Q.Which one of the following will NOT undergo Hofmann bromamide reaction?

Step 1. Hofmann bromamide degradation converts a PRIMARY amide, R-CO-NH2, to a primary amine, R-NH2 (one carbon fewer), via bromination and rearrangement at the free -NH2 nitrogen.

Step 2. Options (b), (c) and (d) -- CH3CH2-CO-NH2 (propanamide), CH3-CO-NH2 (acetamide), and C6H5-CO-NH2 (benzamide) -- are all genuine primary amides, each carrying an unsubstituted -CONH2 group that the reaction mechanism needs.

Step 3. Option (a), CH3-CO-NH-CH3 (N-methylacetamide), is instead an N-SUBSTITUTED (secondary) amide: its nitrogen already carries a methyl group in place of one of the two N-H hydrogens.

Step 4. With only one N-H remaining and that nitrogen already substituted, the bromination/rearrangement pathway that Hofmann degradation depends on cannot proceed the same way, so this is the one amide of the four that does NOT undergo the reaction as written.

Check each option for a genuinely free, unsubstituted -CONH2 group -- Hofmann degradation is defined for a primary amide, and an already N-alkylated amide breaks that requirement.

• Assuming any amide reacts with Br2/KOH the same way regardless of substitution on nitrogen -- the reaction is specifically defined for a PRIMARY amide (free -CONH2); N-substitution removes the very N-H the mechanism operates on.