Identify D:

The correct answer is Option (1) → Aniline (C₆H₅NH₂)

The sequence builds a benzene ring with a side chain, oxidises the side chain to –COOH, turns the acid into an amide and then removes one carbon by Hoffmann degradation. First, benzene is alkylated by propene in acid; the isopropyl carbocation attacks the ring, giving cumene (A):

    \( \displaystyle \ce{C6H6 + CH3CH=CH2} \xrightarrow{\ce{H+}} \ce{C6H5CH(CH3)2} \)

Next, acidified \( \ce{KMnO4} \) oxidises the whole side chain to –COOH, giving benzoic acid (B):

    \( \displaystyle \ce{C6H5CH(CH3)2} \xrightarrow{\ce{KMnO4/H+},\ \Delta} \ce{C6H5COOH} \)

Then ammonia gives ammonium benzoate, which on heating loses water to give benzamide (C):

    \( \displaystyle \ce{C6H5COOH + NH3 -> C6H5COONH4} \xrightarrow{\Delta} \ce{C6H5CONH2 + H2O} \)

Finally, KOBr (\( \ce{Br2} \) + KOH) brings about the Hoffmann bromamide degradation; the amide loses one carbon and gives aniline (D):

    \( \displaystyle \ce{C6H5CONH2 + Br2 + 4KOH -> C6H5NH2 + K2CO3 + 2KBr + 2H2O} \)

Why the other options are incorrect

• Option (2) CH₃CH₂NH₂ — Incorrect. Ethylamine would need an aliphatic amide, propanamide; the benzene ring introduced in the first step stays in every product.

• Option (3) m-Bromobenzamide — Incorrect. KOBr does not brominate the ring here; it converts the –\( \ce{CONH2} \) group into –\( \ce{NH2} \).

• Option (4) CH₃CH(Br)CONH₂ — Incorrect. It has no benzene ring and still carries the amide group, which KOBr removes.

NCERT (2026-27 edition):
From Alkylbenzenes – “Aromatic carboxylic acids can be prepared by vigorous oxidation of alkyl benzenes with chromic acid or acidic or alkaline potassium permanganate. The entire side chain is oxidised to the carboxyl group irrespective of length of the side chain.”

Reaction with Ammonia – “Carboxylic acids react with ammonia to give ammonium salt which on further heating at high temperature give amides.”