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 C₇H₇NO.  

Identify the reaction taking place while converting compound B to compound C in the following reaction sequence:

Reaction Scheme:

C₆H₅COOH  —(i) aq. NH₃, (ii) Δ→  B  —(Br₂ / KOH)→  C

Hofmann Bromamide reaction 

The correct answer is option 3. Hofmann Bromamide reaction.

Step 1:

Benzoic acid reacts with aqueous ammonia and on heating forms benzamide.

Reaction:

$C_6H_5COOH + NH_3 \rightarrow C_6H_5CONH_2 \text{ (Benzamide)}$

So compound $B = \text{Benzamide}$.

Step 2:

Benzamide reacts with $Br_2$ and $KOH$. This reaction converts amide into amine with one carbon less. This reaction is known as Hofmann bromamide degradation or Hofmann rearrangement.

Reaction:

$C_6H_5CONH_2 + Br_2 / KOH \rightarrow C_6H_5NH_2 \text{ (Aniline)}$

So compound $C = \text{Aniline}$.

This reaction involves rearrangement of amide to amine with loss of carbonyl carbon as $CO_2$.

Therefore, the reaction converting $B$ to $C$ is Hofmann bromamide reaction.