See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: This reaction involves the hydrolysis of an ester (ethyl acetate, CH3-C(=O)-O-CH2-CH3) using a labeled nucleophile H-O18^- (hydroxide ion with O-18 isotope). The key is to track where the labeled oxygen (O18) ends up. Step 1: Identify the reaction type. The ester CH3-C(=O)-O-CH2-CH3 reacts with hydroxide ion (H-O18^-). This is base-catalyzed ester hydrolysis (saponification). Step 2: Mechanism of saponification. The hydroxide ion (nucleophile, H-O18^-) attacks the electrophilic carbonyl carbon of the ester in a nucleophilic acyl substitution. The O18 from the hydroxide attacks the carbonyl carbon. Step 3: Tetrahedral intermediate forms with O18 attached to the carbonyl carbon. Then the alkoxide leaving group (CH3-CH2-O^-) departs, breaking the O-C bond between the acyl group and the ethoxy oxygen. Step 4: Products formed. The labeled oxygen (O18) becomes incorporated into the carboxylate product. The carboxylate formed is CH3-C(=O18^-) where the O18 is on the carboxylate. Specifically, the product is CH3-C(=O)-O18^- (acetate ion with the labeled oxygen in the carboxylate anion position, from the attacking hydroxide). Step 5: The other product is ethanol (CH3-CH2-OH) or ethoxide, which retains the unlabeled oxygen from the original ester linkage. Step 6: Evaluate options: - (a) CH3-C(=O)-O-H: This is acetic acid with unlabeled oxygen - incorrect, as the labeled O would be incorporated. - (b) CH3-CH2-O18-H: This would require the labeled O to go to ethanol, but the leaving group is the original ester oxygen (unlabeled), so this is incorrect. - (c) CH3-C(=O)-O18^-: The acetate ion with O18 in the carboxylate - this is correct, as the attacking O18 from hydroxide becomes part of the acetate product. - (d) CH3-CH2-O18^-: The ethoxide with labeled O - incorrect, as ethoxide leaves with unlabeled oxygen. Therefore, the correct answer is C.