Which out of the following show Hoffmann Bromamide Degradation reaction?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(ArCONH_2\)
In the Hoffmann bromamide degradation a primary amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide. The alkyl or aryl group migrates from the carbonyl carbon to the nitrogen, and the product is a primary amine with one carbon atom less than the amide:
\( \ce{RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O} \)
For an aromatic amide such as benzamide:
\( \ce{C6H5CONH2 + Br2 + 4NaOH -> C6H5NH2 + Na2CO3 + 2NaBr + 2H2O} \)
So the compound that shows the reaction is the amide ArCONH2.
Why the other options are incorrect
Option (1) \(ArNO_2\) — Incorrect. A nitro compound is converted to an amine by reduction (Sn/HCl or Fe/HCl, or \( \ce{H2} \) with Pd), not by \( \ce{Br2} \)/NaOH.
Option (2) \(ArCH_2NH_2\) — Incorrect. This is already an amine (benzylamine type); it has no –\( \ce{CONH2} \) group. It is the product of reducing \( \ce{ArCONH2} \) with \( \ce{LiAlH4} \), which keeps all the carbon atoms.
Option (4) \(ArNH_2\) — Incorrect. This is the product of the Hoffmann reaction of \( \ce{ArCONH2} \), not a compound that undergoes it.
NCERT (2026-27 edition):
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.”
Hoffmann Bromamide Degradation Reaction – “… The amine so formed contains one carbon less than that present in the amide.”