See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the transformation: The starting material contains a ketone (C=O) that must be reduced to a methylene (CH2) group, while a secondary alcohol (OH) present elsewhere in the molecule must remain intact. This is a deoxygenation of a ketone, not a simple reduction to an alcohol. Step 2 – Evaluate Clemmensen reduction (option a): Clemmensen reduction uses zinc amalgam in concentrated hydrochloric acid (Zn-Hg / conc. HCl). These strongly acidic conditions would also react with (or at minimum risk damaging) the free hydroxyl group. More critically, the acidic medium is incompatible with substrates bearing sensitive –OH groups that should be preserved. Therefore, Clemmensen reduction is not suitable here. Step 3 – Evaluate Wolff-Kishner reduction (option b): Wolff-Kishner reduction uses hydrazine and strong base (KOH) at high temperature. The strongly basic and high-temperature conditions can cause elimination or other side reactions with secondary alcohols. The OH group in the molecule may not survive these harsh basic conditions cleanly, making Wolff-Kishner unsuitable. Step 4 – Evaluate NaBH4 (option c): NaBH4 is a mild reducing agent that selectively reduces ketones and aldehydes to alcohols. However, the desired transformation is reduction of C=O all the way to CH2 (deoxygenation), not to CHOH. NaBH4 cannot achieve this deoxygenation; it would merely convert the ketone to a secondary alcohol. Step 5 – Conclusion: None of the three listed reagents can selectively reduce the ketone to CH2 while leaving the existing hydroxyl group untouched under mild enough conditions. The correct approach would require a method such as thioketal formation followed by Raney Ni desulfurization, or a similar two-step deoxygenation protocol that is compatible with a free alcohol—none of which are listed among options (a), (b), or (c). Therefore, the correct answer is D.