• This is actually a Hoffman bromamide degradation reaction.

• This reaction occurs by treating an amide with bromine in the presence of the aqueous or ethanolic solution of sodium hydroxide.

• In this reaction, migration of an alkyl or aryl group happens from the carbonyl carbon of the amide to the nitrogen atom of the same amide.

• Hence the amine formed contains one carbon less than that of the reactant amide.

• With other 3 reagents (i) excess H₂/Pt (iii) NaBH₄/methanol (iv) LiAlH₄/ether the same mechanism does not applicable, hence they are wrong answers.