See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify what each conversion requires: (a) The substrate has a trimethylammonium (Me3N+) group on the saturated ring and an acetyl (COCH3) group on the aromatic ring. The product retains the Me3N+ group and the acetyl is reduced to an ethyl group (C=O → CH2). The Me3N+ group is a quaternary ammonium salt — it is acid-sensitive (Clemmensen uses HCl) and base-sensitive (Wolff-Kishner uses strong base/hydrazine). However, LiAlH4 would also reduce the C=O. Clemmensen (Zn(Hg)/HCl) uses acidic conditions — the quaternary ammonium salt would survive acidic conditions since it has no labile proton and is already fully methylated. Wolff-Kishner uses strong base (OH-) and heat, which would cause Hofmann elimination of the trimethylammonium group. LiAlH4 would reduce the ketone to an alcohol, not to CH2. Therefore, Clemmensen (q) is suitable for (a) as it reduces C=O to CH2 under acidic conditions, leaving Me3N+ intact. (b) The substrate has an epoxide on the saturated ring and an acetyl group on the aromatic ring. The product retains the intact epoxide and the acetyl is reduced to ethyl. LiAlH4 would open the epoxide as well as reduce the ketone, so it is not selective here. Wolff-Kishner uses strongly basic conditions — base can open epoxides. Clemmensen uses acidic conditions — epoxides can open in acid too. However, none of the listed reagents (p, q, r) would cleanly reduce the ketone while leaving the epoxide intact. Therefore, the answer is (s) None — no single reagent from the list accomplishes selective ketone-to-methylene reduction without disturbing the epoxide. (c) The substrate has the same epoxide-tetralin with an acetyl group. The product shows the epoxide opened (OH appears at C1 or C2 of the ring) AND the acetyl reduced to a secondary alcohol CH(OH)CH3. LiAlH4 is a strong reducing agent that reduces ketones to secondary alcohols AND opens epoxides (reduces them to alcohols). This matches perfectly. Therefore, (c) → (r) LiAlH4. (d) The substrate has an OH group at C1 of the saturated ring and an acetyl group on the aromatic ring. The product retains the OH and reduces the acetyl to ethyl (C=O → CH2). Clemmensen (Zn(Hg)/HCl) would also reduce the alcohol under harsh conditions. LiAlH4 would reduce C=O to alcohol not CH2. The Wolff-Kishner reduction uses hydrazine/base/heat and selectively reduces ketones/aldehydes to CH2 without affecting secondary alcohols (OH groups are stable under basic conditions). Therefore, (d) → (p) Wolff-Kishner. Summary: (a) → (q) Clemmensen reduction (acid conditions preserve Me3N+, reduces C=O to CH2) (b) → (s) None (no listed reagent selectively reduces ketone without opening epoxide) (c) → (r) LiAlH4 (reduces ketone to alcohol AND opens epoxide) (d) → (p) Wolff-Kishner (reduces ketone to CH2 under basic conditions, OH group preserved) Therefore, the correct answer is {"a": ["Q"], "b": ["S"], "c": ["R"], "d": ["P"]}.