See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: In the acid-catalyzed esterification of a carboxylic acid with an alcohol, the mechanism involves acyl-oxygen cleavage on the acid side. Specifically, the nucleophilic oxygen of the alcohol attacks the carbonyl carbon of the carboxylic acid, and water is eliminated by loss of the OH from the carboxylic acid (not the oxygen from the alcohol). Step 1: Identify the labeled atom. The methanol used is CH3-O18-H, meaning the oxygen in methanol bears the O18 isotope label. Step 2: Determine the mechanism. In Fischer esterification (acid-catalyzed), the carboxylic acid is activated at the carbonyl by protonation, then the alcohol oxygen acts as the nucleophile attacking the carbonyl carbon. The tetrahedral intermediate then collapses by expelling the OH group that originally came from the carboxylic acid as water. Step 3: Trace the labeled oxygen. The O18 from methanol (CH3-O18-H) becomes incorporated into the ester linkage of methyl benzoate as the C-O18-CH3 bond. The water produced comes from the -OH of the original benzoic acid (unlabeled oxygen), not from the methanol oxygen. Step 4: Conclusion. The labeled O18 ends up in methyl benzoate (in the ester oxygen connecting the carbonyl carbon to the CH3 group), and not in the water molecule. Why other options fail: - (a) H2O: The water is formed from the -OH of benzoic acid, which is unlabeled. So O18 is NOT in H2O. - (c) Both (a) and (b): Since O18 is not in H2O, this is incorrect. - (d) Benzoic acid: This is the starting material, not the product; O18 is not retained here. Therefore, the correct answer is B.