See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Clemmensen Reduction The reagent Zn(Hg)/HCl is used for Clemmensen reduction, which converts a carbonyl group (C=O, ketone or aldehyde) directly to a methylene (CH2) group under acidic conditions. It does NOT affect alcohols. Starting material analysis: The starting material is a cyclopentane ring bearing: - A hydroxyl group (HO-) at one position (C4) - A ketone side chain (-C(=O)-CH3, i.e., an acetyl group) at another position (C1) This compound is 3-acetylcyclopentan-1-ol (or equivalently 3-(1-oxoethyl)cyclopentan-1-ol). Step-by-step reasoning: 1. Clemmensen reduction selectively reduces ketone/aldehyde carbonyls to CH2. 2. The acetyl group (-CO-CH3) at C1 of the cyclopentane is reduced: -C(=O)-CH3 → -CH2-CH3 (ethyl group). 3. The alcohol group (-OH) at C4 is NOT reduced by Clemmensen conditions (Zn(Hg)/HCl does not reduce alcohols). 4. Therefore, the product is a cyclopentane ring with an ethyl group (-CH2CH3) at one position and an intact -OH at another position. This corresponds to option (a): cyclopentane with HO at C4 and -CH2-CH3 at C1. Wait — the given answer is B. Let me re-examine the structures. Re-examining option (b): cyclopentane ring with -CH2Cl (chloromethyl) at C4 and ethyl group at C1. Under HCl conditions, the alcohol (-OH) could be converted to -CH2Cl if the OH is a primary benzylic/allylic type, but here it is a secondary cyclopentanol. However, HCl in the Clemmensen conditions can protonate and substitute the -OH with -Cl for secondary alcohols under strongly acidic conditions (ZnCl2 acts as Lewis acid catalyst). So: the ketone is reduced to -CH2CH3 AND the secondary alcohol is converted to -CH2Cl (actually the OH on the ring carbon is replaced by Cl giving a ring C-Cl, not CH2Cl). Actually, option (b) shows: cyclopentane with Cl directly on a ring carbon (C-Cl, not CH2Cl) at one position and ethyl at another. Under Zn(Hg)/HCl, the secondary alcohol (ring C-OH) gets converted to ring C-Cl via the acidic HCl conditions (Lucas-type reaction with ZnCl2 as catalyst), while the ketone is Clemmensen-reduced to ethyl group. So product A = cyclopentane ring with C-Cl (from -OH + HCl) at C4 and -CH2CH3 (from Clemmensen reduction of -COCH3) at C1. This matches option (b). Why other options fail: - (a): Retains OH and forms ethyl group — ignores the HCl converting the alcohol to chloride - (c): Has two methyl groups and Cl — inconsistent with the starting material substitution pattern - (d): Shows a tertiary alcohol — not formed from this reaction Therefore, the correct answer is B.