See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Selectivity of acetylation between -OH and -NH2 groups in the presence of pyridine. Step 1: p-Aminophenol has two nucleophilic groups capable of reacting with acetyl chloride: the phenolic -OH group and the aromatic -NH2 group. Step 2: Relative nucleophilicity comparison. The aromatic amine (-NH2) is a stronger nucleophile toward acyl chlorides than the phenolic -OH group. The nitrogen in -NH2 is more nucleophilic than the oxygen in phenol because nitrogen is less electronegative and its lone pair is more available for nucleophilic attack on the electrophilic carbonyl carbon of acetyl chloride. Step 3: Role of pyridine. Pyridine acts as a base to neutralize the HCl generated during the reaction. It does not significantly alter the selectivity between -NH2 and -OH; the inherently greater nucleophilicity of -NH2 over phenolic -OH governs the selectivity. Step 4: With one equivalent of acetyl chloride, only the more nucleophilic group reacts. The -NH2 group reacts preferentially to form an amide bond (-NHCOCH3), leaving the -OH group unreacted. Step 5: The product is 4-acetamidophenol (paracetamol): a benzene ring with -OH at one para position and -NHCOCH3 at the other para position. This corresponds to option (d). Why other options fail: - Option (a): Acetylation of -OH to give phenyl acetate ester is less favored because -OH is less nucleophilic than -NH2. - Options (b) and (c): These show Friedel-Crafts type acylation on the ring, which does not occur under these mild conditions with acetyl chloride and pyridine. Therefore, the correct answer is D.