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 C₆H₇N.  

Identify (C) from the given reaction. 

None of the above 

The correct answer is option 4- None of the above

1. A is likely benzoyl chloride or benzoic acid, which on reacting with aqueous ammonia and heating gives benzamide (B).

2. Compound B (benzamide) when heated with Br₂ and KOH undergoes Hofmann bromamide reaction, which removes the carbonyl carbon and converts the amide into an amine.

   • So, C = aniline (C₆H₅NH₂), which matches the molecular formula C₆H₇N.

• Option 1 (Benzaldehyde): This is an aldehyde ($C^6{H}^{5}CHO$). The reaction $B{r}^2/KOH$ with an amide specifically produces an amine, not an aldehyde.

• Option 2 (Benzamide): This is Compound (B), the intermediate. You must complete the second step of the reaction to find (C).

• Option 3 (diprop-2-ynylamine): While this has the formula $C^6{H}^{7}N$, it is an aliphatic (non-aromatic) secondary amine. Since the starting material (A) was aromatic, the product must retain the benzene ring.

• Option 4 ( None of above ) is CORRECT