See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: Reduction of a ketone to a secondary alcohol using a deuteride source. The target product is PhC(OH)(D)(CH3), meaning the new C-H bond at the carbinol center must be a C-D bond, and the OH must be a normal O-H (not O-D). Step 1 - Identify what bonds are formed: The carbonyl carbon (C=O) is reduced. The hydride (or deuteride) adds to the carbonyl carbon, and a proton (or deuteron) goes to the oxygen. So: the carbon gets H or D from the reagent, and the oxygen gets H or D from the solvent/workup proton source. Step 2 - Analyze each option: (a) NaBD4 in CH3OH: NaBD4 delivers D⁻ to the carbonyl carbon, giving C-D. The solvent CH3OH provides a normal proton (H) to the alkoxide oxygen, giving O-H. Result: PhC(OH)(D)(CH3). This matches the product exactly. (b) LiAlH4, then D2O: LiAlH4 delivers H⁻ to the carbonyl carbon, giving C-H. Workup with D2O gives O-D. Result: PhC(OD)(H)(CH3). This does NOT match (carbon has H, oxygen has D). (c) NaBD4 in CH3OD: NaBD4 delivers D⁻ to the carbonyl carbon (C-D), and CH3OD provides D to the oxygen (O-D). Result: PhC(OD)(D)(CH3). This does NOT match (oxygen has D instead of H). (d) LiAlD4, then D2O: LiAlD4 delivers D⁻ to the carbonyl carbon (C-D), but workup with D2O gives O-D. Result: PhC(OD)(D)(CH3). This does NOT match (oxygen has D instead of H). Step 3 - Conclusion: Only option (a) gives the correct regiochemistry of deuterium placement: D on carbon and H on oxygen, matching the drawn product PhC(OH)(D)(CH3). Therefore, the correct answer is A.