Hoffmann bromomide degradation of benzamide gives product A, which upon heating with CHCl₃ and NaOH gives product B.
The structures of A and B are :

Solution

Chemistry ma’am · 4 steps

Hoffmann bromamide strips the C=O carbon to give aniline; chloroform and NaOH then make an isocyanide.

1. Hoffmann bromamide. With Br₂ and NaOH, an amide RCONH₂ becomes RNH₂. The carbonyl carbon leaves as CO₃²⁻, so the amine has one carbon fewer and the ring is never brominated.

   C₆H₅CONH₂ + Br₂ + 4NaOH → C₆H₅NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

2. Product A. So A is aniline, C₆H₅NH₂, with no bromine on the ring.
3. Carbylamine reaction. A primary amine heated with CHCl₃ and NaOH (or alcoholic KOH) gives an isocyanide, which has a very foul smell. Dichlorocarbene, :CCl₂, attacks the nitrogen, and the -NH₂ group becomes -N⁺≡C⁻.

   C₆H₅NH₂ + CHCl₃ + 3NaOH → C₆H₅NC + 3NaCl + 3H₂O

4. So. A is aniline and B is phenyl isocyanide, C₆H₅-N⁺≡C⁻. That matches option (a).

The idea to keep. Hoffmann bromamide removes the C=O carbon and leaves a primary amine. A primary amine with CHCl₃ and base gives an isocyanide (carbylamine test), never a ring -CHO.

Answer from the NTA final answer key. Solution by PrepWiser.