Consider the above reaction and identify the product B.

Solution

Chemistry ma’am · 4 steps

First turn the nitro group into an amine, then put an acetyl group on that nitrogen.

1. Reduce the nitro group. H₂ over Pd in ethanol reduces -NO₂ to -NH₂. So A is aniline, C₆H₅NH₂, with the nitrogen still bonded directly to the ring.
2. Acetylate the nitrogen. The lone pair on the NH₂ nitrogen attacks a carbonyl carbon of acetic anhydride. One N–H hydrogen is replaced by an acetyl group, CH₃CO⁻, and acetic acid leaves.

   C₆H₅NH₂ + (CH₃CO)₂O →(pyridine) C₆H₅NHCOCH₃ + CH₃COOH

3. Role of pyridine. Pyridine is a base. It removes the acetic acid as it forms, which pushes the reaction forward and keeps the aniline from being protonated.
4. So. B is acetanilide, C₆H₅NHCOCH₃. The ring is bonded to N, and N is bonded to the C=O of the acetyl group. That is option (a).

The idea to keep. Acetic anhydride or acetyl chloride with pyridine acylates the nitrogen of an amine: it swaps one N–H for N–COCH3. The carbon skeleton around the nitrogen does not change.

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