An aromatic compound (A) on treatment with aqueous ammonia and heating forms compound (B) which on heating with Br2 and KOH forms a compound (C) of molecular formula C6H7N.
Identify (B) from the given reactions.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Benzamide
An amide heated with \( \ce{Br2} \) and an alkali undergoes Hoffmann bromamide degradation and gives a primary amine with one carbon atom less.
Compound (C), \( \ce{C6H7N} \), is aniline (\( \ce{C6H5NH2} \): 6 C, 5 + 2 = 7 H, 1 N). It is formed from (B) by the bromamide reaction, so (B) must be the amide with one carbon more, benzamide:
\( \displaystyle \ce{C6H5CONH2 + Br2 + 4KOH -> C6H5NH2 + K2CO3 + 2KBr + 2H2O} \)
Benzamide itself comes from the aromatic acid (A), benzoic acid: with ammonia it gives ammonium benzoate, which loses water on strong heating. So (B) is benzamide, \( \ce{C6H5CONH2} \).
Why the other options are incorrect
Option (1) Benzylamine — Incorrect. Benzylamine, \( \ce{C6H5CH2NH2} \), is an amine, not an amide, so it does not undergo the bromamide reaction. It is the product of reducing benzamide, in which no carbon is lost:
\( \displaystyle \ce{C6H5CONH2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{C6H5CH2NH2} \)
Option (2) Aniline — Incorrect. Aniline, \( \ce{C6H5NH2} \), is compound (C), the final product with the formula \( \ce{C6H7N} \). It is formed from (B); it is not (B) itself.
NCERT (2026-27 edition):
Reactions Involving Cleavage of C–OH Bond
Preparation of Amines
