See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the target molecule: The product is a bicyclic compound (hydrindanone framework) consisting of a cyclohexane ring fused to a cyclopentanone ring with a double bond at the ring junction. This is 3a,4,5,6,7,7a-hexahydro-1H-inden-2(3H)-one (octahydroindan-2-one with an endocyclic double bond). Step 2 - Retrosynthetic analysis: Working backwards from the bicyclic product, the last step (R) must be an intramolecular aldol condensation (base-catalyzed, HO⊖ with heat Δ) that forms the five-membered ring with the enone. The precursor to this aldol would be a diketone: specifically cyclohexanone bearing a -CH2-CO-CH3 (or -CH2-CH2-CHO) side chain. Given allyl bromide alkylation and Wacker oxidation, the diketone precursor is 2-(3-oxobutyl)cyclohexanone or similar. Step 3 - Step P: Alkylation of cyclohexanone at the alpha position using allyl bromide (H2C=CH-CH2-Br) under base (HO⊖) gives 2-allylcyclohexanone. The base deprotonates the alpha carbon of cyclohexanone, which then attacks allyl bromide. Step 4 - Step Q: Wacker process (PdCl2/CuCl2/O2/H2O) converts the terminal alkene (allyl group, H2C=CH-CH2-) to a methyl ketone. The terminal alkene CH2=CH-CH2- attached to the ring becomes -CH2-CH2-CO-CH3. Wait, Wacker on a monosubstituted terminal alkene H2C=CH- gives a methyl ketone CH3-CO-. So 2-allylcyclohexanone becomes 2-(3-oxobutyl)cyclohexanone... Actually, the allyl group -CH2-CH=CH2: Wacker oxidizes the terminal alkene to give -CH2-CH2-CO-CH3 (Markovnikov, ketone at internal carbon) = 2-(3-oxobutyl)cyclohexanone. This is now a 1,6-diketone. Step 5 - Step R: Intramolecular aldol condensation (HO⊖, Δ) of the 1,6-diketone (2-(3-oxobutyl)cyclohexanone): base abstracts alpha proton adjacent to one ketone, the enolate attacks the other ketone carbonyl intramolecularly, forming the five-membered ring with loss of water (aldol condensation), giving the bicyclic enone product. Step 6 - Sequence: P = allyl bromide with HO⊖ (alpha-alkylation), Q = Wacker process (oxidation of terminal alkene to ketone), R = HO⊖ + Δ (intramolecular aldol condensation). This matches option (b): H2C=CH-CH2-Br(HO⊖), Wacker-process, HO⊖, Δ. Step 7 - Why other options fail: (a) uses HO⊖,Δ before Wacker, which would cause aldol before the side chain ketone is formed - impossible. (c) starts with Wacker on cyclohexanone which has no alkene - nonsensical. (d) also starts with Wacker - same problem. Therefore, the correct answer is B.