An aromatic compound 'X' on treatment with aqueous ammonia and heating forms the compound 'Y' which on heating with $Br_2$ and KOH forms a compound 'Z' of the molecular formula $C_6H_7N$. The structures of X, Y and Z can be

The correct answer is Option (1) → 

The key reaction mentioned in the question is $Br_2$ / $KOH$, which is characteristic of the Hofmann bromamide reaction.

In this reaction, amides on treatment with $Br_2$ and $KOH$ form primary amines with one carbon less.

General reaction:

$RCONH_2 + Br_2 + 4KOH \rightarrow RNH_2 + 2KBr + K_2CO_3 + 2H_2O$

Thus, compound Y must be an amide, and the product Z must be an amine.

The molecular formula of Z is $C_6H_7N$, which corresponds to aniline ($C_6H_5NH_2$).

Therefore:

Z = Aniline

Step 1: Formation of Y

When benzoic acid reacts with aqueous ammonia and heating, it first forms ammonium benzoate, which on heating converts to benzamide.

Reaction:

$C_6H_5COOH + NH_3 \longrightarrow C_6H_5CONH_2 + H_2O$$

Thus:

Y = Benzamide

Step 2: Hofmann Bromamide Reaction

Benzamide reacts with $Br_2$ / $KOH$ to give aniline.

Reaction:

$C_6H_5CONH_2  \rightarrow C_6H_5NH_2$

Thus:

Z = Aniline

Summary

• X = Benzoic acid
• Y = Benzamide
• Z = Aniline