (I) COMPARISON BETWEEN
+ – C–O (II)
C = O GROUP AND
C=C
Similarities: (i)
In both the cases ‘C’ atom is sp2 hybridised.
(ii)
Both consist of a s bond and p bond.
(iii)
Both have planar structure.
(iv)
Both exhibit addition reactions.
BOND
ALDEHYDES AND KETONES
531
Differences: (i)
C = O is polar whereas
C=C
is nonpolar..
(ii)
C = O group undergoes nucleophilic addition reactions whereas
C= C
undergoes electrophilic addition reactions. (iii)
Bond length and bond strength are different in both the cases. Due to small size of oxygen C = O length is shorter (1.23A°) than C = C bond length (1.34A°)
(iv)
Geometrical isomerism is possible with case of
18.5
C=C
whereas no such isomerism in
C = O group.
GENERAL METHODS OF PREPARATION OF ALDEHYDES AND KETONES : 1. From Alcohols (a)
By oxidation:
(i)
Aldehydes and ketones can be prepared by the oxidation of alcohols. Primary alcohols upon oxidation give aldehydes whereas secondary alcohols give ketones. The most commonly used oxidants are KMnO 4 or K 2 Cr 2 O 7 acidified with sulphuric acid.
R– CH2OH
K2Cr2O7 / H2SO4
+ [O]
Primary alcohol CH3– CH2OH + [O]
K2 Cr2O7 / H2SO4
Ethyl alcohol + [O]
CH3
CHOH + [O]
CH3 Isopropyl alcohol
Aldehyde O || CH3 – C – H + H2O Acetaldehyde
CH2 OH Benzyl alcohol R CHOH + [O] R Secondary alcohol
O || R – C – H + H2O
CHO
K2 Cr2O7 + H2SO4
K2Cr2O7 / H2SO4
+ H2O Benzaldehyde R R
K2Cr2O7 / H2SO4
C = O + H2O Ketone
CH3
C=O CH3 Acetone
+ H2O
Aldehydes formed by this process are to be removed from the reaction mixture as soon as they are formed. This is because aldehydes are prone to oxidation further to carboxylic acid. The
532
+2 CHEMISTRY (VOL. - II)
removal of aldehyde from the reaction mixture is usually made by distillation since aldehydes are having lower boiling points than alcohols from which they are formed. (ii)
Ketones can also be prepared from secondary alcohols by an alternate method. The secondary alcohol is refluxed with excess of acetone (solvent) in presence of catalyst aluminium tertiary butoxide [Al(O – tert – Bu) 3]producing ketone. This is known as Oppenauer oxidation. In this case secondary alcohol is oxidised to ketone whereas acetone is reduced to isopropyl alcohol.
R R1
CHOH + Secondary alcohol
CH3 CH3
C=O
[(CH3) 3CO]3 Al
Acetone
CH3
R
C=O + CH3 R1 Ketone
CHOH
In this case secondary alcohol is oxidised to ketone even in the presence of carboncarbon double bond whereas acetone is reduced to isopropyl alcohol. OH
O [(CH 3)3CO]3 Al (CH3)2CO
Cyclohex-2-enol (iii)
Alcohols can be better oxidised to carbonyl compounds by mild oxidsing agents Collin’s reagent (chromium trioxide-pyridine complex, (CrO3.2C5H5N) and PCC i.e.. Pyridinium chlorochromate (CrO3.C5H5N.HCl) which are more advantageous than acidified K2Cr2O7 as using these reagents oxidation can be stopped at the aldehyde stage preventing further oxidation to carboxylic acids. These reagents do not attack double bonds. CH3 CH2 CH2 OH
Collin’s reagent or PCC
H CH3 — CH = CH — C — CH3 OH Pent - 3 - en - 2 - ol (b)
Cyclohex-2-enone
CH3 – CH2 – CHO O
PCC CH3 — CH = CH — C — CH3 or Collin’s reagent Pent - 3 - en - 2 - one
By catalytic dehydrogenation of alcohols. When vapours of alcohols are passed over hot reduced Cu at 573 K, dehydrogenation takes place. Primary alcohols produce aldehydes whereas secondary alcohols produce ketones.
ALDEHYDES AND KETONES
533
Cu
R– CH2 OH
573K
Primary alcohol
Cu
CH3 – CH2 OH Ethyl alcohol R CHOH R Secondary alcohol CH3 CHOH CH3
573K Cu 573K Cu 573K
O || R – C – H + H2 Aldehyde CH3 CHO + H2 Acetaldehyde R C = O + H2 R Ketone CH3 C = O + H2 CH3 Acetone
Isopropyl alcohol 2.
By dry distillation of Ca-salt of fatty acids. Dry distillation of Ca-salts of fatty acids yield aldehydes and ketones. (a)
Aldehydes Formaldehyde is obtained by the dry distillation of calcium formate at 670 K. O || H—C—O H — C —O || O Calcium formate
O || H–C–H
Ca
+
CaCO3
Formaldehyde
A mixture of calcium formate and calcium salt of any other carboxylic acid on dry distillation yields aldehydes. O || CH3— C — O CH3— C — O || O Calcium acetate (b)
Ca + Ca
O || O—C—H O—C—H || O
2CH3 CHO + 2CaCO3 Acetaldehyde
Calcium formate
Ketones Calcium salt of any fatty acid other than formic acid on dry distillation yields ketone.
534
+2 CHEMISTRY (VOL. - II)
O || CH3 — C — O
O || CH3 – C – CH3 + CaCO3
Ca CH3 — C — O || O Calcium acetate
Acetone
Mixed ketones can be obtained by dry distillation of a mixture of calcium salt of fatty acids other than HCOOH. e.g. Ethylmethyl ketone can be obtained by dry distillation of a mixture of calcium acetate and calcium propionate. O || CH3— C — O
Ca + Ca
CH3— C — O || O
O || O — C — CH2CH3
2CH3COC2H5 + 2CaCO3
O — C — CH2CH3 || O
Ethylmethyl ketone
However, this method is not suitable for the preparation of mixed ketones. This is because three products are expected when a mixture of two calcium salts are heated. For example, in the above case the dry distillation of a mixture of calcium acetate and calcium propionate yields a mixture of acetone, diethyl ketone and ethylmethyl ketone. 3.
From carboxylic acid:
When vapours of fatty acid are passed over heated MnO at 573K, aldehydes or ketones are formed. Formic acid gives formaldyhyde whereas acids other than HCOOH produce ketones. A mixture of formic acid and other fatty acids produces aldehydes. O O || || H – C – OH + H – C – OH O O || || CH3 – C – OH + H – C – OH O || CH3— C — OH + CH3— C — OH || O
MnO 573K MnO 573K
O || H–C–H
+ CO2 + H2O
Formaldehyde CH3CHO
+ CO2 + H2O
Acetaldehyde
MnO
CH3
573K
CH3 Acetone
C = O + CO2 + H2O
ALDEHYDES AND KETONES
4.
535
From acid chlorides. Both aldehydes and ketones can be prepared from acid chlorides.
(a)
Aldehydes: Acid chlorides can be reduced to aldehydes with H2 in boiling xylene using palladium as
catalyst supported by BaSO4 and partially poisoned by addition of sulphur or quinoline. O O || || Pd / BaSO4,S R – C – Cl + H2 R – C – H + HCl Acid chloride. Boiling xylene Aldehyde O || CH3 – C – Cl + H2 Acetyl chloride
Pd / BaSO4,S Boiling xylene
O || CH3 – C – H + HCl Acetaldehyde
O C
Cl
CHO
H2 Pd–BaSO4,S
Benzoyl chloride
Benzaldehyde
The partial poisoning of catalyst is made in order to avoid the further reduction of aldehyde to primary alcohol. This reaction is known as Rosenmund reduction. Mechanism: The following mechanism seems to operate. O || R – C – Cl + H2 Acidchloride
Pd
.. ..O–H | R – C – Cl | H
O || R – C – H + HCl
+ –H O
|| R–C | H
+
– Cl
Aldehyde However, formaldehyde can not be prepared by this method since formyl chloride HCOCl is unstable at room temperature. (b)
Ketones: Ketones can be prepared easily by the action of dialkyl cadmium on acid chlorides. O O || || Dry / 2R – C – Cl + Cd R 2 2R – C – R/ + CdCl2 ether Acid chloride Dialkyl cadmium Ketone
536
+2 CHEMISTRY (VOL. - II)
O || 2CH3 – C – Cl +
Cd(C2H5)2
Acetyl chloride
Diethyl cadmium
O || 2CH3 – C – C2H5 + CdCl2
Dry ether
Ethylmethyl ketone.
O 2
O
– C – Cl
Benzoyl chloride
+ Cd (CH3)2
– C – CH3
2
Dry ether
Dimethylcadimium
+ CdCl2
Acetophenone
Dialkyl cadmium can be prepared from Grignard’s reagent by the action of CdCl 2 2RMgBr + CdCl2
5.
R
Cd
+ 2Mg
R Dialkyl cadmium
Br Cl
From alkynes: Hydration of alkynes in presence of HgSO4 and H2SO4 yields enols which readily tautomerise giving aldehydes or ketones. The addition of water to unsymmetrical alkynes is in accordance with Markownikoff’s rule. CH ||| e.g. CH + H – OH Acetylene (Ethyne)
1%HgSO4 42%H2SO4 340K
1%HgSO4
CH3 – C º CH + H – OH 42%H2SO4 Propyne 340K
CH2 || CH – OH
Tautomerise
Vinyl alcohol OH 3 2| 1 CH3 – C == CH2
CH3 | CHO Acetaldehyde
Tautomerise
O || CH3 – C – CH3 Acetone
Prop - 1-en-2-ol Note: However, aldehydes other than acetaldehyde cannot be prepared by this method. 6.
From alkenes:
Alkenes add on a molecule of ozone forming the corresponding ozonides. The ozonides are highly unstable due to the peroxide linkage so that these are shortlived, and explosive in nature. On boiling with zinc dust and water ozonides undergo reductive cleavage producing aldehydes or ketones or a mixture of both depending on the substitution pattern of the alkene. Zinc dust helps in removing H 2O2, else the aldehydes or ketones formed may be oxidised to carboxylic acids.
ALDEHYDES AND KETONES
537
O H H
C=C
H
O3
H
CCl4
Ethylene CH3
C=C
196K H
O3
CH3 CCl4 196K 2-Methylbut -2-ene
CH3
CH3 CH3
C=C
CH3
O3
CH3 CCl4
2,3-Dimethylbut-2-ene 196K
H2C
CH2
O
O || 2H – C – H + H2O+ZnO Formaldehyde
Zn dust H2O
O
Ethylene ozonide CH3 CH3
CH3 CH3
O C
CH
O
O
O
CH3
C
O
O
CH3
CH3
Zn dust H2O
CH3
C=O CH3 Acetone +
CH3 – CH = CH – CH3 But-2-ene 7.
O3
CH3
CCl4
O CH
CH
O
O
C=O
Acetone + CH3CHO Acetaldehyde + ZnO + H2O CH3
C
CH3
Zn dust H2O
CH3 CH3
C=O
Acetone + ZnO + H2O
CH3
Zn dust H2O
196K
2CH3CHO + Acetaldehyde + ZnO + H2O
From gem - Dihalides: A gem-dihalide is a compound containing two halogen atoms attached to the same X carbon. C (X = Cl, Br). The hydrolysis of a gem-dihalide yields a carbonyl group. X Thus, 1.1 dihaloalkanes on hydrolysis with aqueous KOH yield aldehydes whereas 2, 2-dihaloalkanes under similar condition yield ketones.
CH3 — CH
Cl
+ 2KOH(aq)
Cl
(1, 1 -Dichloroethane)
CH3 CH unstable
OH OH
--H2O
CH3CHO Acetaldehyde
538
+2 CHEMISTRY (VOL. - II)
Cl
CH3
OH
CH3
+ 2KOH(aq)
C OH CH3 unstable
C Cl CH3 (2,2 -Dichloropropane)
CH3
--H2O
C=O CH3 Acetone
8. From Grignard Reagent (i) Preparation of Aldehydes (a)
d-
When Grignard reagents react with hydrocyanic acid (HCN) and the products are hydrolysed, aldehydes are formed. For example, d+
Methylmag. bromide
d+
d-
Dry ether
Hydrocyanic acid
[CH3 – CH =NMgBr]
O
+ H3O
Addition product
CH3-C-H Acetaldehyde
+ Mg (OH) Br + NH3
(b)
Aldehydes can also be prepared by the action of Grignard reagents on orthoformic esters. For example,
dry ether Ethylmag. bromide
Ethyl orthoformate
CH3-CH
OC2H5
+ Mg (OC2H5)Br
Diethyl acetaldehyde acetal
O CH3-C-H
(ii) Preparation of Ketones (a) From acid chlorides
OC2H5
H3O+, hydrolysis
+ 2 C2H5OH
Acetaldehyde
Acid chlorides react with Grignard reagents in equimolecular proportion (why ?) and the products formed on hydrolysis with dilute mineral acid yield ketones. For example, O OMgX + R' H3O dry ether R' — C — Cl + RMgX C = O + Mg (Cl) X R' — C — Cl R R (Ketone) Choosing R, R¢ suitable ketones can be prepared using Grignard reagents. For example, methyl magnesium bromide can react with propanoyl chloride (R¢ = C2H5) to give ethyl methyl ketone (butanone) after hydrolysis of the intermediate addition product with dilute mineral acid. O OMgBr + C 2 H5 Dry ether C2H5 – C – Cl + CH3 Mg Br C2H5 – C – Cl H3O C=O CH 3 (Propanoyl chloride) CH3 (Butanone) + Mg (Cl) Br
ALDEHYDES AND KETONES
539
(b) From nitriles Ketones can be prepared by the action of a suitable Grignard reagent on an alkane nitrile followed by acid hydrolysis of the intermediate addition product. For example,
Ethylmag. bromide
Addition product
Methyl cyanide or Ethane nitrile
Butanone
18.6 SPECIAL METHODS OF PREPARATION FOR AROMATIC CARBONYL COMPOUNDS : A.
Preparation of Benzaldehyde
Benzaldehyde may be prepared by the following methods which are applicable to aromatic aldehydes in general (1) Oxidation of Toluene (a) By Etard’s reaction : This reaction involves the preparation of benzaldehyde by the oxidation of toluene. Toluene, when treated with chromyl chloride (CrO2Cl2) in presence of CCl4 or CS2, a brown complex is precipitated as intermediate, which on hydrolysis gives benzaldehyde. Here, chromyI chloride is used as oxidising agent. The reaction is called Etard’s reaction.
CH3 |
2CrO2Cl2 / CCl4
Toluene
OCrCl2OH
CH |
OCrCl2OH
CHO |
+
H3O
(Benzaldehyde)
Brown complex
In case of derivatives of benzene having side chain longer than —CH 3, the end carbon is oxidised by chromyl chloride to —CHO group. For example, CH2CH3 |
CrO2Cl2/CCl4
CH2CHO |
+
Ethylbenzene
H3O
Phenyl ethanal
540
+2 CHEMISTRY (VOL. - II)
(b) Oxidation by Chromium trioxide Other than Etard’s reaction benzaldehyde can also be prepared by the oxidation of toluene with chromium trioxide in acetic anhydride. The gem-diacetate first formed is isolated which upon hydrolysis with dil.alkali or dil. HCl or dil. H2SO4 yield benzaldehyde. CH3
CH
CrO3/(CH3CO) 2O 273-283K
OCOCH3 OCOCH3
toluene
gem-diacetate
+ H3O, D (hydrolysis)
CHO + 2 CH3COOH benzaldehyde
(2) From Acid Chlorides (Rosenmund Reduction) Benzaldehyde may be prepared by passing hydrogen gas through boiling xylene solution of benzoyl chloride in the presence of palladium catalyst supported over BaSO 4 and partially poisoned by the addition of sulphur or quinoline.
O C Cl +
H2
benzoyl chloride
CHO Pd-BaSO4, S Boiling xylene benzaldehyde
+ HCl
(3) From nitriles (Stephen reduction) When an ethereal solution of benzonitrile is reduced with stannous chloride in presence of hydrogen chloride gas at room temperature, benzaldimine hydrochloride is precipitated. This upon hydrolysis with boiling water gives benzaldehyde. CH = NH. HCl C N Dry ether + 2 [H] + HCl 290-295K benzonitrile benzaldimine hydrochloride CHO + H 3O + NH4Cl D benzaldehyde This process is called Stephen reduction. Nitrites can also be selectively reduced by diisobutylaluminium hydride (DIBAL – H) to imines followed by hydrolysis to aldehydes. RCN CH2 = CH – CH2 – CN
1. AlH (i-Bu) 2 2. H2O 1. AlH (i-Bu) 2 2. H2O
RCHO CH2 = CH – CH2 – CHO
ALDEHYDES AND KETONES
541
(4) By side chain halogenation followed by hydrolysis Side chain chlorination of toluene gives benzal chloride which upon hydrolysis gives benzaldehyde. This method is used for the manufacture of benzaldehyde. CH3
CHCl2
2Cl2 , hn (-2HCl)
Toluene
H2O
H2O, 373K (Hydrolysis) (-2HCl)
CH(OH)2
CHO benzaldehyde
(5) By Gattermann – Koch reaction When a mixture of carbon monoxide and HCl gas is passed through a solution of benzene in nitrobenzene at 323K in presence of a catalyst consisting of anhydrous AlCl 3 and a small amount of CuCl, benzaldehyde is formed. [HCOCl]
CO + HCl
Formylchloride (unstable) CHO + HCOCl AlCl3 + CuCl + HCl benzene
benzaldehyde
This reaction is known as Gattermann - Koch reaction. B. (1)
Preparation of Aryl Ketones : From Nitriles : When phenyl magnesium bromide (Grignard reagent) reacts with an alkane nitrile in the presence of dry ether, an intermediate is formed which on acid hydrolysis produces aryl keton. NMgBr
ether CH3 – C º N + C6H5MgBr ¾dry ¾¾¾ ¾® Ethane nitrile
Phenyl magnesium bromide
CH3 — C
H3O+ O CH3 — C Acetophenone
(2)
C 6H 5
C6H5
From benzene or substituted benzene by Friedel-Crafts acylation reaction : Friedel-Crafts acylation reaction involves the conversion of benzene or substituted benzene to aryl ketone on reaction with acid chloride or acid anhydride in the presence of a Lewis acid as catalyst like anhydrous aluminium chloride.
542
+2 CHEMISTRY (VOL. - II)
O O + CH3 – C – Cl Benzene
+
Acetyl chloride
O
Anhydrous Alcl 3
+ C6H5 – C – Cl
C
Benzoyl chloride
Benzene
18.7
CH3
C
CH3
C
+
HCl
Benzophenone
O +
HCl
Acetophenone
O
Benzene
C – CH3
Anhydrous AlCl3
COCH3
Anhydrous Alcl3
O
+ Acetophenone
O Acetic anhydride
CH3 COOH Acetic acid
PROPERTIES OF ALDEHYDES AND KETONES :
(a) Physical Properties: 1. Physical State: Formaldehyde is a gas at room temperature. Other aldehydes and ketones up to C11 are liquids and higher ones are solids. 2. Odour: Lower aldehydes have unpleasant odour whereas higher members have pleasant odour. Ketones are generally sweet smelling. 3. Solubility: Lower members are soluble is water, but solubility decreases with increase in molecular mass. However, all aldehydes and ketones are soluble is organic solvents like alcohol, ether etc. Explanation : The lower members are capable of forming hydrogen bond with water. So they are soluble in water. +d -d +d
-d
R – C = O....H –– O | H
+d -d
+d
H....O = C – R | H Hydrogen bonding of aldehydes with water R R
+d -d +d
-d
C = O....H –– O
+d
d-
+d
H....O = C
R
R Hydrogen bonding of ketones with water
ALDEHYDES AND KETONES
543
As the hydrocarbon portion of alkyl group increases with increase in molecular mass, the solubility rapidly decreases since the hydrogen bonding is prevented. 4. Boiling poiont : Boiling points of aldehydes and ketones are more than those of alkanes but less than those of alcohols of comparable molecular mass. The boiling points of ketones are slightly higher than those of isomeric aldehydes. e.g
Compound CH3CH2CH2CH3 CH3CH2CHO CH3CH2CH2OH CH3COCH3
Molecular mass 58 58 60 58
Boiling point(K) 309 322 371 329
Explanation: Boiling point depends upon molecular association. The stronger the intermolecular forces the more is the boiling point. In case of alkanes the molecules are held together by weak vander Waal’s forces. But in case of aldehydes and ketones the association is due to electrostatic attraction between two opposite ends of carbonyl dipoles. Thus, the boiling points of aldehydes and ketones are more than those of alkanes. In case of alcohols there -dO C +d || is intermolecular hydrogen bonding which is stronger than || +dC O -d dipole-dipole interaction. Hence, the boiling points of alcohols are more than those of aldehydes and ketones. (Dipole-dipole interaction in case of In case of ketones the carbonyl group is attached to aldehydes and ketones) two electron releasing alkyl groups. As a result the carbonyl group in ketones becomes more polar than in H –– O ..... H –– O ..... H –– O | | | isomeric aldehyde. The more the polarity the more is R R R the dipole-dipole interaction. Thus, ketones boil at (Hydrogen bonding in case of alcohols) comparatively higher temperature. (b) Chemical Properties: (1)
Nucleophilic addition reactions: (AdN)
These reactions are due to the presence of a polar carbonyl group. The attacking agent is a nucleophile (nucleus loving) and the reaction is of addition type. Since oxygen atom is more electronegative, there develops an infinitesimal small positive charge over ‘C’ atom and a small -ve charge over ‘O’ atom of the carbonyl group. The nucleophile attacks at the +vely charged carbonyl carbon and as a result the double bond breaks in the direction of ‘O’ atom resulting in the formation of an anion. The anion thus formed takes up a proton from the solvent (usually water) or from the reagent forming the addition product. The reaction may be represented as follows. Nu Nu | | Fast d— d+ Slow — C — O — C — OH C = O + : Nu + + | | E (H ) Addition product Carbonyl Nucleophile compound (Planar) (Tetrahedral) (Tetrahedral)
544
+2 CHEMISTRY (VOL. - II)
Nucleophilic addition reactions are usually carried out in weakly acidic medium in case of weak nucleophiles like NH3 or its derivatives. The protonation of ‘O’ atom of carbonyl group in presence of acids increases the positive charge on the carbon atom thereby facilitating the attack of the weak nucleophile. .. ..
+
C = O + H+
C –O–H
C = O+ – H
(2) Relative reactivities of Aldehydes and Ketones towards nucleophilic addition reactions. Aldehydes are found to be more reactive than the ketones. This can be explained by two effects i.e. electronic effects (resonance effect and inductive effect) and steric effect. A. Aliphatic aldehydes and ketones. (i) Inductive effect: Alkyl groups are electron releasing. The presence of alkyl group at the carbonyl carbon decreases the magnitude of the positive charge over ‘C’ atom thereby decreasing its tendency to accept the nucleophile. Thus, the more the number of alkyl groups attached to the carbonyl group the lesser is its reactivity towards nucleophilic addition reaction. The reactivity basing upon +l effect thus follows the order. HCHO
>
Formaldehyde
>
RCHO Aldehyde
O || R – C – R/ Ketone
Again, presence of electron withdrawing groups at the carbonyl carbon (-l effect) increases the magnitude of the positive charge over ‘C’ atom thereby facilitating the nucleophilic attack. Thus, reactivity of substituted acetaldehyde follows the order, NO2 CH2 CHO
>
Nitroacetaldehyde
ClCH2 CHO
>
Chloroacetaldehyde
CH3CHO Acetaldehyde.
Further, the size of the alkyl groups also affects the reactivity. With increase in size of the alkyl group the reactivity decreases. Thus, the order of reactivity of various ketones is as follows. CH3COCH3 Acetone.
>
CH3 COCH2CH3 Ethylmethyl ketone
>
CH3CH2 COCH2CH3 Diethyl ketone.
(ii) Steric effect: The approach of the nucleophile is prevented by the presence of bulky alkyl groups at the carbonyl carbon. This effect is known as steric effect. The attack of the nucleophile at the carbonyl carbon becomes more and more difficult as the number and size of the alkyl groups increase. Hence, the reactivity of aldehydes and ketones follows the following order,
ALDEHYDES AND KETONES
H H
>
C=O
CH3 H
(CH3)2CH
C=O
(CH3)2CH Di-isopropyl ketone
545
C=O
> >
CH3
C=O
CH3
(CH3)3C (CH3)3C
CH3 CH2
>
C=O
CH3 CH2
>
C=O
Ditert-butyl ketone
B. Aromatic aldehydes and ketone : (i) Aromatic aldehydes and ketones are less reactive towards nucleophilic addition reactions than the aliphatic ones due to negative resonance effect (–R) decreasing the nucelophilic character of the carbonyl carbon atom as carbony group is an electron withdrawing group. This in turn decreases its tendency to be attacked by the nucleophile. R –R
d+
R d-
R -
C=O
C–O
C–O
-
+
+
R C
d-
O
R
R
C=O
C
O
-
+ d+
Resonance hybrid
Resonating structures
(ii) Again aromatic aldehydes are more reactive than aryl ketones due to +I effect of the alkyl group. C6H5CHO > C6H5COCH3 > C6H5COC6H5 3.
Examples of Nucleophilic addition reactions :
(a)
Addition of HCN Both aldehydes and ketones react with HCN forming the corresponding cyanohydrins.
546
+2 CHEMISTRY (VOL. - II)
OH | R — C –– CN | Aldehyde cyanohydrin H OH | CH3 — C — CN | H Acetaldehyde cyanohydrin. OH | R — C –– CN | Ketone cyanohydrin R/ OH | CH3 — C — CN | CH3
O || R – C – H + HCN Aldehyde O || CH3 – C – H + HCN Acetaldehyde. O || R – C – R/ + HCN Ketone O || CH3 – C – CH3 + HCN Acetone
Acetone cyanohydrin O
OH
C
H + HCN
C
OH–
CN
H
Benzaldehyde
Benzaldehyde Cyanohydrin
Since HCN is a poisonous substane the reaction is often carried out by mixing the carbonyl compound with aqueous sodium cyanide and then slowly acidifying the solution by adding mineral acid. This reaction is a base catalyzed reaction proceeding through the following mechanism. HCN + OH R H
+d
--
-CN + H2O
-d
-C = O + CN
Aldehyde.
Nucleophile
CN | -R—C—O | H
H2O
CN | -R — C — OH + OH | H Cyanohydrin
Cyanohydrins are nitriles. They, upon hydrolysis yield a-hydroxy acids which undergo subsequent dehydration forming a, b unsaturated acids. O e.g. CH3CH2 — C — CH3 Ethylmethyl ketone
OH HCN
CH3 — CH2 — C — CN CH3
HOH HCl
ALDEHYDES AND KETONES
547
OH O || b a| CH3–– CH2 –– C –– C –– OH | CH3 a- hydroxy acid (b)
H2SO4 –H2O
b a CH3 –– CH = C –– COOH | CH3 a, b-unsaturated acid
Addition of NaHSO3 Both aldehydes and ketones (specially methyl ketones) react with NaHSO 3 forming
the corresponding crystalline bisulphite compounds. O OH || | CH3 – C – H + NaHSO3 CH3 — C — H | SO3Na Acetaldehyde sodium bisulphite OH | CH3 — C — CH3 | SO3Na
O || CH3 – C – CH3 + NaHSO3
Acetone sodium bisulphite The reaction is carried out by adding carbonyl compound to a concentrated aqueous solution of sodium bisulphite when a crystalline solid bisulphite compound separates out. Machanism: There is initial nucleophilic attack by bisulphite ion on carbonyl carbon followed by proton transfer.
C=O + .. – Na SO3H
C
– + O Na
Proton
SO3H
transfer
C
OH – SO3Na+
Bisulphite compound
The crystalline bisulphite addition product can be decomposed with dilute mineral acids or aqueous alkalies to regenerate the original aldehyde or ketone.
C
OH – SO3Na+
+ HCl
C=O Carbonyl compound
+ NaCl + SO2 + H2O
548
+2 CHEMISTRY (VOL. - II)
OH – SO3Na+
C
C = O + Na2SO3 + H2O
+ NaOH
The reaction is employed for separation and purification of carbonyl compounds from noncarbonyl impurities or compounds. (c)
Reaction with Alcohols Alcohols add to the carbonyl group of aldehydes in the presence of dry HCl gas acid to yield hemiacetals, which further react with alcohols to give a gemdialkoxy compound known as acetal. H H dry HCl | | R¢ – C = O + ROH R¢ – C – OR H+ Aldehyde | OH Hemi acetal (Unstable) H | H+ R¢ – C – OR + ROH | OH Hemiacetal (An alcohol)
H | R¢ – C – OR + H2O | OR Acetal (An ether)
MECHANISM This reaction proceeds through the following steps. H (i)
H
R—C=O
H+
+
+
R—C=O—H
H (ii)
R — C — OH + HO — RI +
H R — C — OH +
H
H
R — C — OH
R — C = OH –H+
+ O—H RI H (iii)
H
R — C — OH OR
I
+
H+
+
R—C—O OR
H H
ORI
ALDEHYDES AND KETONES
H (iv)
H
H
R—C—O—H +
OR
549
+ HO — RI
H
R — C — O — RI
–H2O
+
I
I
OR
–H+ H R — C — ORI ORI For example, acetaldehyde is converted to acetal as follows. The forward reaction is catalysed by dry HCl, while the reverse reaction is catalysed by aqueous HCl. H | HCl C = O + C2 H5 OH CH3 – C – OC2 H5 H | OH Acetaldehyde Hemiacetal CH3
H | CH 3 – C – OC2H5 + C2H5OH | OH Hemiacetal
H | HCl CH3 – C – OC2H5 + H2O | OC2H5 Acetaldehyde diethyl acetal
Ketones would react similarly to form hemiketals and ketals. R R
R '
C = O + ROH
OH
OH Hemiketal
R
OR' C
R
C R
R
OR'
+ R¢ OH
OR' C
OR' R ketals
+ H2O
The above equilibria lie far to the left and are unfavourable to the formation of ketals. However, if the equilibria are tilted to the right by removing water produced during the reaction, ketals are prepared with better yield. This is done by the addition of orthoformic ester.
550
+2 CHEMISTRY (VOL. - II)
OC2H5 OC2H5 + H O H-C 2 OC2H5 ethyl orthoformate
H+
2C2H5OH + HCOOC2H5 ethyl formate
PROTECTION (BLOCKING) OF CARBONYL GROUP FORMATION
BY CYCLIC ACETAL
In the course of organic synthesis it is sometimes necessary to protect or mask a carbonyl group in a molocule by preventing it from reacting during the course of a transformation of some other functional group in the molecule. This is done by converting the carbonyl group into an acetal or cyclic ketal which is regenerated by acid hydrolysis. +
OC2H5 H3O 2C2H5OH C= O C=O C dry HCl OC2H5 –2C2H5OH –H2O (Protected as acetal or ketal)
OH C=O OH
C=O regenerated
–H2O
Cyclic acetals or ketals are obtained by reaction with glycol in the presence of acid and in the absence of alkyl formate. The carbonyl group is regenerated from the cyclic acetals or ketals by acid hydrolysis. R HO – CH2 O CH2 HCl C C=O + R O CH2 HO – CH2 –H2O aldehyde/ketone glycol cyclic acetal/ketal CH2 OH HOH/H + + C=O CH2 OH aldehyde/ketone (d)
Addition of Grignard reagents :
Grignard reagents add to the carbonyl group of aldehydes and ketones forming an adduct which on acid hydrolysis produce alcohols. d+
C=O
d–
+
d–
d+
R – Mg – X
C – O – Mg – X R
+ H 3O
C – OH + Mg R Alcohol
OH X
ALDEHYDES AND KETONES
(i)
551
Formation of primary (I°) alcohol on reaction with formaldehyde. H H
H
C = O + CH3 – Mg – Br
C – O – Mg – Br
H
CH3
Methanal
H+/H 2O H
C – OH + Mg
H
CH3 Ethanol
(ii)
OH X
Formation of secondary (2°) alcohol on reaction with any aldehyde excepting formadehyde. CH3 H
CH3
C = O + CH5 – Mg – Br
H
C – O – Mg – Br CH3
Ethanal
H+ /H2O CH3 H
C – OH + Mg
CH3 Propan - 2-01 (iii)
OH Br
Formation of tertiary (3°) alcohol on reaction with a ketones. H3C H 3C
CH3
C = O + CH3 – Mg – Br
CH3
C — O – Mg – Br CH3
Propanone
H+/H 2O CH3 H3C
C – OH + Mg CH3
2-Methylpropan-2-ol
OH Br
552
+2 CHEMISTRY (VOL. - II)
CH3 CH3
CH3 C = O + C6H5 – Mg – I
CH3 — C — O – Mg – I C6H5 H+ /H2O CH3 CH3
C – OH + Mg C6H5
I OH
3° Alcohol (e)
Nucleophilic addition reactions followed by elimination of water (Addition of ammonia derivaties).
Derivatives of ammonia (H2N–G) such as hydroxyl amine (NH2OH), phenyl hydrazine (C6H5NHNH2), semicarbazide (NH2NHCONH2) etc. react with carbonyl compounds in weakly acidic solution to form compound
C = N – G..
The reaction proceeds through the initial attack of nucleophile on the carbonyl group to form addition product, which subsequently loses a molecule of water to give a compound having a double bond between carbon and nitrogen. The reaction may be represented as follows: C = O + H2N – G Carbonyl
Derivative
compound
of NH3
H+ fast
OH
C
NHG
slow
C = NG + H2O
The above reaction is very much helpful in identifying carbonyl compounds since the compound so formed as final product is an insoluble crystalline solid having sharp melting point. The reactions of carbonyl compounds with ammonia derivatives are carried out at moderate acid strength or pH. The protonation of carbonyl oxygen takes place which makes the carbonyl carbon more prone to nucleophilic attack.
C = O + H+
+ C — OH
.. NH2 — G
H | |+ –C–N –G | | OH H
C = N – G +H2O
H | | – C – N –G | –H+ OH
ALDEHYDES AND KETONES
553
Although the reaction is acid catalysed, the solution should not be too much acidic. This is because the ammonia derivative gets protonated, thereby decreasing its nucleophilicity.
(i)
.. G — NH2 + H+
G – NH3
Free base
Salt
+
Reaction with Hydroxylamine (NH2OH) Aldehydes and ketones react with NH2OH in presence of HCl to form corresponding oximes (aldoximes and ketoximes)
e.g.
H
R–C Aldehyde
O H
H–C Methanal
O
+ H2N – OH
R – CH = N — OH + H2O
Hyxroxylamine
Aldoxime H | H – C = N — OH + H2O
+ H2NOH
Methanaloxime
CH3 – C = O + H2NOH | H Ethanal
CH3 – CH = N — OH + H2O (Ethanal oxime)
R
R
C = O + H2NOH
C = N – OH + H2O R Ketoxime
R Ketone CH3
CH3
C = O + H2NOH CH3 Propanone O || —C–H
C = N – OH + H2O CH3 Propanoneoxime H || — C = N – OH
+ H2NOH
Benzaldehyde (ii)
Benzaldehydeoxime
Reaction with hydrazine ( NH2 – NH2) Aldehydes and ketones react with hydrazine in presence of dilute acid forming corresponding hydrazones. R H
C=O (Aldehyde)
+ H2N – NH2 Hydrazine
H
+
R H
C = N – NH2
+ H2O
(Aldehyde hydrazone)
554
+2 CHEMISTRY (VOL. - II)
R R (iii)
C=O
+ H2N – NH2
H
+
(Ketone)
R R
C = N – NH2
+ H2O
(Ketone hydrazone)
Reaction with phenylhydrazine (C6H5NHNH2) Aldehydes and ketones react with phenylhydrazine in presence of dilute mineral acid forming crystalline phenylhydrazones. R H H
C=O
+ H2N.NHC6H5
Aldehyde
Phenyl hydrazine
R H
C = N.NHC6H5
Aldehyde phenyl hydrazone H
C = O + H2N.NHC6H5
H Formaldehyde CH3
H
H Acetaldehyde R
C = N.NHC6H5 + H2O Acetaldehyde phenylhydrazone
H R
C = O + H2N.NHC6H5
R Ketone CH3
R CH3
C = O + H2N.NHC6H5 CH3 Acetone (iv)
C = N.NHC6H5 + H2O Formaldehyde phenyl hydrazone
CH3
C = O + H2NNHC6H5
+ H2O
CH3
C = N.NHC6H5 + H2O Ketone phenylhydrazone C = N.NHC6H5 + H2O Acetone phenyl hydrazone
O || Reaction with semicarbazide ( H2N – NH – C – NH2) Aldehydes and ketones react with semicarbazide in acid medium to form the corresponding
semicarbazones. R 1
R
C = O + H2N – NH – CO – NH2 Semicarbazide
H+
R 1
R
C = N – NH – CO – NH2+ H2O (Semicarbazone)
R = R1 = H,
Formaldehyde semicarbazone
R = H, R1 = CH3,
Acetaldehyde semicarbazone
R = R1 = CH3,
Acetone semicarbazone
R = H, R1 = – C6 C5
Benzaldehyde semicarbazone
ALDEHYDES AND KETONES
555
Note: The nitrogen derivatives of carbonyl compounds like oximes. hydrazones, phenyl hydrazones, and semicarbazones are used primarily to identify aldehydes and ketones. This is possible because these derivatives are mostly solids with characteristics melting points. 2, 4 - Dinitrophenyl hydrazine (DNPH) is more useful than phenyl hydrazine since the former forms yellow-orange or red precipitate easily with aldehydes and ketones. NO2 O2N
+
NH – NH2
O=C Carbonyl compound
(2, 4 - Dinitrophenyl hydrazine)
H+ — NO2 + H2O
C = N — NH —
NO2 (Orange ppt.)
2, 4 - Dinitrophenyl hydrozone 4.
Reactions due to a - hydrogen
Acidity of a - hydrogens of aldehydes and ketones : Hydrogen atom present on the carbon atom next to the carboxyl group, called a - hydrogen atom is acidic by nature due to the following factors. (i)
Carbonyl group is an electron withdrawing group (– I effect) thus decreasing the electron density from the adjacent carbon - carbon bond making Ca - carbon electron deficient. This in turn results with the decrease of electron density of a - H bond. Hence a - H atom becomes acidic by nature and can be easily abstracted by a base forming enolate anion. d–
O
H
d+
—C— C — d+
a
O +
OH
H 2O
—C— C — Enolate anion
(ii)
Enolate anion i.e. conjugate base is resonancce stablilized. O —C— C —
O — C= C —
Due to acidic nature of a - hydrogen atoms aldehydes and ketones undergo the following reactions.
556
+2 CHEMISTRY (VOL. - II)
A.
Aldol condensation: Two molecules of same or different carbonyl compounds each containing a -H -atom, condense together in presence of dilute alkali like NaOH, KOH, Ba(OH)2 or Na2CO3 to form a syrupy liquid known as Aldol. This aldol which shows the properties of alcohol, aldehyde or ketone is b-hydroxy aldehyde or b-hydroxy ketone.
e.g. CH3 CHO + HCH2CHO
dil NaOH
b a CH3CH(OH)CH2CHO Aldol b -Hydroxy butyraldehyde
Mechanism : (i)
Formation of carbanion due to the abstraction of a - hydrogen by OH O O || || – – CH2 - C - H + OH CH2 - C - H + H2O | Carbanion. H
(ii)
Nucleophilic attack of carbanion on the electron deficient carbonyl carbon of other molecule of acetaldehyde to form an anion O O || || – CH3 - C - H + CH2 - C - H
(iii)
– O | CH3 - C - CH2CHO | H
The anion finally abstracts a proton from water to form aldol. O | CH3 — C — CH2CHO + H — OH | H
OH | CH3 - C - CH2CHO + OH | H Aldol
Restrictions (i)
When concentrated alkali is used the aldol formed undergoes base catalysed dehydration forming a, b-unsaturated aldehyde.
ALDEHYDES AND KETONES
e.g.
557
H b| a CH3 - C - CH2 - C - H | || O OH
H | CH3 - C - HC - C - H + H2O – | || O OH –OH–
– OH –H2O
b a CH3 - CH = CH - CHO
a, b -Unsaturated aldehyde (Crotonaldehyde) (ii)
When a trace of mineral acid is there the aldol formed is no longer stable. It undergoes acid catalysed dehydration to form a, b-unsaturated aldehyde.
e.g.
H | CH3 - C - CH2 CHO | OH
+
H
H | CH3 - C - CH2CHO | + O H
H | CH3— C — CH CHO + | H
– H2O (iii)
--H+
H
b a CH3 - CH = CH - CHO a, b -Unsatd. aldehyde (Crotonaldehyde)
When two different aldehydes each containing a -hydrogen atom condense together in presence of dilute alkali the product formed is a mixture of aldols. This is known as Cross Aldol condensation or Cross Aldolisation. Examples are :– O
I.
CH3 — C — H + H — C — H Acetaldehyde
II.
O
Formaldehyde (No a - hydrogen atom)
O dil.NaOH
b
a
HO — CH2 — CH2 — C — H b-hydroxypropionaldehyde
Here four different aldols are formed, two from condensation of the two aldehydes with like molecules and the other two from condensation of two different aldehydes in two different ways.
558
+2 CHEMISTRY (VOL. - II)
e.g.
CH3CHO + Acetaldehyde
CH3CH2CHO Propionaldehyde OH
–
OH OH OH OH | | | | CH3–CH–CH2CHO+CH3CH–CHCHO+CH3CH2CH–CH–CHO + CH3CH2CH CH2CHO | | CH3 CH3 Simple aldol III.
Cross aldol
Simple aldol
Cross aldol
O O OH O || || | || C6H5 — C — H + CH3 — C — H ¾dil.NaOH ¾ ¾ ¾¾® C6H5 — CH — CH2 — C — H —H2O
O || C6H5 — CH = CH — C — H Cinnamaldehyde
Cross Aldol condensation is also possible in case of two different ketones and in case of condensation reaction between an aldehyde and a ketone. Aldol condensation is case of Ketones: Two molecules of acetone condense together in presence of Ba(OH) 2 to produce diacetonealcohol (aldol of acetone) CH3 – C = O + H.CH2COCH3 | CH3 Acetone
Ba(OH)2
CH3 CH3
Acetone
C
OH CH2COCH3
Diacetone alcohol
Diacetone alcohol upon heating loses a molecule of water to form mesityl oxide. CH3 CH3
C
OH CHCOCH3 | H
– H2O
CH3 CH3
C= CHCOCH3 Mesityl oxide.
ALDEHYDES AND KETONES
B.
559
Iodoform reaction: Compounds which contain CH3CO- group or which produce CH3CO group on oxidation
can undergo iodoform reaction. Methyl ketones or acetaldehyde when treated with sodium hypoiodite (alkaline solution of iodine), iodoform is formed as yellow precipitate along with an acid which contains one C atom less than the parent compound. O || CH3 – C – CH3 + 3I2 Acetone O || CI3 – C – CH3 + NaOH
NaOH
O || CI3 – C – CH3 + 3 HI Tri-iodoacetone CHI3
+ CH3COONa
Iodoform (yellow ppt) Mechanism: NaOH O || CH2 – C – CH3 + OH– | H
Na+ + OH– O || – CH2 – C – CH3 + H2O Carbanion
560
+2 CHEMISTRY (VOL. - II)
Iodoform This test is used for identification of compounds containing CH 3 CO-group or any other group like CH3 –– CH –– OH which can be oxidised to CH3 CO-group by sodiumhypoiodite. | 5.
Oxidation reactions Aldehydes differ from ketones in oxidation reactions. It is a fact that aldehydes can be
easily oxidised to carboxylic acids. O || R – C – H + [O]
O || R – C – OH
Aldehyde
Carboxylic acid The easy oxidation of aldehydes is due to the presence of a H –atom attached to carbonyl group which on oxidation becomes –OH group without breaking any other bond. Aldehydes can also be oxidised with oxidising agents like Cu2+, Ag+ or bromine water to carboxylic acids. On the other hand, ketones can be oxidised to carboxylic acids with less number of carbon atoms under drastic conditions using powerful oxidising agents like acidified KMnO 4 or K2Cr2O7. Thus, aldehydes are treated as good reducing agents. They can reduce Fehling’s solution and Tollens’ reagent whereas ketones can not. Mixed ketones when oxidised under drastic conditions give rise to a mixture of carboxylic acids which involve cleavage of carbon-carbon bonds. When oxidised with acidified KMnO 4 or
K2Cr2O7 and conc.H2SO4, C--CO bond of the ketones breaks and the a- carbon atom containing least number of hydrogen atom is attacked preferentially (Popoff’s rule). Thus,
ALDEHYDES AND KETONES
CH3 – CO – CH2 – CH2 – CH3
561
acid. KMnO4
Pentan -2-one
CH3COOH + CH3 CH2COOH (Acetic acid) (Propionic acid)
If the carbon atoms adjacent to carbonyl group have the same number of hydrogen atoms, the carbonyl group chiefly remains with the smaller alkyl group. Thus, the oxidation of hexan-3one gives propionic acid as the main product with smaller amount of acetic acid and butyric acid. O CrO3 || CH3 – CH2 – C – CH2 – CH2 – CH3 CH3CH2COOH oxdn. Hexan -3-one (Propionic acid) Main product CH3COOH + CH3CH2CH2COOH Small amounts (i)
Reduction of Fehling’s solution. Aldehydes when boiled with Fehling’s solution form a red precipitate of Cu 2O. Fehling’s
solution is alkaline Cu(II) ion complexed with sodium potassium tartrate. It is a mixture of Fehling A and Fehling B. Fehling A contains CuSO4 whereas Fehling B contains NaOH and sodium potassium tartrate (Rochelle salt). CuSO4 + 2NaOH Cu(OH)2
O || R – C – H + 2CuO
Cu(OH)2 + Na2SO4 CuO + H2O O || R – C – OH + Cu2O (Red ppt) Carboxylic acid
Aldehydes thus reduce Cu2+ ions to Cu+ ion and get oxidised to carboxylic acids. Ionically the same may be represented as O || – 2+ R – CHO + 2Cu + 5OH R – C – O + Cu2O + 3H2O Red The function of Rochelle salt is to dissolve insoluble Cu(OH) 2 by complexing it with tartrate ion. i.e. it prevents the precipitation of Cu(OH) 2. Aromatic aldehydes do not reduce Fehling’s solution. (ii)
Reduction of Tollens’ reagent : When an aldehyde is warmed with Tollens’ reagent a silver mirror is formed which appears
on tha walls of the test tube in which the reaction is carried out. Tollens’ reagent is ammoniacal
562
+2 CHEMISTRY (VOL. - II)
solution of AgNO3. It is prepared by adding NH4OH solution to a solution of AgNO3 till the precipitate AgOH formed first again redissolves. AgNO3 + NH4OH
AgOH + NH4 NO3
AgOH + 2NH4OH
[Ag(NH3)2] OH + 2H2O
RCHO + 2[Ag(NH3)2OH
Tollens’ reagent O || – + R – C– ONH4 + 2Ag + 3NH3 + H2O Silver mirror
Here also Ag+ ion is reduced to metallic silver. The reaction is accompanied by oxidation of aldehydes to carboxylic acids. The silver thus deposited shines like a mirror for which this test is known as silver mirror test. This test is exhibited by both aliphatic and aromatic aldehydes. 6.
Miscellaneous reactions:
A.
Reaction with NH3: (i)
Formaldehyde reacts with NH3 to form hexamethylenetetramine [CH 2)6N4], known as Urotropine which is used as urinary antiseptic. 6HCHO + 4NH3
(CH2)6N4 + 6H2O. Hexamethylenetetramine.
Structure of Urotropine:
H2C
N | CH2 | N CH2
CH2
CH2
N
N CH2
Nitration of urotropine under control conditions produces the explosive RDX (Research and Development Explosive)
ALDEHYDES AND KETONES
(ii)
563
Other aldehydes react with NH3 to form aldehydeammonias which on heating undergo dehydration to form aldimine. OH | R — C — NH2 | H Aldehydeammonia
O || R – C – H + HNH2 Aldehyde. O || CH3 – C – H + HNH2 Acetaldehyde. (iii)
D
R — C = NH | H Aldimine
– H2O
OH | D CH3 — C — NH2 – H2O | H Acetaldehydeammonia
CH3 — C = NH | H Acetaldimine
Acetone (2 molecules) react with NH3 to form an addition product diacetone amine, a complex ketonic amine.
CH3 CO CH2 H CH3 CH3
B.
+
CH3
H NH2
CH3
C=O
C
CH2COCH3 NH2
+ H2O
Diacetone amine.
Cannizzaro’s reaction: Aldehydes having no a-hydrogen atom when treated with concentrated alkali undergo
disproportionation i.e. self oxidation-reduction reaction where one molecule of the aldehyde is oxidised to carboxylic acid and other molecule is reduced to alcohol. This reaction is termed as Cannizzaro reaction. e.g. (i) HCHO upon reaction with conc. NaOH undergoes Cannizzaro’s reaction. Out of two molecules of HCHO, one molecule is oxidised to HCOOH and other is reduced to CH3OH. Conc.NaOH
2HCHO Formaldehyde (ii)
— CHO Benzaldehyde
HCOONa
+
Sodium formate Conc.NaOH
CH3OH Methyl alcohol
— COONa + Sodium benzoate
— CH2OH Benzyl alcohol
564
+2 CHEMISTRY (VOL. - II)
Mechanism: (i)
Nucleophilic attack of OH– at the electron deficient carbonyl carbon atom of one molecule of aldehyde. – O | H—C—H | OH
O d|| – H — C — H + OH d+
(ii)
This step involves the hydride transfer from the above intermdiate to the second molecule of aldeyde. – O | H—C—H | OH
(iii)
O || H–C–H
Hydride transfer
– O O | || H – C –OH + H — C — H | H
Since HCOO– ion is more stable than HCOOH, there will be H+ ion shift and as a result formate ion is formed along with methyl alcohol. O || – H — C — O — H + CH3O
H+ transfer
O || – H – C – O + CH3OH Na+ HCOONa
C.
Reaction with Phosphorus pentachloride: PCl5 reacts with simple carbonyl compounds to form dichlorides. C = O + PCl5 (carbonyl comp) H | H — C = O + PCl5 Formaldehyde O || CH3 – C – CH3 + PCl5 (Acetone)
C
Cl
+ POCl3
Cl (dichloroalkane)
H | H — C — Cl + POCl3 | Cl Dichloromethane Cl | CH3 — C — CH3 + POCl3 | Cl (2, 2 -Dichloropropane)
ALDEHYDES AND KETONES
D.
565
Reduction: Aldehydes and ketones yield a variety of products on reduction under different conditions.
(i)
Catalytic reduction: When reduced with hydrogen in presence of a catalyst like Raney Ni, Pt or Pd, aldehydes give primary alcohols whereas ketones give secondary alcohols. R – CHO
H2 /Ni
10alcohol
Aldehyde H – CHO Formaldehyde
H2 /Ni
CH3 OH Methyl alcohol
CH3 – CHO Acetaldehyde
H2 /Ni
CH3 CH2 OH Ethyl alcohol
R
H2 /Ni
R
R|
C=O
R|
CH3 CH3
C=O
CH — OH
20 -alcohol
(Ketone) H2 /Ni
CH3 CH3
Acetone (ii)
RCH2 OH
CH — OH
Isopropyl alcohol
Reduction by chemical reducing agents : Alcohols can also be obtained by reducing cacrbonyl compounds using lithium aluminium hydride (LiAIH 4) or sodium borohydride (NaBH4) or sodium and ethyl alcohol. CH3CHO
LiAlH4
Ethanal
CH3CH2OH Ethanol
NaBH4 is a mild and selective reducing agent, since it reduces only the carbonyl group without affecting the carbon-carbon double bonds in the molecule. Thus, NaBH4 CH3 – CH = CH – CHO CH3 – CH = CH – CH2OH But -2-en-1-al But, CH3 – CH = CH – CHO
H2 /Ni
But -2-en-1-ol CH3 – CH2 – CH2 – CHO Butanal
566
(iii)
+2 CHEMISTRY (VOL. - II)
Reduction to hydrocarbons : The converted to
(I)
C = O group of aldehydes and ketones may be
CH2 by the following methods.
Clemmensen reduction : This involves the reduction of aldehydes or ketone by zinc amalgam in the presence of Conc. HCl H | R — C = O +4H Aldehyde
Zn/Hg &
R – CH3 + H2O alkane
Conc.HCl
R R/
R C = O +4H
CH2 R/ alkane
Zn/Hg & Conc.HCl
Ketone
Acetaldehyde forms ethane whereas acetone forms propane on Clemmensen reduction. C6H5COCH3 + 4[H]
Zn/Hg, Conc.HCl
Acetophenone (II)
C6H5 – CH2CH3 + H2O Ethyl benzene
Wolff - Kishner reduction : Carbonyl group of aldehydes and ketones can be reduced to – CH2– group on reaction with hydrazine followed by heating with sodium or potassium hydroxide in the presence of ethylene glycol.
C=O
NH2NH2
C = NNH2
– H2O
KOH/Ethylene glycol D
CH2 + N2 (g)
For example : CH3CHO Ethanal
(i) NH2NH2 (ii) KOH, Glycol
C6H5COCH3 Acetophenone
D
CH3CH3 Ethane
(i) NH2NH2 (ii) KOH, Glycol D
C6H5CH2CH3 Ethylbenzene
ALDEHYDES AND KETONES
E.
F.
567
Reaction with primary amines : Aldehydes and ketones react with primary amines in the presence of acid as catalyst to form Schiff’s bases. +
CH3CHO + Acetaldehyde
H2NCH3 Methylamine
H D
C6H5CHO + Benzaldehyde
H2NC6H5 Aniline
H D
CH3CH = NCH3 + H2O Schiff’s base
+
C6H5CH = NC6H5 + H2O Benzal aniline (A Schiff’s base)
Electrophilic Substitution Reactions in aromatic aldehyde and ketones :
Aromatic aldehydes and ketones undergo electrophilic substitution reactions such as halogenation, nitration and sulphonation at the meta position. Since carbony group is an elctron withdrawing group, it is a a ring deactivating and m-directing group, it does not undergo FriedelCraft’s reaction. I.
Nitration : Benzaldehyde when treated with concentrated nitric acid in the presence of concentrated sulphuric acid undergoes nitration to form m-nitrobenzaldehyde. The reaction proceeds through electrophilic substitution reaction. The carbonyl group present in benzaldehyde behaves as a ring deactivator and meta-director, towards electrophilic substitution reaction.
+ HNO3 (Conc.) Benzaldehyde
H2SO4
CHO |
|
CHO |
NO2
+ H2O
(m-nitro benzaldehyde)
With acetophenone, m-nitroacetophenone is formed.
+ HNO3 (Conc.) Acetophenone
Conc..H2SO4
COCH3 |
|
COCH3 |
NO2
(m - nitro acetophenone)
Mechanism: Step -1 :
+ H2O
+ Formation of electrophile (NO2) + NO2 + 2HSO4– + H3O+ HNO3 + 2H2SO4
568
+2 CHEMISTRY (VOL. - II)
Step - 2 :
The electrophilic attack and formation of carbocation : CHO | +
CHO | + + NO2 Step : 3
NO2
Loss of proton and formation of m-nitro benzaldehyde. CHO |
CHO | +
HSO 4-
H
+ H2SO4 NO2
NO2
II.
H
m-nitro benzaldehyde
Halogenation : Like nitration, benzaldehyde and acetophenone undergo halogenation. For example, benzaldehyde reacts with chlorine in the presence of FeCl 3 or Anhy. AlCl3 at room temperature, to form m-chlorobenzaldehyde. Here, Cl + ion acts as electrophile. CHO |
CHO | Anh. AlCl3
+ Cl2
+ HCl Cl
Benzaldehyde
m-chloro benzaldehyde
Similarly COCH3 |
COCH3 | Anh. AlBr3
+ Br2
+ HBr Br
Acetophenone III.
m-Bromoacetophenone
Sulphonation : Both aryl aldehydes and ketones undergo sulphonation giving mderivative. COCH3
COCH3 + Conc. H2SO4 Acetophenone
+ H2O SO3H m-acetylbenzenesulphonic acid
ALDEHYDES AND KETONES
569
CHO
CHO + Conc. H2SO4
+ H2O
Benzaldehyde
SO3H m-Benzaldehyde sulphonic acid
18.8
USES OF ALDEHYDES AND KETONES : Aldehydes and ketones are used as solvents starting materials and reagents for the
synthesis of other products in the chemical industry. Some of the important cases of aldehydes and ketones are : (i)
40% aqueous solution of formaldehyde called formalin is used for the preservations of biological or anatomical specimens. It is also used as disinfectant and as a germicide.
(ii)
Formaldehyde is used in the manufacture of bakelite, resins and other polymers.
(iii)
Acetaldehyde is used in the preparation of a number of organic compounds such as acetic acid, ethyl acetate, n-butyl alcohol etc., and in the silvering of mirrors.
(iv)
Benzaldehyde is mostly used as a flavouring agent in perfume industry in addition to its use for the preparation of a number of organic compounds like cinnamic acid, benzoyl chloride etc.
(v)
Acetone is widely known for its use as one of the constituents of the liquid nail polish.
(vi)
Compounds like vanillin, acetophenone, camphor etc. are well known for their use as flavouring and adouring agent.
18.9
SOLVED PROBLEMS :
1.
A ketone ‘A’ which undergoes haloform reaction gives compound ‘B’ on heating with sulphuric acid gives compound’C’ which forms monoozonide ‘D’. ‘D’ on hydrolysis in presence of zinc dust gives only acetaldehyde. Identify A, B, C and write down the reactions involved.
Solution : We know that the ketone which undergoes haloform reaction must contain a O || CH3—C— group. So ‘A’ must be methyl ketone. Again since ‘C’ forms a monoozonide, it must contain a C=C bond. ‘D’ on hydrolysis in presence of zinc dust gives acetaldehyde. Hence the structure of ‘C’ will be CH3—CH = CH— CH3 (C)
570
+2 CHEMISTRY (VOL. - II)
‘B’ on dehydration gives ‘C’. So the structure of ‘B’ will be OH | CH3—CH — CH2— CH3 (B) Again ‘B’ is obtained upon reduction of ‘A’. So ‘A’ has the structure O || CH3—C— CH2— CH3 (A) Butanone The reactions involved may be represented as follows. O OH || | [2H] CH3—C— CH2— CH3 CH3—CH— CH2— CH3 (A)
CH3—CH = CH — CH3
(B) +O3
(C) Zn/H2O 2.
2 CH3 CHO
H2SO4 – H2O
O
CH
3—
CH
CH —
O
O
CH 3
(D)
Compound ‘A’ with molecular formula C 5H12O on oxidation forms compound ‘B’ with molecular formula C 5H10O. The compound ‘B’ gives iodoform test but does not reduce ammoniacal AgNO 3. ‘B’ on reduction with amalgamated zinc and HCl gives compound ‘C’ with molecular formula C 5H12. Identify A, B and C. Write down the chemical reactions involved.
Solution: ‘B’ gives iodoform test but does not reduce ammoniacal AgNO 3 (Tollens’ reagent) Thus O || ‘B’ is not an aldehyde, it must be a ketone having CH3 — C — group i.e. it must be a methyl ketone. So ‘B’ may be O || CH3 – C – CH2 – CH2 – CH3 (B) Pentan -2-one
ALDEHYDES AND KETONES
571
Again ‘B’ is obtained from ‘A’ on oxidation. So ‘A’ is a secondary alcohol. OH | CH3 – CH – CH2 – CH2 – CH3 (A) Pentan -2-ol ‘B’ upon reduction with Zn/Hg and HCl gives ‘C’ having molecular formula C 5H12. The > C=O group is converted to > CH2 group. So it must be a hydrocarbon.
CH3 — CH2 — CH2 — CH2— CH3 n — Pentane (C)
The reactions may be represented as OH | CH3 — CH — CH2 — CH2— CH3
[O]
O || CH3 — C — CH2 — CH2— CH3
(A)
(B) Zn/Hg & HCl CH3 — CH2 — CH2— CH2— CH3 (C)
3.
Compound (A) C5H10O forms phenyl hydrazone and gives negative Tollens’ and iodoform tests. Compound ‘A’ on reduction gives n-pentane. Give the structure of (A). Explain the reaction.
Solution: Compound (A) is a carbonyl compound (Aldehyde or a ketone) since it forms phenyl hydrazone. Again it gives negative Tollens’ and iodoform test, so it is a ketone but not methyl ketone. ‘A’ on reduction gives n-pentane, thus ‘A’ is a straight chain compound. Keeping in view all the above, the formula of ‘A’ is CH3 CH2 — C.CH2— CH3 || O Pentan — 3 — one
(A)
The reaction my be represented as CH3 CH2 — C — CH2— CH3 + H2N. NHC6H5 || O (A)
572
+2 CHEMISTRY (VOL. - II)
CH3CH2
C = N.NHC6H5 + H2O
CH3CH2
Phenyl hydrazone Tollens’ reagent
C2H5
C=O
C2H5
Iodoform test
(A)
No reaction
No reaction
[H] CH3CH2CH2CH2CH3 n — Pentane 4.
An unknown compound of C, H and O contains 69.77% C and 11.63% H and has a molecular weight of 86. It does not reduce Fehling’s solution but forms a bisulphite addition compound and gives a positive iodoform test. What are the possible structures?
Solution : (i)
For finding empirical formula Element
\
%
Relative number of atoms
Simplest ratio
C
69.77
5.76
5
H
11.63
11.63
10
O
19.20
1.2
1
Empirical formula = C5H10O Empirical formula = 5 ´ 12 + 1 ´ 10 + 16 = 86 Also molecular weight = 86
\
Molecular formula = C5H10O The compound forms a bisulphite compound indicates that it is an aldehyde or a ketone. It does not reduce Fehling’s solution, which indicates it is a ketone. It gives +ve iodoform test which indicates that it is a methyl ketone i.e. it must contain (CH 3– CO –) group. Keeping all the above facts in view, the structure of ketone may be
ALDEHYDES AND KETONES
573
CH3 CH2 CH2 C — CH3 || O
or
Pentan — 2 — one 5.
CH3 CH3
CH — C — CH3 || O
3 — Methyl butan — 2 — one
An organic compound ‘A’ C5H8O adds Br2 to form C5H8Br2O. It does not react with Tollens’ reagent but reacts with phenyl hydrazine. ‘A’ on ozonolysis gives acetaldehyde and C3H4O2 which loses CO forming CH3CHO. Identify ‘A’ and explain the reaction. Solution Compound ‘A’ adds one molecule of Br 2, therefore contains a double bond. It forms phenyl hydrazone, hence a carbonyl group is present. Further it does not react with Tollens’ reagent, hence the carbonyl group is ketone and not aldehyde. ‘A’ on ozonolysis gives CH3CHO and C3H4O2. Keeping in view the above facts the structure of ‘A’ is O O O || || Zn/H2O O3 CH — C—CH3 CH3CH=CH—C—CH3 CH3 — CH O O || CH3 CHO + CH3 —C—CHO Pent-3-en-2-one
O
CH3CHO + CO 6.
Compound A (C6H12O2) on reduction with LiAlH4 yielded two compounds B and C. The compound B on oxidation gave D, which on treatment with aqueous alkali and subsequent heating furnished E. The latter on catalytic hydrogenation gave ‘C’. The compound D was oxidised further of give F which was a monobasic acid (mol. weight 60). Deduce the structures of A, B, C, D and E.
Solution
O || F is monobasic acid. The structure is R — C — OH. Mol. wt. is 60 (given) So, MR + 12 + 2 ´ 16 + 1 = 60
Þ
MR + 45 = 60
574
+2 CHEMISTRY (VOL. - II)
Þ
MR = 15, So the alkyl group is CH3. ‘F’ is obtained by oxidation of D. So ‘D’ must be CH 3 CHO. ‘D’ on treatment with aq. alkali undergoes aldol condensation which on heating loses a molecule of water giving a, bunsaturated aldehyde. So ‘E’ must be crotonaldehyde (CH 3CH = CH CHO) CH3CHO+H.CH2 CHO
OH | CH3 —CH—CH2—CHO
aq.NaOH
(D)
Aldol
H2 /Catalyst
CH3—CH2—CH2— CH2 OH
CH3 —CH = CH — CHO
(C) (E) ‘D’ is obtained on oxidation of ‘B’. Since ‘D’ is CH3 CHO, ‘B’ must be CH3CH2OH. Basing on structure of B ( CH3CH2OH) and that of C(CH3—CH2—CH2—CH2—OH), the structure of ‘A’ may in deduced as follows. O || LiAlH4 CH3CH2CH2— C — O CH2 CH3
CH3CH2CH2CH2OH+CH3CH2OH
Ethyl butanoate
(C)
(B) (O)
O || CH3 — C — OH (F) 7.
(O)
CH3CHO (D)
An organic compound (A) contains 40% carbon and 6.7% hydrogen. Its vapour density is 15. On reacting with a concentrated solution of KOH, it gives two compounds (B) and (C). When ‘B’ is oxidised, original compound ‘A’ is obtained. When ‘C’ is treated with conc. HCl it gives a compound D which reduces Fehling’s solution and Tollens’ reagent and also gives effervescence with NaHCO 3 solution. Write the structures of A, B, C and D and explain the reactions.
Solution For empirical formula Element
%
Relative number of atoms
Simplest ratio
C H
40 6.7
3.33 6.70
1 2
O
53.3
3.33
1
ALDEHYDES AND KETONES
\
575
Empirical formula of (A) = CH2O Empirical formula weight = 30 Molecular weight = 2 ´ vapour density = 2 ´ 15 = 30
Hence molecular formula = Empirical formula = CH 2O The reaction path is: CH2O
Conc.KOH
(B)+(C) Conc.HCl
oxdn. A
D Amm. AgNO3
Fehling Na soln. HC O
Cu2O (red ppt)
3
Effervescence
Ag miror
(A) is HCHO, Upon treatment with conc. KOH it undergoes Cannizzaro’s reaction forming CH3OH and HCOOK. Since ‘B’ upon oxidation gives ‘A’, ‘B’ must be CH3OH and ‘C’ is HCOOK. Upon treatment with conc. HCl ‘C’ forms HCOOH HCOOK + HCl
HCOOH + KCl (D)
‘D’ (HCOOH) reduces Fehling soln, Tollens’ reagent and gives effervescence with NaHCO3 HCOOH +2CuO Feh.soln HCOOH + Ag2O Tollens’ reagent HCOOH + NaHCO3 8.
H2O + CO2 +Cu2O red H2O + CO2 + 2Ag Silver mirror HCOONa + H2O + CO2
An alkene ‘A’ on ozonolysis yields acetone and an aldehyde. The aldehyde is easily oxidised to an acid (B). When ‘B’ is treated with Br2 in presence of Phosphorus it yields compound ‘C’ which on hydrolysis gives a hydroxy acid (D). This acid can also be obtained from acetone by the reaction with HCN followed by hydrolysis. Identify A, B, C and D.
576
+2 CHEMISTRY (VOL. - II)
Solution Alkene (A)
O3
CH3
Zn/H2O
CH3
C = O + Aldehyde Oxdn
HOH D C Hydroxy acid CH3
CH3
HCN
CO
CH3
CH3
Br2/P
B
OH
HOH/H+
C
CH3 CH3
CN
OH
C
COOH (D)
Thus ‘C’ is bromo acid, ‘B’ must be an acid formed by the oxidation of the aldehyde. Keeping all these things in view, the reactions may be represented as CH3 CH3
C
H
CH3
[O]
CH3
CHO
C
H
CH3
Br2/P
CH3
COOH
C
COOH HOH
(B) CH3 CH3 So, the structure of alkene (A) must be
CH3 CH3
C = CH.CH (A)
18.9
IMPORTANT CONVERSIONS :
1.
Formaldehyde
O || H—C—H
Br
CH3
C
OH COOH
(D)
CH3
Acetaldehyde.
[H] LiAlH4
CH3OH
PCl5
CH3Cl
KCN
CH3CN
ALDEHYDES AND KETONES
577
CH3CN [H] Na/Alc [O]
CH3CHO
2.
(i)
K2Cr2O7 / H2SO4
Acetaldehyde
CH3CHO
[O]
HNO2
CH3CH2OH
CH3CH2NH2
Formaldehyde NaOH
CH3COOH
CH3COONa
NaOH CaO,
CH4
Cl2 hv [O]
HCHO
(ii)
CH3CHO
[H] LiAlH4
CH3OH
ConcH2SO4
CH3CH2OH
0
170
aq.KOH
CH3Cl
CH2 = CH2
O3 O Zn/H2O
2 HCHO
H2C O
O
Ethylene ozonide
Formaldehyde
3.
CH2
Acetaldehyde to Acetone :
CH3CHO
[O]
CH3COOH
Ca(OH)2
(CH3COO)2Ca
CH3COCH3
578
4.
+2 CHEMISTRY (VOL. - II)
Acetone to Acetaldehyde : Cl | CH3 — C — CH3 | Cl
PCl5
CH3COCH3
AIc KOH
CH3 — C º CH H2
Lindlar’s catalyst
CH3— CH = CH2 Propene O3
O CH3CHO
CH3CH
Zn/H2O
O
Acetaldehyde
5.
Formaldehyde HCHO
6.
O
Propylene ozonide Acetone
as above
Acetone
CH3CHO
as above
CH3COCH3
Formaldehyde as above
CH3COCH3 7.
CH2
Ethylene
CH3CHO as above
HCHO
Acetaldehyde HBr
CH2 = CH2
aqKOH
CH3 — CH2Br
CH2CH2OH
K2Cr2O7 / H2SO4 [O] CH3CHO 8.
Acetaldehyde CH3CHO
9.
Ethylene
[H] Na/Alc
Acetaldehyde CH3CHO
CH3—CH2OH
ConcH2SO4
Ethylalcohol [H]
Na/Alc
CH3CH2OH
0
170
CH2 = CH2
ALDEHYDES AND KETONES
10.
579
Acetone CH3 CH3
Isopropyl alcohol
C=O
LiAlH4
CH3
Reduction
CH3
CHOH
Isopropyl alcohol 11.
Acetaldehyde CH3CHO
Acetylene
as above
CH2 = CH2 + Br2
CH2 — CH2 | | Br Br Alc.KOH CHºCH Acetylene
12.
CH º CH + H2O
13.
Acetaldehyde
Acetylene
1% HgSO4 42% H2SO4 600C
Benzene
OH | CH2 = CH
Tautomerise
Acetaldehyde
vinyl alcohol
n-propyl benzene COCH2CH3
+ CH3CH2COCl
CH3 — CHO
Anhydrous AlCl3
Zn/Hg.Conc.HCL
CH2CH2CH3
580
+2 CHEMISTRY (VOL. - II)
CHAPTER (18) AT A GLANGE
1.
O || The general formula of an aldehyde is R — CHO (R—C—H), whereas that of a ketone is O || RCOR/ (R—C—R/). They are known as carbonyl compounds represented by the general formula CnH2nO.
2.
Aldehydes show chain, functional and position (only for aromatic aldehydes) isomerism whereas ketones show chain isomerism, functional isomerism and metamerism.
3.
The carbon atom of the carbonyl group is sp2 hybridised and the structure is planar.
4.
General Methods of Preparation (a)
Oxidation of alcohols: Chemical (K2Cr2O7 & H2SO4) and catalytic (reduced Cu) R — CH2OH
[O]
R — CHO
(10alcohol) R R
CH — OH
[O]
R
C = O (Also by Oppenauer oxidation)
R/
/
(20alcohol) (b)
From acids: (i)
Dry distillation of Ca-salt of the fatty acids — (HCOO)2Ca
HCHO + CaCO3
(RCOO)2 Ca + (HCOO)2 Ca (RCOO)2 Ca
2R — CHO + 2CaCO3
R–CO–R + CaCO3
(ii) Vapours of fatty acids passed over hot MnO or ThO2 at 560K. HCOOH
MnO 560K
HCHO
RCOOH + HCOOH RCOOH
RCHO RCOR
ALDEHYDES AND KETONES
(c)
581
From acid chlorides:Rosenmund reduction - Acid chlorides can be converted to aldehydes with hydrogen in boiling xylene using Pt or Pd as catalyst supported by BaSO4, poisoned with sulphur. R — COCl
H2
dry ether
1
2R —CO—Cl + Cd R2 (d)
42% H2SO4
CH3CHO
1%HgSO4, 340K
(Acetylene) R — C º CH
RCOCH3
(Propyne)
(Methyl ketone)
From alkenes: Ozonolysis R — CH = CH2 + O3 R R
/
(f)
2R— CO — R1 + CdCl2
From alkynes : CH º CH
(e)
R — CHO
Pd/BaSO4, S
C = CHR// + O3
Ozonide
Ozonide
Zndust
RCHO + HCHO
H2O Zndust R H2O
R/
C = O + R¢¢CHO
From gem-dihalides: Hydrolysis by aq.KOH X
R — CH
X
+ KOH (aq)
R —C—R +KOH(aq)
R—CH
R—C—R/
/
X X
OH OH
OH OH H2O
R —CHO
R—C—R/ || O
Physical Properties 5.
HCHO is a gas at room temperature.
6.
Lower members of carbonyl compounds are soluble is water, but the solubility decreases with increase in molecular mass.
7.
Boiling points of aldehydes and ketones are less than these of the corresponding alcohols.
582
+2 CHEMISTRY (VOL. - II)
Reactions: 8.
Carbonyl compounds undergo nucleophilic addition reactions. The order of reactivity follows the following sequence. HCHO > CH3CHO > CH3 COCH3 > CH3CH2 COCH2CH3 >
O || C6H5CHO > C6H5 – C – CH3
(i)
HCN :-
C = O + HCN
C
(ii)
NaHSO3 :-
C = O+ NaHSO3
C
OH CN
(Cyanohydrin)
OH
(bisulphite compound) SO3Na
(employed for separation and purification of carbonyl compounds) * Note that only methyl ketones undergo bisulphite addition reaction. 9.
Nucleophilic addition followed by elimination of water. Reactent
Product H+
H2N—OH
C=O
H+
H2N—NH2
H+
H2N—NH—C6H5 H2N—NH—CO—NH2 10.
H+
C = N—OH (oxime) C = N—NH2 (hydrazone) C = N—NH—C6H5 (phenyl hydrazone) C = N—NH—CO—NH2 (semicarbazone)
Aldol condensation :- Aldehydes & ketones having a-hydrogen atom undergo aldol condensation in presence of dilute alkali.
2CH3CHO
dil. alkali
Acetaldehyde
CH3COCH3 Acetone
OH | CH3—CH—CH2CHO acetaldol
Ba(OH)2
heat –H2O
OH | CH3—C—CH2COCH3 – H2O | CH3 Diacetone alcohol
CH3—CH=CH—CHO Crotonaldehyde
CH3 CH3
C = CHCOCH3 Mesityl oxide
ALDEHYDES AND KETONES
11.
583
OH | dil.NaOH D C6H5 COCH3 C6H5 — C — CH2 COC6H5 C6H5— C = CHCO C6H5 | | –H2O OH CH CH3 3 | Haloform reaction: Acetaldehyde, methyl ketones and CH3—CH—R (R = H, alkyl or aryl) respond to haloform test. O || CH3—C—R + X2 + NaOH
12.
O || CX3—C—R
– OH
CHX3 + R—COONa
Oxidation Reaction : Aldehydes can reduce both Fehling’s solution (Excepting aromatic aldehydes) and Tollens’ reagent whereas ketones can not. Aldehydes form a red precipitate of Cu2O when warmed with Fehling’s solution and form a silver mirror when warmed with Tollens’ reagent. Fehling’s solution is a mixture of Fehling A(CuSO 4) solution and Fehling B (NaOH and sodium potassium tartrate) Sodium potassium tartrate is called Rochelle salt. Tollens’ reagent is ammoniacal silver nitrate solution
13.
Reduction of Aldehydes and Ketones:- Aldehydes and ketones can be catalytically reduced using Raney nickel or Pt.or by LiAIH 4 to alcohols. NaBH4 can also be used for this purpose. NaBH4 is a mild and selective reducing agent, since it reduces only the carbonyl group and leaves the carbon carbon double bond intact. Clemmensen and Wolff - Kishner reduction converts a > C = O group to —CH 2 group.
14.
PCl5 reacts with simple carbonyl compounds forming dichlorides. C=O
+ PCl5
(Carbonyl comp) 15.
C
Cl
Cl (Dichloride)
Formaldehyde reacts with ammonia forming hexamethylenetetramine or urotropine which is used as urinary antiseptic. And for the synthesis of RDX.
16.
Cannizzaro’s Reaction : Aldehydes having no a-hydrogen atom when treated with conc. alkali undergo Cannizzaro’s reaction. Two molecules of an aldehyde participate in the reaction out of which one molecule is oxidised to a carboxylic acid and the other molecule is reduced to an alcohol. 2R — CHO Aldehyde
Conc.NaOH
RCOONa + RCH2OH Sod. salt of alcohol the carboxylic acid
584
17.
+2 CHEMISTRY (VOL. - II)
Methods of Preparation of Benzaldehyde CH3
1. (a)
COCl
2.
CH3
4.
CHO
CHO
Methods of preparation of Acetophenone. CH3 — C º N
(b)
(i) C6H5 MgBr, dry ether (ii) H3O+
+ CH3 — COCl
O || C6H5 — C — CH3 COCH3
Anh. AlCl3
+ HCl
Reactions of aryl aldehyde and ketone. HCN
Nitration Benzaldehyde
CH
OH
CH
CN
C6H5CH = N – NHC6H5
CHO
20.
CHCl2 (i) Hydrolysis (ii) -HO 2
CO+HCl,(AlCl3 - CuCl) Gattermann - Koch Reaction
(a)
19.
CHO
Stephen Reduction Cl2
5. 18.
CHO
Rosenmund Reduction
CºN
3.
CHO
CrO3 Oxidation
CH3
(b)
CHO
Etard's Reaction
m-nitrobenzaldehyde
Halogenations m-Chlorobenzaldehyde (Cl2), AlCl 3 (Anh.)
Acetophenone undergoes the same reaction. CHO CH2OH Conc. NaOH 2 Cannizzaro Reaction
COONa +
OH COOH
ALDEHYDES AND KETONES
585
QUESTIONS (A)
Very short answer type (1 mark each) (i)
What is Tollen’s reagent ?
(ii)
What is Fehling’s solution ?
(iii)
What is formed if calcium acetate is heated at 400 0C ?
(iv)
To what state does ethanal reduce Cu(II) ?
(v)
What is the IUPAC name of CH3 —CH—C—CH2 —CH3 || | CH O 3
(vi)
The Cannizzaro’s reaction is given by ————
(vii)
Write the IUPAC name of the compound CHO | CH3 —CH—CH2 —CH3
(viii) Write the structural formula of the compound 1-bromo-3-pentanone. (ix)
Aldehydes react with hydroxylamine to produce ———— (i) oxime (ii) alcohol (iii) hydrazone (iv) phenyl hydrazone.
(x)
Name the functional isomer of propanone.
(xi)
The precipitate formed when an aldehyde reacts with Fehling’s solution is——— O
(xii)
IUPAC name of compound
is –––––––– CH3
(xiii) Write the structure of the compound 4-Oxopentanal. (B)
Short answer type (2 marks each) (i)
How would you get oxime from acetaldehyde ? Give equation.
(ii)
What is Fehling’s solution ? How does it react with acetaldehyde ?
(iii)
Write one reaction to distinguish between acetaldehyde and acetone. Give equation
(iv)
What happens when acetaldehyde reacts with HCN ? Give equation.
(v)
What happens when ethylene reacts with ozone ? Explain with equation.
586
+2 CHEMISTRY (VOL. - II)
(vi)
What happens when formaldehyde reacts with HCN. ? Give equation.
(vii)
Show the iodoform reaction in case of acetaldehyde with equation.
(viii) What is Tollens’ reagent ? What happens when it reacts with benzaldehyde ? (ix)
What is Rosenmund’s reduction ? Explain it giving an example.
(x)
What happens when acetone is treated with HCN ? Give equation.
(xi)
What happens when benzaldehyde is treated with Fehling’s solution and why ? Explain.
(xii)
Complete the following equation CH3 CH3
C = O + NaHSO3
(xiii) What happens when HCHO reacts with NH3 ? (xiv) How will you distinguish acetaldehyde and benzaldehyde. (xv)
Which compound can undergo Cannizzaro’s reaction - Give one example
(xvi) What happens when acetaldehyde reacts with iodine in dilute NaOH solution ? Give equation. (xvii) Write with equation what happens when acetone reacts with iodine in presence of sodium hydroxide. (xviii) Identify X & Y in the series. CH3CHO
Acidic KMnO4
X
SOCl2
Y
(xix) How will you prepare phenyl hydrazone of acetone ? Indicate with equation. (xx)
Write with equation, how urotropine is formed
(xxi) Name the compound in IUPAC system having molecular formula C 3H6O, which can reduce Tollens’ reagent. What is Tollens’ reagent ? Δ 2 / KOH (xxii) Calcium acetate ¾¾® [A] ¾I¾ ¾¾ ¾® [B] yellow ppt. Identify the products A & B. Name the reaction in Step-II.
(xiii) Explain why methanal in a gas, where as ethanal is liquid (xiv) Explain why ethanal is having lower boiling point than ethanol.
ALDEHYDES AND KETONES
(C)
587
Short answer type questions (3 marks each) (i)
Distinguish between acetaldehyde and acetone.
(ii)
Distinguish between formaldehyde and acetaldehyde.
(iii)
Distinction between benzaldehyde and acetophenone
(iv)
Name one primary alcohol which gives iodoform test and explain why?
(v)
Distinguish between ethyl alcohol and acetone.
(vi)
Distinguish between acetaldehyde and benzaldehyde.
(vii)
Explain why aldehydes are more reactive than ketones.
(viii) Although both
C=C
and
C = O contain double bond, they differ in their
characteristics. Explain. (ix)
Arrange the following compounds in increasing order of their boiling points with plausible explation. CH3CH2OH, CH3OCH3, CH3CHO, CH3CH3
(x)
Explain Clemmensen’s reduction giving one example.
(xi)
Convert benzene to benzaldehyde.
(xii)
Convert Benzoyl chloride to benzaldehyde.
(xiii) Convert benzaldehyde to m-chlorobenzaldehyde. (D)
Long answer type (7 marks each) (i)
Write notes on
(a) Aldol condensation (b) Cannizzaro reaction
(ii)
How acetone is prepared ? What happens when acetone reacts with
(iii)
(a) HCN (b) NH2OH (c) C6H5NH.NH2 Write notes on (a) Iodoform reaction (b) Reduction of aldehydes and ketones.
(iv)
Describe two general methods of preparation of ketones. State with equations how acetone reacts with (a)
Phenyl hydrazine
(b)
Hydrocyanic acid
(c)
Iodine in presence of sodium hydroxide.
588
+2 CHEMISTRY (VOL. - II)
(v)
How can benzaldehyde be synthesized by the following reaction ? (a)
Etard reaction
(b)
Rosenmund reduction
Explain why it is less reactive than CH3CHO towards nucleophilic addition reaction. (vi)
How is acetaldehyde prepared ? With equation give its reaction with (a) HCN (b) C6H5NH.NH2 (c) NaOH solution
(vii)
(a) What is Haloform reaction ? How will you prepare Iodoform from acetone ? Can you prepare Iodoform from Formaldehyde ? (b) How will you prepare acetophenone by Friedel - Craft’s reaction ?
(viii) How will you differentiate
(ix)
(x)
(a)
Ethyl alcohol and acetone
(b)
Acetaldehyde and acetone
What happens when (give equation) (i)
C2H5OH is passed over hot reduced Cu at 3000C
(ii)
HCHO is treated with NH3
(iii)
Acetone is treated with I2 and NaOH solution
(iv)
CH3CHO reacts with NH2OH
(v)
Calcium acetate is heated
How is acetone prepared ? Give two methods. What happens when acetone reacts with:
(xi)
(a)
Phenyl hydrazine
(b)
Hydroxylamine
(c)
Sodium bisulphite
Explain with equation how aldehydes and ketones can be prepared from calcium salts of organic acids. What happens when acetaldehyde reacts with the following substances ?
(xii)
(a) NH2OH (b) Phenyl hydrazine (c) Hydrogen in presence of nickel catalyst. How acetaldehyde can be prepared from the following ? (a)
Acetyl chloride
(b)
Acetylene
How does it react with (a) Fehling’s solution, (b) Tollens’ reagent ? Give equations. What is formalin ? How is it used ?
ALDEHYDES AND KETONES
589
(xiii) How aldehydes and ketones are prepared from dry distillation of calcium salts ? How aldehydes react with (a) NaHSO3 (b) Phenyl hydrazine ? Describe one test to distinguish between aldehydes and ketones. (xiv) Give two methods for the preparation of ketones. How does acetone react with
(xv)
(i)
Iodine in NaOH solution and
(ii)
Sodium bisulphite ?
(iii)
Hydroxyl amine ?
How are ketones prepared from (a) salts of carboxylic acids (b) acid chlorides? How does acetone react with (i) ammonia (ii) barium hydroxide (iii) hydroxylamine?Explain with equation.
(xvi) Discuss briefly the structure and nature of the carbonyl group. In what respects the C = C and C = O bonds resemble and differ from each other.
ADDITIONAL QUESTIONS I.
Very short answer type (1mark each) (i)
What type of ketones undergo iodoform test ?
(ii)
With which reagent can you distinguish between pentan-2-one and pentan-3-one?
(iii)
What happens when calcium acetate is heated?
(iv)
Which alkene on ozonolysis gives acetone?
(v)
What is Rosenmund’s reduction ?
(vi)
Why HCHO does not undergo aldol condensation ?
(vii)
Which aldehyde can undergo iodoform test ?
(viii) What is the reagent used for conversion of acid chloride to aldehyde ?
II.
(ix)
Name one reagent used to distinguish between acetaldehyde & acetone.
(x)
What is IUPAC name of Formaldehyde ?
Short answer type Questions (2 marks each) (i)
Explain why HCHO is more reactive than acetaldehyde.
(ii)
Explain why diethyl ketone does not exhibit haloform reaction.
(iii)
In what respect, the C = C and C = O bonds resemble each other?
(iv)
Arrange HCHO, CH3CHO and CH3COCH3 in order of increasing reactivity towards HCN.
590
+2 CHEMISTRY (VOL. - II)
(v)
Explain why the boiling points of aldehydes and ketones are lower than those of alcohol.
(vi)
Explain why the lower members of carbonyl compounds are soluble in water ?
(vii)
How is acetaldehyde prepared from ethanol ?
(viii) What is the function of BaSO4 in Rosenmund’s reduction. (ix)
Explain why hydrazones of aldehydes & ketones are not prepared in highly acidic medium.
(x)
Write the structure of Urotropine.
(xi)
How will you detect carbonyl group in an organic compound ?
(xii)
Explain why trichloroacetaldehyde does not undergo aldol condesation.
(xiii) Explain why benzaldehyde does not undergo aldol condensation whereas acetaldehyde does. (xiv) Explain why the bond energy of >C = O is higher than that of >C = C< (xv)
Oximes are more acidic than hydroxylamine. Why ?
(xvi) How does acetaldehyde react with Tollens’ reagent? Give equation. (xvii) How will you distinguish an aldehyde from a ketone? Give one chemical test. (xviii) Which type of aldehydes show Cannizzaro reaction? III.
Long Answer type.
1.
(a)
Give four points of difference between acetaldehyde and acetone
(b)
How can you bring out the conversion of ethanol to 2-hydroxy -3-butanoic acid.
2.
3.
Explain the action of the following reagents on acetophenone (a)
Hydrogen cyanide
(b)
I2, aq.NaOH
(c)
Conc.HNO3 in the presence of Conc.H2SO4
(d)
Phenylhydrazine.
How can you bring out the following conversions ? (a)
Acetaldehyde to acetoxime
(b)
Methanal to ethanal
(c)
Ethanal to 2-hydroxybutanoic acid
(d)
Acetylene to acetone (4 steps)
ALDEHYDES AND KETONES
4.
591
What happens when (a)
Formaldehyde is treated with NH3?
(b)
Acetylene is pased through dilute H2SO4 in presence of HgSO4.
(c)
Formaldehyde is treated with conc H2SO4.
(d)
Acetylchloride is treated with H2 is presence of Pd/BaSO4.
(e)
A mixture of calcium formate and calcium acetate undergoes dry distillation.
5.
Discuss the reaction used to distinguish between aldehydes and ketones.
(F)
Multiple Choice type Questions with answers.
1.
The reagent with which both acetaldehyde and acetone react easily is
2.
3.
4.
5.
6.
7.
(a)
Fehling solution
(b)
Grignard reagent
(c)
Schiff’s reagent
(c)
Tollens’ reagent
Which of the following compounds is oxidised to prepare methyl ethyl ketone ? (a)
2-Propanol
(b)
1-Butanol
(c)
2-Butanol
(d)
Formaldehyde
Cannizzaro reaction is not given by (a)
Trimethyl acetaldehyde
(b)
Acetaldehyde
(c)
Benzaldehyde
(d)
Formaldehyde
The formation of cyanohydrin from a ketone is an example of (a)
Electrophilic addition
(b)
Nucleophilic addition
(c)
Nucleophilic substitution
(d)
Electrophilic substitution
When acetaldehyde is heated with Fehling’s solution, it gives a precipitates of (a)
Cu
(b)
CuO
(c)
Cu2O
(d)
Cu + CuO + Cu2O
Methyl ketones are usually characterised through (a)
Tollens’ reagent
(b)
Iodoform test
(c)
The Schiff’s test
(d)
Benedict’s reagent
Tollens’ reagent is (a)
Ammoniacal cuprous chloride
(b)
Ammoniacal cuprous oxide
(c)
Ammoniacal silver bromide
(d)
Ammoniacal silver nitrate
592
8.
9.
10.
11.
12.
13.
14.
15.
+2 CHEMISTRY (VOL. - II)
Aldehydes can be distinguished from ketones by using (a)
Schiff’s reagent
(b)
Conc. H2SO4
(c)
anhydrous ZnCl2
(d)
resorcinol
Ethyl alcohol on oxidation with K2Cr2O7 gives (a)
acetic acid
(b)
acetaldyhyde
(c)
formic acid
(d)
formaldehyde
Aldehydes can be oxidised by (a)
Tollens’ reagent
(b)
Fehling’s solution
(c)
Benedict’s solution
(d)
all
Isopropyl alcohol on oxidation forms (a)
Acetone
(b)
Ether
(c)
Formalin
(d)
Hydrobenzamide
Formaldehyde reacts with NH3 to give (a)
Hexamethylenetetramine
(b)
Formaldehyde ammonia
(c)
Formalin
(d)
Hydrobenzamide
At room temperature formaldehyde is (a)
gas
(b)
liquid
(c)
solid
(d)
all
Which of the following can reduce Tollens’ reagent ? (a)
2-hydroxy propane
(b)
Acetophenone
(c)
Both
(d)
None
RCH = O + H2
RCH2OH
For this reaction the catalyst is
16.
(a)
Ni only
(b)
Pd only
(c)
Pt only
(d)
any of the above
Which of the following can reduce Tollens’ reagent ? (a)
CH3 — CH = CH2
(b)
CH3COCH3
(c)
CH3CHO
(d)
CH3 — O — CH3
ALDEHYDES AND KETONES
17.
18.
19.
20.
Calcium acetate when dry distilled gives (a)
Formaldehyde
(b)
Acetaldehyde
(c)
Acetone
(d)
Acetic anhydride
Formaldehyde when treated with conc. KOH gives (a)
CH3CHO
(b)
C2H4
(c)
CH3OH, HCOOK
(d)
CH3OH, CH3CHO
Which compound undergoes iodoform reaction (a)
HCHO
(b)
CH3CHO
(c)
CH3OH
(d)
CH3COOH
Identify ‘Z’ in the series CH = CH2
21.
593
HBr
X
Hydrolysis
Y
NaOH I2excess
(a)
C2H5I
(b)
C2H5OH
(c)
CHI3
(d)
CH3CHO
Z
Formaldehyde when treated with conc. KOH gives methanol and potassium formate. The reaction is known as
22.
23.
24.
25.
(a)
Perkin’s reaction
(b)
Claisen’s reaction
(c)
Cannizzaro’s reaction
(d)
Knoevenagel’s reaction
Formalin is an aqueous solution of (a)
Formic acid
(b)
Formaldehyde
(c)
Fluorescein
(d)
Furfuraldehyde
Hexamethylene tetramine is used as (a)
analgesic
(b)
antipyretic
(c)
Urinary antiseptic
(d)
all
Aldol condensation will not occur in (a)
HCHO
(b)
CH3CH2CHO
(c)
CH3COCH3
(d)
CH3CHO
Propyne on hydrolysis in presence of HCl and HgSO4 gives (a)
Acetaldehyde
(b)
Acetone
(c)
Formaldehyde
(d)
None
594
+2 CHEMISTRY (VOL. - II)
26.
When anhydrous acetaldehyde is brought in contact with a drop of conc. H2SO4 at ordinary temperature, the following compound is formed.
27.
28.
29.
30.
31.
32.
33.
(a)
Aldol
(b)
Paraldehyde
(c)
metaldehyde
(d)
acetal
The reaction between formaldehyde and caustic soda to produce methyl alcohol is known as (a)
Cannizzaro reaction
(b)
Kolbe’s reaction
(c)
Rosenmund’s reduction
(d)
Friedel Craft reaction
Rosaniline hydrochloride decolourised by SO2 is called (a)
Benedict’s solution
(b)
Tollen’s regent
(c)
Schiffs reagent
(d)
Magenta solution
Acetaldehyde reacts with phosphorus pentahalide to produce (a)
acetal
(b)
ethyledene chloride
(c)
alcohol
(d)
chloral
Formaldehyde is a raw malerial for the manufacture of (a)
nylon
(b)
terylene
(c)
bakelite
(d)
rayon
Which of the following reagents helps to distingish an aldehyde from a ketone ? (a)
Phenyl hydrazine
(b)
NaHSO3
(c)
neutral FeCI3
(d)
Ammoniacal AgNO3 solution
Which of the following does not show aldol condensation ? (a)
CH3CHO
(b)
CH3CH2CHO
(c)
(CH3)3 C — CHO
(d)
CH3 (CH2)2 CHO
In which of the following reactions, both the oxidised and reduced forms of the same compound are obtained ? (a)
34.
Aldol condensation
(b)
(c) Reimer Tiemann reaction (d) In the Cannizzaro reaction given below
Cannizzaro reaction Kolbe’s synthesis
2 ArCHO OH Ar CH2OH + ArCOO– , the slowest step is (a) The attack of OH– at the carbonyl group (b) The transfer of hydride ion to the carbonyl group (c) The abstraction of a proton from the carboxylic acid (d) The deprotonation of Ar – CH2OH.
ALDEHYDES AND KETONES
35.
36.
37.
38.
39.
595
Which of the following statement is not true about benzaldehyde ? (a) Undergoes Cannizzaro reaction (b) Forms an addition compound with HCN (c) Undergoes aldol condensation (d) Reacts with phenylhydrazine Cannizzaro reaction is given by : (a) C6H5CHO (b) CH3 CHO (c) C6H5CH2CHO (d) C6H5CO CH3 Cannizzaro reaction is an example of : (a) Only oxidation (b) Only reduction (c) Disproportionation (d) None of these Cyanohydrin of benzaldehyde on hydrolysis will give : (a) an optically active hydroxy acid (b) benzoic acid (c) benzyl alcohol (d) Acetophenone In the given reaction ‘X’ will be : H2
C6H5COCl Pd-BaSO ,S ‘X’ 4 (a)
C6H5 COOH
(b)
C6H5CH2OH
(c)
C6H5CHO
(d)
C6H5CH3
(G)
Fill in the blanks (with answers) (1 mark each)
1.
Methanal + Ammonia
2.
Fehling A consists of an aqueous solution of copper suphate while Fehling ‘B’ consists of an alkaline solution of ————
3.
Ethanol vapour is passed over heated copper and the product is treated with aq NaOH.
————
The final product is ———— 4.
40 % aqueous solution of HCHO is ————
5.
Calcium formate on dry distillation gives ————
6.
Carbon atom of carbonyl group is ———— hybridised.
7.
When acetaldehyde is heated with Fehling solution a ———— precipitate of ———— is formed.
8.
IUPAC name of acetaldehyde is ————
9.
When vapour of isopropyl alcohol is passed over hot reduced copper ———— is formed.
10.
Both aldehydes and ketones react with ———— reagent.
11.
Silver mirror test is a test for ————
12.
Acetaldehyde on treatement with ———— gives aldol.
13.
Sodium potassium tartrate is ———— salt.
596
+2 CHEMISTRY (VOL. - II)
14.
Magenta is ————
15.
Conversion of hexan-1-ol to hexanal is carried out by the reagent ––––––––––.
16.
The reagent –––––––––– converts but-2-ene to ethanal.
H.
Complete the following with answers (2 marks each)
1.
(A)
Hg2+ dil.H2SO4
Pd / BaSO4, S
2.
CH3COCl + H2
3.
CH3CHO
4.
2HCHO + NaOH
5.
HCHO + NH3
6.
CH3 CH3
(CH3CO)2O
CH3CHO
dilNaOH
(B)
(A)
(A) _______ + ________ (A)
CHOH
Cu
(A)
0
300
8.
dry (CH3COO)2 Ca + (HCOO)2 Ca (A) O distillation || C2H5 —C—CH3 + I2 + (A) (B) + CHI3 + H2O + NaI
9.
CH3CH2OH
10.
HCHO + HCN
11.
CH3CHO + I2
12.
CH3COCH3 + H2 NOH
7.
(O)
Cl2
A
B
(A) NaOH
(A) (A)
O CH3 — C +
13.
CH3 — C
O
O CH3 (i) CrO 2Cl2 +
(ii) H 2O
14. NO2
A
Anhydrous AlCl 3 CS2
A
CHCl3
ALDEHYDES AND KETONES
15.
(C6H5 – CH2) Cd + 2CH3COCl CO, HCl Anhydrous AlCl 3
16.
O 17. 18.
597
A
A
CH2 — OH
dry, HCl
A
CH3 — C — CH 3 + CH2 — OH CH3CHO + [Ag (NH3)2] OH
A
CHO LiAlH 4
19.
A
CHO D + Conc.NaOH ¾¾® A+B
20.
21.
CH3CO CH2 COO C2H5
22.
C6H5COCH3
(i) NaBH 4 (ii) H +
Zn/Hg, Conc.HCl
A A
ANSWERS Multiple choice type question 1.b
6.b
11.a
16.c
21.c
26.b
31.d
36.a
2.c
7.d
12.a
17.c
22.b
27.a
32.c
37.c
3.b
8.a
13.a
18.c
23.c
28.c
33.b
38.a
4.b
9.b
14.d
19.b
24.a
29.b
34.b
39.c
5.c
10.d
15.d
20.c
25.b
30.c
35.c
598
+2 CHEMISTRY (VOL. - II)
Fill in the blanks 1. (CH2)6N4
2. Sodium potassium tartrate
3.Aldol
4.Formalin
5. HCHO
6. sp2
7.red, Cu2O
8.Ethanal
9. CH3COCH3
10.Grignard
11.Aldehyde 12.dil.NaOH
13.Rochelle
14.Rosaniline hydrochloride
15.PCC
16. O3/Zn–H2O
Complete the following 1.
CH º CH
CH3CH(OCOCH3)2
(A)
(B)
2.
(A) is CH3CHO
3.
(A) is CH3CH(OH) CH2CHO
4.
(A) = CH3OH, (B) = HCOONa
5.
(A) = (CH2)6N4
6.
(A) =
8.
(A) = NaOH (B) = C2H5COONa
10.
(A) = H—CH
12.
(A) =
CH3 CH3
CH3 CH3
C=O
CN OH
7.
(A) = CH3CHO
9.
(A) = CH3CHO (B)CCl3CHO
11.
(A) CHl3
CHO
COCH3
C = N — OH
13.
14. NO2
CHO 15.
C6H5CH2COCH3
16.
17.
CH3
CH2OH 18.
CH3COONH4
CH3
C
O — CH2 O — CH2
19. –
CH2OH
+
COO Na
20.
A=
B=
21.
CH3CHOH CH2 COOC2H5
22.
qqq
C6H5CH2CH3
CARBOXYLIC ACIDS
599
CHAPTER - 19
CARBOXYLIC ACIDS 19.1
GENERAL INTRODUCTION :
O || Organic compounds having carboxyl group (—C—OH) are called carboxylic acids. The carboxyl group is a combination of a carbonyl group ( C=O) and a hydroxyl group (—OH) and hence the name carboxyl (carb from carbonyl and oxyl from hydroxyl). Carboxylic acids are classified as aliphatic or aromatic depending upon whether —COOH group is linked to aliphatic alkyl chain or aryl group respectively. For example, formic acid (HCOOH), acetic acid (CH3COOH), propionic acid (CH3CH2COOH) etc. are aliphatic acids and COOH | Benzoic acid,
; 2-Amino benzoic acid,
COOH | —NH2 ; 3-Methyl benzoic acid,
COOH |
—
etc. are aromatic acids. CH3
These acids are further classified as mono, di, tri carboxylic acids according to the number of carboxyl groups present in their molecules. COOH COOH For example : | | Monocarboxylic acids: HCOOH
CH3COOH
Formic acid
Acetic acid
Dicarboxylic acid : COOH | COOH Oxalic acid
COOH
CH2COOH | CH2COOH
Malonic acid
Succinic acid
H2C
COOH
Benzoic acid
NO2 3-Nitrobenzoic acid
600
+2 CHEMISTRY (VOL. - II)
COOH —COOH —COOH Phthalic acid (1, 2 - Benzenedicarboxylic acid)
COOH Isophthalic acid (1, 3 - Benzenedicarboxylic acid)
COOH
COOH Terephthalic acid (1, 4 - Benzenedicarboxylic acid) The aliphatic monocarboxylic acids are called fatty acids since many higher acids e.g. stearic (C17H35 COOH), palmitic (C15H31COOH), oleic (C17H33COOH) etc. are formed by the hydrolysis of natural fats. The general formula for aliphatic saturated monocarboxylic acids is CnH2n+1 COOH or CnH2nO or RCOOH where R stands for hydrogen or alkyl group. Lower members of monocarboxylic acids occur in various plants and animals. 19.2
NOMENCLATURE :
(1)
Common System : According to common system, lower members are named after the source of the individual acids. For example, formic acid was so named because it was first obtained by the distillation of red ants (Latin: formica meaning ant). Similarly, acetic acid was so named because it was obtained from vinegar (Latin: acetum meaning vinegar). In this system, the positions of the substituents are indicated by Greek letters (a, b, g, d...). Carbon atom adjacent to the carboxyl group is assigned a and the next carbon atom b and so on.
(2)
IUPAC system: (a) Aliphatic carboxylic acids according to IUPAC system are named as alkanoic acids by replacing the terminal ‘e’ of the corresponding alkane by ‘-oic acid’. The positions of the substituents are indicated by Arabic numerals according to the following rules: (i)
The longest carbon chain containing –COOH group is selected.
(ii)
The carbon atom of the carboxyl group is assigned the number 1.
CARBOXYLIC ACIDS
601
Table 19.1: Common and IUPAC names of some aliphatic carboxylic acids Monocarboxylic acid
Common name
IUPAC name
HCOOH
Formic acid
Methanoic acid
CH3COOH
Acetic acid
Ethanoic acid
CH3CH2COOH
Propionic acid
Propanoic acid
CH3CH2CH2COOH
n-Butyric acid
Butanoic acid
Isobutryric acid CH3CHCOOH | CH3
or a-Methyl propionic acid
2-Methylpropanoic acid
b-Methyl butyric acid
3-Methyl butanoic acid
CH3(CH2)3COOH
n-Valeric acid
Pentanoic acid
CH3(CH2)4COOH
n-Caproic acid
Hexanoic acid
HOOC – COOH
Oxalic acid
Ethanedioic acid
HOOC– (CH2)2 – COOH
Succinic acid
Butanedioic acid
CH3 CH3
CHCH2COOH
HOOC – CH2 – CH– CH2 – COOH | COOH (b)
—
Propane-1, 2, 3 –
tricarboxylic acid
Aromatic acids are named by common names or as derivatives of parent benzoic acid,which are also accepted by IUPAC. Examples of this family are : COOH
COOH OH
Benzenecarboxylic acid (Benzoic acid) COOH
2-Hydroxybenzoic acid (Salicylic acid) COOH NH2
COOH Benzene - 1, 3 - dicarboxylic acid (Isophthalic acid)
2-Aminobenzoic acid (Anthranilic acid)
602
+2 CHEMISTRY (VOL. - II)
But the compounds where carboxyl group is present on the side chain of aromatic ring system as in phenyl acetic acid and cinnamic acid are called side-chain aromatic acids showing similar behaviour as aliphatic carboxylic acids. —CH2—COOH
—CH = CH —COOH
Phenylacetic acid
Cinnamic acid
19.3
ISOMERISM :
A.
Fatty acids exhibit chain and functional isomerism.
1.
Chain isomerism: This type of isomerism occurs due to the difference in the nature of carbon chain. For example: CH3 | CH3CH2CH2COOH and CH3—CH—COOH Butanoic acid CH3CH2CH2CH2COOH
2-Methyl propanoic acid and
Pentanoic acid 2.
3-Methyl butanoic acid
Functional isomerism: This type of isomerism occurs due to the presence of different functional groups. Carboxylic acids are the functional isomers of esters. For example, CH3COOH
and
Acetic acid CH3CH2CH2COOH CH3CH2COOH
HCOOCH3 Methyl formate
and
Butanoic acid
CH3CH2COOCH3 Methyl propanoate
and
Propanoic acid B.
CH3 | CH3—CH—CH2—COOH
CH3COOCH3 and HCOOC2H5 Methyl ethanoate
Ethyl methanoate
Position Isomerism : This type of isomerism is exhibited by aromatic acids. For example : Position isomers of benzene dicarboxylic acids are : COOH
COOH HOOC
COOH 1, 2 Benzene dicarboxylic acid (Phthalic acid)
COOH
COOH 1, 3 Benzene dicarboxylic acid (Isophthalic acid)
1, 4 Benzene dicarboxylic acid (Terephthalic acid)
CARBOXYLIC ACIDS
603
19.4
METHODS OF PREPARATION
1.
By the oxidation of Alcohols, Aldehydes or Ketones: Primary alcohols and aldehydes are readily oxidised to carboxylic acids by mild oxidizing agents like potassium permanganate solution in acid or alkaline medium or acidified potassium dichromate solution or chromium trioxide (CrO3) in acidic medium. [O]
R CH2OH Primary alcohol
KMnO4/H+
Aldehyde
[O]
CH3CH2OH
C6H5CH2OH Benzyl alcohol
[O]
Acetaldehyde
[O]
CH3COOH Acetic acid
[O]
C6H5 – CHO
K2Cr2O7/H+
RCOOH Acid
CH3CHO
K2Cr2O7/H+
Ethyl alcohol
[O]
RCHO
Benzaldehyde
C6H5–COOH Benzoic acid
Secondary alcohols on oxidation form ketones which on drastic oxidation gives acids, having less number of carbon atoms. R /
R
R
[O]
CHOH
K2Cr2O7/H+
/
R
Secondary alcohol CH3 CH3
CHOH
NaOI
ketone [O]
CH3
K2Cr2O7/H+
CH3
Isopropylalcohol 2.
[O]
C=O
RCOOH + R/COOH Acids
C=O
[O] NaOI
Acetone
CH3COOH Acetic acid
By the hydrolysis of Alkyl cyanides: Alkyl cyanides are hydrolysed to carboxylic acids in acidic or alkaline medium. R—C º N
+ 2H2O
H+ or OH–
Alkyl cyanide CH3C º N + 2H2O Methyl cyanide
RCOOH + NH3 Acid
H+ or OH–
CH3COOH + NH3 Acetic acid
604
+2 CHEMISTRY (VOL. - II)
H+ or OH–
C6H5 – CN + 2H2O
C6H5 – COOH + NH3
Benzonitrile 3.
Benzolic acid
By the hydrolysis of Esters, Acyl chlorides, Acid amides or Acid anhydrides: Esters, acyl chlorides or acid anhydrides are easily hydrolysed to carboxylic acids by mineral acid or alkali. RCOOR/
+
HOH
H+
Ester CH3COOCH5
+
HOH
Ethyl acetate COOC2H5
H+
HOH
Acid chloride +
HOH
Acetyl chloride RCONH2
+
HOH
Acid amide CH3CONH2
+
HOH
Acetamide CONH2 +
H2O
CH3COOH
+
C2H5OH
+
C2H5OH
+
HCl
+
HCl
+
NH3
CH3—COOH +
NH3
Acetic acid
H+ H+ H+ H+
RCOOH Acid CH3COOH Acetic acid R—COOH Acid
Acetic acid COOH
H+
+ Benzoic acid
RCO RCO Acid anhydride CH3CO
Alcohol
COOH
Benzamide
O
O
CH3CO Acetic anhydride
R/OH
Benzoic acid +
CH3COCl
+
Acid
H3O+
Ethyl benzoate RCOCl
RCOOH
+
HOH
H+
2RCOOH Acid
+
HOH
H+
2CH3COOH Acetic acid
NH3
CARBOXYLIC ACIDS
605
(C6H5CO)2O
+
H2O
2C6H5COOH
Benzoicanhydride 4.
Benzoic acid
By the hydrolysis of Trichloroalkanes: Trichloroalkanes in which all the three chlorine atoms are linked to the same carbon atom are hydrolysed by alkali to carboxylic acids. Cl RC Cl Cl
NaOH
Cl HC Cl Cl
NaOH
OH RC OH OH
HC
–H2O
R—C
O – O H+ Sodium salt of acid O NaOH H—C ONa H+ Sodium salt of Formic acid
O
NaOH
OH Acid
OH –H O 2 OH OH
H—C
O OH
Formic acid 5.
R—C
By Carbonation of Grignard reagent: Alkyl or aryl magnesium halide (Grignard reagent) reacts with solid carbon dioxide in ether to form addition compound which on acidification
gives carboxylic acid. O || H+/H2O dry ether R—C—O MgX RCOOH + Mg O=C=O + RMgX
OH
Carbon dioxide
X
Acid
O = C = O +CH3Mg Br
dry ether
O || H+/H2O CH3 —C—O Mg Br
CH3COOH + Mg Acetic acid
C6H5MgBr Phenylmagnesium bromide
+
O=C=O
dry ether
Br
O || — C — MgBr C6H5 H+/H 2O C6H5 – COOH + Mg Benzoic acid
6.
OH
Br OH
Preparation of aromatic (Benzoic acid) by oxidation of alkylbenzenes (arenes) : Benzoic acid is prepared by the oxidation of alkyl benzenes (irrespective of the length of the side chain) in the presence of strong oxidizing agent like acidified potassium dichromate or alkaline / acidic potassium permanganate or dil. HNO 3. Primary and secondary alkyl groups are oxidized, whereas tertiary group is not at all affected.
606
+2 CHEMISTRY (VOL. - II)
CH3
KMnO4 – KOH
COOK
COOH
H3O+
D Tolune
Benzoic acid CH2 CH2 CH3
COOK
COOH
KMnO4 – KOH
H 3O
+
D 1-Propylbenzene
Benzoic acid
19.5
GENERAL PROPERTIES :
A.
PHYSICAL PROPERTIES:
1.
Physical state: The first three members are colourless liquids having pungent odour. The next six members are oily liquids having faint unpleasant odour. The higher members are colourless and odourless waxy solids.
2.
Solubility: The first four members are soluble in water due to intermolecular hydrogen bonding with water molecules. H—O......H—O—C = O......H—O......H—O—C = O ...... | | | | H R R H The solubility of carboxylic acids gradually decreases with rise in size of the alkyl group. This is due to increase in nonpolar part of the molecule.
3.
Melting point: For the first ten members of the homologous series, the melting points of carboxylic acids having even number of carbon atoms are higher than the nearest lower and next higher acids with odd number of carbon atoms. CH2 CH3
COOH
CH2
(Acid with even number of carbon atoms)
CH2 CH3
CH2
CH2
COOH
(Acid with odd number of carbon atoms)
CARBOXYLIC ACIDS
607
The acids with odd number of carbon atoms have carboxyl group and terminal methyl group on the same side of zig-zag carbon chain and such unsymmetrical molecules fit poorly in the crystal lattice and hence have weaker intermolecular forces. The acids with even number of carbon atoms have carboxyl group and terminal methyl group on opposite side of zig-zag carbon chain and hence they fit better in the crystal lattice resulting in stronger intermolecular forces. This explains why the acids having even number of carbon atoms have melting points higher than the acids with odd number of carbon atoms. Melting points of some Alkanoic acids. Acid
Methanoic
M.P(K) 4.
Ethanoic Propanoic
281
289.6
251
Butanoic
Pentanoic
Hexanoic
265
238.5
270
Boiling points : The carboxylic acids have quite high boiling points due to the presence of intermolecular hydrogen bonding as shown below. R—C
O......H—O O—H......O
C—R
(Dimeric structure) The carboxylic acids have relatively higher boiling points than the corresponding alcohols. This is because, (i)
O—H bond in carboxylic acid is more polar than O—H bond in alcohol due to electron withdrawing effect of carbonyl group on O—H bond.
(ii)
The molecules of carboxylic acids are held together by two hydrogen bonds forming cyclic dimers. Table 19.2 : Boiling points of some Carboxylic acids and Alcohols. Acids/Alcohols
B.P(K)
HCOOH CH3OH
373 338
CH3COOH C2H5OH
391 351
C2H5COOH C3H7OH(Propanol)
424 370
C3H7COOH
437
C4H9OH (Butan-l-ol)
391
608
B.
+2 CHEMISTRY (VOL. - II)
ACIDIC NATURE OF CARBOXYLIC ACIDS Carboxylic acids are relatively weaker acids as compared to mineral acids. The strength
of the acid depends upon the extent of ionisation which mainly depends upon the stability of the carboxylate ion formed. O
R—C (i)
O—H
R—C
O – O
+ H+
A carboxylic acid molecule is a resonance hybrid of the following two resonating
..
O .. .. O—H ..
R—C
.. – O ..
..
structures (I and II) R—C
I
+
O—H .. II
The structure, II indicates that the oxygen atom of the hydroxyl group carries some positive charge. As a result, the electron pair of O—H is drifted towards oxygen atom. This displacement of electron pair causes the formation of a proton and a carboxylate ion. The carboxylate ion which is formed is also a resonance hybrid of the following two resonating structures, III and IV.
..
.. – O ..
..
O .. .. – O ..
R—C
..
O ..
..
R—C
III
IV
Thus we observe that (i) carboxylic acid as well as the carboxylate ion both are stabilised by resonance. (ii)
The contributing structures of carboxylic acid (I & II) are not equivalent and the structure, II involves charge separation. As a result, the structure II contributes less to the stability of the carboxylic acid.
(iii)
The resonating structures, III and IV are equivalent and contribute to the greater stability of the carboxylate ion. Thus, the carboxylate ion being more stable than the undissociated carboxylic acid, the
latter has a tendency to dissociate into proton and the carboxylate ion. Hence, the carboxylic acids show acidic character.
CARBOXYLIC ACIDS
609
The strength of a carboxylic acid is expressed in terms of the dissociation constant (K a)as: RCOOH + H2O
RCOO– + H3O+
[RCOO–][H3O+] Ka = ———————— [RCOOH] Higher the value of Ka, stronger is the acid and vice versa. As the numerical values of K a vary by large magnitude, it is more convenient to express the acidic strength in terms of pKa value, which is defined as pKa = – log Ka Smaller the value of pKa, stronger is the acid and vice versa. The Ka and pKa values of some acids are given below. Table 19.3: Ka and pKa values of some Carboxylic acids
1.
Acid
Ka
pKa
HCOOH
17.7 ´ 10–5
3.75
CH3COOH
1.75 ´ 10–5
4.76
FCH2COOH
260 ´ 10–5
2.59
ClCH2COOH
136 ´ 10–5
2.87
BrCH2COOH
125 ´ 10–5
2.90
I CH2COOH
67 ´ 10–5
3.16
Cl2CHCOOH
5530 ´ 10–5
1.26
Cl3CCOOH
23200 ´ 10–5
0.63
Cl3CH2COOH
1.3 ´ 10–5
4.87
C6H5COOH
6.3 ´ 10–5
4.20
Effect of substituents on acid strength of aliphatic acids : The substituents affect appreciably the acid strength of carboxylic acids.
(i)
Effect of electron releasing groups (+I effect) : An electron releasing group attached to carboxyl group tends to release electrons towards the carbon atom to which it is attached and this effect is transmitted throughout the carbon chain. This increases the electron
610
+2 CHEMISTRY (VOL. - II)
density on the oxygen atom of the – O – H group and as a result the release of H + becomes difficult. This explains why formic acid is a stronger than acetic acid. O O H—C CH3 C O—H O—H Release of H becomes easy Release of H becomes due to absence of electron difficult due to presence of releasing group an electron releasing group (– CH3) Acid strength decreases with increase in electron releasing nature (+I effect) of the substituents. The +I effect of alkyl groups increases in the following order: CH3– < C2H5 – < (CH3)2 CH – < (CH3)3 C – Therefore, the acid strength of carboxylic acids decreases in the following order: HCOOH
>
Formic acid (ii)
CH3COOH
>
CH3CH2COOH
Acetic acid
>
Propionic acid
(CH3)2CHCOOH Isobutyric acid.
Effect of electron withdrawing groups (–I effect): An electron withdrawing group attached to carboxyl group tends to withdraw electrons from the carbon atom to which it is attached and their effect is transmitted throughout the chain. As a result, the electron pair forming the O – H bond is displaced more towards oxygen of O – H group and the release of H+ becomes more easier. Therefore, chloroacetic acid is a stronger acid than acetic acid. H | H —C | H
C
O
CI > Br > I. Hence the acid strength of halogen substituted monocarboxylic acids decreases in the order. FCH2COOH > CICH2COOH > Br CH2COOH
(c)
> I CH2COOH.
Effect of position of halogen atoms: The electron withdrawing effect of halogen atom decreases as its distance from the – COOH group increases. The electron withdrawing effect of halogen is maximum when the halogen atom is linked to a - carbon atom of the carboxylic acid. Thus, a- chloropropionic acid is a stronger acid than b- chloropropionic acid.
2.
Comparison of Relative Acid Strength of Monocarboxylic acid and Alcohol Both carboxylic acid and carboxylate ion are stabilised by resonance and the carboxylate
is more stabilised by resonance as compared to carboxylic acid. Hence the carboxylic acids have tendency to release H+ ions with the formation of stable carboxylate ions. However, resonance is not possible for alcohols and also alkoxide ions. Therefore, carboxylic acids are stronger acids than monohydric alcohols.
R—C
O .. .. – O ..
..
R—C
O .. .. O—H ..
O || – – R – C – O + H3O Conjugate base R—C
+
O—H .. Less stable .. – O ..
..
..
O ..
..
R—C
..
R – OH + H2O
.. – O ..
..
O || R – C – O – H + H2O
– + R – O + H3O Not resonance stabilized
612
3.
+2 CHEMISTRY (VOL. - II)
Acidity of Benzoic acid Aromatic acids are slightly stronger acids than aliphatic acids because of greater value of
Ka. For example, benzoic acid is a stronger acid than acetic acid, because benzoate ion is more resonance stabilized than acetate ion. COOH
COO +
H+
Benzoate ion
Benzoic acid COO |
–
– H+ = 6.3 x 10—5
Ka = COOH |
– O || CH3 —C—O
O || CH3 —C—OH
+ H+
Ka = 1.75 x 10—5 Effect of substitutents on acid strength: Acidity of carboxylic acids is determined by both inductive effect and the resonance effect. Electron releasing substitutents (ortho-para directing groups) tend to decrease the strength of the acid. This is due to intensification of -ve charge on the carboxylate ion making it unstable. Thus with –OH group at the ortho or para position the benzene ring is incapable of delocalising the negative charge on the carboxylate anion. .. +O—H +O—H +O—H O—H O—H | || || || ||
..
(i)
– O
p — Hydroxy benzoate
O
– O
O
– O
| C
|| |
O
| C
|| |
– O
| C
|| |
O
| C
|| |
|| |
| C
O
– O
O
– O
||
| C
O—H
O
– O
| C
O—H – O
|| |
– O
O—H
|| |
O
| C
|| |
.. | – O
O—H
|| |
O
| C
||
.. O—H
|| |
| C
613
||
CARBOXYLIC ACIDS
O
o — Hydroxy benzoate On the other hand electron withdrawing groups like —NO2, –CN etc. tend to increase the strength of the acid. This is due to dispersal of the –ve charge on carboxylate ion making it more stable. In otherwords due to delocalisation of –ve charge on the carboxylate anion it gets stabilised.
Thus, p-nitrobenzoic acid (Ka=40 x10—5)is a stronger acid than benzoic acid. (ii)
The + I or –I effect of the substituents is more pronounced if the substituents are present at the para positions rather than at meta positions.
(iii)
Irrespective of the nature of the substituents (+I or -- I), ortho substituted benzoic acids are found to be stronger than benzoic acid. This is called ortho effect. This is probably due to resultant of steric and electronic factor. However, amino benzoic acids are exceptions. All of them are weaker than benzoic acid. Acidity of amino benzoic acids is in the order. m— > o — > p —.
From the above discussion it can be concluded that the effect of substituents on acidity depends not only on the nature of the group but also on its position with respect to the position of the —COOH group. Some important observations (i)
Strength of o —, m — & p — hydroxy benzoic acids relative to benzoic acid follows the order, COOH |
COOH |
COOH |
|
COOH | OH
>
o-Hydroxy benzoic acid
OH m-Hydroxy benzoic acid
>
> | OH Benzoic acid
p-Hydroxy benzoic acid
614
+2 CHEMISTRY (VOL. - II)
(ii)
Strength of o —, m — & p — nitrobenzoic acids relative to benzoic acid follows the order. COOH |
COOH |
COOH |
|
COOH | NO2
>
>
>
| NO2 o-Nitrobenzoic acid (iii)
NO2
p-Nitrobenzoic acid
m-Nitrobenzoic acid
Benzoic acid
Strength of o —, m — & p — toluic acids relative to benzoic acid follows the order, COOH |
COOH |
COOH |
|
COOH | CH3
>
>
> | CH3
CH3 o-Toluic acid (iv)
Benzoic acid
m-Toluic acid
p-Toluic acid
Strength of o —, m — & p — chlorobenzoic acid reative to benzoic acid follows the order, COOH |
COOH |
COOH |
|
COOH | Cl
>
o-Chloro benzoic acid 4.
Cl m-Chloro benzoic acid
>
> | Cl p-Chloro benzoic acid
Benzoic acid
Comparison of acid strength of aliphatic and aromatic acids (unsubstituted) It is a fact that electron releasing groups tend to decrease the acid strength. Let us compare
the acid strength of HCOOH, CH3 COOH & C6H5 COOH. CH3 —group is electron releasing.
When it is attached to —COOH group the release of H+ becomes comparatively difficult. So it is
a weaker acid than HCOOH, where —COOH group is attached to a H atom. In case of benzoic acid the phenyl group (C6H5—) is attached to —COOH group, phenyl group has an over all electron donating effect when it is attached to a —COOH group. But this effect is weaker than +I
CARBOXYLIC ACIDS
615
effect of alkyl group. Thus, benzoic acid is a stronger acid than acetic acid. The acid strength follows the order, HCOOH > C6H5COOH > CH3COOH But when the phenyl group is attached to —COOH group through one or more CH2— groups, it causes a slight electron withdrawing (—I) effect. Therefore phenyl acetic acid (C 6H5CH2 COOH) is slightly stronger than benzoic acid i.e. C6H5CH2COOH > C6H5COOH. 5.
Comparison of acid strength of Carboxylic acids and Phenols Though phenols are acidic due to the resonance stabilized conjugate base phenoxide ion,
they are less acidic than the carboxylic acids. O
O
R — C — OH + H2O
R — C — O + H 3O
OH
O + H 2O
+ H 3O
The higher strength of carboxylic acid in comparision to phenols can be explained on the basis of higher stability of its conjugate base carboxylate ion which is having two equivalent resonating structures because of the presence of two electronegative oxygen atoms. O
O
R—C
R—C O
O
(Two equivalent resonating structures of carboxylate ion making it highly stable) O
O
So on
(Not equivalent though resonance stabilized.)
616
C.
+2 CHEMISTRY (VOL. - II)
Chemical properties The chemical properties of carboxylic acids may be discussed under the following heads: (a)
Reaction due to H atom of – COOH group.
(b)
Reaction due to – OH part of – COOH group.
(c)
Reaction due to – CO – part of – COOH group.
(d)
Reaction due to – COOH group as a whole.
(e)
Reactions due to alkyl group.
(a)
Reaction due to H atom of – COOH group (acidic property)
1.
Reaction with metals: Carboxylic acids react with strong electropositive metals like Na, Zn, Mg etc. to liberate hydrogen gas with the formation of salts.
2.
2RCOOH + 2Na
2 RCOONa
+ H2
Acid
Sodium of acid
2HCOOH + 2Na
2 HCOONa
Formic acid
Sodium formate
2CH3COOH + Zn
(CH3COO)2 Zn
Acetic acid
Zinc acetate.
+ H2 + H2
Reaction with alkali: Carboxylic acids react with alkali (NaOH or KOH) to form salts and water.
3.
RCOOH + OHNa
RCOONa
+ H2O
HCOOH + OHNa
HCOONa
+ H2O
Foomic acid
Sodium formate
CH3COOH + OHNa
CH3COONa + H2O
CH3COOH + NH4OH
CH3COONH4 + H2O
Reactions with carbonates and bicarbonates: Carboxylic acids react with carbonates or bicarbonates to liberate carbondioxide. 2RCOOH + Na2CO3
2RCOONa +CO2 + H2O
2HCOOH + Na2CO3
2HCOONa +CO2 + H2O
CH3COOH+NaHCO3
CH3COONa +CO2 + H2O
Acitic acid
Sodium acetate
CARBOXYLIC ACIDS
617
This reaction is used to distinguish between carboxylic acids and phenols because phenols do not produce effervescence (evolution of CO2) with aqueous solution of NaHCO3 or Na2CO3. (b)
Reaction due to – OH part of – COOH group.
1.
Formation of amides - Carboxylic acids react with ammonia to form ammonium salts which when heated decompose to give amides. RCOOH + NH3
RCOONH4
RCONH2 + H2O
Acid
Ammonium salt
Acid amide
HCOOH + NH3
HCOONH4
HCONH2 + H2O
Formic acid
Ammonium formate
Formamide
CH3COOH + NH3
CH3COONH4
CH3CONH2 + H2O
Acetic acid
Ammonium acetate
Acetamide
COOH
COONH4
CONH2
+ NH3 Benzoic acid 2.
+ H2O Ammonium benzoate
Benzamide
Formation of esters (Esterification) - Carboxylic acids react with alcohols in presence of a dehydrating agent like anhydrous zinc chloride or concentrated sulphuric acid to form esters. RCO OH + H OR/ Acid
Alcohol
HCO OH + H OH5C2 Formic acid
Methyl alcohol
RCOOR/
+ H2O
Ester H+
Ethyl alcohol
CH3CO OH + H OCH3 Acetic acid
H+
H+
HCOOC2H5 + H2O Ethyl formate CH3COOCH3 + H2O Methyl acetate
618
+2 CHEMISTRY (VOL. - II)
O COOH
C–OC2H5 H+
+ HOH5 C2 Benzoic acid
Ethyl alcohol
+ H2 O Ethyl benzoate
This reaction is reversible and is known as esterification. In order to shift the equilibrium in forward direction water formed has to be removed immediately. The dehydrating agent takes away water. Mecanism : The esterification of carboxylic acid with alcohol is one type of nucleophilic acyl substitution. It involves the following steps. Step-1 : Protonation of the carboxyl oxygen. + O — OH
O R — C — OH + H+
R — C — OH
Step-2 : Nucleophilic attack by the alcohol molecule + O — OH R — C — OH
OH +
H — O — R¢
R — C — OH + O — R¢ H
Step-3 : Loss of a molecule of water and a proton resulting with an ester. OH R — C — OH
OH H+ transfer
+ O — R¢
+ R—C—O
H H
O — R¢
H
— H 2O + O —H
O R — C — OR¢ Ester
— H+
R — C — OR¢
CARBOXYLIC ACIDS
3.
619
Formation of acid chlorides : Carboxylic acids react with phosphorus halides (PCl 5, PCl3) and thionyl chloride (SOCl2) to form acid halides. 3RCOOH + PCl3
3RCOCl
+ H3PO3
Acid chloride
4.
RCOOH + PCl5
RCOCl + POCl3
+ HCl
RCOOH + SOCl2
RCOCl + SO2
+ HCl
3CH3COOH + PCl3
3CH3COCl
+ H3PO3
Acetic acid
Acetyl chloride
CH3COOH + PCl5
CH3COCl + POCl3
+ HCl
CH3COOH + SOCl2
CH3COCl + SO2
+ HCl
Formation of anhydrides : Carboxylic acids on heating with a dehydrating agent like phosphorus pentoxide form acid anhydrides. RCO
R COO H P2O5
R CO OH
RCO
O
+ H2O
Acid anhydride CH3CO
CH3COO H
P2O5
CH3CO OH Acetic acid (c)
O
CH3CO
+ H 2O
Acetic anhydride
Reaction due to – CO – part of – COOH group: Reduction of carboxylic acids: These are reduced to primary alcohols when heated with lithium aluminium hydride (LiAlH4) or with hydrogen and copper chromite (CuCr2O4) (in case of higher fatty acids). R-COOH
LiAlH4 or CuCr2O4 +H2
CH3COOH + 4H
LiAlH4
RCH2OH Primary alcohol
CH3CH2OH +H2O Ethyl alcohol
620
+2 CHEMISTRY (VOL. - II)
(d)
Reaction due to – COOH group as a whole:
1.
Reduction :
(a)
Reaction with HI and red phosphorus (Reduction to alkanes) : Carboxylic acids when heated with red phosphorus and hydrogen iodide at 500K are reduced to alkanes. In this reaction – COOH group is reduced to – CH3 group red P at 500K
RCOOH + 6 HI
RCH3 + 2H2O + 3I2 Alkane
red P at 500K
CH3COOH +6HI
CH3CH3 + 2H2O + 3I2
Acetic acid (b)
Ethane
Reduction to alcohols : Carboxylic acids can be reduced to alcohols on reaction with lithium aluminium hydride (LiAlH4) or with diborane (B2H6). RCOOH
(i) LiAlH4/ether (ii) H3O+
CH3COOH Ethanoic acid
RCH2OH Primary alcohol
(i) LiAlH4/ether
CH3CH2OH Ethanol
(ii) H3O+
COOH
CH2OH (i) LiAlH 4/ether (ii) H 3O +
Benzoic acid 2.
Benzyl alchol
Decarboxylation on Reaction with soda lime: Sodium or potassium salt of a carboxylic acid when heated with dry sodalime (NaOH + CaO) forms alkane with one carbon atom less than the parent acid. NaOH takes part in the reaction and calcium oxide helps in fusion of NaOH. R COONa + H ONa
CaO/D
RH Alkane
+
Na2CO3
CARBOXYLIC ACIDS
621
CH3COONa + HONa
CaO/D
Sodium acetate 3.
CH4
+
Na2CO3
Methane
Kolbe’s Electrolytic reaction (Electrolytic decarboxylation) : Aqueous solution of sodium or potassium salt of a fatty acid on electrolysis gives alkane. 2RCOONa
2RCOO– + 2Na+
H2O
2OH– + 2H+
At anode:
2RCOO–
R – R + 2CO2+ 2e
At cathode:
2H+ + 2e
H2
At cathode both Na+ and H+ are present, but H+ ions are discharged preferentially due to their lower discharge potential thant Na + ions. Hence hydrogen gas is evolved at the cathode. For examples : 2CH3COO– + 2Na+
2CH3COONa Sod.acetate
2OH– + 2H+
2H2O At cathode:
2H+ + 2e
H2
At anode:
2CH3COO– – 2e
CH3 – CH3 + 2CO2 Ethane.
4.
Decomposition of Calcium salts of carboxylic acids : Dry distillation of calicum salts of fatty acids results in the formation of carbonyl compounds. Calcium formate when distilled gives formaldehyde. HCOO HCOO
Ca
CaCO3 + HCHO Formaldehyde.
Calcium formate Calcium formate when distilled with calcium acetate gives acetaldehyde. HCOO HCOO
Ca + Ca
OOCCH3 OOCCH3
2CaCO3 + 2CH3CHO
622
+2 CHEMISTRY (VOL. - II)
Calcium acetate when distilled alone forms acetone. Ca
OOCCH3
CH3
OOCCH3
CH3 Acetone.
C = O + CaCO3
(e)
Reactions due to alkyl or aryl group.
1.
Reaction with halogens : Carboxylic acids having an a-hydrogen atom react with chlorine or bromine in the presence of small amount of red phosphorus to form ahalocarboxylic acids. (i) X2 / Red phosphorus
R — CH — COOH X = Cl, Br —
R — CH2—COOH
a-halocaroxylic acid X When more than one mole of Cl2 or Br2 is used, the second and third hydrogen atoms are replaced. Red P
CH3COOH + Cl2
– HCl
Acetic acid
Cl2/Red P ClCH2 COOH Cl2CHCOOH – HCl Monochloroacetic Dichloroacetic acid acid
Cl2 Red P
–HCl
Cl3CCOOH Trichloroacetic acid CH3CH2COOH
Br2 /Red P –HBr
Propionic acid
Br2 /Red P
CH3CHCOOH | –HBr Br a- Bromo propionic acid
Br | CH3—C—COOH | Br a, a-Dibromo propionic acid
This reaction is known as Hell - Volhard - Zelinsky (HVZ) reaction 2.
Reaction with mild oxidising agent such as H2O2 Carboxylic acids when heated with mild oxidising agent like hydrogen peroxide the alkyl group gets oxidised at the b- position. For example, b a CH3CH2 CH2 COOH n- Butyric acid
[O] H2O2
OH |b a CH3CH CH2 COOH b - Hydroxy butyric acid.
Carboxylic acids can also be oxidised at a - carbon by using oxidising agent such as selenium dioxide, SeO2. Keto acids are obtained in this reaction.
CARBOXYLIC ACIDS
623
O || R—C—COOH + H2O + Se. a - keto acid
RCH2COOH + SeO2 3.
Electrophilic substitution reaction In benzoic acid, the carboxylic functional group is directly linked to the benzene ring and behaves as an electron withdrawing group. Due to the electron-withdrawing character (– R effect), the functional group behaves as a ring deactivator and makes the ortho and para- positions electron deficient.
Hence, the carboxylic acid group behaves as a ring deactivator and meta - director towards electrophilic substitution reactions, like, nitration, sulphonation and halogenation. But benzoic acid does not undergo Friedel crafts reation due to deactivations of the benzene ring by electron withdrawing effect of — COOH (a)
Nitration : Benzoic acid reacts with concentrated nitric acid in the presence of concentrated sulphuric acid to form m-nitro benzoic acid. COOH
COOH
conc. H2SO4
+ HNO3 conc.
NO2
Benzoic acid (b)
+ H2O
m- Nitrobenzoic acid
Sulphonation : Benzoic acid on heating with conc. H2SO4 forms meta sulphobenzoic acid. COOH
COOH + conc. H 2SO4
Benzoic acid
D
+ H 2O SO3H m-sulphobenzoic acid
624
(c)
+2 CHEMISTRY (VOL. - II)
Halogenation :
Benzoic acid reacts with a molecule of halogen in ther presence of FeCl 3 or Anhy. AlCl3 to form m-halobenzoic acid. COOH
COOH + X2
Anhy. AlCl3
+ HX X
Benzoic acid
m - Halobenzoic acid (for X = Cl / Br) COOH
COOH + Cl2
Anhy. AlCl3
+ HCl Cl
Benzoic acid
m - Chlorobenzoic acid
+ HBr
(f)
Unique characteristic reactions of Formic acid: Reducing nature - Formic acid acts as a strong reducing agent since it gets readily oxidised. It reduces ammoniacal solution of silver nitrate (Tollens’ regent), Fehling’s solution and mercuric chloride to metallic silver, red cuprous oxide and mercurous chloride respectively. Acidified potassium permanganate solution is also decolourised by formic acid. This is due to reducing nature of formic acid, which distinguishes it from other carboxylic acid. [O]
HCOOH
CO2
+
H2O
2Ag
+
CO2 + H2O
Formic acid HCOOH + Ag2O (From Tollens’ reagent)
CARBOXYLIC ACIDS
625
HCOOH + 2CuO
Cu2O
+ CO2 + H2O
(From Fehling’s solution)Cuprous oxide (red) HCOOH + 2HgCl2
2HgCl
+ CO2 +
2HCl
Mercurous chloride Reaction of formic acid with acidified KMnO4 solution is represented as : [MnO–4 + 8H+ + 5e
Mn2+ + 4H2O] ´ 2
[HCOOH
CO2 + 2H+ + 2e] ´ 5
2MnO–4 + 6H+ + 5HCOOH
2Mn2+ + 8H2O + 5CO2
19.6
TESTS :
1.
Carboxylic acids react with aqueous sodium bicarbonate to liberate CO 2 with effervescence, which turns lime water milky.
2.
Carboxylic acids react with alcohol in presence of a little conc. H 2SO4 to form ester having fruity smell.
19.7
USES OF CARBOXYLIC ACIDS :
(a)
Methanoic acid is used :
(b)
(i)
as an antiseptic
(ii)
in medicine for teatment of gout
(iii)
in coagulating rubber latex
(iv)
in leather industry
(v)
in textile indusry for dyeing and finishing
(vi)
in the preservation of fuits
(vii)
as a rducing agent
Ethanoic acid used : (i)
as a coagulent for latex in rubber industry
(ii)
as vinegar in cooling and in the preparation pickles
(iii)
as a laboratory reagent
(iv)
in curing meat and fish
(v)
in the manufacture of plastics (polyvinyl acetate), rayon (cellulose acetate) and silk.
626
(c)
+2 CHEMISTRY (VOL. - II)
Benzoic acid is used : (i)
in medicine as urinary antiseptic.
(ii)
for making aniline blue in dye industry.
(iii)
as a preservative. Sodium benzoate is the most useful for this purpose specially for preserving food products like tomato sauce, fruit jams and juices.
(iv) (d)
in ferfumes in the form of its esters.
Higher fatty acids are mainly used for the manufacture of soaps and detergents.
SOME IMPORTANT CONVERSIONS 1.
Formic acid to Acetic acid:
HCOOH Formic acid
Ca(OH)2
HCOO
2H HCHO CH3OH – CaCO3 Formaldehyde Na–Hg/H2O Methyl alcohol
Ca
HCOO
PCl5 CH3COOH Acetic acid 2.
Hydrolysis H+
CH3CN Methyl cyanide
KCN
CH3Cl Methyl chloride
Acetic acid to Formic acid : CH3COOH Acetic acid
NaOH
CH3COONa Sod.acetate
(Na OH + CaO)
CH4 Methane
Cl2
CH3Cl hu Methyl cyanide KOH(aq)
HCOOH Formic acid
[O]
HCHO Formaldehyde
[O]
CH3OH K2Cr2O7+H2SO4 Methyl alcohol
CARBOXYLIC ACIDS
3.
627
Propanoic acid to Propanol : CH3CH2COOH + 4H
LiAlH4
Propanoic acid 4.
CH3CH2CH2OH + H2O Propanol.
Acetic acid to Acetone : CH3COOH
Ca(OH)2
CH3COO
Ca
CH3COO
Acetic acid
D
CH3
– CaCO3
CH3
Ca-acetate 5.
Acetone
Acetic acid to tertiary butyl alcohol.
CH3COOH
Ca(OH)2
CH3COO
Ca
D – CaCO3
CH3 CH3
Ca-acetate
CH3 | CH3— C — OH | CH3
CH3 | CH3— C — OMgBr | CH3
H+ H2O
Acetic acid to Acetic anhydride : 2CH3COOH
D P2O5
Acetic acid 7.
C=O CH3MgBr
CH3COO
Acetic acid
6.
C=O
(CH3CO)2O
+ H2O
Acetic anhydride
Acetic acid to Isopropyl alcohol : CH3COOH Acetic acid
Ca(OH)2
CH3COO CH3COO
Ca
D – CaCO3
CH3 CH3
C=O
H | CH3— C — OH | LiAlH4 CH3 +2H
Isopropyl alcohol
628
8.
+2 CHEMISTRY (VOL. - II)
Acetic acid to Acetaldehyde CH3COOH
9.
+2H
CH3COCl
Acetic acid Acetyl chloride Acetic acid to Ethylene CH3COOH Acetic acid
10.
PCl5
LiAlH4
Pd – BaSO4,S
CH3CH2 OH
+4H
Ethyl alcohol
CH3CHO Acetaldehyde
D
CH2= CH2
conc.H2SO4
Ethylene
at1600C
Acetic acid to Acetylene. CH3COOH
LiAIH4
CH3CH2 OH
Acetic acid
Ethyl alcohol
D
CH2= CH2
conc.H2SO4
Ethylene
at443K
Br2 CH2 Br | CH2 Br
KOH(alc) D
CH º CH Acetylene
Ethylene dibromide 11.
Acetic acid to Ethane. CH3COOH
NaOH
CH3COONa
Acetic acid 12.
Sod. acetate
NaOH
CH3COONa
Acetic acid
Ethane Sodalime D
CH4
Sod. acetate
Methane
Formic acid to Formaldehyde. HCOOH
Ca(OH)2
HCOO
Ca
HCOO Ca-formate
Formic acid 14.
CH3 – CH3
Acetic acid to Methane. CH3COOH
13.
Electrolysis
D – CaCO3
HCHO Formaldehyde
Acetic acid to Acetamide. CH3COOH Acetic acid
NH3
CH3COONH4 Ammonium acetate
D –H2O
CH3CONH2 Acetamide
CARBOXYLIC ACIDS
15.
629
Acetic acid to Diethyl ether. LiAlH4
CH3COOH
+4H
Acetic acid 16.
Excess
NaOH
D
CH3CH2COONa
Propanoicacid
Diethyl ether
Sodalime
Sod. propanoate
CH3–CH3 Ethane
Br2 hv
CH3CH2Br
Ethyl bromide
Acetic acid to Methyl amine. CH3COOH
NH3
Acetic acid 18.
C2H5–O–C2H5
at 413K
Propanoic acid to Ethyl bromide. CH3CH2COOH
17.
conc.H2SO4
CH3CH2OH
CH3COONH4 Ammonium acetate
D
Br2+KOH
CH3 CO NH2
–H2O
Acetamide
CH3NH2
Methyl amine
Acetic acid to Methyl alcohol CH3COOH
NH3
Acetic acid
CH3COONH4 Ammonium acetate
D –H2O
CH3 CO NH2 Acetamide Br2 / KOH
NaNO2+HCl
CH3OH
CH3NH2
Methyl alcohol 19.
Acetic acid to Butane. LiAlH4 CH3CH2OH CH3COOH Acetic acid
20.
Methyl amine
P + I2
CH3CH2I
Ethyl alcohol
Na CH3CH2CH2CH3 Ether Butane (dry)
Propionic acid to n – Butyl amine CH3CH2COOH Propionic acid
LiAlH4
CH3 CH2 CH2OH
P/I2
n – propyl alcohol
CH3 CH2 CH2 I n – propyl iodide KCN(alc)
CH3 CH2 CH2 CH2 NH2 n-Butylamine
Zn–Hg/HCl reduction
CH3CH2CH2CN n-Propyl cyanide
630
21.
+2 CHEMISTRY (VOL. - II)
Benzene to m-Nitrobenzoic acid. Br
MgBr
Br2
Mg/dry Ether
Anh.AlBr3 Benene
Bromobenzene
COOH
COOH
(i) CO 2
conc.HNO3
(ii) H 3 O
conc.H 2SO4 Benzoic acid
22.
Phenylmagnesium bromide
NO2 m-Nitrobenzoic acid
Benzene to p-Nitrobenzoic acid. CH 3
CH 3
CH 3Cl/Anh.AlCl 3
conc.HNO 3
F.C. Alkylation
conc.H 2SO 4
Benene
Toluene
NO 2 p-Nitrotoluene
(i) KMnO4 KOH, D
(ii) H3O+ COOH
NO 2
23.
Benzene to phenylacetic acid
p-Nitrobenzoic acid
CARBOXYLIC ACIDS
631
CHAPTER (19) AT A GLANCE MONOCARBOXYLIC ACIDS
General formula : R – COOH. 1.
Oxidation of alcohols, aldehydes and Ketones (by K2Cr2O7 and conc. H2SO4) [O]
R – CH2OH
R – CHO
[O]
R – CH2 – CH – CH2 – R/ | OH [O]
[O]
R – COOH R – CH2 – CO – CH2 – R/
R – CH2 – COOH + R/COOH or R – COOH + R/CH2COOH
2.
Hydrolysis of alkyl cyanides (by acid or alkali) : H+ R–Cº N
3.
OH
–
Hydrolysis of acid chlorides, acid amides, esters and acid anhydrides(by acid or alkali):
R – CO – X
+
OH
> RCOOH
–
– > RCOO
H
+
RCOOH (X = Cl,NH2, OR/, O.OC.R)
Hydrolysis of trichloroalkanes (by alkali) : R–C
5.
–
> RCOO + NH3
H
4.
+
> RCOOH + NH4
CI CI CI
NaOH
R–C
OH OH OH
– H2O
R–C
O – O
H+
O || R – C – OH
Carbonation of Grignard reagent :
O = C = O + R – Mg – X
R–C
O OMgx
H2O/H+
R–C
O OH
(X = Cl, Br, I)
632
6.
+2 CHEMISTRY (VOL. - II)
Decarboxylation of dicarboxylic acids : R – CH
COOH
heat
COOH
– CO2
COOH | COOH 7.
Glycerol 380K
R – CH2 – COOH
HCOOH + CO2 (Lab method for formic acid)
Heating sodium alkoxide and carbon monoxide : dil.H2SO4
heat R – COONa high press.
R – ONa + CO
When R = H, HONa + CO
High temp. HCOONa High press.
R – COOH + Na2SO4 H
+
HCOOH + Na2SO4
REACTIONS: 1.
Reaction due to H atom of – COOH Group :
(i)
Salt formation (with metals like Na, K, Mg, Zn etc. and with alkali) 1 RCOOM + —H (M = monovalent metal) 2 2 II II (RCOO)2M + H2 (M = divalent metal) I
R – COOH + M/ R – COOH + M//
I
2R – COOH + Na2CO3
2RCOONa + CO2 + H2O
R – COOH + NaHCO3
2RCOONa + CO2 + H2O
R – COOH + NaOH
RCOONa + H2O
2.
Reaction due to –OH part of –COOH group:
(i)
With NH3 :
R – COOH + NH3
R – COONH4 Salt D
R – COONH4
R – CO – NH2 + H2O (acid amide)
(ii)
Esterification : R – CO OH + H O – R/ (acid)
(ii)
Conc.H2SO4
(alcohol)
With PCl3, PCl5 or SOCl2 : 3R – COOH + PCl3
R – COOR/ + H2O (ester) 3RCOCI +H3PO3
R – COOH + PCl5
RCOCl + POCl3 + HCl
R – COOH + SOCl2
R – COCl + SO2 + HCl
CARBOXYLIC ACIDS
633
P2O5
(iii)
Dehydration : 2RCOOH
3.
Reaction due to –COOH group:
(i)
Formation of alkanes : R – COOH R – COONa 2RCOONa + 2H2O
(ii)
D
Red P / HI
RCO – O – CO – R + H2O
R – CH3 + I2 + H2O
500K NaOH + CaO
R – H + Na2CO3
Electrolysis
R – R + 2CO2 (Kolbe’s reaction). + 2H2+ NaOH (at the anode)
Action of heat on the salts of acid : (RCOO)2 Ca
D
R – CO – R
+ CaCO3
(Ketone) (RCOO)2 Ca + (HCOO)2Ca
D
RCHO
+ 2CaCO3
(Aldehyde) (HCOO)2Ca
D
HCHO
+ CaCO3
(formaldehyde) RCOONH4
D
RCONH2 + H2O (amide)
RCOONH4
heat P4O10
R – C º N + 2H2O (alkyl cyanide)
RCOOAg + Br2
R – Br + CO2 +AgBr (Hunsdiecker reaction)
RCOOAg + R/ – X
RCOOR/ + AgX (ester)
4.
Reaction due to alkyl group :
(i)
Reaction with halogens : Hell – Volhard – Zelinsky reaction
634
+2 CHEMISTRY (VOL. - II)
Cl a | Cl /Red P 2 Re d P / Cl 2 RCH 2 - COOH ¾ ¾¾¾¾ ® R - CH - COOH ¾ ¾® R - C - COOH | | Cl Cl (a-chloroacid) (a, a-Dichloroacid) (ii)
Reaction with H 2 O 2 :
OH | R - CH 2 - CH 2 COOH ¾ ¾¾ ¾® R - CH - CH 2 - COOH b (b-hydroxy acid) H 2O2
(iii)
Reaction with SeO 2 : SeO
2 R - CH 2 - COOH ¾ ¾¾ ® R - CO - COOH
(Keto acid) 5.
Special reactions of Formic Acid : Reducing properties: HCOOH + Tollens’ reagent ¾ ¾® Silver mirror
(Ag)
HCOOH + Fehling’s solution ¾ ¾® Red precipitate
(Cu2O)
HCOOH + KMnO4 solution ¾ ¾® Decolourisation
(Mn2+)
HCOOH + HgCl2 solution
(HgCl)
¾ ¾® White precipitate
CARBOXYLIC ACIDS
635
QUESTIONS A.
Very short answer type [1 mark each]
1.
Methyl cyanide on hydrolysis yields ——
2.
What happens when acetic acid is treated with sodium carbonate solution.
3.
How will you prepare acetamide from acetic acid ?
4.
Give any two uses of formic acid.
5.
C6H5COOH is formed by carbonation of ——
6.
Write the structural formula of 3 –Hydroxy Butanoic acid.
7.
Write the IUPAC name of the following compound CH2(Br) – CH2 – COOH.
8.
Write the IUPAC name of the following compounds : (i)
CH3 – CH2 – COOC2H5
(ii)
HOOC – CH2– CH2– CH2– COOH
9.
Write one reaction to distinguish between formic acid and acetic acid. Give equation.
10.
Benzoyl chloride is formed by the action of PCl 5 on ––––– .
B.
Short answer type [2 marks each]
1.
What is the reaction of formic acid with Tollen’s reagent and also with sodium ?
2.
Explain why formic acid has two different C – O bond lengths i.e. 1.23A° and 1.36A°, where in sodium formate two C – O bond are same having bond length 1.27A°
3.
Give two tests to distinguish between formic acid and acetic acid.
4.
How can you convert toluene into m-nitrobenzoic acid ?
5.
How formic acid can be converted to acetic acid ?
6.
What happens when acetic acid is treated with NH4OH and the product is then heated ?
7.
Why m-nitorbenzoic acid is a stronger acid than benzoic acid?
8.
Adentify A and B C6H5COOH
conc. HNO3 conc. H2SO4
A
NaOH/CaO D
B
9.
What happens when acetamide is boiled with NaOH solution ?
10.
Adentify A and B C6H5COOH
SOCl2
A
H2 Pd/BaSO4, S
B
636
+2 CHEMISTRY (VOL. - II)
11.
Distinguish between acetic acid and ethanol
12.
Explain why the bond length of C – O in carboxylic acid is slightly larger than that in aldehydes and ketones.
13. 14.
What happens when sodium propionate is heated with sodalime. What happens when acetic acid reacts with ethyl alcohol in presence of conc. H 2SO4 ?
15.
How acetic acid is prepared from methyl cyanide ?
16.
What happens when sodium acetate is heated with sodalime ?
17.
What happens when formic acid reacts with acidified KMnO 4 solution ?
18.
Explain : acetic acid is stronger acid than ethanol.
19.
What is the reaction of acetic acid with lime water ?
20.
How can you get acetone form acetic acid ?
21.
What happens when formic acid reacts with meruric chloride?
22.
Arrange the following in the increasing order of acidic strength. (i)
ClCH2COOH
(ii)
ClCH2CH2COOH
(iii)
FCH2COOH
(iv)
CH3COOH
[Ans: (iv) < (ii) < (i) < (iii)]
23.
How can you distinguish acetic acid from acetone ?
24.
How would you convert acetic acid to methylamine ?
25.
Explain the reaction of formic acid with Fehling’s solution.
26.
Complete the following equation by writing the structures of A,B and C. Cl / P
NaCN
H O
2 2 CH 3 COOH ¾ ¾¾ ¾® A ¾¾¾ ¾® B ¾ ¾¾ ®C
H+
27.
How can a carboxylic acid be converted to an aldehyde in two steps?
C.
Short answer type (3 marks each)
1.
Convert methyl iodide to acetic acid.
2.
What happens when CH3COONa is electrolysed ? Explain with mechanism.
3.
Explain: Benzoic acid is more acidic than aliphatic acids.
CARBOXYLIC ACIDS
4.
PCl5
Identify A, B and C from the following, C2H5OH (A)
5.
637
KCN
(B)
H3O+
(C)
Acetic acid is weaker than formic acid, but chloroacetic acid is stronger than formic acid. Explain.
6.
How will you distinguish between benzoic acid and phenol ?
7.
How do you prepare 2 - hydroxy propionic acid from acetaldehyde ?
8.
Complete the equation CH3COOH
ClCH2COOH
Give the IUPAC name of
Excess of NH3
?
(i)
CH3COCH2COOH.
(ii)
CH3CH = CHCOOH.
(iii)
HO — CH2CH2CH — COOH —
9.
?
CH3
10.
What are the following reagents ? Give one use of each reagent. (a) Tollens’ reagent (b) Schiff’s base (c) Fehling’s solution.
11.
What is esterification? Explain with mechanism taking the example of benzoic acid.
12.
CH3COOH gives HVZ reaction, where as HCOOH does not. Explain.
D.
Long answer type (7 marks each)
1.
Give any two methods of preparation of monocarboxylic acids. How formic acid reacts with Tollens’ reagent ? How can you obtain acetamide from acetic acid ?
2.
Explain with equation how monocarboxylic acid is obtained from alkyl cyanide. How does acetic acid react with the following ? (a)
3.
4.
C2H5OH
(b)
PCl5
(c)
Sodamide
How do you prepare the followings from acetic acid ? (a)
Acetaldehyde
(b)
Methyl amine
(c)
Methyl cyanide
(d)
Acetone
Describe any two general methods of preparation of monocarboxylic acid. How does formic acid react with. (a)
Tollens’ reagent
(b)
alcohol
(c)
Sodium
638
+2 CHEMISTRY (VOL. - II)
5.
Discuss any two methods for preparing benzoic acid. How benzoic acid is converted to benzoyl chloride?
6.
How is acetic acid prepared ? Give any two methods. Give its reaction with Cl 2 in the presence of Red P, P2O5 and ethyl alcohol in presence of sulphuric acid.
7.
Give any one method of preparation of acetic acid. Write its reaction with
8.
9.
(a)
chlorine
(b)
alkali
(a)
Synthesize benzoic acid from toluene
(b)
How does it react with
(c)
phosphorus pentachloride.
(i)
conc. HNO3 and conc.H2SO4
(ii)
Cl2 in the presence of anhydrous AlCl3
Give any two general methods of preparation of monocarboxylic acids. What is the reaction of formic acid with. (a)
ammonia
(b)
Tollens’ reagent ?
10.
Compare the acid characters of formic acid and acetic acid.
11.
How acetic acid is prepared by using Grignard’s reagent ? Give one method to distinguish acetic acid from formic acid. Mention two uses of acetic acid.
12.
Write notes on — (5marks each)
13.
(i)
Halogenation of benzoic acid
(ii)
HVZ reaction
Explain why carboxylic acids behave as acids ? Discuss bfiefly the effect of electron donating and electron withdrawing substitueents on the acidity of aliphatic carboxylic acids.
MULTIPLE CHOICE TYPE E.
Select the correct answer :
1.
The oils from which soaps are prepared belong to a class of compound known as :
2.
(a)
Amine
(b)
Acid
(c)
Hydrocarbon
(d)
Ester
Formic acid and acetic acid may be distinguished by reaction with (a)
Sodium
(b)
(c)
2, 4 dinitrophenyl hydrazine (d)
Dilute acidified KMnO4 Sodium ethoxide
CARBOXYLIC ACIDS
3.
The end product in the sequence of reaction. R–X
4.
5.
6.
7.
8.
9.
10.
11.
639
KCN
A
NaOH
B is :
(a)
an alkane
(b)
A carboxylic acid
(c)
sodium salt of carboxylic acid
(d)
Saponification
Which of the following reduces HgCl2 to Hg2Cl2 ? (a)
Formic acid
(b)
Ammonia
(c)
Acetic acid
(d)
CCl4
Acetic acid can be halogenated in presence of red P and halogen, but formic acid cannot be halogenated in the same way due to (a)
Presence of a - hydrogen atom in acetic acid
(b)
Presence of – COOH group in formic acid
(C)
Presence of carbonyl group in acetic acid
(d)
None of the above
Among acetic acid, phenol and n-hexanol, which of the compounds reacts with NaHCO 3 solution to give sodium salt and carbon dioxide ? (a)
Acetic acid
(b)
n - Hexanol
(c)
Acetic acid and phenol
(d)
Phenol
Vinegar contains (a)
10 to 20% acetic acid
(b)
10% acetic acid
(c)
6 to 10% acetic acid
(d)
100% acetic acid
Which acid is strongest ? (a)
CCl3 COOH
(b)
Cl2CHCOOH
(c)
ClCH2COOH
(d)
CH3COOH
The acid which does not contain – COOH group is (a)
Ethanoic acid
(b)
Picric acid
(c)
Lactic acid
(d)
Palmitic acid
Which of the following cannot reduce Fehling’s solution ? (a)
Formic acid
(b)
Acetic acid
(c)
Formaldehyde
(d)
Acetaldehyde
When benzoic acid is reacted with LiAlH4, it forms (a)
Benzene
(b)
Benzaldehyde
(c)
Toluene
(d)
Benzylalcohol
640
12.
13.
+2 CHEMISTRY (VOL. - II)
Electrolysis of sodium salt of maleic acid to ethyne is known as (a) Wurtz’s reaction (b) Clemmensen’s reduction (c) Kolbe’s reaction (d) Sabatier Senderen’s reaction. Which of the following reactions is expected to readily give a hydrocarbon product in good yield. (a)
14.
15.
RCOOK
electrolysis oxidation
(b)
I RCOOAg 2
Cl2
(c)
CH3CH3
(d) (CH3)3 C Cl hu Monocarboxylic acids show functional isomerism with (a)
esters
(b)
alcohols
(c)
ethers
(d)
aldehydes
C2H5OH
The boiling point of acetic acid is higher than expected from its molecular weight, because of
16.
17.
18.
19.
20.
21.
(a)
solubility in water
(b)
non-polar character
(c)
strong oxidising character
(d)
association through hydrogen bonding
Acids are obtained as a result of reaction between a Grignard reagent and (a)
Oxygen
(b)
CO2
(c)
CH3COCl
(d)
CH3CHO
Which acid is weaker than benzoic acid? (a)
p-Methyl benzoic acid
(b)
p-chlorobenzoic acid
(d)
p-Nitrobenzoic acid
(d)
o-Chlorobenzoic acid
Carboxylic acids are more soluble in (a)
Ether
(b)
C6H6
(c)
Na2CO3 solution
(d)
CHCl3
Stinges of bees and wasps contain (a)
formalin
(b)
formaldehyde
(c)
acetic acid
(d)
formic acid
Formic acid is obtained by the hydrolysis of (a)
HCN
(b)
CH3CN
(c)
(COONa)2
(d)
CO + CO2
What is the main reason for the fact that carboxylic acids can undergo ionisation? (a)
Absence of a-hydrogen
(b)
Resonance stabilisation of the carboxylate ion
(c)
High reactivity of a-hydrogen
(d)
Hydrogen bonding.
CARBOXYLIC ACIDS
22.
RCOOH RCH2OH. This mode of reduction of an acid to alcohol can be affected by (a) Zn/HCl (b) Na/Alcohol (c) Aluminium isopropoxide and isopropyl alcohol. (d)
23.
24.
25.
26.
641
LiAlH4
Calcium acetate on heating yields (a)
CaO, CO2 and H2O
(b)
CaCO3 and H2O
(c)
Acetaldehyde and CaCO3
(d)
CaCO3 and acetone
(b) (d)
Butanoic acid Benzaldehyde
n-Butyl benzene on oxidation will give (a) Benzyl alchohol (c) Benzoic acid
Benzoic acid may be converted into ethyl benzoate by reaction with (a)
Ethyl chloride
(b)
DryHCl – C2H5OH
(c)
Ethyl alcohol
(d)
Sodium ethoxide
C6H5MgBr
(i) CO2 + (ii) H 3O
P
In the above equation, Product ‘P’ is
27.
28.
29.
30.
(a)
Benzaldehyde
(b)
Benzoic acid
(c)
Phenol
(d)
Benzophenone
Which of the following does not give benzoic acid on hydrolysis ? (a)
Phenyl cyanide
(b)
Benzoyl chloride
(c)
Benzyl chloride
(d)
Methyl benzoate
Which of the followings is the strongest acid ? (a)
o-Nitrobenzoic acid
(b)
p-Nitrobenzoic acid
(c)
p-Chlorobenzoic acid
(d)
Benzoic acid
Zn
CH 3Cl
Alk
Phenol dust X Anhyd.AlCl Y KMnO Z, the product ‘Z’ is 3 4 (a)
Benzaldehyde
(b)
Benzoic acid
(c)
Benzene
(d)
Toluene
Which has the highest pKa value ? (a)
Benzoic acid
(b)
p-Nitrobenzoic acid
(c)
m-Nitrobenzoic acid
(d)
o-Nitrobenzoic acid
642
+2 CHEMISTRY (VOL. - II)
31.
Which of the following compounds will have the smallest pK a value ?
32.
(a)
Benzoic acid
(b)
Formic acid
(c)
Acetic acid
(d)
Phenylacetic acid
Among the following compounds, most acidic is : (a)
p-Nitrophenol
(b)
p-Hydroxybenzoic acid
(c)
o-Hydroxybenzoic acid
(d)
p-Toluic acid
ANSWERS A. 1. 2.
Acetic acid, 2CH3COOH + Na2CO3
2CH3COONa + CO 2+ H2O sodium acetate
3.
CH3COOH + NH3
CH3COONH4 Amm.acetate
4.
CH3CONH2 + H2O Acetamide
Formic acid is used in leather industry in tanning for removing lime, from the hides. It is also used for preparing certain medicines and antiseptics.
5.
Phenyl magnesiumbromide
6.
CH3 – CH – CH2 –COOH | OH
7.
3 Bromo propanoic acid.
8.
(i) Ethylpropanoate (ii) Pentanedioic acid
9.
Formic acid reduces Tollen’s reagent or Fehling solution.
10.
Benzoic acid E.
1. (d) 2. (b) 3. (c) 4. (a) 5. (a) 6. (a) 7. (c) 8. (a) 9. (b) 10. (b) 11.(b) 12.(c) 13.(a) 14.(a) 15.(d) 16.(b) 17. (a) 18. (c) 19. (d) 20.(a) 21.(b) 22.(d) 23.(d) 24.(c) 25.(b) 26.(b) 27.(c) 28.(a) 29. (b) 30.(a) 31. (b) 32.(c)
qqq
AMINES
643
UNIT - XIII
ORGANIC COMPOUNDS CONTAINING NITROGEN CHAPTER - 20
AMINES 20.1
INTRODUCTION : The amines are considered as alkyl / aryl derivatives of ammonia in which one or more
hydrogen atoms in an ammonia molecule have been replaced by alkyl and / or aryl groups. 20.2
TYPES OF AMINES :
20.2.1 ALIPHATIC AMINES : Aliphatic amines are the derivatives of ammonia in which one or more hydrogen atoms in ammonia molecule are replaced by alkyl groups. These are classified as primary, secondary and tertiary amines depending on whether one, two or three hydrogen atoms of ammonia molecule are replaced by alkyl groups. H | H—N—H or NH3 Ammonia
H | R—N—H or RNH2 Primary amine
R | R—N—H or R2NH Secondary amine
R | R—N—R or R3N Tertiary amine
Primary, Secondary and Tertiary amines : (i)
Primary or 10 amines : The primary amines have the general formula, RNH 2 and are characterised by — NH2 group linked to an alkyl group. The —NH2 group in primary amine is called the amino group.
(ii)
Secondary or 20 amines : The secondary amines have the general formula, R 2NH and are characterised by NH group linked to two same or different alkyl groups. The NH group in secondary amine is called the imino group.
(iii)
Tertiary or 30 amines : The tertiary amines have the general formula, R 3N and are characterised by — N group linked to three same or different alkyl groups.
644
+2 CHEMISTRY (VOL. - II)
Quaternary ammonium salts : If all the four hydrogen atoms of an ammonium salt are replaced by same or different alkyl groups, the compound is called quaternary ammonium salt or tetra- alkyl ammonium salt. + H | H — N — H Br— | H
+ R | R — N — R Br— | R
NH4Br
R4NBr
Ammonium bromide
Tetra-alkyl ammonium bromide
20.2.2 ARYL AMINES : Arylamines are those having the amino group (one or more) directly attached to an aromatic NH2
nucleus. For example : aniline NH2 NH2
methylaniline
NH-CH3
O-phenylenediamine
etc. Like aliphatic amines, arylamines may be primary (1 0), secondary (20) or tertiary (30)
according as one, two or all the hydrogen atoms of ammonia are replaced by phenyl or aryl radicals. In case of secondary and tertiary amino compounds, all the hydrogen atoms may not be replaced by aryl groups. There may be both alkyl and aryl groups. Such amines are of mixed aliphatic – aromatic type secondary and tertiary amines. A few examples of both pure and mixed amines are given below.
NH2
NH2
NH2
Primary :
CH3 Aniline m- toluidine Secondary :
NH Diphenylamine
Tertiary :
NH2 p-phenylenediamine NHCH3
N- Methylaniline (mixed) CH3
N N
Triphenylamine
N, N -Dimethylaniline (mixed)
CH3
AMINES
645
Aromatic compounds with amino groups present in the side chain are called arylalkyl amines. For example, benzylamine
CH2–NH2
CH2– CH2– NH2
b - phenylethylamine
,
etc. These are
aryl substituted aliphatic amines and behave very much like aliphatic amines. NH2
Aminobenzene or Benzenamine
or
C6H5 – NH2
Aniline [C6H5NH2] is the simplest primary arylamine. 20.3
NOMENCLATURE :
1.
Common system : The common names of aliphatic amines are derived by listing the alkyl groups on the nitrogen in order of increasing size ending with amine. If same type of alkyl group occurs twice or thrice on the nitrogen atom, the prefix di- or tri- is placed before the name of the alkyl group. CH3NH2
CH3NHCH3
Methylamine Dimethylamine 2.
CH3 | CH3 — N — CH3
CH3NHCH2CH3
Trimethylamine
Methyl ethylamine
IUPAC system : In IUPAC system primary amines are called alkanamines; secondary amines, alkyl alkanamines and tertiary amines, dialkyl alkanamines. The longest alkyl group is considered as the parent chain. CH3NH2
CH3CH2NH2
CH3CH2CH2NH2
Methanamine
Ethanamine
Propanamine
CH3CH2NH2CH3
CH3 — CH2 — CH2 — N
N – Methyl ethanamine
N, N – dimethyl propanamine
CH3 CH3
In the higher and more complex compounds the amino group is treated as a substituent and is assigned the locant keeping in view the lower number rules and the order of priority of different functional groups. Names of a few amines are given below. NH2 | CH3 — CH — CH2 — CH3 2-Butanamine
CH3CH2CH2CH2CH2NH2 1-Pentanamine
NH2 | CH3— CH—CH3 2-Propanamine
646
+2 CHEMISTRY (VOL. - II)
NH2 | CH3 — C — CH3 | CH3
CH3 | CH3 — CH — CH — CH2— CH3 | NH2
2 – Methyl – 2 –propanamine 3.
3–Methyl –2– pentanamine
Nomenclature of some common aryl amines are given below : Structure
Common name
IUPAC name
Aniline
Aniline or Benzenamine
o-Toluidine
2-Methylanine
p-Cloroaniline
4-Chlorobenzenamine or 4-Chloroaniline
N, N-Dimethylaniline
N, N – Dimethyl - benzenamine
NH 2
NH 2
CH3
NH 2
Cl N(CH 3) 2
20.4
ISOMERISM : Amines exhibit (i) Chain isomerism (ii) Position isomerism (iii) Functional isomerism
(iv) Metamerism. (i)
Chain isomerism :
CH3 | CH3 — CH — NH2 Isopropylamine (common name)
CH3 — CH2 — CH2 — NH2 n-Propylamine (common name) (ii)
Position isomerism :
NH2 | CH3 — CH — CH2 — CH3 sec-Butylamine
CH3 — CH2 — CH2 —CH2 — NH2 n-Butylamine NH 2
NH 2
NH 2 CH3 CH3
1-Methylaniline
2-Methylaniline
CH3
3-Methylaniline
AMINES
(iii)
(iv)
647
Functional isomerism : CH3 — CH2 — NH2
CH3 — NH — CH3
Ethylamine
Dimethylamine
C6H5CH2NH2
C6H5 — NH — CH3
Benzylamine
N-Methylaniline
Metamerism : C2H5 C2H5
Diethylamine C6H5 C2H5
CH3
NH
NH
C3H7 Methyl propylamine
C6H5CH2
NH
CH3
N-Ethylaniline
CH3 — NH — CH
CH3 CH3
Methyl isopropylamine NH
N-Methylbenzylamine
20.5
GENERAL METHODS OF PREPARATION :
1.
Ammonolysis of Alkyl halides (Hofmann’s method) Primary amines may be prepared by heating an alkyl halide with excess of alcoholic solution of ammonia in a sealed tube. R– X
+
H – NH2
Alkyl halide
Ammonia
CH3 – I + H – NH2 Methyl iodide
RNH2
+
HX
10 amine CH3NH2 + Methylamine
HI
The primary amine reacts with another molecule of alkyl halide to form secondary amine. R – X + H HN R
R2NH
+
HX
20 amine The secondary amine can combine with another molecule of alkyl halide of form a tertiary amine which in turn can combine with another molecule of alkyl halide to form quaternary ammonium salt. R X + H N R2 R X + R3 N
R3N
+
HX
30 amine
R4NX Tetra - alkyl ammonium halide Commercially, the amines are separated by fractional distillation.
648
+2 CHEMISTRY (VOL. - II)
Simple aryl halides i.e. when the halogen atom (X) is directly attached to the aromatic ring are ordinarily inert towards ammonia. This is because, in aryl halide the C – X bond is stronger than C – X bond in alkyl or benzyl halides. However, aryl halides react with ammonia in presence of catalyst (copper salts) at high temperature and under high pressure to form the corresponding amino compounds. Ar – X + NH3
Cu2 O
Ar – NH2 + HX
D, pressure
Aryl halide
1°-Amine
C6H5 – X + NH3
Cu2 O D, pressure
Halobenzene
C6H5NH2 + HX Aniline
Like aliphatic primary amines, aniline react with alkyl halides (RX) giving successively secondary, tertiary amines and quaternary ammonium salts. C6H5NH2
RX
C6H5NHR
RX
2° Amine 2.
C6H5Nr2
RX
3° Amine
+
–
C6H5N R3X
Quaternary ammonium salt
Ammonolysis of alcohols and phenol : When vapours of an alcohol and ammonia are passed over a dehydrating catalyst such as thoria or alumina at 300 0C, a mixture of primary, secondary and tertiary amines is formed. R – OH + H – NH2
ThO2 3000C
RNH2 + H2O
R – OH + H – H NR
R2NH + H2O
R – OH + H – NR2
R3N + H2O
Phenols react with ammonia in the presence of zinc chloride at about 300°C to form the corresponding amines. Ar – OH + HNH2 C6H5 – OH + NH3 Phenol
ZnCl 2 Ar – NH2 + H2O 300°C ZnCl 2 C6H5 – NH2 + H2O 300°C Aniline
AMINES
3.
649
Reduction of nitroalkanes and nitroarenes : When a nitro group is reduced it is easily converted to amino group. The reduction can be brought about by tin and hydrochloric acid or iron and hydrochloric acid or hydrogen in presence of nickel or platinum or lithium aluminium hydride. R NO2 Nitroalkane
Sn + HCl
+
6H
R NH2 + 2H2O
CH3NO2 + Nitromethane
6H
CH3NH2 + 2H2O Methylamine
C6H5NO2 + Nitrobenzene
6H
C6H5NH2 + 2H2O Aniline
or Fe + HCl
Catalytic reduction : Nitrobenzene can also be catalytically reduced by hydrogen under pressure in presence of Platinum or Raney nickel. Raney Ni or
C6H5NO2 + 3H2 Nitrobenzene 4.
C6H5NH2 + 2H2O
Pt or Pd
Aniline
Hofmann bromamide reaction : The action of bromine on acid amides in an alkaline (NaOH or KOH) solution produces an amine containing one carbon atom less than the number present in the amide. This reaction was developed by Hofmann and used in degradation of carbon chain.
RCONH 2 + Br2 + 4 KOH
® R - NH 2 + 2 KBr + K 2 CO 3 + 2 H 2 O
CH 3CONH 2 + Br2 + 4 KOH Acetamide
® CH 3NH 2 + 2 KBr + K 2 CO 3 + 2 H 2 O Methylamine
C6H5CONH2 + Br2 + 4KOH ® C6H5NH2 + 2KBr + K2CO3 + 2H2O Benzamide 5.
Aniline
Reduction of alkyl and aryl cyanides : When alkyl cyanides are reduced by sodium and alcohol (Mendius reaction) or lithium aluminium hydride, 10amines are formed Na+ Alcohol RCN + 4H or LiAlH4
RCH2NH2
CH3CN + 4H
CH3CH2NH2
C6H5CN + 4H
C6H5CH2NH2
650
6.
+2 CHEMISTRY (VOL. - II)
From Grignard reagent : Alkyl magnesium halides react with chloramine to form primary amines. RMgX
+
ClNH2
X
RNH2 + Mg
Cl
Grignard reagent 7.
From isocyanates : Acyl azides on heating produce isocyanates. When alkyl isocyanates are boiled with alkali, they undergo hydrolysis to form primary amines. This is known as Curtius reaction. RCON3 + NaCl
RCOCl + NaN3 heat
RCON3
R—N=C=O
Acyl azide
Alkyl isocyanate
R—N=C=O
H2O
RNH2 + CO2
Carboxylic acids react with hydrazoic acid to form primary amines. This is called Schmidt reaction. RCOOH + HN3 8.
H2SO4
RNH2 + N2 + CO2
heat
Gabriel’s phthalimide synthesis : Phthalimide reacts with alcoholic potash to form potassium phthalimide which reacts with alkyl halide to give N - alkyl phthalimide. This on hydrolysis with 20% HCl gives primary amine CO
CO
NH KOH
CO Phthalimide
NK RI
CO Potassium phthalimide
CO
NR
2H2O
CO N — alkyl phthalimide
COOH COOH Phthalic acid
+ RNH2 10Amine
Phthalic acid can be utilised for preparing phthalimide. Aromatic primary amines cannot be prepared by this method because aryl halides will not react with potasium phtholimide to form N – aryl phthalimide.
AMINES
20.6
651
GENERAL PROPERTIES : Basicity of aliphatic amines : Amines are derivatives of ammonia and hence like ammonia
these are basic in character. The basicity of amines and ammonia is due to the presence of unshared electron pair on nitrogen atom which accepts a proton. The readiness with which the lone pair of electrons is available for coordination with a proton determines the relative basicity of amines. .. H—N—H | H
.. R—N—H | H
.. R—N—R | H
.. R—N—R | R
Ammonia
Primary amine
Secondary amine
Tertiary amine
Alkyl groups are electron releasing groups. As a result the nitrogen atoms become electron rich. The lone pair of electrons is more readily available than ammonia. Thus amines are stronger bases than ammonia. Accordingly the basicity should increase with number of alkyl groups. However, the basicity follows the order, [in case of methylamine (1 0), dimithelamins (20) and trimethyl amine (30)] Secondary > Primary > Tertiary > Ammonia The reason for this is the steric factor. In tertiary amines three alkyl groups prevent the proton for coordination with lone pair. Hence tertiary amines are less basic than even primary amine. Ethyl amine is more basic than methyl amine since the +I effect of ethyl group is more than that of methyl group. Since chlorine is electron withdrawing group, chloramine is less basic than ammonia. C2H5NH2 > CH3NH2 NH3 > ClNH2 Basic Character of aromatic amines : Aniline is weakly basic and forms crystalline salts with acids. It also forms double salt with PtCl 4 and AuCl3 + – C6 H5 – NH2 + HCl [C6H5 – NH3 ] Cl Aniline hydrochloride C6H5 NH2 + H2SO4
[C6H5 – NH3] HSO4 Aniline hydrogensulfate
2 C6H5 NH2 + 2HCl + PtCl4
[C6H5 NH3Cl]2 PtCl4 Aniline chloroplatinate +
Basicity of a compound is its ability to accept a proton (H ). We have seen that the amines behave as bases due to the capacity of the lone pair of electron on nitrogen to form a
652
+2 CHEMISTRY (VOL. - II)
coordinate bond with the proton. Thus aliphatic amines are basic and more basic than ammonia, because the alkyl group pushes the electron towards nitrogen (+I effect). Hence, electron density on nitrogen increases. Aniline is a resonance hybrid of the following canonical structures:
Due to the resonance interaction between the p – electron of the benzene ring and the lone pair of electron on nitrogen atom of the amino group, the lone pair of electron on the nitrogen atom responsible for the basicity of aniline is less available to be donated to a proton. Further, a small amount of positive charge on nitrogen atom tends to repel proton. Hence, aniline is a weak base. Aniline may accept a proton to form a small amount of anilinium ion. +
..
NH3
NH2 + H+
(Anilinium ion) But anilinium ion less stable since there is no resonance in this ion. Therefore, the equilibrium shifts to the other side ie. towards aniline which is more stable. The low basicity of aniline is, therefore, due to the fact that aniline is stabilized by resonance to a greater extent than anilinium ion. Electron releasing substituents, like – CH3 increase the basicity of aniline whereas electron withdrawing substituents, like Cl, Br, I, NO2 decrease the basicity. This is due to the fact that electron release stabilizes the anilinium ion relative to aniline by dispersing the positive charge of the anilinium ion. Thus, monomethyl and dimethylanilines are stronger bases than aniline, CH3
Again, o-toluidine
NH2
CH3
and p –toluidine
would be more basic than aniline. It NH2
is observed that o-toluidine has a Kb= 2.5 x 10-10 and p-toluidine, Kb = 12 x 10-10 in comparison to the Kb of aniline (4.2 x 10-10). The anomalous behaviour of o-substituted anilines is, however, possibly due to ortho effect. Thus o-substituted anilines tend to have Kb lower than that of
AMINES
653
aniline irrespective of the nature (electron – releasing or electron withdrawing) of the substituent. The base –weakening effect of electron –withdrawing substituents is most marked when present in ortho position of the amino group. Physical Properties : (1)
State : The lower members of amines are gases. Those of three to eleven carbon atoms (except trimethyl amine) are liquids at ordinary temperature. Higher members are solids.
(2)
Odour : Amines of lower molecular mass possess odour similar to that of ammonia. As the molecular mass increases these have characteristic unpleasant odour of a decaying fish. In general it can be stated that aliphatic amines possess fishy or fish -like odour.
(3)
Density : These are lighter than water.
(4)
Solubility : The lower members (less then six carbon atoms) of the family are highly soluble in water as all amines including tertiary amines are capable of forming hydrogen bonds with water.
(5)
Boiling point : The boiling points are higher than those of the corresponding alkanes. This is due to the polar nature and intermolecular hydrogen bonding in amines. The boiling points of amines are lower compared to alcohols and carboxylic acids of comparable molecular mass. It is due to weaker hydrogen bonding as oxygen is more electronegative than nitrogen.
(6)
Physical properties of Aniline : Freshly distilled aniline is a colourless liquid with 0
unpleasant odour, b.p. 184 C. On exposure to light and air, it undergoes oxidation and turns brown. It is heavier than water and is slightly soluble in it, but it is highly soluble in alcohol, ether and benzene. Aniline is a polar molecule and its dipole moment is 1.48D. Chemical Properties : 1.
Reaction with water : Amines combine with water to form alkyl ammonium hydroxides which ionise to give protonated amines and hydroxide ion. RNH2 + H2O Amine
2.
RNH3OH
RNH3+
+
OH—
Alkyl ammonium hydroxide Protonated amine
Reaction with acids : Amines react with strong acids to form ammonium salts. Methyl amine combines with hydrochloric acid to form methylammonium chloride. CH3NH2
+ HCl
+
CH3NH3 Cl– Methylammonium chloride
654
+2 CHEMISTRY (VOL. - II)
RNH2
+ HCl
RNH3Cl
1° Amine
Alkylammonium chloride
2C2H5NH2
+ H2SO4
(C2H5NH3)2SO4
C6H5NH2
+ HCl
[C6H5 – N H3] Cl (Crystalline salt)
+
Aniline
Aniline hydrocloride
R2NH
+ HCl
R2NH2Cl
20Amine
Dialkyl ammonium chloride
R3N
+ HCl
R3NHCl
30Amine
3.
–
Trialkyl ammonium chloride.
Reaction with alkyl halides : Primary amines take up three molecules of alkyl halide to form secondary and tertiary amines and quaternary ammonium salts. CH3NH2
+ CH3I
(CH3)2NH
Methyl amine (CH3)2NH
+ HI
Dimethyl amine (CH3)3N
+ CH3I
+ HI
Trimethyl amine (CH3)3N
+ CH3I
(CH3)4NI Tetramethyl ammonium iodide.
Like aliphatic primary amines, aniline can react with alkyl halides giving successively secondary, tertiary amine and quaternary ammonium salts. These are all typical nucleophilic substitution reactions. .. C6H5– NH2 +
C6H5– NH2R X
alkyl halide
Aniline
.. C6H5– NH – R + R – X .. C6H5– NR2 +
heat
R–X
R–X
heat
heat
C6H5– NHR2 X
OH
OH
C6H5 – NHR N-alkylaniline (Sec. amine)
C6H5 NR2 N,N-dialkylaniline (tertiary amine)
C6H5– NR3 X Quaternary ammonium salt
This process of introducing alkyl group in the molecule is called alkylation. N-arylation is usually more difficult and purely aromatic quaternary ammonium salt of the type +
[(C6H5)4N)] X is unknown.
AMINES
4.
655
Reaction with acid chloride and acid anhydride (Acylation) : Primary and secondary amines react with acid chlorides and anhydrides to form alkyl derivatives of amides (substituted amides). O || RNH – H + Cl — C — R/
O || R — NH — C — R/
10 Amine R2NH
N - Alkylamide O || R2N — C — R/ + R/ COOH
R/CO O R/CO Acid anhydride
+
20 Amine
N, N - dialkylamide
Aniline reacts with acetyl chloride in presence of a base like pyridine or sodium hydroxide to neutalise the acid formed. The product is a substituted acid amide. O O || || base CH3 – C – Cl + H2N — CH3– C – NH – – HCl (Acetyl chloride) (Acetanilide or (aniline) N-phenylethanamide) Substituted acid amides like acetanilide can also be considered as acyl derivatives of 0
amines. Thus, acetanilide is acetyl derivative of aniline. It is to be noted that only 1 and 0
=
2 amines containing hydrogen atom attached to nitrogen can undergo acylation reaction. O Acylation is a process of introducing R – C – group into the molecule. Introduction of
=
CH3 –CO – group is called acetylation. Similarly, introduction of benzoyl group O (C6H5– C –) is called benzoylation. Benzoylation of aniline is carried out by treating aniline with benzoyl chloride in presence of NaOH. C6H5– NH2 + Cl – CO – C6H5 (Aniline)
NaOH
(Benzoyl chlonde)
C6H5– CONHC6H5 (Benzanilide)
In this reaction, H atom of – NH2 group is substituted by benzoyl group. (C6H5 – CO –). This reaction is known as Schotten- Baumann reaction. 5.
Reaction with nitrous acid : (Distinction): Amines are distinguished by this reaction. Nitrous acid is formed by the reaction by sodium nitrite with dilute mineral acids. (a)
Primary amines react with nitrous acid to form primary alcohol with the liberation of nitrogen.
656
+2 CHEMISTRY (VOL. - II)
NaNO2
+
CH3CH2NH2 +
HCl
NaCl +
HONO
CH3CH2OH + N2 +H2O
Ethyl amine
HNO2
Ethyl alcohol
Important : Methylamine does not give methyl alcohol with NaNO2 & dil. HCl. Instead it gives methyl nitrite and dimethyl ether. CH3NH2 + 2HONO
CH3 — O — N = O + N2 + H2O
2CH3NH2 + 2HONO (b)
CH3 — O — CH3 + 2N2 + 3H2O
Secondary amines react with nitrous acid to form nitrosoamine (yellow oil). (CH3)2NH
+
(CH3)2N — N = O + H2O
HONO
Dimethylamine (c)
Nitrosodimethylamine
Tertiary amines dissolve in nitrous acid to form addition product which decomposes to form nitrosoamine and alcohol. HNO2
(CH3)3N
(CH3)3 NHNO2
heat
(CH3)2NNO + CH3OH
Nitrosodimethylamine Aryl primary, secondary and tertiary amines react differently with nitrous acid. Since nitrous acid is unstable, it is prepared in the reaction medium by the action of dilute hydrochloric acid on sodium nitrite at low temperature. (a)
Primary amines : Aromatic primary amines react with nitrous acid at 273 – 278 K (0 – 5 0C) to form arenediazonium salts. Thus, NaNO2 + HCl
HNO2 + NaCl
NH2 + HNO2 + HCl
+ N NCl
273-278K
+ 2H2O
Benzenediazonium chloride This reaction of converting primary aromatic amines into diazonium salts by treatment with a cold (273 - 278K) solution of nitrous acid is called diazotisation. (b)
Secondary amines : Secondary amines (both aromatic and aliphatic) react with nitrous acid to give Nnitrosoamine which being insoluble in dilute mineral acids separate out as yellow oily compounds.
AMINES
657
CH3
CH3
NH + HONO
N – N = O + H2O
N-Methylaniline
N-Nitroso-N-methylaniline (Yellow oil)
These N-nitrosoamines on warming with a crystal of phenol and a few drops of conc.H 2SO4 form a green solution which when made alkaline with aqueous NaOH, turns deep blue and then red on dilution. This reaction is called Liebermann nitroso reaction and is used as a test for secondary amines. (c)
Tertiary amines : Aromatic tertiary amines undergo electrophilic substitution with nitrosonium ion at pposition of the phenyl ring to form green-coloured p-nitrosoamines. CH3 CH3
N
CH3
+ HONO
N, N-Dimethylaniline
N
CH3
N = O + H2O
p-Nitroso-N, N-dimethylaniline (green)
Thus, the reaction with nitrous acid can be used for the distinction of 1 0, 20 and 30 amines. Uses : Aniline is largely used for preparation of dyes and drugs. It is also used in the manufacture of accelerators used in rubber industry. Aniline also finds its use in calico printing. 6.
Carbylamine reaction : This reaction is responded only by primary amines. Primary amines on heating with chloroform and alcoholic caustic potash give isocyanides (carbylamines) with very unpleasant odour. RNH2 + CHCl3 + 3KOH
RNC + 3KCl + 3H2O Alkylisocyanide (Carbylamine)
CH3NH2 + CHCl3 + 3KOH
CH3NC + 3KCl + 3H2O
Primary arylamines like aliphatic primary amines undergo carbylamine reaction. When heated with chloroform and alcoholic KOH, aniline give obnoxious smelling carbylamine or phenyl isocyanide. C6H5NH2 + CHCl3 + 3KOH (Aniline)
C6H5NC
+ 3KCl + 3H2O
(Phenyl isocyanide)
In general, Ar NH2 + CHCl3+ 3KOH (Arylamine)
Ar NC
+
Aryl isocyanide or carbylamine
3KCl + 3H2O
658
7.
+2 CHEMISTRY (VOL. - II)
Reaction with benzene sulphonyl chloride (Distinction): Primary amines react with benzene sulphonyl chloride to form alkyl benzene sulphonamides, soluble in caustic alkali. Secondary amines form dialkyl benzene suphonamides insoluble in caustic alkali. Tertiary amines do not react.
8.
RNH– H + Cl – SO2 C6H5
RNHSO2C6H5 + HCl
R2N – H + Cl – SO2 C6H5
R2NSO2C6H5 + HCl
Reaction with Grignard reagent : Primary amines react with Grignard reagent to form hydrocarbons. RNH2 + R/MgX
9.
R/H + RNHMgX
Formation of Schiff’s bases : Primary amines react with aldehydes to form imine or Schiff’s bases. This is catalysed by acids. H+
RN H2 + O HCR/ 10.
R — N = CH — R/ + H2O Schiff’s base
Electrophilic Substitution reaction of Aryl amines : We have seen that due to resonance, the ortho – and para – positions with respect to – NH2 group in aniline become electron rich. Therefore, electrophilic substitution in aniline becomes easier, and the substituents enter the ortho – or / and para-positions in the ring with respect to – NH2 group. (a)
Halogenation : The presence of – NH2 group in aniline activates the nucleus so much that all the H – atoms of o – and p – positions are readily replaced by bromine or chlorine atoms. Thus, aniline reacts with bromine – water to form 2, 4, 6 – tribromoaniline. Bromine enters para and both the ortho positions. NH2 + 3Br
H2O
Br
NH2
Br
2
(Aniline)
Br (2, 4, 6 tribromoaniline)
Chlorine, however, reacts with aniline in presence of water – free solvent like chloroform to form 2 , 4, 6 – trichloroaniline. NH2 + Cl (Aniline)
Chloroform
Cl
NH2
Cl
2
Cl (2, 4, 6 trichloroaniline)
AMINES
659
For the preparation of a monobromo compound, aniline is first acetylated before bromination. Finally, the acetyl group is removed through hydrolysis.
(aniline)
NHCOCH3 | Hydrolysis
NHCOCH3 | Br2
base + CH3COCl (acetyl chloride)
CH3COOH (Acetanilide)
|
NH2 |
H2O,H
Br (p- bromoacetanilide) Major product
NH2 | +CH3COOH | Br (p- bromoaniline)
The reactivity of the nucleus is decreased by acetylating – NH 2 group, since activating influence of – NHCOCH3 group is much less than that of – NH2 group. In that case, only mono substituted product is formed. (b)
Sulphonation : When aniline is heated with excess of conc. H2SO4 or fuming sulphuric acid at about 1800 – 2000C for 3 – 4 hours, p – amino benzene sulphonic acid or sulphanilic acid is formed as the main product. Aniline hydrogen sulphate, formed loses a molecule of water to form phenyl sulphamic acid which rearranges to sulphanilic acid.
NH2 |
H2SO4
(Aniline)
– NH3HSO4 |
NH.SO3H |
heat
1800 – 2000 (aniline hydrogen sulphate)
NH3 |
NH2 |
| SO3H (Phenyl Sulphamic (Sulphanilic acid) acid)
| SO3 (Zwitter ion)
Sulphanilic acid contains one acidic group (SO3H) and one basic group (-NH2). The H ion of – SO3H group combines with – NH2 group forming –NH3. This is a special type of salt in which an acid group and a basic group of the same molecule +
–
neutralize each other. Thus, sulphanilic acid molecule contains both N H3 and SO3
ions. So, it is a dipolar ion or Zwitterion. Sulphanilic acid does not show basic character. The acidic property predominates in the molecule. (c)
Nitration : Aniline can not be nitrated directly by the nitrating mixture (conc. H2SO4 + Conc. HNO3), because the amino group is oxidized to form various tarry products. To prevent this oxidation, - NH2 group is first protected by acetylating the group with acetyl chloride or acetic anhydride and acetanilide formed is then nitrated. – NHCOCH3 group is o–, p– directing. So, o–nitro and p–nitroacetanilides
660
+2 CHEMISTRY (VOL. - II)
are formed. The two isomers are separated and hydrolysed separately with dilute acid, when o–nitroaniline and p–nitroaniline are formed. NH2 |
(aniline)
(CH3CO)2O or (CH3COCl
NHCOCH3 | HNO3
+
H2SO4 (Acetanilide)
NHCOCH3 |
NHCOCH3 | NO2
(o-nitroacetanilide)
H2O
| NO2 (p- nitroacetanilide)
HCl
NH2 |
H2O HCl NH2 |
NO2
(o-nitroaniline)
| NO2 (p- nitroaniline)
20.7
TESTS :
(1)
The aqueous solution of water soluble amines turn red litmus blue. All three classes of amines (1°, 2°. 3°) being basic dissolve in mineral acids like HCl, H 2SO4 etc. to form salt.
(2)
On addition of a concentrated aqueous solution of sodium nitrite to a solution of amine in dil HCl, primary amines give alcohols with rapid effervescence of nitrogen, secondary amines form water-insoluble yellow oil and tertiary amines form nitrite addition salts. Aromatic primary amines, however, react with HNO2 at O – 5°C to from arenediazonium salts which couple with an alkaline solution of b-naphthol to form orange or red coloured azo dyes. (distinction form primary alkyl amines).
(3)
Carbylamine test (For primary amines) : When a primary amine (aliphaticor aromatic) is heated with chloroform and alcoholic KOH, an alkyl isocyanide is formed which has a foul smell.
(4)
Liebermann nitroso reaction (For secondary amines) : R2NH + (NaNO2 + HCl)
R2N — N = O + H2O (Nitrosoamine) Insoluble yellow oil
AMINES
661
Nitrosoamines on warming with phenol and conc. H2SO4 give a brown or red colour and when the mixture is poured into alkaline solution, a blue or violet colour is observed. This is known as Liebermann’s nitroso reaction and often used to detect a secondary amino group. 20.8
SEPARATION OF PRIMARY, SECONDARY AND TERTIARY AMINES : Hinsberg’s method — The mixture of amines is treated with Hinsberg’s reagent (Benzene sulphonyl chloride). Primary and secondary amines react and tertiary amines do not react. — HCl R.NH.SO C H R — NH + C H SO Cl 2 0 1 amine
6 5
2 6 5
2
N– alkyl benzene sulphonamide — HCl
R2NH + C6H5SO2Cl 20 amine
R2N.SO2C6H5 N, N– dialkyl benzene sulphonamide
To the above mixture some NaOH solution and ether are added. Sulphonamide from primary amine forms its sodium salt. RNHSO2 .C2H5 + NaOH
R.N (Na)SO2C6H5 Sodium salt (in aqueous layer)
Sulphonamide from secondary and unreacted tertiary amine dissolves in ether and forms the upper layer. The two layers are separated. (1)
The aqueous layer is acidified with dilute hydrochloric acid and then hydrolysed with conc. HCl. It is then distilled over NaOH when primary amine passes over. R.N(Na)SO2.C6H5
+ HCl
R.NH SO2C6H5 + NaCl
R.NHSO2.C6H5
+ HCl+H2O
R.NH2HCl + C6H5SO2OH
R.NH2HCl
+ NaOH
R.NH2 + NaCl + H2O Primary amine
(2)
Ether layer is subjected to fractional distillation when tertiary amine distils over. The remaining sulphomamide of secondary amine is treated with conc. HCl R2NSO2C6H5 + HCl + H2O
C6H5SO2OH + R2NH.HCl
Dialkylamine hydrochloride is then distilled over NaOH when secondary amine is produced. R2NH.HCl + NaOH 20.9
R2NH + NaCl + H2O
USES : 1. Methyl and ethylamines are used in leather industries for dehairing of hides . 2. Methylamine is used as a refrigerant. 3. Methylamine and ethylamine are used in preparing dyes and medicines. 4. Amines are used in organic synthesis as condensing agents and catalysts.
662
+2 CHEMISTRY (VOL. - II)
5.
Aromatic amines such as aniline are widely used in the manufacture of dyes and drugs. Used as additives (antioxidants) and vulcanization accelerators in rubber industry. It is used for the preparation of phenyl isocyanate needed for the manufacture of polyurethane plastics. It is also used for the preparation of arenediazonium salts which in turn is used in the synthesis of wide variety of aromatic compound.
20.10 CONVERSIONS : 1.
Methyl amine to ethyl amine HNO2 CH3NH2 CH3OH KCN
2.
P + I2
Reduction 4H Ethyl amine to Methyl amine : [O] HNO2 CH3CH2NH2 CH3CH2OH [O]
CH3CN
CH3COOH
NH3
CH3COONH4
heat
CH3I CH3CH2NH2
CH3CHO CH3CONH2
Acidified K2Cr2O7 Br2+KOH
CH3NH2
20.11 DISTINCTION BETWEEN PRIMARY, SECONDARY AND TERTIARY AMINES : Reaction 1. Action of nitrous acid
2. Action of CH3I
Primary Alcohol and N2are formed RNH2 + HONO ROH + N2 + H2O (Aromatic primary amines undergo diazotisation) at 0 to 5° Three molecules CH3I react to form quaternary salt
3. Action of CH3COCl
Forms N– alkyl amide
4. Carbylamine reaction
Forms isocyanide having a foul smell Forms N-alkyl benzene sulphonamide soluble in aqueous NaOH
5. Action of Hinsberg’s reagent(C6H5SO2Cl)
Secondary
Tertiary
Nitrosoamine is formed. R2NH+ HNO2 R2N—N=O+H2O
No reaction (Only salt is formed.)
Two molecules of CH3I react to form quaternary salt
One molecule of CH3I reacts to form quaternary salt. No reaction.
Forms N,N– dialkyl amide No reaction Forms N,N– dialkyl benzene sulphonamide insoluble in aqueous NaOH
No reaction. No reaction.
20.12 CYANIDES AND ISOCYNIDES : These classes of organic compounds are not dealt separately. Their methods of preparation and chemical reactions have been discussed at relevant places in the text.
AMINES
663
CHAPTER (20) AT A GLANCE General formula: R–NH2(10),
R
R R/ R//
NH (20)
R/
/ // N (30) (R, R , R may be same or different alkyl groups)
Methods of Preparation : 1.
Hofmann’s method: NH3 +
alc.soln.
RX
RNH2
+
HX
R NH2 + RX
R2NH
+
HX
R2NH + RX
R3N
+
HX
1000C
(10amine) (20amine)
(30amine) R3N
+ – R4 N X
+ RX
Quaternary ammonium salt 2.
Ammonolysis of Alcohols: ThO2 orAl2O3
R — OH + H NH2
3000C
R — OH + H NHR
R–NH2
+
H2O
R2NH
+
H2O
R3N
+
H2O
+
2H2O
(10amine) (20amine)
R — OH + H NR2
(30amine) 3.
Reduction of nitroalkanes : R – NO2
4.
+
6H
Sn/conc.HCl
R–NH2
(10amine)
Hofmann bromamide reaction: RCONH2 + Br2
+ 4KOH
RNH2 + 2KBr + K2CO3 + 2H2O (10amine)
664
5.
+2 CHEMISTRY (VOL. - II)
Reduction of alkyl cyanides : Mendius reaction R – CN
6.
+
+
(10amine)
ClNH2
R – NH2
+
Mg(X)Cl.
(10amine)
From Acids and Acid derivatives : Curtius reaction :
RCOCl + NaN3 RCON3
heat
(Acid azide) Schmidt reaction :
8.
R – CH2 – NH2
or LiAlH4
From Grignard reagent : RMgX
7.
Na/C2H5OH
4H
RCON3 + NaCl NaOH boil (alkyl isocyanate)
RNCO
R–NH2 (10amine)
RCOOH +
HN3
RNH2 + N2 + CO2
(Acid)
(Hydrazoic acid)
(10amine)
Gabriel’s Phthalimide synthesis :
CO
NH
CO
KOH
CO (Phthalimide)
NK
CO RI
CO CO
NR
HCl
RNH2+ (10amine)
COOH COOH (Phthalic acid)
General Properties : 1.
Basicity : Amines are more baisc than ammonia due to the alkyl group(s) attached to nitrogen. Aromatic amines are less basic than aliphatic amines.
2.
Boiling points : Amines have higher boiling points than the corresponding alkanes due to intermolecular hydrogen bonding.
AMINES
3.
665
Chmical reactions : (i)
(ii)
Basic nature : +
H2O
RNH3 OH
RNH2
+
HX
RNH3 X
R2NH
+
HX
R2NH2 X
R3N
+
HX
+ – R3NHX
+
–
–
+
–
Free bases (amines) can be obtained from the salts by adding dilute base solution.
Alkylation : RNH2 (10amine)
(iii)
+
RNH2
RX(heat) Pressure (–HX)
RX –HX
R2NH (20amine)
RX
R3N (30amine)
R4NX
Quarternary salt
Acylation : By acid chloride and acid anhydride. R/COCl
RNH2 +
RNHCOR/
+
HCl
(N–Alkyl amide) R/COCl
R2NH +
R2NCOR/
+
HCl
(N, N–Dialkylamide) No – NH group
R3N
No acylation.
Acid anlydrides react in a similar way as above. (iv)
With HNO2 (NaNO2 and cold dilute mineral acid) R – NH2
+ HONO
(10amine) R2NH
R – OH
+ N2 + H2O
(alcohol) + HONO
(20amine)
R2N– N = O + H2O (Nitrosoamine) (Insoluble yellow oil)
R2N–N=O + phenol + conc.H2SO4 (Liebermann’s nitroso reaction) R3N
+
HNO2
heat
Brown or red colour
NaOH
Blue or Violet colour
+ – R3NHNO2
Aryl primary amines, however, react with HNO2 to form arenediazonium salts.
666
+2 CHEMISTRY (VOL. - II)
(v)
Carbylamine reaction : For 10 amines only. heat R–NH2 + CHCl3 + 3KOH RNC + (alcoholic)
3KCl +3H2O
(Carbylamine) unpleasant odour
R2NH and R3N do not respond to this reaction. (vi)
With benzene sulphonyl chloride : KOH R–NH2 + C6H5SO2Cl C6H5SO2NHR (10amine) R2NH + C6H5SO2Cl (20amine)
Excess soluble KOH (benzene sulphonamide) KOH KOH C6H5SO2NR2 Insoluble
No reaction with 30amine (vii)
With aldehydes : Formation of Schiffs base. H
R – NH2 + R/– CHO
+
R – N = CH – R/ + H2O
(1 amine) (viii)
(Schiff’s base)
Aryl amines undergo electrophlic substitution reactions. NH2
X
3X2
(a)
NH2
X
[X = Cl or Br] X 2,4,6 - trihaloaniline NH2
NH2 Sulphonation
(b)
SO3H NH2 (c)
NH2
Nitration
NO2
NH2 and NO2
(d)
C6H5NH2 Diazotisation [ C6H5 N N ]+X
(e)
10, 20, 30 Amines
HNO 2
React differently
AMINES
667
4.
Distinction between 10, 20 and 30 amines :
(i)
By the action of nitrous acid : 10 amine 20 amine
alcohol + N2 gas. (except methylamine) nitrosoamine
responds to Liebermann nitroso reaction.
30 amine salt formation (dissolves in HNO2 solution). Aryl primary amines form arenediazonium salts which couple with alkaline solution of b- naphthol to red or orange colour azo dyes. (ii)
Alkylation : 10 amine takes up 3 moles of CH3I for quaternisation. + – RNH2 + 3RI R4 N I 20 amine takes up 2 moles of CH3I for quaternisation. + – R4 N I R2NH + 2RI 30 amine takes up 1 moles of CH3I for quaternisation. + – R3N + RI R4 N I
(iii)
(iv)
Acylation: 10 amine
RCOCl
N–alkyl acid amide
20 amine
RCOCl
N, N–dialkyl acid amide
30 amine
RCOCl
No reaction.
Carbylamine reaction : 10 amine 20 amine 30 amine
(v)
5.
CHCl3 alc.KOH
Isocyanide (unpleasant odour) No reaction No reaction
Hinsberg’s reagent (C6H5SO2Cl) 10 amine
N–alkyl benzene sulphonamide. (soluble in alkali)
20 amine
N, N–dialkyl benzene sulphonamide. (insoluble in alkali)
30 amine
No reaction.
Separation : A mixture of 10, 20 and 30 amines can be conveniently separated by Hinsberg’s method.
668
+2 CHEMISTRY (VOL. - II)
QUESTIONS A.
Short questions (one mark each)
1.
Which of the following is basic ? (a) CH3 CH2 OH (c) CH3 NH2
2.
(b) CH3 COOH (d) CH3 OCH3
Complete the following reaction and give the names of the products ? CH3CH2NH2 + HNO2
——— + —— ?
What happens when an alkyl cyanide is reduced by sodium metal in ethanol ? 3.
Why aniline is soluble in aqueous HCl.
4.
Write the IUPAC name for C6H5N (CH3)2
5.
Identify the products in the following
6.
(i)
C6H5NO2
(ii)
C6H5NO2
Sn/HCl
(iii)
Raney Ni or Pt or Pd H2 C6H5NH2 RX A RX B RX
(iv)
C6H5NH2
(v)
C6H5NH2
(vi)
C6H5NH2 Nitration
(vii)
C6H5NH2
(viii)
C6H5NHCH3
(ix)
C6H5N(CH3)2
C (identify A, B & C)
Br2 water
Sulphonation
Diazotsation
HNO 2 HNO2
Fill in the Blanks : (i)
Benzyl amine is
(ii)
Among the isomeric o- m- and p- anisidine
(iii)
The reaction of aniline with NaNO2 and HCl at 00 C to from benzenediazonium chloride is called
(iv)
Aniline on treatment with bromine water gives
(v)
Tetra-alkyl ammonium salts are called
Answer :
basic than aniline is the weakest base.
1. (more), 2. (m-Anisidine), 3. (diazotisation), 4. (2,4,6 - tribromoaniline), 5. (quaternary ammonium salts)
AMINES
669
B.
Short questions (Two marks each)
1.
How will you prepare ethyl amine from methyl iodide ?
2.
Give a method of preparation of primary amine.
3.
Why methylamine is more basic than aniline ?
4.
What is carbylamine reaction ? Give equation.
5.
How will you obtain methanol from methyl amine ?
6.
How can you get methylamine from ethylamine?
7.
Give the IUPAC name of the following compounds: (a)
CH3CH2CH2NH2
(b)
CH3NHCH2CH3
(c)
(CH3)2CHNH2
(d)
(CH3)3 C.NH2
(e)
(CH3)2 NCH2CH2CH3
(f)
CH3CH2CH (NH2) CH2CH3
8.
Give the structural isomers of C3H9N and C4H11N and give their names. Classify each as primary, secondary and tertiary amines.
9.
Name the functional groups in the following compounds :- CH3CH2NH2 and CH3CONH2
10.
What happens when nitromethane is reduced ?
11.
Give a chemical test and reagents used to distinguish between ethylamine and diethylamine.
12.
A compound molecular formula CH5N on treatment with HNO2 liberates a colourless and odourless gas. What is the name of the compound and the gas liberated ? Write the equation.
13.
How you will distingush C2H5NH2 for C6H5NH2
14.
What is carbylamine test?
15.
Why aniline is less basic than ammonia?
16.
Why ethyl amine is more basic than ammonia?
17.
How will you carry out the conversion of benzene to p-nitroaniline ?
18.
Direct nitration of aniline is not carried out at all. Explain why ?
19.
Give an example of a Zwitterion.
+ [Ans. H3 N
SO3– Sulphanilic acid]
670
+2 CHEMISTRY (VOL. - II)
C.
Short questions (Three marks each) :
1.
Arrange the following compounds in decreasing order of basicity, Give reason. Methyl amine, dimethylamine, aniline, N-methyl aniline
2.
How will you prove that amines are basic compounds ?
3.
Distinguish between primary and secondary amines Give one test only with equation.
4.
Complete the following equation and balance. C2H5OH
5.
A
KCN
B
H3O+
C
NH3
D
How will you prepare ethylamine from (a)
6.
PCl5
Methyl cyanide
(b)
Propanamide
Name a tertiary amine in IUPAC system which is isomeric with CH3 | CH3—C—NH2 | CH3
and
C2H5 – CH2 – CH2 – NH2
7.
Describe a test to distinguish between aniline, N-methylaniline and N-ethyl-Nmethylaniline
8.
How will you convert the following (i) Benzene to aniline and (ii) Aniline to N,N-dimethylaniline ?
9.
Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
10.
Write the chemical reaction of aniline with benzoyl chloride and mention the name of the product.
11.
Convert aniline to 1,3,5 – tribromobenzene.
12.
Account for the following. (i)
Aniline is less basic than methylamine
(ii)
p-Nitroaniline is less basic than aniline
(iii)
p-Toluidine is more basic than aniline.
13.
Describe a method for the identification of primary, secondary and tertiary amines.
14.
Accomplish the conversion of aniline to p-bromoaniline.
15.
Write the reactions of aromatic and aliphatic primary amines with nitrous acid.
16.
How you will prepare benzonediazonium chloride?
17.
How you will obtain 2,4,6-trichloroaniline from aniline ?
18.
How will you convert nitrobenzene to (i) acetanilide and (ii) benzamide ?
AMINES
671
D.
Long questions :
1.
Give any two methods for the preparation of a primary amine. How does it react with (a) HNO2
2.
(b) CH3CHO
(c) CH3COCl
Give any two methods for the preparation of a an aromatic primary amine. How does it react with (a) CH3COCl
3.
(b) HNO2
Write notes on : (a) Hofmann reaction (b) Hofmann bromamide reaction (c) Hinsberg’s method for distriction of 1 0, 20 & 30 amines ?
4.
What are amines ? Explain three types of aliphalic amines giving one example from each. Write with equations how amines react with (a) alkyl halide and
(b) acid chloride
5.
Give any two methods of preparation of primary amine. How does it react with CHCl 3 presence of alcoholic KOH?
6.
What are different types of amines ? Give two methods for the preparation of primary amine. How does it react with methyl iodide? C2H5NH2 is more basic than CH3NH2; Explain.
7.
What are different types of amines ? How primary amines are prepared from (a) nitroparaffins
(b) cyanides
(c) alkyl halides ?
What happens when methylamine reacts with methyl iodide ? 8.
What are different types of aliphatic amines? How aliphatic primary amines are prepared from (i) acid amides, (ii) alkyl cyanides. What happens when methylamine reacts with nitrous acid?
9.
Describe the methods of preparation of aromatic monoamines, taking example of aniline. Discuss the relative basic character of aniline and methyl amine Explain the weak basic character of aniline.
10.
Give the general chemical reactions of aryl amines.
11.
How is aniline prepared on a laboratory scale ? Summarise its chemical reactions.
12.
How is basicity of aniline affected by substituents on the benzene ring ? How do you explain the ortho and para directive influence of –NH2 group ?
13.
Explain the action of nitrous acid on primary, secondary and tertiary amines. How aniline differs from methyl amine in its reaction with nitrous acid ?
14.
How does aniline react with the following reagents ? (i) Acetic anhydride, (ii) benzoylchloride (iii) Sodium nitrite / HCl.
15.
How will you distinguish between aniline and benzylamine ? How N, N-dimethylaniline is prepared ? How will you distinguish it from aniline ?
672
+2 CHEMISTRY (VOL. - II)
E.
Multiple Choice Questions :
1.
Most basic among the following is C6H5NH2,
(C2H5)2 NH
I (a) I 2.
(b) II
(c) III
(C2H5)3 N,
II
III
C2H5NH2 IV
(d) IV
Which of the following compounds gives dye test ? (a) Aniline (b) Methylamine (c) Diphenylamine (d) Ethylamine
3.
In the following reaction, X is X Bromination Y NaNO2/HCl
Z
Boiling C2H5OH
Tribromobenzene
(a) Benzoic acid (b) Salicylic acid (c) Phenol (d) Aniline 4.
The correct increasing order of basic strength for aniline (I), p-nitroaniline (II), and ptoluidine (III) is..... (a) II < III < I
5.
(b) NO+
(d) II < I < III
(c) NO2+ (d) NO2–
Reduction of nitrobenzene by which of following reagent gives aniline ? (a) Sn/HCl
7.
(c) III < II < I
In the nitration of benzene using a mixture of conc. H 2SO4 and conc. HNO3, the species which initiate the reaction is..... (a) NO2
6.
(b) III < I < II
(b) LiAlH4
C6H5NO2 Sn/HCl
(c) Zn/NH4OH
(d) SnCl2
C6H5X
In the above reaction ‘X’ is +
(a) Cl (b) NH2 (c) NH3+Cl– (d) N2 Cl 8.
Which of the following is the weakest Bronsted base ? (a)
9.
10.
11.
NH2 (b)
N– H (c) CH3
NH2 (d)
CH2 – NH2
Which of the following will react with CH3COCl ? (a) Trimethylamine (c) Dimethylethylamine (b) Dimethylamine (d) Trimethylamine Acetamide is treated separately with the following reagents. Which of these would give methylamine? (a) PCl5 (c) NaOH + Br2 (b) Sodalime (d) Hot Conc. H2SO4 Which of the following shows optical activity? (a) Butanamine - 1 (c) Isopropylamine (b) Butanamine - 2 (d) Ethylmethylamine
AMINES
12.
13.
14.
673
Number of saturated isomeric primary amines possible for the molecular fomula C 3H5N is (a) Zero
(c) 2
(b) 3
(d) 4
Primary amines on reaction with alcoholic KOH yields : (a) Isocyanide
(c) Cyanide
(b) Aldehyde
(d) Alcohol
The compound on reaction with aqueous HNO 2 at low temperature produces oily nitrosoamine is (a) Methylamine
(c) Diethylamine
(b) Ethylamine
(d) Triethylamine
Answer :
1. (b), 2. (a), 3. (d), 4. (d), 5. (c), 6. (a), 7. (c), 8. (a), 9. (b), 10. (c), 11. (b), 12. (a), 13. (a), 14. (c).
qqq
674
+2 CHEMISTRY (VOL. - II)
CHAPTER - 21
ARYL DIAZONIUM SALTS When primary arylamines or their salts react with nitrous acid in ice-cold solution, an important class of compounds called the diazonium salts are formed. They are so called, because N (Di means two, azo from French word azote, meaning nitrogen). they contain diazo group –N + – These salts have the general formula Ar N 2 X where Ar is an aryl group ( e.g. benzene ring) – – – – and X is any anion like Cl , Br , HSO 4 etc. Structurally, they may be represented as
21.1
|
|
|
[Ar – N N:] X and are named by adding diazonium to the parent aromatic compound to which they are related, followed by the name of the anion. For example – + – N N Cl is benzenediazonium chloride, + – H3C N N Br is p-toluenediazonium bromide etc., PREPARATION OF BENZENEDIAZONIUM CHLORIDE :
Aniline is dissolved in dil. hydrochloric acid and the solution is cooled to 0 0 – 50C. Aqueous solution of sodium nitrite is added to the cooled solution in small portions with stirring. The temperature is maintained at 0 0 – 50C. Addition of sodium nitrite solution is discontinued when a drop of the solution turns starch-iodide paper blue indicating the presence of slight excess of nitrous acid. Benzenediazonium chloride is formed in the solution. 0 0 0 –5 C NaNO2 + HCl NaCl + HNO2 0 0 + – C6H5 – NH2 + HNO2 + HCl 0 – 5 C C6H5 N2 Cl + 2H2O (Aniline) (Benzenediazonium chloride) This reaction between a primary arylamine and nitrous acid in presence of a mineral acid to produce aryldiazonium salt is called diazotisation. Nitrous acid is unstable and it is always generated in situ from NaNO2 and mineral acid like HCl or H2SO4. The reaction is carried 0 0 out at a low temperature (0 – 5 C), because both nitrous acid and diazonium salt would decompose at higher temperature.
ARYL DIAZONIUM SALTS
21.2
675
PROPERTIES :
Physical : Dry aryldiazonium salts are unstable crystalline solids and readily explode when in dry state liberating nitrogen gas. Therefore, they are not isolated and are used in the solution in which they are prepared. Chemical : Aryldiazonium salts are very reactive and undergo a large numbers of substitution reactions in which the diazo group is replaced by different univalent atoms or groups. These are all unimolecular nucleophilic substitution reactions. (SAr N1) H2O Ar N2X i.
Ar – N2 + X
slow – N2
+ fast Ar
X CN
Ar OH Ar X Ar CN
Replacement by hydrogen : Synthesis of benzene : When benzene diazonium salt solution is treated with hypophosphorus acid, the diazo
group is replaced by hydrogen and benzene is formed. + _ C6H5N2 Cl + H3PO2 + H2O (Benzenediazonium chloride)
C6H6 + H3PO3 + HCl + N2 (Benzene)
The reaction is carred out simply by dissolving the amine in hypophosphorus acid and cold NaNO2 solution is added. The reaction takes place immediately. The net result is removal of NH2 group from the aromatic ring and hence the process is known as de-amination. Nitro group can also be removed after converting it to – NH 2 group. ii.
Replacement by halogens : Synthesis of chlorobenzene, bromobenzene and iodobenzene
Sandmeyer’s Reaction : When benzenediazonium chloride solution is added to a solution of cuprous chloride dissolved in HCl or cuprous bromide dissolved in HBr, the diazo group is replaced by chlorine or bromine respectively resulting in the formation of chlorobenzene or bromobenzene.
+ _ C6H5 N2 Cl
CuCl, HCl
C6 H5 Cl
+ N2
(Chlorobenzene) _ C6H5 N+ Cl 2
CuBr, HBr
C6 H5 Br
+ N2
(Bromobenzene)
676
+2 CHEMISTRY (VOL. - II)
The reaction is known as Sandmeyer’s reaction. Gatterman used a modified method in which cuprous halides were replaced by copper powder to prepare chlorobenzene and bromobenzene. The reaction is known as Gattermann reaction. + – C6H5 N2 Cl
+ C6H5 N2 Br–
Cu-powder HCl
C6 H5– Cl + N2 (Chlorobenzene)
Cu-powder C6 H5– Br + N2 HBr (Bromobenzene)
The best method for introducing iodine atom in benzene nucleus is to add a saturated solution of potassium iodide to benzene diazonium chloride solution. It is not necessary to add copper salt. + – C6H5 N2 Cl + KI
C6 H5– I + N2 + KCl (Iodobenzene)
iii.
Replacement by cyano group:Synthesis of cyanobenzene or benzonitrile or phenyl cyanide When diozonium salt solution is treated with cuprous cyanide dissolved in aqueous
potassium cyanide, the diazo group is replaced by –CN group and aryl cyanide is formed. This is also called Sandmeyer’s reaction. Benzene diazonium chloride with cuprous cyanide in aqueous KCN yields phenyl cyanide or cyanobenzene. + _ CuCN, KCN C6H5 N2 Cl
C6 H5 – CN + N2 (Phenyl cyanide or cyanobenzene)
This is a method for introducing – CN group and hence – CH2 – NH2 and – COOH groups in the benzene nucleus (how?) C6H5 CH2 NH2 Benzylamine
H2 / Ni
C6 H5–CN Phenyl cyanide
H2 O/ H+
C6H5–COOH
SnCl2 / Conc. HCl C6H5–CHO (Benzaldehyde)
Benzoic acid
ARYL DIAZONIUM SALTS
677
iv.
Replacement by hydroxyl group : Synthesis of phenol When benzenediazonium salt solution is boiled with water containing dil H 2SO4, it undergoes hydrolysis and diazo group is replaced by – OH group. Phenol is formed. + C6H5 N2 Cl + H2O v.
[H+]
C6 H5 — OH + N2 + HCl
boil
(Phenol)
Replacement by fluorine : Synthesis of fluorobenzene
When an aqueous solution of a diazonium salt is treated with fluoroboric acid (HBF 4), diazonium fluoroborate gets precipitated. It is filtered and dried. Unlike other diazonium salts, diazonium fluoroborates are fairly stable. When dry diazonium fluoroborate is heated, it composes to aryl fluoride and boron trifluoride. + HCl
ArN2+ BF4–
ArN2+Cl– + HBF4
Diazonium fluoroborate +
Ar N NBF4
D
Ar – F + N2 + BF3 Aryl fluoride
This reaction is known as Balz - Schiemann reaction. vi.
Replacement by nitro (–NO2) group : Synthesis of nitrobenzene.
Nitro compounds are generally prepared by treating an arenediazonium fluoroborate with an aqueous solution of sodium nitrite in presence of copper powder.
+ N NCl + HBF4
+ N HCl
Benzenediazonium chloride vii.
NO2
NBF4 NaNO2 / Cu
+
N2 + NaBF4
D Benzenediazonium fluoroborate
Nitrobenzene
Coupling reactions :
Arene diazonium salts react with electron rich (highly reactive) aromatic compounds such as phenols and amines to form brightly coloured azo compounds, Ar – N = N – Ar. This reaction is known as Coupling reaction. Azo compounds are used as dyes. Coupling takes place almost exclusively at the para position if it is unoccupied. If it is not, then coupling takes place at ortho position. Coupling with phenols occurs in basic medium (pH 9-10) whereas with amines it occurs in faintly acidic medium (pH 4 – 5) at 273 – 278 K.
678
+2 CHEMISTRY (VOL. - II)
+ N NCl + H Benzenediazonium chloride
– OH
Base 273-278K
N=N
Phenol
p-Hydroxyazobenzene
OH
OH N=N
Base 273-278K
+ N NCl +
+ HCl
CH3
CH3 2-phenylazo-4-methylphenol
p-cresol (Para position occupied)
+ C6H5 – N NCl + H
OH + HCl
N
CH3
C6H5 – N = N
CH3
N
CH3
CH3 (Butter yellow) + HCl
p-N,N-Dimethylazobenzene Azo dyes almost always contain one or more than one SO3– Na+ groups so that they are soluble in water.
+ N NCl + H
+ NaO3S
N
CH3 CH3
+ NaO3S
N
N=N
Methyl orange 21.3
SOME IMPORTANT CONVERSIONS :
1.
Toluene to m- nitrotoluene
Nitration (Toluene)
CH3
CH3
CH3
Reduction
CH3
CH3 CH3COCl – OH
Sn / HCl NO2
CH3
NH2
NHCOCH3
ARYL DIAZONIUM SALTS
679
CH3
CH3
CH3 NaNO2 / HCl
hydrolysis
Nitration
NO2
NO2
O0C
_ Cl N2
NH2
NHCOCH3
NO2
CH3 H3PO2
NO2
H2O
m- nitrotoluene
2.
Benzene to 1, 3, 5 - tribromobenzene NH2
NO2 Nitration (Benzene)
Sn / HCl
Br2/H2O
Br
(Aniline)
(Nitroben-
NH2
Br
Br
zene)
+
NaNO2/HCl
Br
_
N2Cl
Br
Br
H3PO2 H2O
00C Br
3.
Br
Br (1, 3, 5 - tribromobenzene)
Toluene to p-hydroxybenzoic acid CH3
CH3 Nitration (Benzene)
COOH oxdn Na2 Cr2O7 / H2SO4
NO2
COOH Sn HCl
NO2
NH2
680
+2 CHEMISTRY (VOL. - II)
COOH
COOH
NaNO2/HCl
H2O/H boil
0C 0
OH (p- hydroxybenzoic acid)
N2Cl
4.
Benzene to m-bromophenol NO2
NO2 Br2/Fe
Nitration
Br
Sn/HCl
NaNO2/ HCl Br 00C
(nitrobenzene)
(Benzene)
_ N+2Cl
OH H2O / H+
Br
5.
NH2
Br (m-bromophenol)
Aniline to p-toluidine
NH2
NHCH3 CH3Cl
(Aniline)
NH2
NH.CH3HCl HCl
heat CH3 (p-toludine)
ARYL DIAZONIUM SALTS
6.
681
p - Toluidine to m - bromotoluene CH3
CH3
CH3
Acetylation
Br2 NHCOCH3
NH2 (p- toluidine)
Br NHCOCH3
CH3
i. H3+O ii. NaNO2 / H+ iii. H3 PO2
Br (m- bromotoluene)
CHAPTER (21) AT A GLANCE Methods of Preparation + NaNO2/HCl C6H5 N NCl 273-278K
C6H5 NH2
Synthetic applications (Properties) H3PO2+H2O
ArH
CuCl, HCl or Ar – Cl or Ar – Br Sandmeyer’s reaction CuCl, HBr
Ar
HX Ar – X Cu KI Ar – I
+
N
[X = Cl or Br] Gattermann reaction
CuCN, KCN Ar – CN H2O, D HBF4 /D
Ar – OH Ar – F
(i) HBF4 (ii) NaNO2, Cu
Coupling
Schiemann reaction
Ar – NO2
Phenol (Basic)
Ar – N = N
OH
Aniline (Acidic)
Ar – N = N
NH2
682
+2 CHEMISTRY (VOL. - II)
QUESTIONS A. Short Answer questions : (1 mark) 1.
Write the structure of benzenediazonium chloride.
2.
Why benzenediazonium salts are soluble in water ?
3.
What should be the nature of medium (neutral/acidic/basic) for diazotisation of arylamines.
4.
Give only the equations for the synthesis of the following from benzenediazonium chloride. (i) Benzene (ii) Phenol (iii) Chlorobenzene (iv) Fluorobenzene (v) Phenyl cyanide (vi) p-Hydroxyazobenzene
B.
Short Answer Questions (2 marks)
1.
Suggest reasons why excess mineral acid is used in diazo reaction.
2.
Complete the following equations (i) C6H5N2Cl + H3PO2+H2O (ii) C6H5N2Cl (i) HBF4 (ii) NaNO2/Cu, D (iii) C6H5N2Cl + H2O CuCl (iv) C6H5N2Cl HCl (v) C6H5OH
ArN+2Cl OH
(vi) C6H5N2Cl + ArNH2 3.
D
ice cold pH 4-5
Give the chemical equation for the following. (i) Diazotisation reaction (ii) Coupling reaction (iii) Sandmeyer reaction
4.
How will you convert ? (a) Aniline to nitrobenzene, (b) Aniline to iodobenzene
C.
Short Answer Questions (3 marks)
1.
Why the diazonium salts of aromatic amines are more stable than those of aliphatic amines ?
2.
How you will obtain p-hydroxyazobenzene from aniline ?
3.
What is coupling reaction ? give two examples.
4.
How will you carry out the conversion of aniline to m-bromonitrobenzene ?
5.
Give the structure of A, B and C in the following reactions.
ARYL DIAZONIUM SALTS
(i) C6H5N2Cl (ii) C6H5NO2
683
CuCN Fe/HCl
A A
H2O/H
+
HNO2 273-278K
B B
NH3 C6H5OH
C C
D.
Long Answer Questions
1.
Discuss in detail the preparation of benzenediazonium chloride from aniline. Why it is not separated in solid state and used as soon as produced in solution ? How can you convert benzenediazonium chloride to fluoro and iodobenzene.
2.
Describe the synthetic applications of aryl diazonium salts.
3.
How will you obtain (a) Iodobenzine from nitrobenzine (b) m-Dichlorobenzene from benzene
4.
How will you synthesise the following ? (a) Azobenzene from benzene (b) Benzene from azobenzene (c) Azobenzene from nitrobenzene (d) Nitrobenzene from azobenzene
5.
Write notes on : (a) Sandmeyer reaction (b) Coupling reaction
6.
How can benzene diazonium chloride be prepared from nitrobenzene ? Starting from a diazonium salt how can you prepare (a) iodobenzene (b) benzoic acid ?
E.
Multiple Choice Questions
1.
The diazonium salts are the reaction products of the reaction of nitrous acid with (a) primary aliphatic amines (b) primary aromatic amines (c) secondary aliphatic amines (d) secondary aromatic amines Preparation of a diazonium salt from a primary aromatic amine is known as : (a) Coupling reaction (b) Sandmeyer reaction (c) Diazotisation (d) Corey-House synthesis Which of the following reagents is used to prepare benzenediazonium chloride from aniline ? (a) NaN02 + HC1 (b) LiAlH4 (c) NH2NH2 + KOH (d) NaOH Which of the following are optimum temperature conditions for making benzonediazonium chloride from aniline ? (a) 0°C to 10°C (b) 20°C to 25°C (c) 30°C to 40°C (d) 45°C to 50°C
2.
3.
4.
684
5.
6.
7.
8.
9.
10.
11.
12.
+2 CHEMISTRY (VOL. - II)
Benzenediazonium chloride reacts with warm water to give (a) Aniline (b) Phenol (c) Benzene (d) Chlorobenzene Bromobenzenes can be prepared by treating aniline with (a) Conc. HBr (b) Br2/FeBr3 (c) CuBr (d) Nitrous acid then CuBr Chlorobenzene can be prepared by treatment of aniline with (a) Cuprous chloride (b) Chlorine in the presence of UV light (c) Nitrous acid followed by treatment with CuCl (d) Chlorine in the presence of FeCl3. lodobenzene can be prepared by (a) treating chlorobenzene with I2 using FeCl3 catalyst. (b) treating phenol with I2 in NaOH solution. (c) treating benzenediazonium chloride with KI (d) treating benzene with CH3I using A1C13 catalyst. Benzene diazonium chloride reacts with Kl to form : (a) Benzene diazonium iodide (b) m-Diiodobenzene (c) lodobenzene (d) o-, plus p-Diiodobenzene What is the major product of the following reaction ? NH2 NaNO2 CuCN HC1, 0°C (a) Benzonitrile (b) Benzoic acid (c) Nitrobenzene (d) Benzene diazonium chloride Benzene diazonium chloride reacts with hypophosphorus acid to form : (a) Phenol (b) Benzaldehyde (c) Aniline (d) Benzene Benzene diazonium chloride reacts with phenol to form : (a) p-chlorophenol (b) Chlorobenzene (c) p-Hydroxyazobenzene (d) DDT ANSWERS
1. (b) 2. (c) 3. (a) 4. (a) 5. (b) 6. (d) 7. (c) 8. (c) 9. (c) 10. (a) 11. (d) 12. (c)
qqq
BIOMOLECULES
685
UNIT - XIV
CHAPTER - 22
BIOMOLECULES 22.1
INTRODUCTION :
The branch of Chemistry dealing with the structure, composition and the chemical changes which take place in the living system is called biochemistry. Living cells, the fundamental unit of life are composed of some complex biologically important organic molecules otherwise termed as biomolecules. Important biomolecules constituting the ultimate structure of living cells are carbohydrates, fats, proteins, lipids, nucleic acids etc. In this chapter elementary ideas on such biomolecules are described in brief which are related to the living organism in the following sequence. Living organisms ® Organs ® Tissues ® Cells ® Organelles ® Biomolecules. 22.2
CARBOHYDRATES :
Introduction : The glucose, cellulose, starch and glycogen, all belong to the class of organic compounds known as carbohydrates. Carbohydrates are the main source of energy for our body in the form of food. The carbohydrates provide us with three necessities of life, that is food, clothing and shelter. The term “carbohydrate” was originally given to the compounds with general formula, Cx(H2O)y where x and y may be same or different and they were considered to be hydrates of carbon. However, this definition could not hold ground for long due to following reasons: (i)
Compounds like formaldehyde (HCHO), acetic acid (CH3COOH) etc. have the general formula Cx(H2O)y, but they do not show the characteristic properties of carbohydrates.
(ii)
Some carbohydrates, such as deoxyribose (C5H10O4) do not have the required ratio of hydrogen to oxygen.
To accommodate wide variety of compounds, the carbohydrates are now-a-days broadly defined as polyhydroxy aldehydes or ketones and their derivatives or as substances that yield one of these compounds on hydrolysis.
686
+2 CHEMISTRY (VOL. - II)
A.
CLASSIFICATION OF CARBOHYDRATES :
I.
CLASSIFICATION ON THE BASIS OF HYDROLYSIS PRODUCTS :
Carbohydrates are classified into following classes depending upon their behaviour towards hydrolysis. (i)
Monosaccharides
(ii)
Oligosaccharides
and
(iii)
Polysaccharides.
(i)
Monosaccharides : (Greek: Mono = one; Sakcharon= sugar):
Monosaccharides are simple sugars and can not be hydrolysed to still simpler compounds. Monosaccharides contain 3 to 7 carbon atoms each. Monosaccharides are generally sweet to taste, soluble in water and crystalline. Monosaccharides may be further subdivided into trioses, tetroses, pentoses, hexoses, heptoses etc. depending upon the number of carbon atoms they possess. Monosaccharides which contain an aldehyde (–CHO) group are called aldoses. Since the aldehyde group is monovalent, it is present at one end of the carbon chain. Similarly, monosaccharides which contain a keto group are called ketoses. As keto group is divalent, it is present anywhere along the chain. In all, the naturally occurring ketoses, keto group is present at the carbon atom next to the terminal carbon, that is, at C–2. Some important examples of monosaccharides are : Name
Formula
Aldoses
Ketoses
(Aldo Sugars)
(Keto Sugars)
Trioses
C3H6O3
Glycerose
Dihydroxyacetone
Tetroses
C4H8O4
Erythrose
Erythrulose
Pentoses
C5H10O5
Ribose
Ribulose
Hexoses
C6H12O6
Glucose
Fructose
Heptoses
C7H14O7
Glucoheptose
Sodoheptulose
Both these characters, that is, number of carbon atom and the nature of functional group may also be combined into one. For example, glucose is an aldohexose and fructose is a ketohexose
BIOMOLECULES
687
CHO | H—C—OH | HO—C—H | H—C—OH | H—C—OH | CH2OH Glucose (Aldohexose)
CH2OH | C=O | HO—C—H | H—C—OH | H—C—OH | CH2OH Fructose (Ketohexose)
(ii)
Oligosaccharides : (Greek : Oligos = a few): There are carbohydrates which on hydrolysis give two to ten molecules of the same or different monosaccharides on hydrolysis. Depending upon the number of monosaccharide molecules obtained on hydrolysis, they are further classified as di, tri, tetrasaccharides etc. (a)
Disaccharides : These carbohydrates on hydrolysis give two molecules of monosaccharides. The general formula of disaccharides is C n(H2O)n–1 The exmaples of disaccharides are sucrose, lactose, maltose, cellobiose etc.
(b)
Trisaccharides : These carbohydrates on hydrolysis give three molecules of monosaccharides. For example, raffinose, C18H32O16.
(c) (iii)
Tetrasaccharides : These carbohydrates on hydrolysis give four molecules of monosaccharides. Example : Stachyose, C24H42O21.
Polysaccharides : (Poly = many) : These are carbohydrates which on hydrolysis give a large number of monosaccharide molecules. The most commonly occurring polysaccharides are starch, glycogen, cellulose and gums. Their general formula is (C6H10O5)n where n = 100 to 3000.
II.
Sugars and Non-sugars : Carbohydrates can be classified on the basis of their taste as follows :
(a)
Sugars : All the monosaccharides and oligosaccharides are crystalline solids, soluble in water and sweet in taste and hence are termed as sugars.
(b)
Non-sugars : Polysaccharides are amorphous in nature, insoluble in water and tasteless and hence are called non-sugars.
688
III.
+2 CHEMISTRY (VOL. - II)
Reducing and Non-reducing sugars : On the basis of the reducing character of carbohydrates, they can be classified into two types.
(a)
Reducing sugars : All those carbohydrates which contain free aldehydic or ketonic group, thus reducing Tollen’s reagent and Fehling’s solution are called reducing carbohydrates or sugars. Examples are glucose, fructose, maltose, lactose.
(b)
Non-reducing sugars : Those carbohydrates which are incapable of reducing Tollens’ reagent or Fehling’s solution are called non-reducing sugars. Examples : sucrose.
B.
FUNCTIONS OF CARBOHYDRATES : Carbohydrates have a number of diverse functions in the living organisms. These are,
(i)
Structural material : The chief structural material in the cell walls of all plants are cellulose, which is a polysaccharide. We make furnitures from cellulose in the form wood and clothe ourselves with cellulose in the form of cotton fibre. Cellulose is the raw material for industries like textiles, paper etc.
(ii)
Reserve food materials: Some polysaccharides act as reserve food materials for plants and animals. For example, starch is a major food reserve in plants. In seeds, starch acts as a reserve food material for the tiny plant till it is capable of preparing its own food by the process of photosynthesis. Similarly, in animals glycogen acts as a reserve food material, which is present in liver cells. Glycogen is a source of glucose for the blood.
(iii)
As a biofuel : Carbohydrates, such as glucose, fructose, starch, glycogen etc. provide energy for the functioning of living organisms and act as a biofuel. In the living systems, polysaccharides such as starch and glycogen present are hydrolysed by enzymes to glucose. Glucose is then transported to the various cells, where it is oxidised to CO 2 and H2O by a series of enzyme catalysed reactions. The energy thus released during oxidation, provides energy for the functioning of cells. C6H12O6 + O2
6CO2 + 6H2O + energy.
(glucose) (iv)
In nucleic acids : Monosaccharides like ribose and 2-deoxyribose are the essential components of RNA and DNA respectively. Nucleic acids play an essential role in the biosynthesis of proteins.
BIOMOLECULES
689
C.
MONOSACCHARIDES
1.
GLUCOSE
Glucose, the most common monosaccharide is also known as Dextrose because it occurs in nature as the optically active dextrorotatory isomer. Glucose is found in most of the sweet fruits especially grapes and honey. (a)
Preparation : (i)
From starch : Glucose is produced commonly by the hydrolysis of starch with dilute hydrochloric acid at high temperature under pressure. atm (C6H10O5)n + n H2O
HCl 393K, 2 - 3 atm
Starch (ii)
n C6H12O6 Glucose
From sucrose : Sucrose i.e. cane sugar on acid hydrolysis produces an equal amounts of glucose and fructose. C12H22O11 + H2O Sucrose
(a)
H+ D
C6H12O6 + C6H12O6 Glucose
Fructose
Structure of Glucose : The structure of glucose can be elucidated basing on its following characteristics
I.
Open Chain Structure of Glucose : (i)
Molecular formula : Elemental analysis and molecular weight determination have established the molecular formula of glucose to be C 6H12O6
(ii)
Presence of 6-carbon unbranched chain : The complete reduction of glucose with concentrated hydrogen iodide and red phosphorus gives n-hexane proving that glucose molecule is made of an unbranched i.e. a straight chain of six carbon atoms. Conc. HI/Red P C6H12O6 CH3 – CH2 – CH2 – CH2 – CH2 – CH3 D Gluclose n-Hexane
(iii) Presence of 5-OH groups : Glucose forms a pentacetyl derivative when refluxed with acetic anhydride showing the presence of five hydroxyl groups. Since it is a stable compound, no two - OH groups are attached to the same carbon and hence the five -OH groups are on different carbons. Reflux C6H12O6 + 5 (CH3CO)2O ¾¾ ¾¾® C6H7O (OCOCH3)5 + 5 CH3COOH
Glucose
Glucose penta-acetate
690
+2 CHEMISTRY (VOL. - II)
(iv) Presence of C = O Group : Presence of a carbonyl group in glucose is proved by the following observations. (1)
With HCN, it forms an addition product, called glucose cyanohydrin. C6H12O6 + HCN Glucose
(2)
It forms an oxime with hydroxylamine. C6H12O6 + NH2OH Glucose
(v)
C5H111O5CH (OH) CN Glucose cyanohydrin H+
C5H111O5CH = NOH + H2O Glucoxime
Presence of – CHO (aldehyde) group : On mild oxidation with bromine water, glucose is converted to gluconic acid which when reduced with excess of HI yields n-hexanoic acid. HI (O) C5H11O5 × CHO C5H111O5 × COOH CH3 (CH2)4 COOH Br2(aq) Glucose
n-Hexanoic acid
This shows that glucose contains a straight chain of six carbon atoms with – CHO at one end, which has been oxidised to – COOH (vi) Presence of a primary alcoholic group (–CH2OH) : On heating with nitric acid, glucose forms dicarboxylic acid (glucaric acid) with the same number of carbon atoms confirming the presence of a terminal primary alcohol group.
HNO3 OHC – (CHOH)4 – CH2OH ¾¾ ¾¾ ® HOOC – (CHOH)4 – COOH Glucose
Glucaric acid
(vii) Open chain structure of glucose : Basing on the above observations the open chain structure of glucose can be represented as : CHO | *CH (OH) | *CH (OH) | *CH (OH) | CH2OH The above structure having four chiral carbon atoms correspond to 16 ( = 2 4) optically active isomers, out of which one isomer is Glucose. II.
Configuration of Glucose :
Basing on its preparation from lower aldose and its properties Fischer put forth the exact arrangement of different –OH groups in space around the chiral centres and the open chain structure of glucose with correct canfiguration can be reprsented as :
BIOMOLECULES
691
CHO | H — C — OH | HO — C — H | H — C — OH | H — C — OH | CH2OH D (+) Glucose Here D stands for configuration, where as (+) represents the dextrorotatory nature of glucose. This open chain structure is able to explain all the above reaction. CH3–CH2–CH2–CH2–CH2–CH3 n–Hexane
/ HI
Re
,D dP
O AC 2
CHO | H — C — OH | HO — C — H | H — C — OH | H — C — OH | CH2OH
CHO O | (CH–O–C–CH3)4 O | CH2O–C–CH3 Glucose penta acetate CN | CH–OH | (CHOH)4 | CH2OH Glucose cyanohydrine
HCN
NH 2 OH
D (+) Glucose
q) (a Br 2
O3 HN
COOH | (CHOH)4 | COOH Glucaric acid
CH=N–OH | (CHOH)4 | CH2OH Glucoxime COOH | (CHOH)4 | CH2OH Gluconic acid
692
III.
+2 CHEMISTRY (VOL. - II)
Evidence against open chain structure :
The open chain structure of glucose explains most of the properties, but unable to explain the followings (i)
It does not react with NaHSO3 and also with ammonia, although it contains a free–CHO group.
(ii)
Penta acetyl derivative of glucose is not oxidized by Tollen’s reagent or Fehling’s solution, indicating the absence of free –CHO group.
(iii)
Two stereo isomeric forms (a- and b-) are found to exist with different value of specific rotation. Crystallization from alcohol produces a- Glucose (+110°) whereas from pyridine results with b- Glucose (+19°). This behaviour could not be explained by open structure.
(iv)
Mutarotation : The phenomenon involving spontaneous change in specific rotation of the aqueous soluton of an optically active compound without any change in other properties is known as muta-rotation. Muta rotation is observed in glucose. The optical rotation of the aqueous solution of aGlucose (+110°) and b- Glucose (+19°) change gradually with time till finally a constant value of +53° is reached, which represents the state of equilibrium between a-D-glucose and b-D-glucose. a-D (+) Glucose [a]D = + 110°
IV.
Equilibrium mixture [a]D = + 53°
b-D (+) Glucose [a]D = + 90°
Cyclic structure of D-Glucose :
Failure of open chain structure to explain some of the important characeristics of glucose suggests that the –CHO gr is not free, rather it combines with one of the –OH grs to form a cyclic hemiacetal structure. The stable cyclic structure expected is either a six-membered ring (pyranose structure) or a five-membered ring (furanose structure), out of which the pyranose structure for glucose is confirmed explaining the existence of two cyclic forms in equilibrium with open chain structure. 1*CHOH
Furan
Pyran
| H—2C—OH | HO—3C—H | H—4C—OH | H—5C | CH 6 2OH
O
Pyranose structure of glucose
BIOMOLECULES
693
During hemiacetal formation C1 - aldehydic group combines with C5 – OH group resulting with a chiral centre at C1 and thus it has two possible arrangements of H and OH groups around it. Hence D-glucose exists in two stereoisomeric forms i.e. a-D-glucose and b-D-glucose. Such pair of stereoisomers which differ in configuration only around C 1 are called anomers and C1 carbon is called anomeric carbon.
H—1C—OH | H—2C—OH | HO—3C—H | H—4C—OH | H—5C | 6CH OH 2
O || H—C | H—C—OH | HO—C—H | H—C—OH | .. H—C—OH .. | CH2OH
O
a-D- (+) Glucose
D (+) Glucose (Open chain form)
HO—1C—H | H—2C—OH | HO—3C—H | H—4C—OH | H—5C | 6CH OH 2
O
b-D- (+) Glucose
The cyclic pyranose structure of glucose can be more correctly represented by Haworth structures as shown below. 6CH OH 2
CH2OH
5
O H
H
H
OH OH H
1 H OH OH
a-D- (+) Glucose 2.
H
CH2OH OH .. H
H
4 OH OH 3 H
O
H
C=O 1
2 OH
D (+) Glucose (Open chain form)
H
OH
H
1 OH OH H
H
H
OH
b-D- (+) Glucose
FRUCTOSE
Fructose, the important ketohexose is otherwise termed as levulose bcause the naturally occurring form of fructose is laevorotatory. It is found along with glucose in the juices of ripe fruits and in honey. In the combined state it occurs in sucrose. (a)
Preparation
Fructose is obtained from sucrose (cane sugar by warming with dilute sulphuric acid or with enzymes inverstase.
694
+2 CHEMISTRY (VOL. - II)
H2SO4 C6H12O6 + C6H12O6 or invertase Sucrose Fructose Glucose Structure of fructose : Open Chain Structure of fructose : (i) Molecular formula : As per elemental analysis and molecular weight determination, the molecular formula of fructose is C 6H12O6. (ii) Presence of a straight chain of six carbon atoms : On complete reduction with HI and red P, it forms n-Hexane thus proving the presence of a straight chain of six carbon atoms. (iii) Presence of 5 –OH groups : It gives pentaacetyl derivative on reaction with acetic anhydride indicating the presence of five hydroxy groups, each being attached to a separate carbon as fructose is a stable compound. (iv) Presence of carbonyl ( C = O) group : On oxidation with nitric acid fructose forms a mixture of trihydroxy glutaric aicd, tartaric acid and glycollic acid, all containing fewer carbon atoms than six. Thus, the carbonyl group is fructose must be ketonic. (vi) Carbonyl is on C-2 : Fructose forms cyanohydrin which upon hydrolysis and subsequent reduction with HI and red P yields 2-methylhexanoicacid. Since the COOH has appeared in place of CN, it is bonded to C-2 in the cyanohydrin confirming the position of carbonyl group at position 2 on the six carbon chain.
C12H22O11 + H2O (a) I.
1CH OH 2
1CH OH 2
Fructose
Cyanohydrin
| 2C = O ¾HCN ¾¾ ¾® | (CHOH)3 | 6CH OH 2
| 2C CN | OH (CHOH)3 | 6CH OH 2
CH2OH CH3 | | COOH 2 C CHCOOH [H] HO ¾¾ ¾® | ¾¾2¾® | OH (CHOH)3 (CH2)3 | | 6CH OH 6CH 2 3 2-Methylhexanoic acid
(vii) Open-chain structure : From the above observations, the open chain structure of fructose can be represented as : 1CH OH 2 | 2C = O | 3*CHOH | 4*CHOH | 5*CHOH | 6*CH OH 2 This structure has three chiral centres with 8 ( = 2 3) optically active forms out of which one is D (–) fructose.
BIOMOLECULES
II.
695
Configuration of Fructose :
In the presence of excess phenyl hydrazine fructose forms the osazone which is identical with that obtained from glucose. This shows that the configuration of asymmetic carbon atoms C–3, C–4 and C–5 in D-fructose is the same as in D-glucose. Thus, CHO CH = NNHC6H5 CH2OH | | | H—C—OH C = NNHC6H5 C=O Excess Excess | | | HO—C—H HO—C—H HO—C—H C H NHNH2 C H NHNH 2 ¾¾6 ¾5 ¾ ¾ ¾ ¬¾6¾5¾ ¾ ¾ ¾¾ ¾® | | | H—C—OH H—C—OH H—C—OH | | | H—C—OH H—C—OH H—C—OH | | | CH2OH CH2OH CH2OH D - Glucose III.
Osazone
Evidence against open chain structure :
D-fructose Open chain structure with correct configuration
Open chain structure of fructose fails to explain the following : (i)
It does not form addition product with NaHSO3, although it contains a ketonic group.
(ii)
Since an aqueous solution of fructose shows the properties of mutarotation, it shows the existence of two stereo-isomers of fructose (a and b varieties)
(iii) The a - and b - isomers of fructose are also confirmed as it forms two stereoisomeric methyl fructosides. IV.
Ring or cyclic structure of fructose :
To explain the objection led against open chain structure, a five membered cyclic structure i.e. furanose structure was proposed for fructose which is confirmed. HOH2C—C—OH | HO—C—H O | H—C—OH | H—C | CH2OH a-D- (–) Fructofuranose
CH2OH | C=O | HO—C—H | H—C—OH | .. H—C—OH .. | CH2OH Open chain structure of D (–) Fructose
HO—C—CH2OH | O HO—C—H | H—C—OH | H—C | CH2OH b-D- (–) Fructofuranose
696
+2 CHEMISTRY (VOL. - II)
The furanose structures can be better represented by Haworth structures.
O
HOH2C H
CH2OH
H
HO
OH
H
HOH2C
OH
H
a-D- (–) Fructofuranose
O C
H
HO
OH
H
CH2OH
Open chain structure of D (–) Fructose O
HOH2C H
.. OH ..
H
OH HO
CH2OH
OH H b-D- (–) Fructofuranose
D.
DISACCHARIDES
Disaccharides (C12H22O11) are carbohydrates that produce two monosaccharides on acid hydrolysis.
H O/H+ Sucrose ¾¾2¾ ¾ ¾® Glucose + Fructose H O/H+ Maltose ¾¾2¾ ¾ ¾® 2 Glucose H O/H+ Lactose ¾¾2¾ ¾ ¾® Glucose + Galactose Disaccharides are composed of two units of monosaccharides joined by a glycosidic linkage with the elimination of a water molecule. I.
SUCROSE : Sucrose is ordinary table sugar obtained from cane sugar.
(i)
Sucrose is composed of a-D-glucose and b-D-fructose unit being joined by a, b - glycosidic linkage between C–1 of the glucose unit and C–2 of the fructose unit.
BIOMOLECULES
697
6 CH OH 2 H
CH2OH O
5
O H
H
4
H
H
H
1
OH OH H
H
3
1 OH OH
OH
2 OH
– H2O
H
H
a-D- Glucose 6 HOCH2 5 H H
O HO
4 OH
3
O OH 2 CH2OH 1
H
(iii)
O
HOCH2
2 H
H
HO
OH
H
b-D- Fructose (ii)
a, b - linkage
OH
CH2OH
Sucrose
The structure of sucrose explains the following characteristics. (a)
Sucrose does not form an osazone with phenyl hydrazine.
(b)
It does not reduce Tollens’ reagent or Fehling’s solution.
(c)
It does not exhibit mutarotation.
Sucrose is dextrorotary but after hydrolysis gives dextrorotary glucose (+52.5°) and laevorotatory fructose (– 92.4°) and hence the mixture is laevorotatory. Thus hydrolysis of sucrose is associated with change in the sign of rotation, for which the product is termed as invert sugar. + CH O + CH O C H O + H O ¾H ¾ ¾® 12 22 11
2
6 12 6
6 12 6
Sucrose
D (+) Glucose
D (–) Fructose
[a]D = 66.5°
+ 52.5°
– 92.4°
Equimolar mixture [a]D = –19.85° 2.
MALTOSE :
Maltose is obtained from starch, being composed of two a-D-Glucose units joined by an a-glycosidic linkage between C–1 of one unit and C–4 of the other unit.
698
+2 CHEMISTRY (VOL. - II) CH 2OH
CH 2OH O
H
H
H OH OH
OH H
O
H
H
H
1 +4
H
CH2 OH
OH
OH
HO
OH
H
O
H
H
H
1
H
CH2 OH
– H 2O
OH
1
H
OH
H
4
OH
H OH
OH
Two a-D-Glucose units
H
H
O
OH
O
H
H
OH
a-Linkage Maltose
(i)
C–1 of the second Glucose unit in the maltose structure is a hemiacetal carbon and hence can exist in a - and b forms.
(ii)
Since it has a potential aldehyde group, it shows mutarotation.
(iii)
It forms osazone.
(iv)
It reduces Fehling’s solution and Tollen’s reagent. Thus maltose in a reducing sugar.
3.
LACTOSE :
Lactose is found in the milk of all animals, and hence commonly known as milk sugar. Lactose is composed of b-D-Galactose unit and a-D-Glucose unit joined by b-D-Glycosidic linkage beween C–1 of the galactose and C–4 of the glucose unit. CH 2OH
CH 2OH O
OH H OH
H
H
OH 1 H
H
OH
b-D-Galactose
O
H +
CH2 OH
4
H OH
H
H
H
OH
a-D-Glucose
O
OH
– H 2O
OH
O H
OH
O
H
H
1 OH
HO
CH2 OH
OH
H OH
H H
OH
H
H
H
OH
b-Linkage Lactose
(i)
Because of potential aldehyde group present lactose exists in both a - and b - forms.
(ii)
It is a reducing sugar reducing Tollen’s reagent and Fehling’s solution.
(iii)
It shows mutarotation.
E.
POLYSACCHARIDES :
Polysaccharides are made of a large number of monosaccharide units joined together by glycosidic linkages. The most common naturally occurring polysaccharides mainly acting as the food storage or structural materials are :
BIOMOLECULES
(a)
699
STARCH :
Starch is the main contributor of carbohydrates in our diet. It exists exclusively in plants, stored in seeds, roots and fibres as food reserve. The chief sources of starch are cereals, potatoes, corn and rice. Structure : Starch is a polymer of a-glucose and consists of two components. (i)
Amylose : The water soluble component of starch is amylose. The amylose molecule is made up of D-glucose units joined by a - glycosidic linkages between C-1 of one glucose unit and C–4 of the next glucose unit. CH 2OH O
H 4 O
CH 2OH H
H H
H
OH
O
H 1
OH
CH 2OH H
H
4 O
1
OH
H
H
OH
O
H 4 O
a- link
H
H
1
OH
H
H
OH
O
a - link Amylose
(ii)
Amylopectin : Amylopectin has a branched chain structure and insoluble in water constituting about 80-85% of starch. It is composed of chains of 25 to 30 D-glucose unit joined by a-glycosidic linkages between C-1 of one glucose unit and C-4 of the next glucose unit. These chains are in turn connected to each other by 1, 6 - linkages. CH 2OH O
H 4 O
CH 2OH H
H OH
H
H
OH
O
H 1
4 O
H
H OH
H
H
OH
1
a - link O
CH 2OH
4 O
6 CH 2 O
H
Branch at C6
H
H 1
OH
H
H
OH
5
O
OH
H
H
OH
H
Amylopectin
4 O
H
H 1 O
700
(b)
+2 CHEMISTRY (VOL. - II)
CELLULOSE :
Cellulose is the main structural material of trees and other plants. Wood is 50% cellulose, while cotton wool is almost pure cellulose. Other sources of cellulose are straw, corncobs, bagasse and other agricultural wastes. (i)
Structure : Cellulose is a straight-chain polysaccharide composed of D-glucose units being joined by b - glycosidic linkages between C–1 of one glucose unit and C–4 of the next glucose unit. The number of D-glucose units in cellulose ranges from 300 – 2500. H CH2 OH O
H H
CH 2OH O
H
OH
H
O
H
H
CH2 OH O
H
O
OH
H
H
OH
H
O
H OH
H
H
O
OH
H H
OH
b - links Cellulose
(ii)
The grazing animals like, cow, deer etc. can digest cellulose of grass and plants as the enzyme cellulase is present in their stomach, whereas the humans can not digest cellulose because of the absence of cellulase.
(C)
GLYCOGEN :
Glycogen is present in liver and muscles and is known as the reserve carbohydrate of animals or animal starch. When energy i.e. glucose is needed for any work by the body, glycogen is broken down to glucose by enzymes. The structure of glycogen is similar to that of amylopectin and it has 1,6 as well as 1,4 glycosidic linkages, but it is much more branched than amylopectin. 22.3
AMINO ACIDS :
Proteins are linear polymers of a-amino acids. Therefore amino acids are regarded as “building blocks of proteins.” The general structure of a-amino acid is represented as, NH2 | a R—C—COOH | H
BIOMOLECULES
701
Here amino group is present at a-carbon atom, R may be a hydrogen atom or any alkyl group or an aromatic ring or a heterocyclic ring. Hence each amino acid is a nitrogeneous compound containing both an acidic carboxyl and a basic amino group. Except in glycine, a-carbon atom in all the amino acids is asymmetric in nature. About 20 commonly occurring a-amino acids are known, which are obtained by the hydrolysis of proteins. These amino acids differ from one another due to the different nature of their side chain group, R. The properties of side chain present in the amino acid determine the properties of proteins they constitute. A.
Structure of amino acids :
According to the general formula, amino acids contains both an acidic (—COOH) group and a basic (-NH2) group. In fact, these two groups interact resulting in the transfer of a proton from acidic carboxylic acid group to basic amino group, thereby resulting in the formation of an internal salt as shown below. NH3 NH2 R - CH - COO a| R—CH—COOH The dipolar structure of internal salt is known as zwitter ion. The a- carbon atom of all amino acids (except glycine) is asymmetric. Hence, except glycine all amino acids are optically active. They give two stereoisomers, which are mirror images of each other. However, all the naturally occurring amino acids have L-configuration having– NH2 group on the left as –OH group in L-glyceraldehyde. COOH H 2N
CHO
H
HO
R
CH2OH
L-aminoacid
L-glyceraldehyde
COOH | H2N—C—H | R L (–) a - Amino acid
H
COOH | H—C—NH2 | R Mirror
D (+) a - Amino acid
702
B.
+2 CHEMISTRY (VOL. - II)
Classification of Amino acids : Amino acids can be classified mainly in two ways;
(a)
(a)
Neutral, acidic and basic amino acids.
(b)
Essential and non-essential amino acids.
Neutral, acidic and basic amino acids : Amino acids are classified as neutral, acidic and basic according to the relative number of amino and carboxyl groups in the molecule. (i) Neutral amino acids contain one amino group and one carboxyl group. (ii) Acidic amino acids contain one amino group and two carboxyl groups. (iii) Basic amino acids contain two amino groups and one carboxyl group. Examples of each type are given in Table 22.1 Table 22.1 Amino acids derived from proteins Common name
Abbreviated name
Structure
NEUTRAL AMINO ACIDS: Glycine
Gly
Alanine
Ala
Valine
Val
Tyrosine
Tyr
Proline
H2N – CH2 – COOH NH2 | CH3 — CH — COOH CH3 NH2 | | CH3 — CH — CH — COOH NH2 | HO — — CH2 — CH — COOH
Pro N H
ACIDIC AMINO ACIDS :
Aspartic acid
Glutamic acid
CH - COOH
Asp
NH2 | HOOC — CH2 —CH COOH
Glu
NH2 | HOOC—CH2—CH2—CH—COOH
BIOMOLECULES
Common name
703
Abbreviated name
Structure
BASIC AMINO ACIDS :
Arginine
Histidine
Arg
His
HN = C —NH —(CH2)3— CH — COOH | | NH2 NH2 NH2 | HC===C — CH2 — CH — COOH | | N NH C H
Lysine
(b)
Lys
NH2 | H2N — (CH2)4 — CH — COOH
Essential and non-essential amino acids : Amino acids can also be classified as essential and non-essential amino acids.
C.
(i)
Those amino acids that can not be synthesised by the body and must be supplied in the diet are called essential amino acids. For example; valine, leucine, isoleucine etc.
(ii)
But the amino acids that can be synthesised by the human body are called nonessential amino acids. Examples are glycine, alanine, tyrosine etc.
PEPTIDES & POLYPEPTIDES Peptides are the condensation products of two or more a-amino acids. The bond between
two adjacent amino acids is a special type of amide bond, known as the peptide bond and the chain thus formed, is called a peptide chain. Peptides can be classified as dipeptides, tripeptides and polypeptides depending upon the number of amino acid molecules taking part in the condensation. When two amino acids condense together, the resulting product is called a dipeptide. Similarly, when theree amino acids combine, the product is called a tripeptide. When four or more amino acids combine in this way, the product is called a polypeptide. Proteins are polypeptides containing atleast 100 or more amino acids, but there is no clear demarcation between polypeptides and proteins.
704
+2 CHEMISTRY (VOL. - II)
As for example, when glycine and alanine condense together, a dipeptide results. O || H2N — CH2— C — OH (glycine)
+
O H | || H — N — CH — C — OH | CH3 (Alanine)
A peptide linkage
– H2O O O H || | || H2N — CH2 — C — N —CH— C — OH | CH3 Glycil - alanine (A dipeptide) 22.4
PROTEINS :
Proteins are essential compounds for living cells. The name protein (Greek : Proteios = pre-eminent or first) was suggested by Berzelius. According to him, proteins are complex organic nitrogeneous substances found in the cells of the living beings and essential to cell structure and cell function. Thus, proteins are vital chemical substances essential for the growth and maintenance of life. Proteins are present in almost all living cells of plants and animals. Chemically, proteins are linear unbranched polymers of a-amino acids. In a protein molecule, a-amino acids are linked together by peptide bonds formed between amino and carboxylic groups of successive amino acids. Hence, proteins are polypeptides with more than 100 amino acids. 1.
CLASSIFICATION OF PROTEINS :
(a) On the basis of structures, the proteins are classified into three main groups namely simple, conjugated and derived proteins. (i)
Simple proteins give aminoacids on hydrolysis, carried out by acid, alkali or enzyme. Examples are globulins, collagens, elastins etc.
(ii)
Conjugated proteins are simple proteins which are bonded with non-proteinous prosthetic groups. Examples are nuclein, nucleohistone, haemoglobin, chlorophyll, cytochrome, albumin, casein, serum proteins etc. Albumin contains glycoprotein as protein and carbohydrates as prosthetic group. Casein has phosphoprotein and phosphoric acid other than nucleic acid. Serum protein has lipoprotein and lipid as prosthetic group.
(iii)
Derived proteins are obtained from simple proteins by regulated hydrolytic process. (b) According to solubility, proteins are classified as fibrous proteins and globular proteins.
BIOMOLECULES
705
(i)
Fibrous proteins are insoluble in water and possess long threadlike structures. They are found in nails, horn, hair (keratin), wool, silk and feathers.
(ii)
Globular proteins are soluble in water, acid and alkali. These are highly branched and cross-linked. These proteins perform various functions related to maintenance and regulation of life process. They include all enzymes, many hormones (insulin), albumin in egg, haemoglobin, antibodies responsible for allergies and for defence against foreign organisms.
2.
STRUCTURE OF PROTEINS :
The complete structure of a protein is quite complex. Therefore, their structures are usually discussed in four different levels such as primary, secondary, tertiary and quaternary structures of the protein. (i)
Primary structure : The primary structure of proteins refers to the sequence in which various amino acids are linked together by peptide linkages. The primary structure of a protein is determined by its successive hydrolysis with acids, alkalies or enzymes. Successive hydrolysis of proteins yield different products having decreasing molecular masses as shown below. Proteins
Proteoses
Peptones
Polypeptides
Simple Peptides
a - amino acids
In primary structure, the peptide bonds form the backbone and the side chains of amino acids project outside the peptide backbone. H
O
H
1.24A° N C H
° 3A 1. 5
R
C
1 .3 2A °
N O
C ° 7A 1. 4
H
R
N C O
Fig : 22.1 Primary structure of protein The amino acid sequence of a protein determines its function and is critical to its biological activity. The amino acid sequence in proteins can be determined by taking the following generalisations. They are (a) Proteins are made up of L-amino acids only; (b) Sequence of amino acids along the protein chain is random; and (c) Even a change of just one amino acid can drastically alter the properties of the entire protein molecule. For example a disease called sickle cell anaemia is caused when one a-amino acid is replaced in haemoglobin. (ii)
Secondary structure : The secondary structure of proteins refers to the conformation which the polypeptide chains assume as a result of hydrogen bonding between the carboxylic
706
+2 CHEMISTRY (VOL. - II)
acid and amino groups. Two different secondary structures are possible depending on the size of the side chain (R-) (a)
a-Helix structure : When the size of the R- group is very large, intermolecular hydrogen bonding occurs between C = O group of one amino acid unit with N — H group of the fourth amino acid unit present in the chain. As a result, the polypeptide chain coils up into a spiral structure, called right handed a- helix structure. a- Helix structure are seen in most of fibrous proteins like a-keratin in hair, nail and myosin in muscles.
H | N
C
||
R
H | N
H C
C
||
O
O
C
||
H
O
|
||
R C
N
C
H
C
N | H
R
H
O
C
R
H
H | N
C
N | H
||
C
O
||
O
C
||
O
H | C N || O H H | | N C N
||
O
H | N
||
O C
R
H
N |
H
C
C
||
O
O
||
C C
R
H
Fig : 22.3 b-Pleated sheet structure
Fig : 22.2 a-Helix structure (b)
b-pleated sheet structure: When the size of R-group is small, the polypeptide chains lie side by side in a zig-zag manner with alternate R- groups on the same side situated at fixed distances. As a result of which intermolecular hydrogen bonding occurs between two neighbouring chains. This results in the formation of a flat sheet structure.
(iii)
Tertiary structure : An a- helix may be considered to be a piece of a rope which is free to bend, twist and fold. The tertiary structure of a protein refers to the final three dimensional shape that results from the twisting, bending and folding of the protein helix with main forces operating are hydrogen bonding, disulphide linkage, and van der Waal’s forces of attraction. Tertiary structures of proteins results with fibrous and globular molecular shapes.
(iv)
Quaternary structure : Complex proteins are formed by the combination of two or more polypeptide chains. Each chain is a complete protein with a characteristic primary, secondary and tertiary structure. The quaternary structure refers to the way in which these polypeptide chains of a complex protein are associated with each other.
BIOMOLECULES
3.
707
BIOLOGICAL ROLES OF PROTEINS :
Proteins are of great importance to biological systems. These are most essential to life and perform various functions. Some of their important roles are given below:
4.
(i)
Many proteins act as catalysts, which enhance the rate of chemical reactions to an appriciable extent as required by the living cells and are known as enzymes.
(ii)
The fibrous proteins act as “components of the tissues” holding together the skeletal elements. For example, collagen is a structural unit of connective tissues.
(iii)
Protein hormones regulate the growth of plants and animals, besides controlling many other physiological functions.
(iv)
Interferon is a cellular protein, which is naturally produced by the body in response to virus infections.
(v)
Blood plasma, which is a solution of proteins in water is used for the treatment of shock produced by serious injuries and operations.
DENATURATION OF PROTEINS :
The process after which proteins lose their physiological activity and certain other properties is called denaturation which is caused by
(a)
(b)
(i)
action of bacteria
(ii)
action of heat
(iii)
by shaking with alcohol and
(iv)
by treatment with acids.
Examples of denaturation : (i)
Boiling of an egg : Coagulation of the white of an egg by the action of heat is a very common example of denaturation.
(ii)
Formation of cheese : The coagulation of milk in the presence of an acid (lemon juice) to form cheese is an example of denaturation.
Cause of denaturation :
Denaturation is due to the disruption of the secondary and tertiary structure due to the breaking of hydrogen bonding and salt bridges in proteins causing uncoiling of the protein molecules from an ordered and specific confirmation into a more random conformation resulting into its coagulation from solution.
Denaturing Agent
Natural protein
Denatured protein
708
(c)
+2 CHEMISTRY (VOL. - II)
Renaturation :
The reversible process by which the original form of the denatured protein can be regained is known as renaturation. Protein denaturation may or may not be reversible. Coagulation of white of an egg on heating is an irreversible process. But in case of reversible denaturation, the protein can be coagulated from their colloidal solution by saturating the solution with soluble salts like ammonium sulphate, or by the addition of water or alcohol thus rejoining the original form. 22.5
ENZYMES :
Enzymes are defined as the complex nitrogenous organic compounds produced by living plants and animals. They are polypeptides i.e. proteins of high molecular mass responsible for catalyzing natural processes prevailing in the bodies of animals and plants, for which the enzymes are also termed as biochemical catalysts and the phenomenon is termed as biochemical catalysis. Some examples of enzyme catalysis are : (a)
Inversion of cane sugar : The enzyme invertase converts cane sugar into glucose and fructose. C12H22O11(aq) + H2O(l) ¾Invertase ¾ ¾ ¾¾® C6H12O6(aq) + C6H12O6(aq) Glucose Fructose
(b)
Conversion of milk into curd : The enzyme lacto bacilli is responsible for the conversion of milk into curd.
(c)
Conversion of glucose into ethyl alcohol : The zymase enzyme converts glucose into ethyl alcohol and carbon dioxide. C6H12O6(aq) ¾Zymase ¾ ¾ ¾® 2 C2H5OH(aq) + 2 CO2(g) Glucose Ethyl alcohol
(d)
Conversion of starch into maltose : The diastase enzyme converts starch into maltose. 2(C6H10O5)n (aq) + n H2O(l) ¾Diastase ¾¾¾ ¾® n C12H22O11 (aq) Starch Maltose
(e)
Conversion of maltose into glucose : The maltase enzyme converts maltose into glucose. C12H22O11 (aq) + H2O (l) ¾Maltose ¾¾¾ ¾® 2 C2H12O6(aq) Maltose Glucose
(f)
Decomposition of urea into ammonia and carbon dioxide : The enzyme urease catalyses the decomposition of urea. NH2CONH2 (aq) + H2O (l) ¾Urease ¾ ¾¾® 2NH3 (g) + CO2(g)
BIOMOLECULES
709
(g)
In stomach the pepsin enzyme coverts proteins into peptides while in intenstine, pancreatic trypsin converts proteins into amino acids by hydrolysis.
1.
NATURE OF ENZYMES : (i)
Enzymes are globular proteins.
(ii)
Like proteins enzymes are amphoteric by nature.
(iii) Enzymes also get denatured like proteins. (iv) When purified enzymes are injected in the body, specific antibodies are produced. (v)
They are generally named after the compound or class of compounds on which they are effective. For example, the enzyme is named maltase which catalyzes hydrolysis of maltose. Maltose ¾Maltase ¾¾¾ ¾® 2 Glucose.
(vi) Enzymes are also named after the reactions. For example the enzymes are named as oxido reductase enzymes, when they catalyze the oxidation of one substrate with simultaneous reduction of another substrate. The name of the enzyme always ends with suffix-ase. 2.
PROPERTIES OF ENZYMES : (i)
Catlytic efficiency : Enzymes are very efficient catalysts. A very small quantity of any enzyme is needed to catalyse a reaction. For example, the enzyme rexin coagulate over a million times its weight of milk during cheese formation.
(ii)
Specificity : Enzymes are highly specific in nature. An enzyme catalyses only a particular reaction. Invertase can break up sucrose into glucose and fractose, but fails to breake up a similar disaccharide, maltose, which can only be broken by another enzyme maltase. Similarly urease hydrolyses urea to ammonia and carbon dioxide but does not hydrolyse N-methyl urea which has a similar structure. NH2CONH2 + H2O ¾Urease ¾ ¾¾® 2NH3 + CO2 CH3NHCONH2 + H2O ¾Urease ¾ ¾¾® No action
(iii) Effect of temperature : Enzymes are most effective near body temperature i.e. the rate of enzyme reaction becomes maximum at the optimum temperature with the range 298 - 310K. In general all the chemical reactions proceed faster with increase in temperature, bu the rate of enzyme catalyzed reaction first increases, become maximum at about 35°C and then decreases at higher temperature.
710
+2 CHEMISTRY (VOL. - II)
(iv) Effect of pH : The rate of an enzyme-catalysed reaction is maximum at a particular pH called optimum pH, which is between pH values 5-7. (v)
Effect of co-enzymes and activators : The enzymatic activity is increased in the presence of small organic molecules called co-enzymes which are derived from vitamins such as thiamine, riboflavin etc. Metal ions like Na+, Mn2+, Co2+, Ca2+ etc when weekly bonded to enzyme molecules increase their catalytic activity, for which there are called activators. For example amylase in presence of sodium chloride are catalytically very active.
(vi) Influence of inhibitors : The enzyme activity can be reduced or inhibited by the presence of certain compounds known as enzyme inhibitors. 22.6 HORMONES : A hormone is a chemical released by one or more cells that affects cells in other part of the organism. Only a small amount of hormone is required to alter cell metabolism. All multicellular organisms produce hormones. Plant hormones are called Phytohormones. Hormones in animals are often transported in the blood. Cells respond to a hormone when they express a specific receptor for that hormone. The hormone binds to the receptor protein, resulting in the activation of a signal transduction mechanism that ultimately leads to cell type - specific responses. Endocrine hormone molecules are secreted by ductless glands directly into the blood stream whereas exocrine hormone molecules are secreted directly into a duct and from the duct they flow into the blood stream. Then they move to different parts of the body and exert strong regulatory influence on the chemical processes taking place there. Chemical classes of Hormones Vertebrate hormones fall into three chemical classes. (i)
Amine derived hormones : Examples are catecholamines and thyroxine.
(ii)
Peptide hormones : They are protein hormones and include insulin and growth hormone. More complex protein hormones bear carbohydrate side chains and called glyocoprotein hormones. Luteinizing hormone, follicle stimulating hormones and thyroid stimulating hormones are glycoprotein hormones.
(iii)
Lipid and phospholipid - derived hormones : They are the derivatives of linoleic acid and arachidonic acid and phospholipids. The main classes are the steroid hormones that derive from cholesterol and cortisol. Calcatriol is a sterol hormone. The adrenal cortex and the gonads are the main sources of steroid hormones.
BIOMOLECULES
711
Some hormones in man are given in the following table. Hormone
Source
Function
1.
Adrenalin -
Adrenal medulla -
Increases pulse rate & blood pressure, releases glucose from glycogen and fatty acids from fats.
2.
Testosterone -
Testis -
Normal functions of male sex organ.
3.
Estrone or Estradiol
- Ovary -
Normal functioning of female sex organ.
4.
Insulin -
Pancreas -
Metabolism of Glucose
5.
Cortisone -
Adrenal cortex -
Metabolism of water, fats, proteins and carbohydrates.
6.
Pituitary hormone -
Pituitary glands -
Stimulation of thyroid gland, testis, ovary and mammary glands.
Effects of Hormones : A few important effects of hormones are 1.
Stimulation or inhibition of growth.
2.
Mood swing
3.
Induction or suppression of apoptosis (Programmed cell death)
4.
Activation or inhibition of immune system.
5.
Regulation of metabolism.
6.
Preparation of the body for fighting sex, mating and other activity.
7.
Preparation of the body for a new phase of life.
8.
Control of reproductive cycle.
A hormone may also regulate the production and release of other hormones. Plant Hormones : An example of plant hormone is a series of substances called auxins. Auxins increase the length of most plant cells and thereby contribute to the growth and elongation of the plant. Another plant hormone is abscisic acid, which is produced in mature leaves and inhibits growth in developing leaves and germinating seeds. Another hormone ethylene encourages ripening and dropping of leaves and fruits from the trees. Two important growth regulating hormones are the gibberellins and cytokinins. Gibberellins affect plants by stimulating their growth via rapid stem elongation. Cytokinins induce plant cells to undergo mitosis, therefore they encourage increased growth in the roots and stems in plants. They also enhance flowering and stimulate some type of seeds to germinate.
712
22.7
+2 CHEMISTRY (VOL. - II)
VITAMINS :
Vitamins are defined as the naturally occurring organic compounds required in the diet in very small quantities to maintain the normal health and development of the organism. Vitamins, themselves do not supply much energy to the body, but in small quantities perform specific and vital functions such as energy transformation reactions in the body. Dificiency of a particular vitamin causes specific disease. Excess of vitamins is also harmful and hence vitamin pills should not be taken without the advice of Doctor. Classification of vitamins : Vitamins can be categorized into two groups depending on their solubility in fat or water. These are : A.
Fat soluble vitamins :
Vitamins like A, D, E and K are soluble in fat and oils, but isoluble in water. These are found to be stored in liver and adipose tissues. B.
Water soluble vitamins :
Water soluble vitamins are vitamin C and vitamin B complex. These vitamins must be supplied regularly in diet to our body as they are readily excreted in urine and can not be stored in our body excepting B12. The sources, properties and functions of some of the important vitamins are discussed here. 1.
Vitamin A (Retinol) : This belongs to a class of organic compounds called carotenoids to which carotenes belong. Vitamin A is a carotene derivative, insoluble in water, but soluble in fats and oils. It is not easily destroyed on heating. It is available in cod liver oil, egg yolk, carrots, milk, butter and green vegetables. Function : Deficiency of vitamin A results in the retarded growth and a change in epithelial cells. Deficiency of this vitamin causes nightblindness, xerosis in which the skin becomes dry and xerophthalmia in which the corneas of eyes become opaque.
2.
Vitamin B Complex : This consists of a number of complex substances like vitamin B 1 or thiamine, vitamin B2 or riboflavin, vitamin B6 and vitamin B12 or cyanocobalamine. (i)
Vitamin B1 (Thiamine) : It is a water-soluble white crystalline compound and stable up to 1000C in dry condition. Stability to heat decreases in moist condition. Thiamine is available in milk, green vegetables, meat, egg, yeast, cereals, nuts and rice polishings. Function : Its deficiency in diet causes loss of appetite, and a severe disease called beriberi.
BIOMOLECULES
(ii)
713
Vitamin B2 (Riboflavin) : It is an orange yellow water-soluble crystalline compound and quite stable to heat, but sensitive to light. Vitamine B 2 is available in milk, green vegetables, egg, liver and rice polishings. Function : Deficiency of vitamin B2 causes skin diseases, sore tongue and anaemia.
(iii) Vitamin B6 : It consists of three similar components pyridoxal, pyridoxine and pyridoxamine and is water-soluble. It is easily destroyed by heat and ultraviolet radiation. The chief source of vitamin B6 is cereal grain, egg yolk, meat, fish, milk and cabbage. Function : Deficiency of vitamin B6 causes nervous disturbances and convulsions. (iv) Vitamin B12 (Cyanocobalamine) : It is a large complex molecule which has a central cobalt atom coordinated to four nitrogen atoms. Vitamin B 12 is water-soluble red crystalline solid. It is present in meat, fish, egg and curd. Function : Lack of this vitamin causes a bad form of anaemia. 3.
Vitamin C (Ascorbic acid) : It is insoluble in fats and oils, but soluble in water. Vitamin C is destroyed in contact with oxygen of air even at room temperature since it undergoes reversible equilibrium. It loses effectiveness upon open storing and therefore, vitamin C tablets are kept in sealed strip foils. A major part of this vitamin is lost when vegetables are cooked in open pans in contact with air. Therefore, cooking of vegetables in closed pans or pressure cookers is considered desirable to avoid loss of vitamin C. It is present in fresh vegetables, citrus fruits (lemon, orange), tomatoes, chillis, amla, papita etc. Function : Absence of vitamin C is a diet causes scurvy and pain in joints.
4.
Vitamin D (Calciferols) : It is a white crystalline solid soluble in fats and oils only and present in milk, egg, liver, cod liver oil etc. Morning sun rays falling on the skin form vitamin D from the sterols present under the skin. Function : Vitamin D helps in the development of bones and teeth in growing children. Its deficiency causes rickets in which the bones and legs get curved and the teeth get deformed.
5.
Vitamin E (a-Tocopherol) : This vitamin, also known as ‘‘antisterility’’ vitamin is a light yellow oily substance soluble in fats and oils. It is present in animal and vegetable oils, cotton seed oil, maize oil, pea-nut oil, wheat germ oil, sunflower oil etc. Function : Deficiency of vitamin E causes damage to the reproductive system of both males and females, and increased fragility of RBCs and muscular weakness.
6.
Vitamin K : This vitamin is a mixture of vitamin K 1 and K2. It is fat soluble and stable to heat. Vitamin K is present in leafy vegetables, fish, eggyolk and liver. Function : It plays a vital role in coagulation properties of blood. Its deficiency leads to profuse bleeding even from small wounds and causes disturbances in bleeding time and clotting time of blood.
714
22.8
+2 CHEMISTRY (VOL. - II)
NUCLEIC ACIDS :
Nucleic acid was first isolated in 1871 by a Swiss physician and chemist Friedrich Miescher. He isolated a substance from the nuclei of the pus cells and named it as nuclein. This substance was quite different from the carbohydrates and proteins. Later, it was found that the nuclein had acidic properties and it was renamed by Altman in 1889 as nucleic acid. Nucleic acid is a substance, which is responsible for the transmission of hereditary characters. Nucleic acids are long chain biopolymers present in most living cells either in the free state or bound to proteins as nucleoproteins. The nucleic acids are biopolymers of high molecular mass with nucleotide as their repeating units. As regards their elemental composition, the nucleic acids contain carbon, hydrogen , oxygen , nitrogen and phosphorous. (a)
Composition of Nucleic acids :
We know, nucleotide is the monomeric unit of nucleic acid. A nucleotide is further made up of a phosphoric acid unit and a nucleoside. A nucleoside is made up of two components, that is a pentose sugar and a nitrogen containing heterocyclic base. Thus, in general, a nucleotide and a nucleic acid can be represented as follows: Base O || Sugar — O — P — OH |– O (A nucleotide) Base
Base
O O || || — Sugar — O — P — O — Sugar — O — P — O — | | –O –O (A nucleic acid) Two types of sugars are present in nucleic acids. They are b-D ribose and b-D-2deoxyribose. Both these pentose sugars are present in furanose form. The bases present in nucleic acids are purine and pyrimidine. 5 HOH 2C 4 H
O H
OH
3 OH
1
H 2 OH
b-D ribose
H
5 HOH 2C 4 H
O H
OH
3 OH
1
H 2
H
H
b-D-2- deoxyribose
BIOMOLECULES
715
The bases derived from purine are adenine (A) and guanine (G). Similarly, the bases derived from pyrimidine are thymine (T), uracil (U) and cytosine (C). The nucleic acid which contains only ribose is known as ribonucleic acid (RNA) and the nucleic acid which contains deoxyribose is known as deoxyribonucleic acid (DNA). Nucleic acids are long chain polymers of nucleotides and therefore they are known as polynucleotides. DEOXYRIBONUCLEIC ACID (DNA) : DNA is composed of two strands of polynucleotides coiled around each other in the form of a double helix resembling a gently twisted ladder. This information was given by James Dewey Watson who received Nobel prize along with Francis Crick and Maurice Wilkins with the foundation of a new field of Molecular Biology. The bases (specific pyrimidine groups) on one strand of DNA are paired with the bases (purine groups) on the other stand with the help of hydrogen bonding, which is highly specific because the structures of bases allow only one mode of pairing. Adenine (A) pairs with only thiamine (T) via two hydrogen bonds while guanine (G) pairs with cytosine (C) through three hydrogen bonds. It is because the sequence of bases in one strand automatically determines that of other.
Double strand helix structure of DNA
The complimentary pairing of nucleotides bases explains how indentical copies of parental DNA pass on to two daughter cells thus transmitting inherent characters (heredity). Har Gobind Khurana, an Indian Scientiest shared the Nobel prize for Medicine and Physiology for cracking the genetic code while working at Cambridge, UK with Marshal Nirenberg and Robert Holley. DNA FINGER - PRINTING : Each person has unique finger prints which occur at the tips of the fingers being used for indentification. But unfortunately now fingerprints can be easily altered by surgery. A sequence of bases on DNA is also unique for a person and information regarding this is called DNA finger printing. It is same for every cell and can not be altered by any treatment. For this reason DNA fingerprinting is now used : (i)
to determine paternity of an individual
(ii)
in forensic laboratories for identification of criminals.
(iii)
to identify dead bodies in any accident by comparing the DNA’s of their parents or children.
(iv)
to identify racial groups to rewrite biological evolution.
716
+2 CHEMISTRY (VOL. - II)
RIBONUCLEIC ACID (RNA) : The process of translating coded message into action requires ribonucleic acid (RNA). The chemical constitution of RNA is almost similar to that of DNA except that the sugar in RNA is ribose and the base is uracil (U) in place of thymine (T) as in DNA. The structure of RNA in most organisms is single stranded, but appears helical because at some places it is wound back upon itself. There are three kinds of RNA (i)
Messenger RNA (m-RNA) : This carries the genetic information from that template of DNA chain to the site of protein.
(ii)
Transfer RNA (t-RNA) : This is the smallest of all the RNAs, single stranded and occurs free in the cytoplasm of the cell, one for each aminoacid found in protein.
(iii)
Ribosomal RNA (r-RNA) : This is present in ribosome and constitutes about 80% of total RNA. r-RNA helps in protein synthesis and carries no message of DNA.
BIOLOGICAL FUNCTIONS OF NUCLEIC ACIDS : The two important biological functions of nucleic acids are replication and protein synthesis. (i) Replication is a process by which a single DNA molecule produces two identical copies of itself. Replication is an enzyme catalysed process. The two identical copies of DNA produced from the original DNA are then passed on to the two new cells resulting from cell division. In this way hereditary characters are tranmitted from one cell to another. DNA is double stranded. (ii) Another important function of RNA is the synthesis of proteins. The RNA molecule present in the cell nucleus is responsible for the synthesis of all the proteins present in a cell. Protein synthesis is a very fast process.
CHAPTER (22) AT A GLANCE 1.
Carbohydrates : Carbohydrates are polyhydroxy aldehydes or ketones and their derivatives or substances that yield one of these compounds on hydrolysis.
2.
Aminoacids : Aminoacids are regarded as building blocks of proteins.
3.
Proteins : Proteins are linear or branched polymers of a-amino acids linked by peptide bonds. Enzymes, many hormones and antibiotics are also proteins in nature.
4.
Nucleic acids : Nucleic acids contain long chain of alternating sugar and phosphate units with bases attached to the sugar residue. DNA contains deoxyribose sugar and bases are adenine, guanine, cytosine and thiamine. RNA contains ribose suagr and bases like adenine, guanine, cytosine and uracil. While DNA is present in the molecule of the cell, RNA is
BIOMOLECULES
717
present outside the nucleus in the surrounding fluid called cytoplasm. DNA is the chemical bases of heredity and have the coded message for the proteins to be synthesised in the cell. The three types of RNA i.e. m-RNA, t-RNA and r-RNA carry out the synthesis of proteins in the cell. RNA is single stranded whereas DNA is double stranded. 5.
Vitamins : Vitamins are specific organic compounds required by animals, bacteria and micro-organisms in small amounts for the maintenance and normal growth of life. Vitamin A, D, E and K are fat soluble while vitamin B and C are water soluble. Deficiency of these vitamins causes diseases.
6.
Enzymes : They are considered as conjugated proteins. They are globular proteins (soluble in water, acid and alkali).
7.
Hormones : Chemicals released by one or more cells that affect cells in other part of the organism are called hormones. They may be amine-derived hormones (catecholamines, thyroxine etc), peptide hormones or protein hormones, (insulin, TRH and growth hormones), lipid and phospholipid derived hormones - steroid hormones (testoterone, cortisol, calcitriol etc). Plant hormones are called auxins. Other plant hormones are gibberellins and cytokinins, ethylene, abscisic acid etc.
QUESTIONS A.
VERY SHORT ANSWER QUESTIONS (1 MARK EACH)
1.
Enzymes are
2.
The carbohydrate present in DNA molecule is
3.
The general formula of a carbohydrate is
4.
DNA stands for
5.
RNA stands for
6.
Name the nucleic acids which are used in protein synthesis
7.
Name the bases present in both RNA and DNA.
8.
Which vitamin is Ascorbic acid ?
9.
Which vitamin is obtained from sunlight ?
10.
Which elements are generally present in proteins ?
11.
What are the hydrolysis products of sucrose ?
12.
Name two carbohydrates which act as biofuels.
13.
What causes the disease sickle cell anaemia ?
14.
_________ is the chemical basis of heredity.
15.
Name the building blocks of proteins.
16.
Which polysaccharide is stored in the liver of animals.
type of proteins
718
+2 CHEMISTRY (VOL. - II)
[Answers. 1. Conjugated, 2. Deoxyribose, 3. Cx(H2O)y, 4. Deoxyribonucleic acid, 5. Ribonucleic acid, 6. DNA & RNA, 7. Adenine, guanine and cytosine, 8. Vitamin C 9. Vitamin D, 10. C, H, O, N, S & P., 11. Glucose & Fructose, 12. Starch and glycogen, 13. Defective Haemoglobin, 14. DNA, 15. a - Amino acids, 16. Glycogen] B.
SHORT ANSWER QUESTIONS (2 MARKS EACH) :
1.
What are oligosaccharides ? Give one example.
2.
What are zwitter ions?
3.
What is meant by inversion of sugar?
4.
What are polysaccharides? Give one example.
5.
What happens when protein is denatured?
6.
What is a prosthetic group?
7.
What are proteins?
8.
What are monosaccharides ? Give one example.
9.
Why can not vitamin C be stored in our body?
10.
What are the different types of RNA found in the cell?
11.
What is the difference between a nucleoside and a nucleotide?
12.
What are enzymes?
13.
What are anomers?
14.
What are nucleic acids?
15.
How does glucose differ from fructose?
C.
SHORT ANSWER QUESTIONS (3 MARKS EACH) :
1.
Classify Carbohydrates giving one example of each class.
2.
What are the important units of protein ? What is the name of the bond between these units ?
3.
Give two biological functions of nucleic acids.
4.
Differentiate between DNA and RNA.
5.
Where does the water present in the egg go after boiling the egg?
6.
How are the vitamins classified?
7.
Explain mutarotation of D-glucose.
8.
Name four important vitamins of vitamin B complex. Deficiency of which of these vitamins causes beriberi ?
9.
Name three hormones from three different classes of vertebrate hormones.
10.
Name two plant hormones giving one function of each of them.
BIOMOLECULES
719
11.
What are reducing and non-reducing sugars?
12.
What is the basic differenc between starch and cellulose?
13.
How can you explain the absence of aldehyde group in the pentaacetate of D-glucose.
14.
Differentiate between globular and fibrous proteins.
15.
Explain why glucose and fructose give the same osazone, on treatment with excess phenyl hydrazine.
D.
LONG QUESTIONS :
1.
What are proteins ? How are they classfied ? Discuss their primary, secondary and tertiary structures.
2.
What are vitamins ? Classify the vitamins along with the roles of various vitamins in our body.
3.
How glucose is synthesized? Discuss the ring structure of D-Glucose.
4.
What are monosaccharides? Elucidate the structure of D-Fructose.
5.
What are hormones? How are they classified. Give one example of each along with its functions.
6.
What are plysaccharides? Discuss the structure of starch and cellulose.
E.
MULTIPLE CHOICE QUESTIONS :
1.
Which one of the following is a disaccharide ? (a) glucose (b) starch (c) sucrose (d) cellulose
2.
Ascorbic acid is the name of (a) a vitamin (b) an enzyme (c) an aminoacid (d) a hormone
3.
Which of the following is a monosaccharide ? (a) maltose (b) fructose (c) sucrose (d) lactose
4.
Ribose is an example of (a) ketohexose (b) disaccharide (c) pentose (d) polysaccharide
5.
Enzymes are (a) proteins (b) minerals (c) oils (d) fatty acids
6.
The main structural feature of protein is (a) the ester linkage (b) the ether linkage (c) the peptide linkage (d) all of the above
7.
In nucleic acids, the sequence is (a) phosphate - base - sugar (b) sugar - base - phosphate (c) base-sugar - phosphate (d) base- phosphate - sugar
8.
a- and b- glucose differ in the orientation of –OH group around (a) C1 (b) C2 (c) C3 (d) C4
720
9.
+2 CHEMISTRY (VOL. - II)
Vitamin A deficiency leads to a disease known as (a) Beriberi (b) Night blindness (c) Scurvy (d) All
10.
Insulin is (a) Antibiotic (b) Antiseptic (c) Hormone (d) Vitamin.
11.
Which of the following is known as antiscurvy factor ? (a) Vitamin A (b) Vitamin C (c) Vitamin D (d) Vitamin E
12.
The fat -soluble vitamin is (a) Retinol (b) Thiamine (c) Riboflavin (d) Pyridoxine
13.
Vitamin B1 is called (a) Riboflavin (b) Thiamine (c) Pyridoxine (d) Cyanocobalamine
14.
Which of the following contains cobalt ? (a) Chlorophyll (b) Haemoglobin (c) Vitamin C (d) Vitamin B 12
15.
Hair, Finger nails, hoofs etc. are all made of (a) Fat (b) Vitamins (c) Proteins (d) Iron
16.
Proteins are polymers of amino acids. Which of the following is not a protein ? (a) Wool (b) Nails (c) Hair (d) DNA
17.
The chemical messengers produced in ductless glands are (a) Vitamins (b) Hormones (c) Lipids (d) Antibiotics
18.
The function of DNA is (a) to synthesise RNA (b) to contain message to carry out specific protein synthesis (c) to carry hereditary characteristics from generation to generation (d) all of the above.
19.
Adenosine is an example of (a) Nucleoside (b) Nucleotide (c) Purine base (d) Pyrimidine base
20.
Two purine bases present in both DNA & RNA are (a) Guanine & Adenine (b) Guanine and Uracil (c) Adenine & Thiamine (d) Cytosine and Uracil
21.
A mixture of amylose and amylopectin is called (a) starch (b) cellulose (c) lactose (d) sucrose
22.
Which vitamin is water soluble? (a) Vitamin-K (b) Vitamin-B (c) Vitamin-E, (d) Vitamin-A
23.
The number of chiral carbons in b- D(+) glucose are (a) three (b) four (c) five (d) six
24.
Which hormone contains iodine. (a) Thyroxine (b) Insulin (c) Adrenaline (d) Testosterone
BIOMOLECULES
25.
721
Sugars have the suffix (a) –ol (b) –al (c) –ose (d) –one
26.
27.
28.
The function of enzymes in the living system is to (a) provide immunity
(b) provide energy
(c) catalyse biochemical reaction
(d) transport oxygen
The helical structure of protein is stabilised by (a) Hydrogen bonds
(b) Dipetide bonds
(c) Peptide bonds
(d) van der Waals forces.
Vitamin A is called (a) Ascorbic acid
(b) Retinol
(c) Calciferol
(d) Tocopherol ANSWERS
1. (c)
2. (a)
3. (b)
4. (c)
5. (a)
6. (c)
7. (c)
8. (a)
9. (b)
10. (c)
11. (b)
12. (a)
13. (b)
14. (d)
15. (c)
16. (d)
17. (b)
18. (d)
19. (a)
20. (a)
21. (a)
22. (b)
23. (c)
24. (a)
25. (c)
26. (c)
27. (a)
28. (b)
qqq
722
+2 CHEMISTRY (VOL. - II)
UNIT - XV
CHAPTER - 23
POLYMERS 23.1
INTRODUCTION :
Polymers (Greek words, poly = many and meros = parts) are defined as the large molecules having very high molecular mass consisting of a very large number of repeating structural units joined together through covalent bonds in a regular fashion. The simple molecules from which the repeating structural units are derived are called monomers and the process of formation of polymers from their monmers is called polymerization. For example, (i)
In case of polythene, polymer which is obtained by the polymerization of ethene
CH2 – CH2
molecules the repeating structural unit
is derived from the
nonomers ethene.
( CH – CH
Polymerization
n CH2 = CH2 ¾¾ ¾ ¾ ¾ ¾¾®
2
Ethene (Monomer) (ii)
2
n
Polythene (Polymer)
In case of Nylon 6, 6, two different types of monomers used are hexamethylene diamine and adipic acid interacting with each other with the repeating structural unit H H O O N – CH2)6 – N – C – (CH2)4 – C n NH2 (CH2)6 NH2 + n HOOC (CH2)4 COOH Hexamethylene Adipic acid Polymerisation diamine H
H O
O
N – CH2)6 – N – C – (CH2)4 – C Nylon 6, 6
n
+ (2n–1) H2O
POLYMERS
723
Polymers because of their large size are also termed as macromolecules. But the reverse is not true i.e. all polymers are macromolecules but all macromolecules are not polymers. For example, proteins and nucleic acids are regarded as macromolecules but not polymers as these molecules do not contain repeating structural units. Where as polythene can be regarded as both a macromolecule and a polymer as it contains a number of repeating structural units. 23.2
CLASSIFICATION OF POYMERS : Polymers can be classified into different categories basing upon different factors. These
are : A.
Classification based upon the sources :
Depending on the source from which they are derived, poymers can be categorized into three classes, such as; (a)
Natural polymers : Polymers which are found in nature i.e. in animals and plants are called natural polymers. Examples are : (i)
Proteins, silk, wool etc.
(ii)
Polysacharides : Starch (potatoes, rice etc.), cellulose of cotton and wood.
(iii) Natural rubber (b)
Semi-synthetic polymers : These are chemically modified natural polymers. For example cellulose on acetylation with acetic anhydride in presence of con. H 2SO4 gives cellulose diacetate which is used for making threads of acetate rayon and other materials like films, glasses etc. Other examples are gun cotton and vulcanized rubber.
(c)
Synthetic polymers : Man made polymers like plastics (polyethylene, polypropylene), fibres (nylon and terylene), rubbers (neoprene) adhesives, paints etc. are called synthetic polymers.
B.
Classification based on structure : On the basis of structures, polymers can be classified into three types. These are:
(a)
Linear polymers : In linear polymers the monomers are joined together to form long straight chains of polymers molecules [Fig. 23.1(a)]. The various polymeric chains are then stacked over one another to give a well packed structure resulting with high melting point, high densities and high tensile strength. Examples of linear polymers are high density polythene, nylon, polyester etc.
(b)
Branched chain polymers : In these polymers the manomer units not only combine to produce the linear chain called the main chain, but also form branches of different lengths along the main chain [Fig. 23.1(b)]. Due to branching these polymers do not pack well for which they have lower melting points, densities and tensile streangth than that of
724
+2 CHEMISTRY (VOL. - II)
linear polymers. Examples of this class are low density polythene, starch, glycogen, amylopectin etc. (c)
Cross-linked or three demensional network polymers : In these type of polymers the initially formed linear polymer chains are joined together to form a three-dimensional network structure [Fig. 23.1(c)]. Only two cross-links per polymer chain are required to join together all the long chain polymer molecules to form a giant molecule. Because of the presence of cross-links, these polymers are called cross-linked polymers, which are usually formed from bifunctional and trinfunctional monomers containing strong covalent bonds between various linear polymer chains. Examples are bakelite, melamine etc.
(a) Linear structure
(b) Branched chain structure
(c) Three dimensional network or cross-linked structure Fig. 23.1 : Different structures of polymers. C.
Classification based on the nature of repeating structural units : Depending on the nature of repeating structural units, polymers are of two types:
(a)
Homopolymers : Polymers containing only one type of monomers are called homopolymers. Polymerization
n CH2 = CH2 ¾¾ ¾ ¾ ¾ ¾¾® Ethene (Monomer)
( CH – CH 2
2
Polythene (Polymer)
n
POLYMERS
725
Other examples are polyvinyl chloride (PVC), polyacrylonitrile (PAN), polypropylene etc. (b)
Copolymers : Polymers whose repeating structural units are derived from two or more types of monomers are called copolymers. Nylon-66 is a copolymers of hexamethylene diamine and adipic acid. n H2N – (CH2)6 – NH2 + n HOOC – (CH2)4 – COOH Hexamethylene diamine (Monomer)
Adipic acid (Monomer) Polymerization
HN – (CH2)6 – NH – CO – (CH2)4 – CO
n
+ (2n–1) H2O
Nylon 6, 6 (Copolymer) Other examples are Buna–S–rubber, bakelite, polyethylene terephthalate (PET) etc. D.
Classification based on mode of polymerization :
Polymerization i.e. the process of synthesis of polymers can be carried out in two modes. These are addition polymerization and condensation poymerization. Based on these two methods of synthesis, the poymers can be classified into two categories. (a)
Addition polymers : Addition polymers are those polymers which are formed by the repeated addition of monomers molecules possesing double or triple bonds without elemination of simple molecules like water. These are also referred to as chain growth polymers. For examples : (i)
ethene polymerises to from polyethylene. Polymerization
n CH2 = CH2 ¾¾ ¾ ¾ ¾ ¾¾® Ethelene (ii)
2
2
n
Polyethylene
Vinyl chloride polymerises to form polyvinylchloride (PVC). Polymerization
n CH2 = CH ¾¾ ¾ ¾ ¾ ¾¾® ½ Cl Vinyl chloride (b)
( CH – CH ( CH – CH 2
2
n
½ Cl PVC
Condensation polymers : Polymers which are formed by the codensation reaction of molecules having more than one functional group resulting with the removal of simple molecules like H2O, NH3 or CH3OH. These are also termed as step growth polymers. For examples :
726
+2 CHEMISTRY (VOL. - II)
(i)
Dacron (polyester) is produced by successive condensation of ethylene glycol molecules and terephthalic acid molecules. n HO – CH2 – CH2 – OH + n HOOC –– Ethylene glycol
–– COOH
Terephthalic acid Polymerization O
O
O – CH2 – CH2 – O – C –––
–––– C
n O –– + (2n – 1) H2O
Dacron (polyethylene terephthalate) (ii)
Similarly Nylon – 6, 6 is a condensation polymer. O O n HO – C – (CH2)4 – C – OH + n H2N – (CH2)6 – NH2 Adipic acid Hexamethylenediamine Polymerization O (2n – 1) H2O +
O
C – (CH2)4 – C – NH – (CH2)6 – NH
n
Nylon – 6, 6 E.
Classification based on molecular forces :
Depending on the magnitude of intermolecular forces like hydrogen bond, van der Waals forces and dipole–dipole interactions existing between the adjacent polymer chains, the polymers can be classfied into four categories. These are : (a)
Elastomers : Polymers in which the intermolecular forces of attraction between the polymer chains are the weakest resulting with high degree of elasticity are called elastomers. Elastomers are amorphous polymers consisting of randomely coiled molecular chains of irregular shape having a few cross links. Under the application of an external force, these randomly coiled chains straighten out and the polymer is stretched. But as soon as the force is withdrawn, because of very weak van der Waal forces of attraction, the polymers return to its original randomly coiled state. Thus, whereas weak van der Waals forces of attraction permit the polymer chains to be stretched, the cross links help the polymer to come back to the original position when the force is withdrawn. (Fig. 23.2)
POLYMERS
727
Cross lines Cross links
Strech Relax Cross links
Unstretched
Stretched
Fig. 23.2 : Unstretched and stretched forms of an elastomer Most important examples of elastomers are buna-S, buna-N, neoprene etc. (b)
Fibres : Fibres are the thread forming polymers possessing very high tensile strength, high modulus, but least elasticity which are attributed due to very strong inter molecular forces of attractions like hydrogen bonding or dipole interactions (Fig. 23.3) Most important examples are polyamides (nylons), polyesters (terylene, dacron) and polyacrylonitrile (orlon, acrilan). H C
C O
O
N
N
H
C
C
O
H
O
H
O
H
N
N
C
H
C
N
N
O
Fig. 23.3: Hydrogen bonding in Nylon-6,6 (c)
Thermoplastic polymers : Thermoplastic polymers are the linear or slightly branched long chain molecules which can be easily softened repeatedly on heating and hardened on cooling with little change in their properties because of the inter molecular forces being intermediate between elastomers and fibres. When heated these polymers melt and form a fluid which can be moulded into any desired shapes and then cooled to get the desired product. Some common examples of thermoplastics are polythene, polypropylene, polystyrene, polyvinyl chloride, teflon, polyacrylonitrile etc.
728
+2 CHEMISTRY (VOL. - II)
(d)
Thermosetting polymers : Thermosetting polymers are semi-fluid substances with low molecular masses which when heated in a mould undergo change in chemical composition to give a hard, infusible and insoluble mass due to extensive cross-linking between different polymer chains to give a three-dimensional network solid. (Fig. 23.4).
heat
Uncrosslinked polymer
Highly cross-linked polymer
Fig. 23.4 : Conversion of uncrosslinked polymer into highly cross-linked thermosetting polymer on heating. In contrast to thermoplastic polymers thermosetting polymers can be heated only once as these undergo prermanent change on melting and set into a new solid which can not be remelted. Thus these can not be reused. Common examples are : phenol-formaldehyde resins (bakelite), urea-formaldehyde resins etc. F.
Classification based on the basis of degradability of polymers by micro-organisms:
Polymers may be classified into biodegradable polymers and non-degradable polymers on the basis of their degradability by micro-organisms. (a)
Biodegradable polymers : These are the polymers which are degraded by microorganisms during a certain period thus not posing any threat to our environment. Some new biodegradable synthetic polymers have been designed and developed containing functional groups similar to the functional groups present in biopolymers. Biopolymers like starch, cellulose etc. disintegrate themselves in biological systems during a certain period of time by enzymatic hydrolysis and to some extent by oxidation and hence are biodegradable. Hence they do not cause any pollution. Examples of biodegradable polymers are : (i)
Poly b-hydroxybutyrate-co-b-hydroxyvalerate (PHBV) : It is a copolymer of 3-hydroxybutanoic acid and 3-hydroxypentanoic acid in which two monomer units are connected by ester linkages. OH
OH
n CH3 – CH – CH2 – COOH + n CH3 – CH2 – CH – CH2 – COOH 3-Hydroxy butanoic acid 3-Hydroxypentanoic acid
POLYMERS
729
O – CH – CH2 – C – O – CH – CH2 – C – O CH3
O
CH2CH3
n
+ (2n – 1) H2O
O
PHBV PHBV is used in special packaging, orthopaedic devices and in controlled drug release. When a drug is enclosed in a capsule of PHBV, it is released only when the polymer is degraded in the body. (ii)
Poly-Glycollic acid - Poly Lactic acid : This is a copolymer of glycollic acid and lactic acid, commercially called dextron. This was the first biodegradable polyester used as sutures, that is for stitching of wounds after operation. n HO – CH2 – COOH Glycollic acid
+
n HO – CH – COOH CH3 Polymerization
O – CH2 – C – O – CH – C – O O
n
Lactic acid
+ (2n – 1) H2O
CH3 O
(iii) Nylon 2 - Nylon 6 : It is an alternating copolymer of glycine and e-aminocaproic acid and is biodegradable. n H2N – CH2 – COOH Glycine
+
n H2N – (CH2)5 – COOH e-amino caproic acid
HN – CH2 – C – NH – (CH2)5 – C
n
+ (2n – 1) H2O
O O Nylon 2 - Nylon 6 All the above polymers are plyesters and are therefore, susceptible towards hydrolysis of their ester links. Hence these are completely degraded and absorbed by the body within 15 days to one month for which these are mainly used for medical goods such as surgical sutures, tissue in growth materials or for controlled drug release devices, plasma substitutes etc.
730
+2 CHEMISTRY (VOL. - II)
(b)
Non-degradable polymers : The polymers which are not degraded by micro-organisms within a suitable period and seriously affect the environment leading to environmental pollution are called non-degradable polymers. Most of the commercially synthesized polymers like polythene, PVC, teflon, nylons, polyesters, bakelite etc are the examples of non-degradable polymers.
23.3
GENERAL METHODS OF POLYMERIZATION : The process of polymerization can be carried out by two methods.
A.
ADDITION POLYMERIZATION :
The type of polymerization involving successive addition of monomer units to the growing chain with a reactive intermediate such as a free radical, a carbocation or a carbanion is termed as addition polymerization. This can also be termed as chain growth polymerization as after each act of addition of monomer the reactive intermediate is regenerated leading to increase in chain length. The monomers used are unsaturated compounds like alkenes, alkadienes, and their derivatives with free radical governed addition polymerization is the most common method. (a)
Free radical addition polymerization : Addition polymerization of alkenes or their derivatives involves the chain reactions in which the chain carrier is a free radical formed in the presence of a free radical generating initiator (catalyst) like benzoyl peroxide, acetyl peroxide etc. This vinyl polymerization involves three steps. (i)
Initiation : The initiation of polymerization is a two step process. The first step is the decomposition of the initiator to produce a free radical. O
O D
C6H5 – C – O – O – C – C6H5 ¾¾® Benzoyl peroxide O 2 C6H5 – C – O
·
Bezoyloxy radical
®
·
2 C6H5 + 2 CO2 Phenyl radical
In the second step, the free radical interacts with the vinyl monomer to give rise the initiated monomer radical. ·
·
C6H5 + CH2 = CH2 ® C6H5 – CH2 – CH2 Ethene
POLYMERS
731
(ii)
Propagation : This step involves the addition of the new monomer radical to the monomer molecules in rapid succession to form a growing polymer chain. ·
·
C6H5 – CH2 – C H2 + CH2 = CH2 ® C6H5 – CH2 – CH2 – CH2 – C H2 + (n – 1) CH2 = CH2 C6H5
( CH – CH 2
·
nCH2 – C H2
2
(iii) Termination : In the termination step, the growing chain radical deactivates either by combination or by disproportionation to form a dead polymer. Combination : 2 C6H5 C6H5
( CH – CH 2
·
n CH2 – C H2
2
( CH – CH 2
2
n CH2 – CH2 – CH2 – CH2
( CH – CH 2
2
n C6H5
Polythene (Dead polymer) Disproportionation : 2 C6H5 C6H5
(
·
CH2 – CH2
( CH – CH 2
2
CH2 – C H2
n n
CH = CH2 + C5H6
( CH – CH 2
2
n
CH2CH3
Dead polymers (b)
Cationic polymerization : Cationic polymerization takes place by the addition of a positively charged ion to a vinyl monomer to form a carbocation as the reaction intermedicate. The initiators used for the cationic polymerization include strong protic acids like H2SO4 or Lewis acids like AlCl3, BF3 in the presence of water. The reaction proceeds through three steps such as initiation, propagation and termination. (i)
HB ® H+ + B–
Initiation :
Protic acid +
H+ + CH2 = CH ® CH3 – CH CH3 Propene
CH3
732
+2 CHEMISTRY (VOL. - II)
(ii)
+
+
Propagation : CH3 – C H + CH2 = CH ® CH3 – CH – CH – C H CH3
CH3
CH3
CH3
n CH2 = CH CH3 CH3 – CH
(
CH3
CH2 – CH
+
n
CH2 – C H
CH3
CH3
(iii) Termination : Transfer to counter ion CH3 – CH
( CH – CH 2
CH3
CH3 – CH
+
– n CH2 – C H + B
CH3
( CH – CH 2
CH3
CH3
n
CH3
CH = CH + HB CH3
Polypropene Transfer to monomer : CH3 – CH
2
CH3
CH3
CH3 – CH
( CH – CH
CH3 (c)
( CH – CH
2
CH3
+
n CH2 = C H + CH2 = CH CH3
CH3
+
n CH = CH + CH3 – C H CH3
CH3
Anionic polymerization : Anionic polymerization involves the successive addition of vinyl monomers to a negatively charged species known as carbanion as reaction intermediate formed by the reaction of a negative ion or group to the monomer molecule. The initiators used for the anionic polymerization are strong bases like potassium amide or lithium amide. The steps involved are :
POLYMERS
(i)
733
Initiation : +– + K NH2 + C H2 = CH ® H2N – CH2 – C H K
Cl (ii)
Cl
Propagation : +
-
-
+
H2N – CH2 – C H K + CH2 = CH ® H2N – CH2 – CH – CH2 – C H K Cl
Cl
Cl
Cl
+ n CH2 = CH Cl H2N – CH2 – CH
( CH – CH
-
n CH2 C H K
2
Cl
+
Cl
Cl
The most important aspect of the anionic plymerization is that the termination is absent in this case. If inert solvents and pure reactants are used, the system will result in the formation of carbanion end-group. Hence the anionic polymers are also termed as living polymers. B.
CONDENSATION POLYMERIZATION :
Condensation polymerizaton is the type of polymerization involving the condensation between two bifunctional monomers with the loss of simple molecules like H 2O, NH3, HCl etc. leading to the formation of high molecular mass polymers. The reaction regenerates a bifunctional species for which the sequence of condensation goes on till any one of the reactants is totally consumed. This type of polymerization is therefore termed as step growth polymerization. The simplest example of the condensation polymer is the polyethylene terephthalate (PET) formed by the condensation of ethylene glycol and terephthalic acid with the elimination of water molecules. –– COOH
n HOH2C – CH2OH + n HOOC–– Ethylene glycol
Terephthalic acid – (2n –1) H2O O
O – CH2 – CH2 – O – C––
O –– C
n Polyethylene terephthalate (Terylene or Dacron)
734
23.4
+2 CHEMISTRY (VOL. - II)
COPOLYMERIZATION :
A polymer consisting of two or more chemically different types of monomer units in the chain is called a copolymer and the process of synthesis of copolymer is called copolymerization which may be a chain growth polymerization or a step growth polymerization. For example, a mixture of 1, 3 – butadiene and styrene can form a copolymer. CH = CH2 n CH2 = CH – CH = CH2 + n 1, 3 – Butadiene
Styrene
CH2 – CH = CH – CH2 – CH – CH2
n
Butadiene-styrene copolymer The properties of copolymers are quite different from that of homopolymers. For example, butadiene-styrene copolymer is quite tough and is a good substitute for natural rubber and is used for the manufacture of floor tiles, cable insulation, autotyres etc. Copolymers can be classified into four types basing on the distribution of different monomers in the chain. Suppose A and B are two monomers. (i)
Random copolymers : These are the copolymers in which the monomers are randomly arranged. –A–B–A–A–A–B–B–A–
(ii)
Alternating copolymers : In case of alternating copolymers the monomers are arranged alternately in the chain. –A–B–A–B–A–B–
(iii)
Block copolymers : Block copolymers contain long sequence of monomers in a linear chain. –A–A–A–A–B–B–B–B–A–A–A–A–
(iv)
Graft copolymers : Graft copolymers consist of main homopolymer chain with branches of another type of homopolymer.
POLYMERS
735
–A–A–A–A–A–A–A | | B B | | B B | | B B | | B B | | 23.5
SOME IMPORTANT POLYMERS :
A.
POLYTHENE :
It is a polymer of ethylene. Two types of polythenes can be synthesized depending on the conditions of synthesis with different types of intiators having widely different properties. These are : (a)
Low density polythene : The low density polythene is manufactured by heating pure ethylene to 350 - 570 K under pressure of 1000 - 2000 atmospheres and in presence of a trace of oxygen or peroxide following free radical mechanism. It consists of highly branched chain molecules, for which these molecules do not pack well resulting with low density and low melting point. Low density polythene is a transparent polymer of moderate tensile strength and high toughness. It is mainly used (i) as a packing material in the form of thin plastic film bags (ii) for insulating wires and cables, and (iii) in manufacturing toys, squeeze bottles and flexible pipes. n CH2 = CH2 Ethene
(b)
350 - 570 K 1000 - 2000 atm
~~~ CH2 – CH – CH2 – CH2 ~~~ | CH2 | CH2
Low density polythene High density polythene : High density polythene is prepared by co-ordination polymerization where ethene is heated to 330 – 350 K under a presure of 6–7 atmospheres in the presence of a catalyst consisting of triethyl aluminium and titanium tetrachloride (Ziegler-Natta catalyst). n CH2 = CH2
330 – 350 K, 6 – 7 atm. (Et)3 Al + TiCl4
( CH – CH 2
2
n
High density polythene
736
+2 CHEMISTRY (VOL. - II)
This polymer consists of linear chains, for which the molecules can be closely packed in space resulting in very high density and high melting point. It is quite harder, tougher and has greater tensile strength than low density polythene. It is used for (i) manufacturing containers like buckets, dustbins, bottles etc and (ii) manufacturing different housewares, pipes etc. B.
POLYACRYLONITRILE (PAN) OR ORLON :
Vinyl cyanide (acrylonitrile) on polymerization in the presence of a peroxide catalyst results in the formation of polyacrylonitrile. CN Polymerization n CH2 = CHCN CH2 – CH n Peroxide catalyst
(
Acrylonitrile
Polyacrylonitrile
PAN also known as acrilon or orlon is used for (i) making blankets, sweaters, bathing suits etc. (ii) for making synthetic carpets. C.
POLYTETRAFLUOROETHYLENE (PTFE) OR TEFLON :
Telfon is an addition polymer of tetrafluoroethylene formed on heating in the presence of a free radical initiator at high pressure. n CF2 = CF2
Catalyst Heat, High pressure
Tetrafluoroethene
( CF – CF 2
2
n
Teflon
Teflon is a very tough meterial and is resistant towards heat, action of chemicals such as acids and bases. it is used: (i)
as material resistant to heat and chemical attack
(ii)
for coating articles and cookware to make them non-sticky as non-sticky utensils.
(iii) for making gaskets, pump packings, valves, seals etc. D.
POLYAMIDES OR NYLONS
Polyamides with amide linkages (– CO – NH –) in the chain are formed by the condensation polymerization of diamines with dibasic acids and also of amino acids and their lactams. Polyamides popularly known as nylons are of different types as follows : (a)
Nylon - 6, 6 : Nylon - 6, 6 can be prepared by heating equimolar mixture of hexamethylene diamine and adipic acid under high pressure. It is given the name nylon - 6, 6 since both hexamethylene diamine and adipic acid contain six carbon atoms each.
POLYMERS
737
n H2N – (CH2)6 – NH2 Hexamethylene diamine
+ n HOOC – (CH2)4 – COOH Polymerization Adipic acid 553K High pressure O
O
( NH – (CH ) – NH – C – (CH ) – C 2 6
2 4
n + (2n – 1) H2O
Nylon – 6, 6 Nylon - 6, 6 is used for making tyre cord, fishing nets, climbing ropes, strings for sports rackets and bristles for brushes. (b)
Nylon - 6 : Nylon -6 can be synthesized by heating caprolactam with water at a high temperature. N | H H2C
C=O
H2C H2C
( C – (CH ) – N 2 5
533 – 543K
CH2
H
O
H2O
O–
Nylon - 6
CH2
Caprolactam Nylon - 6 is used for the manufacture of ropes, tyre cords and fabrics. E.
POLYESTERS :
O
Polyesters are the polymers with ester linkages (– C – O –) which can be formed by the condensation of dicarboxylic acids with diols. Dacron or terylene is the best example of polyester, being prepared by the condensation of ethylene glycol and terephthalic acid with elimination of water at 420 – 460 K in the presence of a catalyst consisting of a mixture of zinc acetate and antimony trioxide. O O n HO – CH2 – CH2 – OH
+
n HO – C ––
Ethylene glycol
–– C – OH
Terephthalic acid
Polymerization 420 – 460 K O O
( O – CH – CH – O – C –– 2
2
–– C
Terylene or Dacron
n + (2n – 1) H2O
738
+2 CHEMISTRY (VOL. - II)
It is used for (i) making textile, sarees, dress materials and curtains. (ii) making sails of sail-boats. (iii) making water houses for fire-fighting operations. (iv) making conveyor belts. F.
PHENOL-FORMALDEHYDE RESIN (BAKELITE) :
When phenol reacts with formaldehyde in the presence of either an acid or a base catalyst, polymer of high molecular mass is obtained. The process starts with the initial formation of ortho and para hydroxymethyl phenol derivatives which further react with phenol to form compounds having rings joined to each other through methylene bridges resulting initially with a linear polymer called Novolac. OH OH OH CH2OH
H+ + HCHO or OH–
+ o-Hydroxybenzyl alcohol
CH2OH p-Hydroxybenzyl alcohol
OH
OH CH2OH
OH CH2
Polymerization
CH2
Novolac Novolac an heating with formaldehyde undergoes cross linking to form an infusible solid mass called Bakelite. D Novolac + n HCHO – n H2O OH H2C
OH CH2
CH2
H2C
CH2
CH2
CH2 OH
OH CH2
CH2
CH2 OH Bakelite
CH2 OH
POLYMERS
739
Bakelite is used for (i)
making combs, fountain pen barrels, gramophone records etc.
(ii)
making electrical goods like switches, plugs etc.
G.
RUBBER
(a)
Natural Rubber : Rubber is a natural linear polymer of isoprene (2-methy - 1, 3 butadiene) CH3 CH3 Polymerization n CH2 = C – CH = CH2 ¾¾ ¾ ¾ ¾ ¾¾®
( CH – C = CH – CH 2
Isoprene
2
n
Poly isoprene
In polyisoprene each repeating unit contains a double bond, it exhibits cis-trans isomerism. Natural rubber has all cis-configuration having weak van der Waals interactions between various chains thus resulting with a coiled structure. Hence natural rubber can be stretched like a spring and exhibits elastic properties. CH3 CH2
CH2
CH2
C=C
C=C CH3
H
CH2
C=C
CH2
H
CH2
CH3
H
Cis-polyisoprene (Natural rubber) In contrast synthetic rubber (gutta percha) obtained by free radical plymerization isoprene has all trans-configurations. CH2
H CH2
C=C CH3
H CH2
C=C
CH2 CH3
C=C
CH2 CH3
Trans-isoprene (Gutta-percha)
H
CH2
740
+2 CHEMISTRY (VOL. - II)
Vulcanization of rubber : The process of heating natural rubber with sulphur and an appropriate additive like zinc oxide at temperature 373K to 415K to improve its properties is called vulcanization. The vulcanized rubber has excllent elasticity, low water absorption tendency and resistant to the action of organic solvents and oxidising agents in contrast to the natural rubber which is soft and sticky, having high water absorption capactiy, low tensile strength and resistance to abrasion. During vulcanization linear chains of isoprene units present in natural rubber get crosslinked through sulphur bridges at the allylic positions, which make rubber hard and stronger removing sticky quality of the natural rubber since the individual chains can no longer slip over the other because of locking together in a giant size molecule. The structures of vulvanized rubber can be represented as : CH3 CH2
CH2
CH3
C
CH
S
S
C
CH
CH2
CH
C = CH
CH2
C = CH
CH2
S CH2
CH3
CH
CH3
The extent of hardness or toughness depends on the amount of sulphur added. For making tyre rubber 5% sulphur is used, whereas for ebonite 20 – 25% sulphur and for making battery case rubber 30% sulphur is used. (B)
Synthetic Rubber :
(i)
Cis-polybutadiene : Polymerization of 1, 3-butadiene in the presence of Zeigler - Natta catalyst results with cis-polybutadiene having properties similar to that of natural rubber. It is used for making footwear, tyres and toys etc. - Natta catalyst n CH2 = CH – CH = CH2 ¾Zeigler ¾ ¾¾ ¾ ¾¾ ¾ ¾¾® 1, 3 - Butadiene (C2H5)3 Al + TiCl3
CH2
CH2 C=C
H
H
Cis - 1, 4 - Polybutadiene
n
POLYMERS
(ii)
741
Neoprene : Neoprene is prepared by the free radical polymerization of chloroprene (2chloro -1, 3- butadiene). Cl Cl Polymerization n CH2 = C – CH = CH2 ¾¾ ¾ ¾ ¾ ¾¾® CH2 – C = CH – CH2 n Chloroprene Neoprene
(
Neoprene is used (i)
for manufacturing hoses, shoes heel, stoppers etc as it is resistant to oils, gasoline and other solvents and stable towards aerial oxidation.
(ii)
as an insulator.
(iii) for making conveyer belts and printing rollers. (iii)
Buna-N : Buna-N is a copolymer of 1, 3-butadiene and acrylonitrile prepared by free radical polymerization. n CH2 = CH – CH = CH2 + n CH2 = CH Acrylonitrile 1, 3- Buadiene CN Polymerization
( CH – CH = CH – CH – CH – CH 2
2
2
n
CN Buna - N is used for making oil seals, manufacture of hoses and tank linings. (iii)
Buna-S : It is a copolymer of 1, 3 butadiene and styrene. n CH2 = CH – CH = CH2 + n CH2 = CH 1, 3 - Butadiene
Styrene Polymerization
( CH – CH = CH – CH – CH – CH 2
2
2
n
Buna - S It is used for the manufacture of automobile tyres, rubber soles, water proof shoes, belts etc.
742
+2 CHEMISTRY (VOL. - II)
CHAPTER (23) AT A GLANCE 1.
Polymers are macromolecules built up by repeating structural units joined by the covalent bonds, whereas all macromolecules are not polymers.
2.
Monomers are the simple raw materials (molecules) from which polymers are synthesized.
3.
Polymers can be classfied on different basis i.e. natural or synthetic, homopolymer or copolymer, linear or branched chain polymer etc.
4.
The process of polymerization may be addition polymerization or chain growth polymerization and condensation polymerization or step growth polymerization.
5.
Polythene, polystyrene, teflon, orlon etc are examples of addition polymers where as Nylon, Bakelite and Dacron are the examples of condensation polymers.
6.
Bifunctionality is the minimum criterion for a molecule to act as monomer.
7.
Natural rubber is a cis-1,4-polyisoprene the properties of which can be modified by vulvanization process on heating with sulphur. Neoprene, Buna-S & Buna-N are the examples of synthetic rubbers.
8.
In order to avoid the hazardous effect of polymeric wastes, biodegradable polymers like PHBV, Nylon-2, Nylon-6 have been synthesized successfully.
QUESTIONS A.
VERY SHORT QUESTIONS (1 mark each) : (a) Bakelite is formed by the chemical combination of phenol and ––––. (b) Name the linkage (bond) prevailing in the proteins. (c) Write any two uses of Neoprene. (d) Mention two uses of Buna-S. (e) What is Buna-N? (f) What is neoprene? (g) Name the monomer used for the preparation of Teflon. (h) What is PHBV? (i) What is a copolymer? (j) Name the monomer of Nylon-6 (k) What are thermosetting polymers? (l) Name the monomers used for Novolac. (m) What are the monomers required for terylene? (n) Define the term polymerization.
POLYMERS
B.
C.
743
SHORT ANSWER QUESTIONS (2 marks each) : (a)
What are polymers ? Give two examples.
(b)
What is Tefflon ? Give one of its uses.
(c)
What is natural rubber? Why it can not be used for making foot-ball bladders?
(d)
What is terylene? How is it prepared?
(e)
What are biodegradable polymers? Give two examples.
(f)
What is Nylon-6, 6? How is it prepared?
(g)
What are natural and synthetic polymers? Give one example of each.
(h)
What do you mean by vulvanization of rubber?
(i)
Define copolymerization and give one example.
(j)
What do you mean by free radical polymerization ? Name one initiator used for it.
SHORT ANSWER QUESTIONS (3 marks each) : Distinguish between
D.
(a)
Thermoplastic and thermosetting polymers.
(b)
Homopolymers and copolymers.
(c)
Natural rubber and synthetic rubber.
(d)
Addition polymerization and condensation polymerization.
(e)
Nylon-6 and Nylon-6, 6
(f)
Buna-N and Buna-S
(g)
Novolac and Bakelite
LONG QUESTIONS (7 marks each) : (a)
What are polymers ? What are the different methods of their preparation? Name an important polymer and describe a method of its preparation with chemical equation.
(b)
What are polymers? How are they classified on the basis of (i) structure (ii) synthesis and (iii) molecular forces involved. Give one example of each kind.
(c)
What are natural rubbers and what are synthetic rubbers? What is vulcanisation? How does it improve the characteristics of rubbers?
(d)
What is free radical addition polymerization? What are the intiators used for it? Discuss the mechanism of free radical polymerization taking one example.
744
+2 CHEMISTRY (VOL. - II)
E.
MULTIPLE CHOICE QUESTIONS :
1.
Chloroprene is the repeating unit in (a) PVC (b) Polythene (c) Neoprene (d) Polystyrene Orlon is a polymer of (a) Vinyl chloride (b) Styrene (c) Butadiene and adipic acid (d) Acrylonitrile Which of the following is a condensation polymer? (a) Teflon (b) Bakelite (c) Buna-S (d) Neoprene Natural rubber is a polymer of (a) Acrylonitrile (b) Isoprene (c) Vinyl chloride (d) Tetrafluoroethylene Which of the following is not a condensation polymer? (a) Bakelite (b) Nylon (c) Dacron (d) Teflon Teflon is a type of (a) Condensation polymer (b) Synthetic rubber (c) Addition polymer (d) None of these Which of the following is synthetic rubber? (a) Buna-S (b) Polyisoprene (c) Dacron (d) None of these Which is naturally occurring polymer? (a) Rayon (b) Nylon (c) Protein (d) PVC Which of the followings is called a polyamide? (a) Rayon (b) Terylene (c) Nylon (d) Teflon A raw material used in making nylon is (a) Adipic acid (b) Butadiene (c) Ethylene (d) Methyl methacrylate Teflon is a polymer of (a) Monofuoroethene (b) Difluoroethene (c) Trifluoroethene (d) Tetrafluorethene Which of the following is used to make non-stick cookware? (a) PVC (b) PET (c) Teflon (d) Nylon
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
POLYMERS
13.
14.
15.
16.
745
Nylon-6,6 is obtained from (a) Propene and adipic acid (c) Hexmethylene diamine and adipic acid Nylon-6 is made from (a) 1, 3 - Butadiene (c) Adipic acid Baby feeding bottles are generally made up (a) Polystyrene (c) Polyamide Teflon, styrene and neoprene are all (a) Monomers (c) Copolymers
(b) (d)
Phenol and formalehyde Adipic acid and phthalic acid
(b) (d)
Chloroprene Caprolactam
(b) (d)
Polyurethane Polyester
(b) (d)
Homopolymers Condesation polymers
ANSWER FOR MULTICPLE CHOICE QUESTIONS 1. (c)
2. (d)
3. (b)
4. (b)
5. (d)
6. (c)
11. (d)
12. (c)
13. (c)
14. (d)
15. (a)
16. (b)
qqq
7. (a)
8. (c)
9. (c)
10. (a)
746
+2 CHEMISTRY (VOL. - II)
UNIT - XVI
CHAPTER - 24
CHEMISTRY IN EVERYDAY LIFE 24.1
INTRODUCTION :
Life without Chemistry is totally impossible, because the basic needs of human being i.e. food, cloth and shelter involve the basic principle of Chemistry. Thus Chemistry has played a vital role in everyday life. For example, clothes like wool, silk, cotton, nylon etc. we wear, soaps and detergents used to wash them are all organic compounds. The food (carbohydrates, proteins, oils and fats etc.) we eat to nourish and build up our body are organic in nature. The house where we live uses cement, paints, adhesives, marbles etc., all involving the chemical processes. The medicines protecting us from diseases like antibiotics, sulphadrugs etc. are also boon of Chemistry. So the basic principles of Chemistry have influenced every sphere of human life. 24.2
CHEMICALS IN MEDICINE :
Chemical substances of natural or synthetic origin used for curing diseases and reducing suffering from pain are called medicines or drugs. The branch of science which deals with the treatment of diseases using suitable chemicals is known as chemotherapy. Though all drugs are medicines and all medicines are drugs, medicine is safer to use having negiligible toxicity without causing addiction, whereas drug causes addiction with serious side effects. Medicines are generally classified basing on the purpose for which they are used. Different classes of medicines are : A.
ANALGESICS
An analgesic may be defined as a drug that selectively relieves pain by acting on the central nervous system and on peripheral (external) pain mechanisms without significantly altering consciousness. Simply analgesics may be referred to the medicines used for getting relief form pain. Analgesics are categorized into two classes :
CHEMISTRY IN EVERYDAY LIFE
747
(a) Narcotics : Drugs which produce sleep and unconsciousness are called narcotics, and are mainly used for the relief of post-operative pain, cardiac pain, pains of terminal cancer and in child birth. Narcotic analgesics are mostly opium products which contains morphine and codeine. These are very effictive analgesics, but regular doses cause addiction. Other examples are heroin and marijuana. HO
H3CO
NCH3
NCH3 H
HO
HO
MORPHINE
CODEIN
(b) Non-narcotics : Regular doses of this type of analgesics do not cause addiction. They give temporary relief from pain and possess anti-inflammatory property. Examples of nonnarctoic analgesies are aspirin, analgin and paracetamol. H3C OCO CH3
HN
COOH
C
N
CH3
C
N
C6H5
+
CH2SO3Na
C O
ASPIRIN
ANALGIN
HO
NHCOCH3
PARACETAMOL
There have antipyretic properties also. Aspirin is the commonly used analgesic which gives relief from headache, muscular pain and toothache. The toxic side effect of using aspirin is gastric irritation leading to ulceration. Thus it should not be administered in empty stomach. But its calcium salt i.e. DISPRIN which is water soluble, is better than Aspirin having few side effects. Because of the side effects other analgesics like ibuprofen, naproxen are now being used. However because of anti blood clotting action af aspirin, it is still the drug of choice for prevention of heart attacks and reliever of angina pains. B.
TRANQUILIZERS
Tranquilizers may be defined as the neurologically active drugs which help in reducing stress and fatigue by inducing a sense of well being.
748
+2 CHEMISTRY (VOL. - II)
These affect the message transfer mechanisms from nerve to receptor for which they are used for the treatment of mild or even severe mental diseases. They form an essential component of sleeping pills. Tranquilizers are of different types. Such as : (a) Sedatives : These drugs are suitable for violent patients who are mentally upset. Examples are calmpose and equanil. CH3
O
N
O
C
O
H2N –– C –– O –– CH2 –– C –– CH2 –– O –– C –– NH2
N
Cl
CH3
C3H7 EQUANIL
CALMPOSE (b) Antidepressants : These drugs are also called mood elevators, which are given to those patients who lose self-confidence and are much depressed, thus improving their efficiency. Examples are amphetamine and methamphetamine. CH2 –– CH –– NH2
CH2 –– CH –– NHCH3
CH3 AMPHETAMINE (Benzedrine)
CH3 METHAMPHETAMINE (Methedrine)
(c) Hypnotics : Luminal and Seconal are some important hypnotics producing sleep which are the derivative of barbituric acid called barbiturates. These are habit forming drugs.
C2H5
O
H
C
N
C C2H5
H2C = HC – CH2
N
O
H
LUMINAL
H
C
N
C
C=O C
O
H3C – CH2 – CH2 – CH CH3
C=O C
N
O
H
SECONAL
CHEMISTRY IN EVERYDAY LIFE
C.
749
ANTISEPTICS AND DISINFECTANTS :
Antiseptics and disinfectants are the chemicals which either kill or prevent the growth of micro-organisms. (a) Antiseptics : Antiseptics are applied to the living tissues such as wounds, cuts, ulcers and diseased skin surface in order to kill or prevent the growth of micro-organisms in the form of antiseptic creams like furacin, soframycin etc. Antiseptics has the ability to reduce the odour produced as a result of bacterial decomposition of organic matter on the body or in the mouth for which they are added to face powders and breath purifiers. More examples of antiseptics are : (i)
Dettol, most commonly used antiseptic, is a mixture of chloroxylenol and terpineol. CH3 Cl
H3C
CH3
OH
CHLOROXYLENOL (ii)
H3C
OH CH3
TERPINEOL
Bithional is added to soaps to impart antiseptic properties to reduce the odour produced by bacterial decomposition of organic matter on the skin.
(iii) Iodine is a very powerful antiseptic used as tincture of iodine which is 2–3% solution of iodine in alcohol and water. (iv) Very dilute aqueous solution of boric acid is used as a mild antiseptic for eye wash. (v)
Iodoform is used as antiseptic powder for wounds.
(vi) 2–5% mercurochrome solution is a very good antiseptic for skin, mucous surfaces and wounds. (vii) Salol is used as an intenstial antiseptic for throat ailments. (b) Disinfectants : Disinfectants are the chemical substances which kill micro-organisms but are not safe to be applied to the living tissues, but can be applied to abiotic objects such as floors, drains, toilets, instruments etc. Examples are : (i)
Solution of cresols : (i.e. o–, m– and p– methyl phenols) in soapy water is called lysol which is used as a disinfectant.
750
+2 CHEMISTRY (VOL. - II)
(ii)
Chlorine in the concentration of 0.2 to 0.4 ppm in aqueous solution is a disinfectant.
(iii) SO2 in very low concentration acts an disinfectant. (iv) One percent solution of phenol is disinfectant whereas 0.2 percent solution acts as an antiseptic. D.
ANTIMICROBIALS
Antimicrobials are defined as the drugs used to destroy or prevent development or inhibit the pathogenic action of micro-organisms such as bacteria, viruses, fungi or other parasites. Microbial diseases can be controlled by the following ways : (i)
By using a bactericidal drug which kills the organisms in our body.
(ii)
By checking the growth of organisms using a bacteriostatic drug.
(iii) By improving the resistance power of the body towards infection i.e. increasing the immunity power. Examples of antimicrobial drugs are (i)
antibiotics- Penicillin, Tetracycline, Ofloxacin
(ii)
Sulpha drugs - Sulphanilamide, Sulphapyridine.
O || C6H5 – CH2 – C – NH
H
H
CH3 CH3
O
N
Pencillin - G E.
S
H2N
SO2NH2
COOH H
Sulphanilamide
ANTIBIOTICS
The term antibiotic has been derived from the word "antibiosis" which means survival of fittest i.e. a process in which one organism may destroy another to preserve itself. That means antibiotics are chemical substances produced by or derived from living cells which is capable in small concentration of inhibiting the life proceses or even destroying the micro-organisms. But with development of synthetic methods producing compounds which are same as derived from micro-oganisms, the definition of antibiotics needs modification. An antibiotic is now defined as the chemical substance produced wholly or partially by chemical synthesis, which in low concentration either kills or inhibits the growth of micro-ogranisms by intervening in their metabolic processes.
CHEMISTRY IN EVERYDAY LIFE
I.
751
Conditions for a chemical substance to be an antibiotic :
All chemical substances produced by or derived from living cells can not be antibiotics. These have to satisfy the following conditions. (i)
Antibiotics must be a product of metabolism although it might have been synthesized.
(ii)
If the antibiotic is a synthetic product, it should be a structural analogue of naturally occurring antibiotic.
(iii) It must be effective at low concentrations. (iv) In order to act as therapeutic agent, it must have the following characteristics.
II.
(a)
It must be effective against a pathogen.
(b)
It should not have any toxic side effects.
(c)
It should be stable and can be stored for a longer time.
Classification of anibiotics : (a)
(b)
Basing on their action on micro-organism antibiotics are of two types: (i)
Bactericidals which kill the organisms in the body. Examples : Penicillin, Ofloxacin.
(ii)
Bacteriostatic which checks the growth of micro-organisms. Examples : Erythromycin and Tetracycline.
The broad based classification of antibiotics classifies them into two types : (i)
Broad spectrum antibiotics : These include the antibiotics which are effective for a number of microbial diseases. Examples are penicillin, chloramphenicol etc.
(ii) (c)
III.
Narrow spectrum antibiotics : These are the antibiotics which are highly specific in their action. Examples are bacitracin and nystatin.
On the basis of the type of bacteria the antibiotic can destroy, the antibiotics are of two types. These are : (i)
Gram positive bacteria and
(ii)
Gram negative bacteria.
Examples of antibiotics : (i)
Penicillin : The first antibiotic penicillin was discovered by Alexander Fleming in 1929, who along with H.W. Florey received Nobel Prize for medicine in 1945 for their contribution to the development of penicillin.
752
+2 CHEMISTRY (VOL. - II)
Penicillin can not be taken orally. To persons who are allergic, it can cause fatal coma. The general structure of pencillins can be represented as : O || R – C – NH
H
H
S
CH3 CH3
O For Penicillin - G and Penicillin - V
N
COOH R = – CH2 – C6H5 R = – CH2 – O – C6H5
Penicillin is very much effective for pneumonia, bronchitis, sore throat, and other infectious diseases caused by various cocci and some gram positive bacteria. (ii)
Streptomycin : Streptomycin is a very effective drug for the treatment of tuberculosis, meningitis, pneumonia and also for common infections like throat, lungs etc.
(iii) Tetracyclines : Terramycin and Auromycin are the best examples of tetracyclines, which are highly effective against a number of bacteria, viruses, protozoa, parasites etc. Unlike penicillin and streptomycin tetracyclines can be taken orally as these are absorbed from the gastrointestinal tract. Due to this advantage streptomycin is largely replaced by tetracyclines. (iv) Chloramphenicol (Chloromycetin) : As an antibiotic chloramphenicol is having certain advantages over others. These are : (a)
Chemically it is the first antibiotic to be synthesized on a commercial scale.
(b)
It can be taken orally and very effective in the treatment of typhoid, dysentry, acute fever, urinary infections, meningitis and pneumonia.
(c)
It is the first natural compound found to contain a nitro group and the presence of –CHCl2 group is most unusual. NHCOCHCl2 O2N
CH –– CH –– CH2OH OH CHLORAMPHENICOL
CHEMISTRY IN EVERYDAY LIFE
F.
753
ANTIFERTILITY DRUGS
Antifertility drugs may be defined as the chemical substances which are used to check pregnancy in women having control over the female menstrual cycle and ovulation. Rising social problems like environmental pollution, insufficient food resources, unemployment etc. due to high population growth have become serious global problems which gave birth to the idea of family planning which is possible by birth control pills or oral contraceptives. These antifertility drugs contain a mixture of synthetic estrogen and a progesterone derivative which are more potent than the natural hormones. For example, a common brand name Enovid E contains norethindrone (a progesterone derivative) and mestranol (an estrogen). However all these drugs have side effects and should be used only under doctor's advice. CH3 H
OH
CH3
C º CH
H
OH C º CH
H
H
H
O
H3CO Norethindrone
Mestranol
Mifepristone the synthetic steroid is now widely used as "morning after pill" that blocks the effects of progesterone. CH3 H3C
N CH3
OH C º C – CH3
H H O Mifepristone G.
ANTACIDS
Antacids are defined as the chemical substances which neutralize the acid secreted in the stomach and raise the pH to an appropriate level.
754
+2 CHEMISTRY (VOL. - II)
The most common disease now the people are suffering is gastritis, which is due to the presence of excess hydrochloric acid in the gastric juice. Over productions of acid in the stomach causes irritation and pain with the development of ulcers. The various antacids used to check the gastritis are : (i)
Sodium hydrogen carbonate. But excessive use of it may make the stomach alkaline and accelerates the production of more acid.
(ii)
Metal hydroxides:- a mixture of aluminium and magnesium hydroxide.
(iii) The drug cimetidine (Tegamet) was a better invention in the field of antacids as it prevents the interaction of histamine which actually stimulates the secretion of pepsin and hydrochloric acid in the stomach, with the receptors present in the stomach wall resulting in the production of low amount of acid. (iv) Now another drug ranitidine (Zantac) has now prevailed in the medicinal field as a very effective antacid. (v)
Recently omeprazole and lansoprazole have been used as antacids preventing the formation of acid in the stomach. CH3 HN N – CN
HN
S N
N
NH2
Histamine
N
C NH CH3
H
Cimetidine
CHNO2
H3C N
S
H3C
N
O
C NH CH3
H
Ranitidine H N
O
N
S
CH3
N Omeprazole
CH3
OCH3 H N
O
N
S N Lansoprazole
CH3
OCH2CF3
CHEMISTRY IN EVERYDAY LIFE
H.
755
ANTIHISTAMINES
Antihistamines are defined as the antiallergy drugs used to treat allergy such as skin rashes, inflammation of tissues, asthma and itching of hives due to the release of histamine is the body. Histamine is a potent vasodilator with various functions. These are : (i)
It contracts the smooth muscles in the bronchi and gut and relaxes other muscles, such as those in the walls of fine blood vessels.
(ii)
It is also responsible for the nasal congestion associated with common cold.
Antihistamines work on different receptors for which these are completely different from antacids. They interfere with the natural action of histamine by competing with histamine for binding sites of receptor where histamine has its impact. These are mainly used for treatment of (i)
hay fever
(ii)
conjuctivitis
(iii) seasonal rhinitis (inflammation of nasal mucosa) i.e. sneezing, nasal discharge (iv) itching of eyes, nose and throat (v)
nausea in pregnancy
(vi) post operative vomiting. Commonly used antihistmine drugs are : diphenylhydramine, cetrizine, promethazine, brompheniramine (Dimetapp) and terfenadine (Seldane). CH3 CH3 C6H5 C6H5
CH2 – CH – N
CH3 CH – O – CH2 – CH2 – N Diphenylhydramine
CH3
N CH3 S Promethazine
N N N Br Brompheniramine (Dimetapp)
Ph HO
OH Ph
Terfenadine (Seldane)
756
24.3
+2 CHEMISTRY (VOL. - II)
CHEMICALS IN FOOD :
All those chemicals which are added to food to improve its shelf life, taste, appearance, odour and nutritive (food) value are termed as food additives. Main types of food additives are: (i)
preservatives
(ii)
food colours
(iii) flavouring agents (iv) artificial sweetners (v)
antioxidants
(vi) fortifiers (vii) emulsifiers (viii) antifoaming agents (ix) nutritional supplements such as minerals, vitamins and amino acids. I.
PRESERVATIVES
Preservatives may be defined as the chemicals which are capable of inhibiting or arresting the process of fermentation, acidification or other decomposition of food caused by bacteria, yeasts and moulds. (a) Preservatives used in a common man's house are salt, sugar, spices and oil to check the growth of micro-organisms in food. (i)
The preservation of food like raw mango, amla, fish, meat etc by adding excess salt is called salting which prevents the availability of water for bacterial growth.
(ii)
Sugar syrup is used for the preservation of fruits like mango, apples, carrot etc, which checks the growth of micro-organisms by dehydrating them.
(iii) Oil and spices are commonly used as preservaties for pickles, jams, squashes etc. (b) Sodium benzoate : Sodium benzoate is used as preservative in soft drinks and acidic foods. It is metabolized by conversion into hippuric acid, C 6H5CONHCH2COOH which is ultimately excreted in urine. (c) Sodium metabisulphite (Na2S2O5) : Sodium or potassium metabisulphite is used as a preservative for pickles, jams and squashes. Its preservative action is due to SO 2 which dissolves in water to give sulphurous acid. Na2S2O5 ® Na2SO3 + SO2 SO2 + H2O ® H2SO3 Sulphurous acid inhibits the growth of yeasts, moulds and bacteria.
CHEMISTRY IN EVERYDAY LIFE
757
(d) Epoxides : Epoxides such as ethylene oxide and propylene oxide are used to preserve low moisture foods such as spices, dried fruits and nuts. These foods are expossed to these gaseous chemicals in closed chamber for sufficient time when all types of micro-organisms including spores and viruses get killed. (e) Sodium and Calcium propionate : These are used in bread and cakes to inhibit the growth of moulds. (f) Sorbic acid and its salts : Sodium and potassium sorbates are now used for controlling the growth of yeasts and moulds in processed cheese, pickles, baked food items, certain meat and fish products. (g) p-Hydroxy benzoate esters : The methyl, ethyl, propyl and heptyl esters of p-hydroxy benzoic acid are used to preserve baked food, beer, soft drinks and syrups. They are highly effective in inhibiting the growth of moulds and yeasts but less effective on bacteria. II.
ARTIFICIAL SWEETENING AGENTS :
Most commonly used natural sweetners are sucrose and fructose which not only add to our calorie intake, but also causes tooth decay. Particularly for diabetic patients who want to control their sugar intake, calorie free artificial sweetners have been synthesised. Some commonly used artificial sweetners are : (a) Saccharin : Ortho-sulphobenzoimide also called saccharin is the first most popular sweetening agent which is about 550 times as sweet as cane sugar. It is insoluble in water but its soidum salt is soluble in water. Chemically it is 1, 2 benzisothiazolin - 3 - one - 1, 1 - dioxide (o-sulphobenzoimide) and accurs as white crystals. O O +
NH
N Na
SO2
SO2 Sodium salt of Saccharin (Soluble in water)
Saccharin (Insoluble in water)
It is not biodegradable and is excreted as such in urine. So it does not possess any calorific value and is used as sweetening agent by diabetic patients. (b) Cyclamate : Cyclamate is seven times sweeter than sucrose. However a mixture of cyclamate and saccharin in the ratio 10:1 is found to be sweeter than either of the two individual compounds. Its structure is: H N SO3H Cyclamate or N - cyclohexylsulphamate
758
+2 CHEMISTRY (VOL. - II)
Its use has been banned because of its suspected link with cancer. (c) Aspartame : Aspartame is the most efficient and widely used artificial sweetner which is the methyl ester of the dipeptide derived from phenylalanine and aspartic acid. Å NH3
COOCH3 O
C6H5 – CH2 – CH – NH – C – CH – CH2 – COO Aspartame It is approximately 100 times as sweet as cane sugar, but it has short shelf life. It decomposes at baking or cooking temperature and its use is limited to cold foods and soft drinks. Again people suffering from the disease phenylketone urea are advised not to use aspartame since their metabolism converts it to phenylpyruvic acid, accumulation of which is harmful especially to infants due to mental retardation and brain damage. (d) Sucralose : Sucralose is a trichloro derivative of sucrose which is roughly 600 times as sweet as sucrose. It tastes and looks like sucrose and is quite stable at baking and cooking temperature. It neither provides calories nor causes tooth decay. CH2OH
Cl H
O
H
HO
H
H OH OH
H
O
H CH2Cl
H
ClH2C
OH O
H
Sucralose (e) Alitame : Alitame is 2000 times as sweet as sucrose. It is a high potency sweetner, but it is very difficult to control the sweetness of the food to which it is added. O
NH2 O
CH3 O
H3C
HO – C – CH2 – CH – C – NH – CH – C – NH –– Alitame III.
ANTIOXIDANTS
CH3
CH3 S CH3
Antioxidants are defined as the chemicals which are added to processed foods like potato chips, biscuits, breakfast cereals etc. to prevent the oxidation of fats and subsequent spoilage of the food.
CHEMISTRY IN EVERYDAY LIFE
(i)
759
The most common antioxidants are butylated hydroxytoluene (BHT), butylated hydroxyanisole (BHA) and several esters of gallic acid such as propyl gallate.
OH (CH3)3C
OH C(CH3)3
C(CH3)3
CH3
OCH3
Butylated hydroxytoluene (BHT)
Butylated hydroxyanisole (BHA)
COOH
HO
(ii)
COO CH2 CH2CH3
OH
HO
OH
OH
OH
Gallic acid
Propyl gallate
Antioxidants are also termed as sacrificial materials as they are more reactive towards oxygen than the materials they are protecting. They also reduce the rate of involvement of free radicals in the ageing process.
(iii) The addition of BHA to butter increases its shelf-life from months to years. (iv) Sulphur dioxide and sodium metabisulphite which are used as preservatives, are also used an antioxidants for wine, beer, sugar syrups, dried fruits and vegetables. (v)
24.4
Ascorbic acid prevents browning caused by enzymatic oxidation of phenolic compounds.
CLEANSING AGENTS :
I. (a) Soaps : Soaps are the sodium or the potassium salts of higher fatty acids like stearic acid, palmitic acid, oleic acid etc. They are prepared by the alkaline hydrolysis i.e. saponification of oils and fats. Soaps may also be obtained by the direct neutralisation of higher fatty acids, like stearic acid, palmitic acid and oleic acids with alkali, such as NaOH or KOH.
760
+2 CHEMISTRY (VOL. - II)
O (i)
CH2 – O – C – C17 H35 O CH – O – C – C17 H35
+ 3 NaOH ® 3C17H35COONa
O
Sodium stearate (Soap)
CH2 – O – C – C17 H35 Glyceryl ester of stearic acid
CH2 – OH +
CH – OH CH2 – OH Glycerol
(ii)
neutralization C17H35 – COOH + NaOH ¾Direct ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾® C17H35 COONa + H2O
Stearic acid
Soap (Sodium stearate)
The saturated fats and oils give hard soaps than the unsaturated one. Also sodium salts of a given fat or oil is harder and less soluble than the potassium salts. Hence, soft soaps are particularly of potassium salts of higher unsaturated fatty acids and hard soaps are sodium salts of higher saturated falty acids. (b) Types of soaps : Saponification of oils or fats produces soap. However, different soaps are prepard by varying diffeent raw materials. (i)
Washing soaps : These are made from cheaper fats like mohwa oil, rosin etc using caustic soda for their saponification. Sodium silicate is added to these soaps as a filler. Some cheap starch or white clay is added to increase the weight and reduce the cost.
(ii)
Toilet soaps : Toilet soaps are made from best animal or vegetable fats or mixture of both with removal of excess alkali. Then colour, perfume and germicide etc. are added to the soap during crutching or shredding.
(iii) Transparent soaps : These are made by dissolving the soap in ethanol followed by evaporation of the excess solvent. (iv) Shaving soaps : During preparation of shaving soaps glycerol is added to prevent rapid drying. Rosin, a gum is added while making them which forms sodium rosinate responsible for lather. (v)
Laundry soaps : These soaps contain fillers like sodium silicate, sodium rosinate, borax and sodium carbonate.
CHEMISTRY IN EVERYDAY LIFE
761
(c) Advantages and disadvantages of soaps : Soap is an effective cleaning agent and is completely biodegradable. Micro-organisms present in sewage water can oxidise soap completely to carbon dioxdes and hence does not create any pollution. However soaps are having the following limitations. (i)
Ordinary soaps are unsuitable for washing of silk, wool etc. as the alkali present causes harm to the fibre.
(ii)
Soaps can not be used in acidic medium because they change the soaps into free carboxylic acids which strick to the fabrics being incapable of removing oil and grease from fabrics.
(iii) Soaps can not be used in hard water because hard water contains calcium and magnesium ions which form insoluble calcium and magnesium soaps when sodium or potassium soaps are dissolved in hard water. 2 C17H35 COONa + CaCl2 ® 2NaCl + (C17H35 COO)2Ca Sodium stearate Calcium stearate (Soap) (Soluble) (Insoluble) Insoluble soaps separate as scum in water and are hindrance to good washing. This causes wastage of soap and also discolours and hardens the fabric being washed. II.
SYNTHETIC DETERGENTS :
Synthetic detergents are generally sodium salts of long-chain alkyl hyrogen sulphate or sodium salts of long chain alkyl benzene sulphonic acids which act as cleansing agents having all the properties of soaps, but do not contain any soap. These are also referred to as syndets or soapless soaps. In contrast to soaps, synthetic detergents can be efficiently used even with hard water, because calcium and magnesium salts of synthetic detergents like their sodium and potassium salts are also soluble in water. Hence they do not form curdy white precipitates with hard water. Again they can also be used for acidic solutions. However, unlike soaps they are not completely biodegradable and hence result in water pollution. Synthetic detergents are of three types. These are : (a) Anionic detergents : These are so called because a large part of their molecules are anions. These are of two types: (i)
Sodium alkyl suphates : These are the sodium salts of sulphonated long chain alcohols formed by the treatment of long chain alcohol with concentrated sulphuric acid followed by neutralisation with sodium hydroxide. Examples are sodium lauryl sulphate and sodium stearyl sulphate.
762
+2 CHEMISTRY (VOL. - II)
H SO 4 ¾¾2¾¾ ®
CH3(CH2)10CH2OH Lauryl alcohol
–H2O
CH3(CH2)10CH2OSO3H Lauryl hydrogen sulphate –H2O ¯ NaOH(aq) CH3(CH2)10 CH2O SO–3Na+ Sodium lauryl sulphate
These sodium alkyl sulphates are 100% biodegradable. (ii)
Sodium alkyl benzene sulphonates : These are obtained by Friedel-Craft's alkylation of benzene with a long chain alkyl halide or an alkene or an alcohol followed by sulphonation and neutralization with NaOH. Most widely used detergent is sodium n–dodecylbenzenesulphonate (SDS), which is also referred to as LAS detergents (linear alkyl sulphonates) where as an example of ABS detergents (alkyl benzene sulphonates) with a branched chain alkyl groups is CH3
CH3 ––– SO–3Na+
CH3 ––– CH – CH2––– CH ––– 3
Hard or non-biodegradable ABS detergent CH3 – (CH2)11 –––
––– SO–3Na+
Soft or biodegradable LAS detergent. (Sodium n-dodecylbenzene sulphonate) Sodium n-dodecylbenzene sulphonate can be prepared as follows: CO(CH2)10CH3 + CH3 (CH2)10COCl
(CH2)11CH3
Anhydrous AlCl3 – HCl
Zn/Hg–HCl
(CH2)11CH3
NaOH (aq)
ConcH2SO4
– H2O
– H2O
SO–3 Na+ Sodium dodecyl-benzenesulphonate
SO3H Dodecylbenzene sulphonic acid
(CH2)11CH3
Dodecylbenzene
CHEMISTRY IN EVERYDAY LIFE
763
(b) Cationic detergents : Cationic detergents are quarternary ammonium salts of amines with chlorides, bromides or acetates as anions containing one or more long chain alkyl groups and a positive charge on nitrogen atom, for which these are called cationic detergents. Being more expensive than the anionic detergents, they find limited use. However they are extensively used as germicides. Cetyltrimethylammonium bromide is a popular cationic detergent and is used in hair conditioners. CH3 + CH3(CH2)15 ––––– N ––––– CH3
Br–
CH3 Cetyltrimethyl ammonium bromide (germicide) Other examples of cationic detergents are: CH3 + ––– CH2 ––––– N ––––– C12H25 Cl–
(CH3)3 N+ (C18H37) Br– Trimethyl stearyl ammonium bromide
CH3 Benzalkonium chloride (an antibacterial)
(c) Neutral or non-ionic detergents : Non-ionic detergents do not contain any ion in their structure. One most widely used neutral detergent is formed when stearic acid reacts with polyethylene glycol. n HO – CH2 – CH2OH Ethylene glycol
+
CH2 – CH2 ® HO (CH2CH2O)n CH2CH2OH
O Ethylene oxide
Polyethylene glycol
CH3 (CH2)16 COOH + HO (CH2CH2O)n CH2CH2OH Stearic acid
Polyethylene glycol ¯ – H2O
CH3 (CH2)16 COO (CH2CH2O)n CH2CH2OH Non-ionic detergent
764
+2 CHEMISTRY (VOL. - II)
Another example of non-ionic detergent is lauryl alcohol ethoxylate which is 100% biodegradable. This can be synthesized as follows : CH3 (CH2)10 CH2OH
+
n – Lauryl alcohol
8 CH2 – CH2 Ethylene oxide
¯ Base
CH3 (CH2)10 CH2 (OCH2CH2)8OH Lauryl alcohol ethoxylate Non-ionic detergents are liquid dish washing detergents. Advantages and disadvantages of synthetic detergents over soaps : (a) Advantages : Synthetic detergents find wide application these days as cleasing agents having the following advantages over soaps. (i)
Synthetic detergents can be used even in hard water.
(ii)
Synthetic detergents can be used even in acidic medium because these are the salts of strong acids and are not decomposed in the acidic medium.
(iii) These are more soluble in water than soaps. (iv) They have stronger cleansing power than soaps. (v)
Synthetic detergents are prepared from the hydrocarbons obtained from petroleum. This saves vegetable oils used as food.
(b) Disadvantages of synthetic detergents over soaps : Synthetic detergents having branched chain hydrocarbon chains are not biodegradable and cause water pollution making it unfit for use by the aquatic life. However, this difficulty has been overcome now by using a straight chain hydrocarbon in the detergent instead of branched chain hydrocarbon making it biodegradable. But all soaps are biodegradable. Difference between Soaps and Detergents : Main points of difference between soaps and synthetic detergents are given in Table 24.1.
CHEMISTRY IN EVERYDAY LIFE
765
Table 24.1 Difference between soaps and synthetic detergents. Synthetic detergents
Soaps 1
These are alkali metal salts of long chain fatty acids
1. There are sodium salts of long chain alkyl sulphates or long chain alkylbenzene sulphonates.
2. These can not be used in hard water.
2. These can be used even in hard water
3. These can not be used in acidic solutions.
3. These can be used even in acidic solutions.
4. These are obtained from vegetable oils.
4. These are prepared from hydrocarbons obtained from petroleum.
5. These are biodegradable.
5. Some of the synthetic detergents are not biodegradable.
III.
CLEANSING ACTION OF SOAPS AND DETERGENTS :
The cleaning action of a soap or a detergent is called its detergent action or detergency. All the soaps and detergents contains two characteristics groups i.e. (i)
Non-polar hydrocarbon chain which is oil soluble (lyophilic or lipophilic) and is called hydrocarbon tail or oil-soluble non-polar part of the soap or detergent molecule.
(ii)
Polar end which is water soluble (hydrophilic) and is called negative end or water soluble polar part of the soap or detergent molecule. O CH3 – CH2 – (CH2)15 Non polar hydrocarbon chain (Oil soluble)
CH3 – (CH2)10 – CH2 Oil soluble non-polar part
Å C – O Na (Soap) Polar part (Water soluble) Å O SO2 O Na (Detergent) Water soluble polar part
Cleansing action of soaps and detergents takes place in two steps : (i)
Wetting : This results from lowering of surface tesnsion of water by mixing soap with it. As a result the oil droplets of the dirt are wetted on the surface and form a colloidal suspension in the medium of water. The soap forms a protective film on their surface to prevent coagulation. Such colloidal particles are then washed away with water.
766
+2 CHEMISTRY (VOL. - II)
(ii)
Emulsification : When dirty cloth is immersed into the mixture of soap and water, the non-polar part of the soap or detergent dissolves in oil or grease and the polar part of the soap or detergent is held by the surrounding water as shown in Fig. 24.1. The grease layer is then dislodged from skin by rubbing or from garment by tumbling and strirring. Na+
Oil or grease droplet (Dirt particle)
Polar end Non-polar end
Fig. 24.1 : Cleansing action of soap and detergents Each oil droplet is surrounded by an ionic atmosphere with negative heads outwards in water. Due to repulsion between similar charges, oil or grease particles change into very small drops and a stable emulsion of oil in water is obtained. The emulsified grease particles bearing dirt can be easily washed with water. Thus soap and detergent clean by emulsifying the fat and grease containing the dirt.
CHAPTER (24) AT A GLANCE 1.
Chemicals in medicine : Chemical substances used for curing diseases and reducing suffering from pain are called medicines or drugs. The branch of science which deals with the treatment of diseases using suitable chemicals is known as chemotheraphy. (a)
Analgesics : Analgesics decrease pain. These are of two types. (i) Narcotics : Examples are morphine and heroin. (ii) Non-narcotics : Examples are aspirin and phenacetin.
(b)
Tranquilizers : These are medicines used to treat mental diseases. Examples are equanil, seconal and luminal.
CHEMISTRY IN EVERYDAY LIFE
2.
3.
767
(c)
Antiseptics : These are applied to skin, on wounds and cuts to kill or prevent the growth of micro-organisms. Examples are dettol, savlon and acriflavin.
(d)
Disinfectants : These are the chemicals which kill micro-organisms, but not safe for human beings.
(e)
Antimicrobials : Drugs used to cure diseases caused by fungi, bacteria and viruses are called antimicrobials. Examples are penicillin, streptomycin and AZT widely used against AIDS.
(f)
Antifertility drugs : These are used to check pregnancy in women. Example is Enovid E.
(g)
Antibiotics : These are chemical substances produced by some micro-organisms and can be used to kill other micro-organisms which cause infections. Examples are tetracyclin, chloramphenicol, penicillin.
(h)
Antacids : These are the chemicals used to remove the excess acid and raise the pH to on appropriate level in stomach. Examples are omeprazole and lansoprazole.
(i)
Antihistamines : Allergy is caused by liberation of histamines. The drugs used to treat allergy are called antihistamines. Examples are diphenylhydramine and promethazine.
Chemicals in food : Food additives are those chemicals which are added to food to improve its keeping qualities, appearance, taste, colour, odour and food value. These are : (a)
Preservatives : These are used to protect food against bacteria, yeasts and moulds. Examples are: Sodium metabisulphite, Sorbic acid and Sodium benzoate.
(b)
Artificial sweetening agents : Aspartame, Alitame and Saccharin are the artificial sweetners having no calorie intake for which used by the diabetic patients.
(c)
Antioxidants : These are used to prevent oxidation of fats in processed foods such as potato chips, biscuits etc. Examples are BHT and BHA.
Cleansing Agents (Soaps and Detergents) (a)
Soaps are sodium and potassium salts of higher fatty acids like stearic acid, palmitic acid, oleic acid etc. Washing soaps contain sodium salts of fatty acids while soft soaps contain potassium salts.
(b)
Synthetic detergents are generally sodium salts of long chain alkyl hydrogen sulphates or sodium salts of long chain alkyl benzene sulphonic acids.
768
+2 CHEMISTRY (VOL. - II)
QUESTIONS A.
VERY SHORT QUESTIONS (One mark each) : (a)
Give the name of the first antibiotic.
(b)
Name a substance which can be used as an antiseptic as well as disinfectant.
(c)
What are tranquilizers?
(d)
What type of medicine chloramphenicol is?
(e)
Name two antipyretic drugs.
(f)
Name two analgesic drugs.
(g)
Aspirin is an –––––– .
(h)
Streptomycin is a –––––– spectrum antibiotic.
(i)
What is the name given to medicines used for getting relief from pain?
(j)
Name the antibiotic used specifically for treatment of typhoid fever.
(k)
Name one broad spectrum antibiotic.
(l)
What is the difference between a preservative and an antioxidant?
(m) Name an antacid which prevents the formation of acid in the stomach.
B.
(n)
What structural unit makes detergents non-biodegradable?
(o)
What type of detergents are used for dish washing?
(p)
Give an example of a bactericidal antibiotic.
(g)
In transparent soaps –––––– is used.
(h)
Name two artificial sweetening agents.
(i)
–––––– is the sweetening agent used in the preparation of sweets for a diabetic patient.
SHORT QUESTIONS (Two marks each) : (a)
What are detergents? Name one from each type of detergent.
(b)
Write down the difference between soap and detergent.
(c)
What are broad spectrum antibiotics? Give two examples.
(d)
What are antipyretics? give two examples.
(e)
Whay soap is not used for hard water?
(f)
What are antifertility drugs? Give one example.
(g)
What are the main constituents of Dettol?
(h)
What are food preservatives? Give two examples.
(i)
Why are Cimetidine and Ranitidine better antiacids than sodium hydrogen carbonate or magnesium hydroxide?
CHEMISTRY IN EVERYDAY LIFE
C.
D.
E.
769
(j)
Why is the use of aspartame limited to cold foods and drinks?
(k)
What problem arises in using alitame as artificial sweetener?
SHORT QUESTIONS (Three marks each) : (a)
How are antiseptics different from disinfectants ? Give one example of each of them.
(b)
What are soaps and detergents?
(c)
Discuss the cleansing action of soaps?
(d)
Why does a detergent act even with hard water? Is it always biodegradable?
(e)
Give the composition of one oral contraceptine.
(f)
What are detergents? Give their scheme of classification.
(g)
What are antioxidants? How do they differ from preservatives.
(h)
What is tincture of iodine? What is its use?
(i)
What are biodegradale and non-biodegradable detergents? Give one example of each.
(j)
Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs.
LONG QUESTIONS (7 marks each) : (a)
What are food additives? Discuss briefly the various types of food additives.
(b)
What are soaps and synthetic detergents? In what respects detergents preferred over soaps? Discuss the cleansing action of soaps and detergents.
(c)
Write notes on : (i)
Antibiotics
(ii)
Antacids
(iii)
Analgesics
MULTIPLE CHOICE QUESTIONS : 1.
2.
3.
The oils from which soaps are prepared belong to a class of compounds known as (a) Amine
(b) Acid
(c) Hydrocarbon
(d) Ester
Novalgin is a common (a) Analgesic
(b) Antibiotic
(c) Antipyretic
(d) Antimalarial
Equanil is a drug to control (a) Pneumonia
(b) Malaria
(c) Ordinary fever
(d) Mental disease
770
+2 CHEMISTRY (VOL. - II)
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
Medicine which is an antibiotic is (a) Ampicilin
(b) Aspirin
(c) Calmpose
(d) Chloroquine
A detergent is a (a) Drug
(b) Catalyst
(c) Surface active agent
(d) Soap
Which of the following is used as a preservative to protect processed food? (a) Saccharin
(b) BHT
(c) Sodium sulphate
(d) Sodium metabisulphite
The drugs used to get relief from pain are called. (a) Antibiotics
(b) Antipyretics
(c) Analgeics
(d) Antiseptics
Which of the following is an antihistamine drug? (a) Ciprofloxacin
(b) Chloroquine
(c) Chloramphenicol
(d) Chlorpheniramine maleate
Which of the following is a tranquilizer? (a) Morphine
(b) Seconal
(c) Phenacetin
(d) Streptomycin
2-Acetoxy benzoic acid is called (a) Mordant dye
(b) Aspirin
(c) Antiseptic
(d) Antibiotic
Morphine is (a) Antiseptic
(b) Antibiotics
(c) Analgesics
(d) Antimalarial
Heroin is a derivative of (a) Nicotine
(b) Morphine
(c) Caffeine
(d) Cocains
Which of the followings is used as germicide? (a) Sodium lauryl sulphate
(b) Cetyltrimethylammonium chloride
(c) Lauryl alcohol ethoxylate
(d) Sodium-2-dodecylbenzene sulphonate
Which of the followings is an anionic detergent? (a) C6H5 – SO3Na
(b) CH3 (CH2)16 N+ (CH3)3 Cl–
(b) CH3 (CH2)16 CH2OSO3Na
(d) CH3 (CH2)16 COO (CH2CH2O)n – – CH2CH2–OH
CHEMISTRY IN EVERYDAY LIFE
15.
16.
17.
18.
19.
20.
771
The oxidant which is used as an antiseptic is (a) KMnO4
(b) KBrO3
(c) CrO3
(d) KNO3
Which of the followings is used as a "morning after pill"? (a) Norethindrone
(b) Mifepristone
(c) Promethazine
(d) Bithional
Tincture of iodine is (a) aqueous solution of I2
(b) alcoholic solution of I2
(c) aqueous solution of KI
(d) solution I2 in aqueous KI
Pick out the wrong statement. (a) BHT is an antioxidant
(b) Alitame is an artificial sweetener
(c) Sodium alkyl sulphate is a cationic detergent
(d) Tetrazine is a harmful edible colour.
Pick out the wrong statement for a detergent molecule. (a) It is not easily biodegraded.
(b) It is sodium salt of fatty acid.
(c) It is a source of active agent.
(d) It has both non-polar organic part and a polar group.
Detergents can be made biodegradable by taking (a) Cyclic hydrocarbon chain
(b) Unbranched hydrocarbon chain
(c) Benzenoid with hydrocarbons
(d) Hydrocarbon with more branching
ANSWERS FOR MULTIPLE CHOICE QUESTIONS 1. (d)
2. (a)
3. (d)
4. (a)
5. (c)
6. (d)
7. (c)
8. (d)
9. (b)
10. (b)
11. (c)
12. (b)
13. (b)
14. (c)
15. (a)
16. (b)
17. (b)
18. (c)
19. (b)
20 (b)
qqq
772
+2 CHEMISTRY (VOL. - II)
I. LOGARITHMS
LOGARITHMS
773
II. LOGARITHMS
774
+2 CHEMISTRY (VOL. - II)
III. ANTILOGARITHMS
ANTILOGARITHMS
775
IV. ANTILOGARITHMS