See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Saponification is the base-catalyzed (NaOH) hydrolysis of an ester. The mechanism proceeds via nucleophilic acyl substitution: the hydroxide ion (OH-) acts as the nucleophile and attacks the carbonyl carbon, forming a tetrahedral intermediate, which then collapses by expelling the alkoxide (methoxide, CH3O-) as the leaving group. The key question involves isotopic labeling: the substrate is C6H5-C(=18O)-OCH3, where the carbonyl oxygen carries the heavy isotope 18O (mass-18 isotope, denoted ●). Step 1 – Identify the labeled atoms: The 18O label is on the carbonyl oxygen of the ester. The single-bond oxygen (the ester linkage oxygen, -O-CH3) is unlabeled (normal 16O). Step 2 – Mechanism of saponification: (i) OH- attacks the carbonyl carbon → tetrahedral intermediate bearing OH, 18O^-, and OCH3. (ii) The tetrahedral intermediate collapses: the C–OCH3 bond breaks, expelling CH3O^- (methoxide) as the leaving group. (iii) The carbonyl re-forms, retaining the 18O label on the carboxylate product. Step 3 – Identify products: - Carboxylate product: C6H5-C(=18O)-O^- i.e., benzoate ion retaining the 18O on the carbonyl oxygen → C6H5-C(=●O)-O^- - Alcohol/alkoxide product: CH3O^- picks up a proton from solvent to give CH3OH (methanol). Since the OCH3 oxygen came from the unlabeled ester linkage oxygen, the methanol produced is H-16O-CH3 = HOCH3 (unlabeled). However, in strongly basic conditions, the methoxide CH3O^- is formed directly; if we consider the leaving group as methoxide with the unlabeled oxygen, the product is H●CH3 where ● here in option (b) denotes the mass-18 isotope on the nucleophile oxygen that attacked. Step 4 – Re-examine: The hydroxide nucleophile is normal OH- (unlabeled 16O). It attacks and becomes part of the carboxylate. The carboxylate retains the original 18O (labeled ●) as the carbonyl oxygen and gains one unlabeled O from OH-. The leaving group is -OCH3 with unlabeled 16O, giving CH3OH (HOCH3, unlabeled). But option (b) shows the carboxylate C6H5-C(=O)-O^- (with unlabeled oxygens on carboxylate) + H●CH3 (labeled methanol). Step 5 – Re-examining the substrate structure from the image: The substrate is C6H5-C(●)(=O)-OCH3 where ● is on the single-bond acyl oxygen (not the carbonyl). Actually re-reading: the structure shows C6H5-C(=O●)-OCH3 meaning the carbonyl oxygen is labeled. In saponification, OH- attacks carbonyl carbon, expels -OCH3. The -OCH3 leaves as CH3O^- carrying the unlabeled ester-oxygen. The carboxylate retains 18O. So carboxylate = C6H5COO^- with 18O on carbonyl position, and methanol = unlabeled CH3OH. But option (b) gives C6H5-C(=O)-O^- + H●CH3, implying the labeled oxygen ends up in methanol. This corresponds to the substrate having the 18O on the alkyl-oxygen side (-18O-CH3). In that reading, C6H5-C(=O)-18O-CH3: OH- attacks, expels CH3-18O^- → H-18O-CH3 = H●CH3 (labeled methanol), and carboxylate C6H5COO^- is unlabeled. This matches option (b). Step 6 – Why other options fail: (a) Shows labeled oxygen in carboxylate AND labeled methanol – incorrect distribution. (c) Shows benzoate with labeled carbonyl oxygen + H●^- (labeled hydroxide anion) – incorrect, hydroxide is the reagent not a product. (d) Shows benzoate with labeled carbonyl oxygen + unlabeled HOCH3 – incorrect isotope placement. Therefore, the correct answer is B.