The compound which will undergo Hoffmann Bromamide degradation
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Benzamide
The Hofmann bromamide degradation converts a primary amide into a primary amine with one carbon atom less.:
\( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)
Of the given compounds only benzamide, \( \ce{C6H5CONH2} \), is an amide. It gives aniline:
\( \displaystyle \ce{C6H5CONH2 + Br2 + 4NaOH -> C6H5NH2 + Na2CO3 + 2NaBr + 2H2O} \)
Why the other options are incorrect
Option (1) Aniline — Incorrect. Aniline is an amine, the product of the Hofmann degradation of benzamide, not a starting material for it.
Option (3) Nitrobenzene — Incorrect. Nitrobenzene has no –\( \ce{CONH2} \) group; it is converted to aniline by reduction (Sn/HCl or Fe/HCl), not by Hofmann degradation.
Option (4) Benzylamine — Incorrect. Benzylamine, \( \ce{C6H5CH2NH2} \), is an amine; it is what benzamide gives on reduction with \( \ce{LiAlH4} \), where no carbon is lost.
NCERT (2026-27 edition):
Hoffmann Bromamide Degradation Reaction
