The compound which will undergo Hoffmann Bromamide degradation

Benzamide

The correct answer is Option (2) → Benzamide

Hofmann bromamide degradation (Hofmann rearrangement) converts a primary amide ($R\text{--}CONH_2$) into a primary amine ($R\text{--}NH_2$) with one carbon less using $Br_2$ and $KOH/NaOH$. The reaction specifically requires the $\text{--}CONH_2$ group.

Explanation of Options:

1. Aniline

Aniline is an aromatic amine ($C_6H_5NH_2$), not an amide. It does not contain the carbonyl ($C=O$) group attached to $\text{--}NH_2$. Since Hofmann degradation acts on amides, aniline will not undergo this reaction.

2. Benzamide

Benzamide ($C_6H_5CONH_2$) is a primary amide, the exact functional group required. On treatment with $Br_2/\text{alkali}$, it forms aniline ($C_6H_5NH_2$) with loss of one carbon as $CO_2$. Hence, benzamide undergoes Hofmann bromamide degradation.

3. Nitrobenzene

Nitrobenzene contains a $-NO_2$ group, not an amide. The nitro group does not participate in Hofmann rearrangement. Thus, it cannot undergo this reaction.

4. Benzylamine

Benzylamine ($C_6H_5CH_2NH_2$) is already a primary amine, not an amide. Hofmann degradation reduces carbon count of amides, not amines. Therefore, benzylamine does not undergo this reaction.