Reaction of propanamide with Br₂/KOH(aq) produces :

Solution

Chemistry ma’am · 3 steps

Amide plus Br₂ and KOH is Hofmann degradation: the C=O carbon leaves, so count one carbon less.

1. Name the reaction. A primary amide heated with Br₂ and aqueous KOH undergoes Hofmann bromamide degradation. It gives a primary amine with one carbon fewer than the amide.
2. Where carbon goes. The amide is first brominated on nitrogen. The ethyl group then shifts from the carbonyl carbon to nitrogen, giving an isocyanate, C₂H₅-N=C=O. Hydrolysis removes the carbonyl carbon as CO₂, which ends up as K₂CO₃.

   CH₃CH₂CONH₂ + Br₂ + 4KOH → CH₃CH₂NH₂ + K₂CO₃ + 2KBr + 2H₂O

3. So. Propanamide has 3 carbons, so the product is the 2-carbon amine CH₃CH₂NH₂, ethylamine, option (d).

The idea to keep. Br₂/KOH on an amide means Hofmann degradation: you get RNH₂ with one carbon less. Reducing the amide with LiAlH₄ keeps every carbon instead.

Answer from the NTA final answer key. Solution by PrepWiser.