Given below are two statements :
Statement I : In Hofmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom.
Statement II : The group is migrated in Hofmann degradation reaction to electron deficient atom.
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Ask two things: which groups can move, and where do they land? Benzamide answers the first one.

1. The reaction. Hofmann degradation turns an amide into an amine with one carbon less: RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O. The group R moves from the carbonyl carbon onto nitrogen.
2. Where it lands. The amide first becomes an N-bromoamide. Base removes the N–H proton, and the anion loses Br⁻. That leaves nitrogen with only six electrons, so it is electron deficient. R shifts onto it with its bonding pair and gives an isocyanate, R-N=C=O, which hydrolyses to RNH₂. So Statement II is correct.
3. Which groups move. The migrating group does not have to be alkyl. Benzamide, C₆H₅CONH₂, gives aniline, C₆H₅NH₂, so an aryl group migrates just as well. The word "only" makes Statement I incorrect.
4. So. Statement I is incorrect and Statement II is correct, which is option (d).

The idea to keep. In Hofmann degradation, an alkyl or an aryl group moves from carbon to an electron-deficient nitrogen and keeps its configuration. Watch for "only" in a statement: one counterexample breaks it.

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