See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material. The starting material is 1-(4-hydroxycyclohexyl)acetophenone: a benzene ring bearing an acetyl group (–C(=O)CH3) and attached at its para (or meta, as drawn) position to C-1 of a cyclohexane ring that has an –OH at C-4. Step 2 – Reaction A: Clemmensen reduction [Zn(Hg) / HCl]. The Clemmensen reduction converts a carbonyl group (C=O) directly to a CH2 group under strongly acidic conditions. The ketone –C(=O)CH3 on the benzene ring is reduced to –CH2CH3 (an ethyl group). The cyclohexane ring and its –OH substituent are NOT affected by acidic Clemmensen conditions — alcohols are stable to Zn(Hg)/HCl. Therefore Product (A) retains the –OH on the cyclohexane ring and now has an ethyl group on the benzene ring. This matches structure Q? Let us re-examine: Q has Cl, not OH. Structure P has OH and an ethyl group on the benzene ring with a fully saturated cyclohexane — but wait, P shows OH at C-4 of cyclohexane and a 3-ethylphenyl group. However, under Clemmensen (HCl), an allylic/alcoholic –OH can sometimes be replaced by Cl, but a simple secondary alcohol on a cyclohexane ring under Zn(Hg)/HCl can undergo substitution to give the chloride. Actually, Clemmensen conditions (concentrated HCl) can convert the secondary alcohol –OH to –Cl (since HCl can substitute –OH → –Cl in secondary alcohols, especially under heating). Therefore the –OH at C-4 of the cyclohexane is converted to –Cl, and the ketone is reduced to –CH2CH3. This gives a product with –Cl on the cyclohexane and –CH2CH3 on the benzene ring — which is structure Q (fully saturated cyclohexane with Cl and ethylphenyl). So A = Q. Step 3 – Reaction B: Wolff–Kishner reduction [NH2NH2 / HO^- / Delta]. The Wolff–Kishner reduction converts a carbonyl group (C=O) to CH2 under basic conditions. The ketone –C(=O)CH3 → –CH2CH3. The –OH on the cyclohexane ring is NOT affected under basic conditions (alcohols are stable to base). Therefore Product (B) has –OH on the cyclohexane and an ethyl group on the benzene ring, which matches structure P. So B = P. Step 4 – Verify why other options fail. - Option (a) A = P, B = Q: Wrong, because Clemmensen (acidic) should convert –OH to –Cl giving Q not P, and Wolff–Kishner (basic) preserves –OH giving P not Q. - Option (b) A = Q, B = R: Wrong, because Wolff–Kishner does not introduce a double bond into the cyclohexane ring; R has a ring double bond which is not expected. - Option (d) A = R, B = P: Wrong, because Clemmensen does not produce a cyclohexene ring; R has an endocyclic double bond unexplained by Clemmensen. Therefore, the correct answer is C.