CBSE Class 12 Chemistry: Amines - Marking Scheme Solution

(i) Why does aniline not give Friedel-Crafts reaction?

Step 1 (Governing Principle): [0.5 Mark] Aniline is a Lewis base due to the lone pair of electrons on the nitrogen atom.

Step 2 (Reaction Mechanism): [0.5 Mark] Friedel-Crafts reagents (like $\text{AlCl}_3$) are Lewis acids. Aniline reacts with $\text{AlCl}_3$ to form a salt:

Step 3 (Conclusion): [0.0 Mark] The nitrogen atom acquires a positive charge, which acts as a strong deactivating group, preventing further electrophilic substitution.

💡 EL10 Concept Insight: Think of the Lewis acid as a "bully" that grabs the lone pair on the nitrogen, turning the ring "shy" and preventing it from reacting further.

(ii) Arrange in increasing order of $\text{pK}_b$ values: $\text{C}_6\text{H}_5\text{NH}_2, \text{NH}_3, \text{C}_2\text{H}_5\text{NH}_2, (\text{CH}_3)_3\text{N}$

Step 1 (Concept): [0.5 Mark] $\text{pK}_b$ is inversely proportional to basic strength. Stronger bases have lower $\text{pK}_b$ values.

Step 2 (Comparison): [0.5 Mark] Basic strength order: $\text{C}_2\text{H}_5\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3 > \text{C}_6\text{H}_5\text{NH}_2$.

Step 3 (Final Answer): [0.0 Mark] Increasing order of $\text{pK}_b$:

(iii) Distinguish between $\text{CH}_3\text{CH}_2\text{NH}_2$ and $(\text{CH}_3\text{CH}_2)_2\text{NH}$ by Hinsberg test

Step 1 (Reagent): [0.5 Mark] Hinsberg reagent is Benzenesulphonyl chloride ($\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$).

Step 2 (Observation): [0.5 Mark] Primary amine ($\text{CH}_3\text{CH}_2\text{NH}_2$) forms N-ethylbenzenesulphonamide, which is soluble in alkali due to the acidic H attached to N. Secondary amine ($(\text{CH}_3\text{CH}_2)_2\text{NH}$) forms N,N-diethylbenzenesulphonamide, which is insoluble in alkali as it lacks an acidic H.

(iv) (a) Structures of A and B

(I) [0.5 Mark]