CBSE Class 12 Chemistry 2014 Outside Delhi Set 1 Solved Paper

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Question : 25
Total: 30
Account for the following:
(i) Primary amines (R−NH2) have higher boiling point than tertiary amines (R3N).
(ii) Aniline does not undergo Friedel-Crafts reaction.
(iii) (CH3)2‌NH is more basic than (CH3)3N in an aqueous solution.
OR
Give the structures of A, B and C in the following reactions:
(i) C6H5NO2
Sn+HCl
───▸
A
NaNO2+HCl
───────▸
B
H2O
───▸
C

(ii) CH3‌CN‌
H2O∕H+
───▸
‌A
‌
NH3
───▸
B
Br2+KOH
─────▸
C
Solution:  
(i) Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule. This intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation. Therefore, the order of boiling points of isomeric amines is as follows:
Primary > Secondary > Tertiary
(ii) Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst. Due to this, nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group for further reaction
(iii) In the aqueous phase, the substituted ammonium cations get stabilised not only by electron releasing effect of the alkyl group (+I) but also by solvation with water molecules. The greater the size of the ion, lesser will be the solvation and the less stabilised is the ion. Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base. Thus, the order of basicity of aliphatic amines should be:
primary > secondary > tertiary.
OR
(i) C6H5NO2
Sn+HCl
───▸
C6H5NH2
Aniline
NaNO2+HCl
───────▸
C6H5−
+
N2
−
Cl
Benzeediazonium‌chloride
H2O
───▸
C6H5‌OH
Phenol

(ii)
CH3‌CN
Methyl‌cyanide
H2O∕H+
───▸
CH3‌COOH
Ethanoic‌acid
NH3
───▸
CH3‌CO‌
+
N
‌
−
H2
Ethanamide
Br2+KOH
─────▸
CH3NH2
Methylamine
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