See image — Amines Chemistry Question

💡 Solution & Explanation

Concept: Acetic anhydride (Ac2O) reacts with both alcohols (to form esters) and amines (to form amides). However, amines are more nucleophilic than alcohols under normal conditions, so when Ac2O is added in limited (one equivalent) amount to a compound containing both an -OH and an -NH2 group, it selectively acylates the more nucleophilic amine first. Step 1 - First reaction (starting material + Ac2O → P): The starting material is 4-(aminomethyl)cyclohexanemethanol (a cyclohexane with -CH2OH at C1 and -NH2 at C4, or equivalently trans-4-(hydroxymethyl)cyclohexanemethanamine). When one equivalent of Ac2O is added, the more nucleophilic -NH2 reacts preferentially to form an amide (-NH-C(=O)-CH3), while the -OH remains unreacted. Therefore, P = cyclohexane ring with -CH2OH at C1 and -NH-C(=O)-CH3 at C4. Step 2 - Second reaction (P + Ac2O → Q): Now P has only the -OH group remaining as a nucleophile (the amine has already been acetylated). The second equivalent of Ac2O reacts with the -OH to form an acetate ester (-CH2-O-C(=O)-CH3). Therefore, Q = cyclohexane ring with -CH2-O-C(=O)-CH3 at C1 and -NH-C(=O)-CH3 at C4. Why other options fail: (a) P shows the alcohol acetylated first (ester formed) while amine is free - this contradicts the selectivity; amines react faster with Ac2O than alcohols. (b) Both P and Q show a -CH2-C(=O)-O-CH3 group which would imply oxidation of the CH2OH to a carboxylic acid then esterification - not what Ac2O does. (c) Q shows a -CH2-C(=O)-O-CH3 group (same issue as b - this is not an acetate ester of the primary alcohol but rather a methyl ester of a carboxylic acid, which is not the product of Ac2O reacting with -CH2OH). (d) P = cyclohexane with -CH2OH and -NH-C(=O)-CH3 (amine acetylated first, correct), Q = cyclohexane with -CH2-O-C(=O)-CH3 and -NH-C(=O)-CH3 (alcohol then acetylated, correct). This matches the expected selectivity and correct product structures. Therefore, the correct answer is D.