Q.Which one of the following will NOT undergo Hofmann bromamide reaction?

The molecular formula $C_7H_7ON$ corresponds to benzamide, which undergoes Hofmann bromamide degradation with Br$_2$/KOH to give aniline, which then reacts with phosgene to give phenyl isocyanate.

Identifying A: The molecular formula $C_7H_7ON$ (one N, one O, degree of unsaturation matching a benzene ring plus a carbonyl) corresponds to benzamide, $C_6H_5-CO-NH_2$ (MW = 121; C:7, H:7, O:1, N:1) — consistent.

Step 1 — Hofmann bromamide degradation ($A \rightarrow B$):

Treating an amide with bromine in the presence of concentrated/aqueous KOH brings about the Hofmann bromamide (Hofmann rearrangement) reaction, converting the amide into a primary amine with one carbon less (the carbonyl carbon is lost as carbonate), with retention of configuration at the migrating group:

$C_6H_5CONH_2 \xrightarrow[KOH]{Br_2} C_6H_5NH_2\ (B,\ \text{Aniline}) + K_2CO_3 + KBr + H_2O$

Mechanism (outline): N-bromination of the amide, then base-induced loss of HBr to form an $N$-bromoamide anion (nitrene-like), then a 1,2-migration of the phenyl group from carbon to nitrogen with simultaneous loss of the carbonyl as isocyanate, then hydrolysis of the isocyanate under the basic aqueous conditions to the primary amine.

Step 2 — Reaction with phosgene ($B \rightarrow C$):

Aniline reacts with phosgene ($COCl_2$, carbonyl chloride) to give phenyl isocyanate, with loss of two equivalents of HCl:

$C_6H_5NH_2 + COCl_2 \rightarrow C_6H_5-N=C=O\ (C,\ \text{Phenyl isocyanate}) + 2HCl$

(This proceeds via an intermediate phenyl carbamoyl chloride, $C_6H_5NHCOCl$, which loses HCl to give the isocyanate.)