In the above reactions, product A and product B respectively are :

Solution

Chemistry ma’am · 4 steps

Same amide, two reagents: KOBr throws away the C=O carbon, while LiAlH4 keeps it and turns it into CH₂.

1. Reaction A. KOBr, which is Br₂ with KOH, carries out the Hofmann bromamide degradation. The aryl group moves from the carbonyl carbon to nitrogen, and the carbonyl carbon leaves as CO₃²⁻. So -CONH₂ becomes -NH₂ directly on the ring, with one carbon fewer.

   Ar-CONH₂ + Br₂ + 4KOH → Ar-NH₂ + K₂CO₃ + 2KBr + 2H₂O

2. Reaction B. LiAlH₄ followed by H₃O⁺ reduces the C=O of the amide to CH₂. So -CONH₂ becomes -CH₂NH₂, and the carbon count stays the same.
3. Ring bromine. Neither reagent touches the C-Br bond on the benzene ring. The aryl bromide is not attacked by hypobromite, and LiAlH₄ does not reduce it under these conditions. So Br stays at the meta position in both products.
4. So. A is 3-bromoaniline and B is 3-bromobenzylamine (3-Br-C₆H₄CH₂NH₂). Both keep the bromine, which is option (d).

The idea to keep. Hofmann degradation gives an amine with one carbon fewer than the amide. LiAlH₄ reduction gives an amine with the same number of carbons. Groups not involved, like a ring halogen, just ride along.

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