Identify the product C

The correct answer is Option (2) → 3,5-dibromoaniline

This is a three-step conversion of a carboxylic acid into an amine with one carbon atom less:

• SOCl₂ converts –COOH into the acid chloride –COCl (with \( \ce{SO2} \) and HCl as gases).

• NH₃ converts the acid chloride into the amide –\( \ce{CONH2} \).

• NaOBr (\( \ce{Br2} \) + NaOH) carries out the Hofmann bromamide degradation: the aryl group migrates from the carbonyl carbon to nitrogen, the C=O carbon is lost as carbonate, and a primary amine with one carbon fewer is formed.

    \( \displaystyle \ce{ArCOOH + SOCl2 -> ArCOCl + SO2 ^ + HCl ^} \)

    \( \displaystyle \ce{ArCOCl + 2NH3 -> ArCONH2 + NH4Cl} \)

    \( \displaystyle \ce{ArCONH2 + Br2 + 4NaOH -> ArNH2 + Na2CO3 + 2NaBr + 2H2O} \)

Here Ar = 3,5-dibromophenyl. The two Br atoms are not touched, so:

• A = 3,5-dibromobenzoyl chloride

• B = 3,5-dibromobenzamide

• C = 3,5-dibromoaniline (\( \ce{NH2} \) in place of the COOH, meta to both Br atoms), which is structure 2.

Why the other options are incorrect

• Option (1) 3,5-dibromobenzamide — Incorrect. This is intermediate B; NaOBr converts it further into the amine.

• Option (3) 1,3,5-tribromobenzene — Incorrect. The Hofmann reaction puts an \( \ce{NH2} \) group, not a Br atom, in place of the carbonyl group.

• Option (4) 3-amino-5-bromobenzoic acid — Incorrect. The COOH carbon is the one lost in the Hofmann step, and the ring bromines are not replaced; this structure keeps the COOH and loses a Br.

NCERT (2026-27 edition):
Hoffmann Bromamide Degradation Reaction