The reaction of R-CO-NH2 with bromine and KOH gives RNH2 as the end p…

The correct answer is Option (3) → R–N=C=O

This is the Hoffmann bromamide degradation. The amine formed has one carbon less than the amide, because the alkyl group R moves from the carbonyl carbon to nitrogen. First, \( \ce{Br2} \)/KOH brominates the nitrogen:

    \( \displaystyle \ce{R-CO-NH2 + Br2 + KOH -> R-CO-NHBr + KBr + H2O} \)

Then the base removes the N–H proton; as \( \ce{Br-} \) leaves, R migrates from carbon to nitrogen, giving the isocyanate, R–N=C=O:

    \( \displaystyle \ce{R-CO-NHBr + KOH -> R-N=C=O + KBr + H2O} \)

Last, the isocyanate is hydrolysed by KOH to the primary amine and carbonate:

    \( \displaystyle \ce{R-N=C=O + 2KOH -> R-NH2 + K2CO3} \)

The net reaction is \( \ce{R-CO-NH2 + Br2 + 4KOH -> R-NH2 + K2CO3 + 2KBr + 2H2O} \). The isocyanate is the intermediate that gives the amine.

• Option (1) R-CO-NH-Br — Incorrect. The N-bromoamide is only the first brominated species; R is still on the carbonyl carbon, and it must rearrange to the isocyanate before any amine can form.

• Option (2) R-NH-Br — Incorrect. This would need the carbonyl carbon to leave before the rearrangement. In fact the C=O carbon stays bonded to nitrogen in the isocyanate and leaves only at the end, as carbonate.

• Option (4) R-CO-NBr₂ — Incorrect. The N-bromoamide loses its remaining N–H proton and rearranges at once; a second bromine is not taken up on the way to the isocyanate.

Note: The question is from beyond the NCERT textbook.

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. 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.”