See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: Selective reduction of a ketone in the presence of other functional groups (nitrile and alkene). Step 1: Identify the functional groups in the starting material. - Nitrile group (N≡C–) - Ketone (C=O) - Alkene (CH=CH2) Step 2: Identify the transformation required. The ketone (C=O) is converted to a secondary alcohol (C–OH), while the nitrile and the alkene remain untouched. This is a selective reduction of only the ketone. Step 3: Evaluate each reagent. (a) NaBH4: NaBH4 is a mild, selective hydride reducing agent. It selectively reduces ketones and aldehydes to alcohols. It does NOT reduce nitriles, and it does NOT reduce isolated alkenes. Therefore, it will reduce the ketone to the alcohol while leaving the nitrile and alkene intact. This matches the desired transformation. (b) LiAlH4: LiAlH4 is a powerful reducing agent. It reduces ketones, aldehydes, nitriles, esters, carboxylic acids, and more. It would also reduce the nitrile group (N≡C–) to a primary amine (–CH2–NH2), which is not desired. This does not match the required selective transformation. (c) 1 mole H2, poisoned catalyst, low pressure: This reagent is used for selective hydrogenation of alkenes (and alkynes to alkenes with Lindlar's catalyst). It would likely reduce the alkene rather than the ketone, which is not the desired transformation. (d) H3O+: This is an acid (aqueous acid), not a reducing agent. It cannot reduce a ketone to an alcohol. Step 4: Conclusion. Only NaBH4 selectively reduces the ketone to the alcohol without affecting the nitrile or the alkene. Therefore, the correct answer is A.