The correct structures of A and B formed in the following reactions are :

Solution

Chemistry ma’am · 4 steps

Reduce the nitro group first, then let the more nucleophilic NH₂ grab the one acetyl group.

1. Find A. H₂/Pd in ethanol reduces only the -NO₂ group to -NH₂. The -OH is untouched, so A is p-aminophenol, HO-C₆H₄-NH₂.
2. Compare nucleophiles. A now has two groups that can attack acetic anhydride, -NH₂ and -OH. Nitrogen is less electronegative than oxygen, so its lone pair is more available and -NH₂ is the stronger nucleophile.
3. Only one equivalent. With just 1.0 eq of (CH₃CO)₂O, there is only enough reagent for one acylation, and the faster -NH₂ uses it up. The product is the amide HO-C₆H₄-NHCOCH₃, which is paracetamol.
4. So. A is p-aminophenol and B is N-(4-hydroxyphenyl)acetamide, with the -OH still free. That is option (d).

The idea to keep. When a molecule has both -NH₂ and -OH and you add one equivalent of an acylating agent, the amine reacts first. You get the amide, not the ester.

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