See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The starting material is 5-hydroxy-2-hexanone (contains both a secondary alcohol group at C5 and a ketone group at C2). The product is 2-hexanol (contains only a secondary alcohol at C2, with the ketone fully reduced to CH2, i.e., deoxygenated). This means the ketone C=O must be reduced to CH2 (complete deoxygenation), not just to CHOH. Step 1: Identify what transformation is needed. The ketone (C=O at C2 in the starting material counting from the methyl end) is converted to a methylene (CH2) in the product. This is a deoxygenation of the carbonyl, not a simple reduction to alcohol. The OH group at the other end of the chain is retained. Step 2: Identify which reagent achieves C=O --> CH2 (deoxygenation of ketone) while leaving an existing alcohol intact. - Wolff-Kishner reduction (hydrazine + KOH/heat) converts C=O to CH2 under basic conditions. It does NOT affect existing alcohols. This fits perfectly. - Clemmensen reduction (Zn-Hg/HCl) also converts C=O to CH2 but uses strongly acidic conditions, which could affect the alcohol group (though it generally tolerates alcohols, the harsh acidic conditions are a concern, and more importantly Clemmensen is not selective for ketone in presence of alcohol under acid). - LiAlH4 reduces C=O to CHOH (gives alcohol), not CH2. It would give a diol, not the desired product. - NaBH4 also reduces C=O to CHOH (gives alcohol), not CH2. It would give a diol. Step 3: Wolff-Kishner reduction specifically deoxygenates ketones/aldehydes to CH2/CH3 under basic conditions (NH2NH2, KOH, ethylene glycol, heat). The pre-existing secondary alcohol at the other carbon is unaffected by basic conditions. Step 4: Why other options fail: - Clemmensen: acidic conditions, and converts C=O to CH2 but may not cleanly preserve the alcohol; Wolff-Kishner is the standard choice for base-stable substrates with existing OH. - LiAlH4: reduces ketone to secondary alcohol, giving a diol product, not the desired mono-alcohol with CH2. - NaBH4: same issue as LiAlH4, gives diol. Therefore, the correct answer is A.