What is the product formed when benzamide undergoes Hoffmann bromamide degradation?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Aniline
In the Hoffmann bromamide degradation, an amide is treated with bromine and sodium hydroxide. The alkyl or aryl group moves from the carbonyl carbon to the nitrogen, the carbonyl carbon leaves as carbonate, and a primary amine with one carbon atom less is formed:
\( \displaystyle \ce{C6H5CONH2 + Br2 + 4NaOH -> C6H5NH2 + Na2CO3 + 2NaBr + 2H2O} \)
Benzamide (7 C) gives an aromatic primary amine with 6 C, with \( \ce{-NH2} \) directly on the ring: aniline.
Why the other options are incorrect
Option (2) Nitrobenzene — Incorrect. The reaction gives an amine (–NH2), not a nitro compound (–NO2).
Option (3) Benzylamine — Incorrect. \( \ce{C6H5CH2NH2} \) still has 7 carbon atoms; it would need the carbonyl carbon to be kept (as in reduction of the amide), but this reaction removes it.
Option (4) 2-Methylaniline — Incorrect. It also has 7 carbon atoms, and benzamide has no methyl group on the ring to give it.
NCERT (2026-27 edition):
9.4 Preparation of Amines – “6. Hoffmann bromamide degradation reaction
Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.”
“… (ii) Benzamide is an aromatic amide containing seven carbon atoms. Hence, the amine formed from benzamide is aromatic primary amine containing six carbon atoms.”