10. (i) In aniline, the lone pair of the electrons on the N – atom is delocalized over the benzene ring.
As a result, electron density on the nitrogen decreases.
In contrast in CH 3 NH 2 +I effect of CH 3 increases the electron density on the Natom.
Therefore, aniline is a weaker base than methylamine and hence its pK b value is higher than that of methylamine.
(ii) Methylamine being more basic than water, accepts a proton from water liberating OH – ions.
: CH NH + H OH 3 2 CH NH + OH 3 2
+ _
These OH – ions combine with Fe 3+ ions present in H 2 O to form brown ppt. of hydrated ferric oxide.
3
FeCl3 ®Fe++ 3Cl -
3
Hydrated ferric oxide 3 (Brown ppt.)
2 Fe + +6 OH- ® 2 Fe(OH) .
(iii) Aniline being a Lewis base reacts with Lewis acid AlCl 3 to form a salt.
C 6 H 5 NH 2 + AlCl 3 ® C 6 H 5 N + H 2 AlCl 3 – .
As a result, N of aniline acquires positive charge and hence it acts as strong deactivating group for electrophilic substitution reaction.
Consequently, aniline does not undergo Friedel Crafts reaction.
11. (i) Loss of proton from amine gives amide ion whereas
loss of a proton from alcohol gives an alkoxide ion.
RNH NH + H 2 R
– +
Amine Amide ion
ROH O + H R – +
Alcohol Alkoxide ion
Since O is more electronegative than N, therefore, RO – can accommodate the –ve charge more easily than RNH – . Hence amines are less acidic than alcohol.
(ii) Due to resonance in aniline, the lone pair of electron on nitrogen gets delocalised over the benzene ring and becomes less available for protonation. In alkyl amine, alkyl group releases electrons and increase density of electron of nitrogen, making it stronger base
: +
NH 2 NH 2 NH 2 + NH 2 +
NH 2
NH 2 R N NH : 2
12. (i) The increasing order of basic strength in water of the given amines and ammonia follows the following order.
C 6 H 5 NH 2 < NH 3 < (C 2 H 5 ) 3 N < (C 2 H 5 ) 2 NH
(ii) The increasing order of basic strength in gas phase of the given amines follows the order.
CH 3 NH 2 < C 2 H 5 NH 2 < (C 2 H 5 ) 2 NH < (C 2 H 5 ) 3 N.
13. (i) Hofmann bromamide reaction : Primary amines when heated with Br 2 and (aqueous or ethanoic solution of) NaOH lose a carbon atom and are converted to the corresponding amines. It is an example of stepdown reaction.
CH CONH + Br + 4NaOH 3 2 2
CH NH + 2NaBr + Na CO + 2H O 3 2 2 3 2 D
Acetamide
(ii) Gabriel phthalimide synthesis : In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic potassium hydroxide. Then potassium phthalimide is heated with an alkyl halide to yield an Nalkylpthalimide which is hydrolysed to phthalic acid and primary amine by alkaline hydrolysis.
CO CO
NH –NK +
CO CO
KOH (alc.) C H I 2 5
–H O 2 –KI
CO NC H 2 5 CO
NaOH (aq ) (Hydrolysis)
COOH
+ C H NH 2 5 2 COOH Ethyl amine Phthalic acid
This synthesis is very useful for the preparation of pure aliphatic primary amines. However, aromatic primary amines cannot be prepared by this method.
14. (i) Aniline gives white or brown precipitate with bromine water.
NH 2
+ 3Br 2
NH 2
+ 3HBr
Br Br
Br
Ethyl amine does not react with bromine water (ii) Each compound is warmed separately with few drops
of chloroform and alcoholic KOH. Aniline (primary amine) gives offensive smell of isocyanide while N
methyl aniline (secondary amine) does not give carbylamine test.
C H NH + CHCl + 3KOH 6 5 2 3 (alc.) Warm C 6 H 5 NC
CHCl + KOH (alc.) 3
C H NHCH 6 5 3 No reaction.
15. (i) R C NH 2 R CH NH O
LiAlH 4
H O 2 2 2
(ii) C 6 H 5 N 2 Cl + H 3 PO 2 + H 2 O
C 6 H 6 + N 2 + H 3 PO 3 + HCl
(iii) C 6 H 5 NH 2 + Br 2(aq)
NH 2
Br Br Br
+ 3 HBr
16. (i) C 6 H 5 NH 2 + CH 3 COCl C H NHCOCH + HCl 6 5 3
Acetanilide
(ii) C 2 H 5 NH 2 + C 6 H 5 SO 2 Cl C 6 H 5 SO 2 NHC 2 H 5 (iii) C H NH + HNO 2 5 2 2 [C H N Cl ] 2 5 2 + –
H O 2
C H OH + N + HCl 2 5 2 17. (i) Solubility of ethylamine in water is attributed to its ability to form hydrogen bonds with water molecules.
In aniline the non polar hydrocarbon part is relatively larger and has no interaction with polar water molecules.
(ii) Refer Ans. 10 (i).
18. (i) Increasing order of basic strength :
C 6 H 5 NH 2 < C 6 H 5 N(CH 3 ) 2 < CH 3 NH 2 < (C 2 H 5 ) 2 NH (ii) Decreasing order of basic strength :
ptoluidine > aniline > pnitroaniline (iii) Increasing order of pK b values :
(C 2 H 5 ) 2 NH < C 2 H 5 NH 2 < C 6 H 5 NHCH 3 < C 6 H 5 NH 2 . 19. (i) C 6 H 5 6N 2 5Cl + C 2 6 6H 5 5NH 2 2 OH
–
NH + Cl + H O 2 – 2 N N
(ii) C 6 H 5 N 2 Cl + CH 3 CH 2 OH
+ N 2 + HCl + CH 3 CHO (iii) R NH + CHCl + KOH 2 3 R NC + KCl + H O 2
20. (i) Aniline gives white or brown precipitate with bromine
(ii) Aniline and benzylamine : Benzylamine reacts with nitrous acid to form a diazonium salt which is unstable at low temperature, decomposes with evolution of N 2 gas
C H CH NH NaNO HCl 2 Benzylamine
[C H CH N NCl ] + – unstable
6 5 2 2 6 5 2
H O 2
C H CH OH + N + HCl - Benzene alcohol
6 5 2 2
Aniline reacts with HNO 2 to form benzene diazonium chloride which is stable at 273 278 K and hence does not evolve N 2 gas. It forms orange dye with 2naphthol.
N NCl + –
21. (i) CH CH Cl NaCN
Chloroethane
22. (a) Due to electron releasing nature, the alkyl group (R) pushes electrons towards nitrogen in alkyl amine and
NaNO /HCl 2 HBF 4 273 – 278 K
(ii) Refer Ans. 15 (ii).
(iii) Refer Ans. 15 (i).
24. (i) In aniline, the lone pair of electrons on Natom are delocalised over benzene ring due to resonance. As a result, electron density on the nitrogen atom decreases.
In contrast, in methylamine, +Ieffect of CH 3 group increases electron density on the nitrogen atom.
Therefore, aniline is a weaker base than methylamine, hence its pK b value is more than that for methylamine.
(ii) Ethylamine is soluble in water due to formation of intermolecular hydrogen bonds with water molecules. Hbonding. As a result, primary amines have higher boiling points than tertiary amines.
25. (i) Refer Ans. 13 (ii).
(ii) Coupling reactions :
Diazonium salts react with aromatic amines in weakly acidic medium and phenols in weakly alkaline medium to form coloured compounds called azo dyes by coupling at pposition of amines or phenols. The mechanism is basically that of electrophilic aromatic
(iii) Refer Ans. 13(i).
26. (i) NaNO / HCl 2
273 278 K 273 278 K Nitrobenzene
NO 2 NH 2 Sn / HCl
N Cl 2 + – CuCN
Benzoic acid
CN COOH
H O / H 2 +
(ii) KCN LiAlH 4
Benzyl Chloride
CH Cl 2 CH CN 2 CH CH NH 2 2 2
2phenylethanamine
(iii) NaNO /HCl 273 278 K 2 Aniline
NH 2 N Cl 2 CN
CuCN + –
NaNO /HCl 2 273 278 K LiAlH 4
Benzyl alcohol CH NH 2 2 CH OH 2
27. (i) Carbylamine reaction : Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is carbylamine reaction or isocyanide test and is used as a test for primary amines.
R NH + CHCl + 3KOH 2 3 Heat R NC + 3KCl + 3H O 2 Carbylamine
(foul smell)
(ii) Refer Ans. 13(i).
28. (i) Refer Ans. 15 (ii).
(ii) C 6 H 5 NH 2 + Br 2(aq)
NH 2
Br Br Br
+ 3 HBr
JJJ
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK) 1. Describe the following giving example for each.
(i) Glycosidic linkage. [AI 2008]
2. Name the bases present in DNA. Which one of these is not
present in RNA? [AI 2009]
3. What are the expected products of hydrolysis of lactose?
[AI 2009]
4. What is the biological effect of denaturation of protein?
[AI 2009]
5. Explain the following terms.
(i) Invert sugar
(ii) Polypeptide. [AI 2010]