See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: When an ester undergoes hydrolysis (saponification) with a base (HO^-), the mechanism proceeds via nucleophilic acyl substitution at the carbonyl carbon of the acetate (OAc) group. The key stereochemical consideration is whether the reaction proceeds with retention, inversion, or racemization at the chiral center. Step 1: Identify the reaction type. The substrate is (d)-CH3-CH(OAc)-Et, a secondary alkyl acetate ester. Treatment with HO^- (base/nucleophile) causes ester hydrolysis (saponification). Step 2: Determine the mechanism. In base-promoted ester hydrolysis (saponification), the hydroxide attacks the carbonyl carbon of the ester (the acyl carbon), NOT the carbon bearing the chiral center. The C-O bond that breaks is the acyl-oxygen bond (between carbonyl carbon and the oxygen attached to the chiral center), leaving the alkyl-oxygen bond intact at the chiral center. Step 3: Stereochemical outcome. Since the nucleophile (OH^-) attacks the carbonyl carbon of the acetate group and the C-O bond at the chiral carbon is NOT broken, the configuration at the chiral center is RETAINED. The (d) or dextrorotatory configuration is preserved. Step 4: Product identification. The product is CH3-CH(OH)-Et with retention of configuration, giving the (d) enantiomer: (d)-CH3-CH(OH)-Et, which corresponds to option (a). Why other options fail: - Option (b): Shows (l) configuration, which would require inversion at the chiral center. This would only occur in an SN2 mechanism at the chiral carbon, which does not happen in saponification. - Option (c): Shows racemization giving both (d) and (l). This would require breaking the C-O bond at the chiral center via a mechanism that allows racemization, which does not occur here. - Option (d): Shows an elimination product (alkene), which is not the expected product under mild base hydrolysis conditions for an ester. Therefore, the correct answer is A.