See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The transformation requires converting a cyclic ketone (cyclohexanone ring in the perhydropyrene framework) into an acetyl group (-COCH3) attached to the ring carbon. This means the C=O of the ring must be removed (reduction of ketone to CH2) and an acetyl group must be introduced via Friedel-Crafts acylation on the aromatic part of the pyrene system. Wait - more carefully: the starting material has a ketone on the alicyclic ring, and the product has an acetyl (-COCH3) substituent on the aromatic pyrene ring. So the two steps are: (1) Friedel-Crafts acylation to introduce the acetyl group on the aromatic ring, and (2) Clemmensen reduction (Zn(Hg)/HCl) to reduce the existing alicyclic ketone to CH2. Step-by-step reasoning: 1. The starting material is a perhydropyrene with a ketone (cyclohexanone moiety) fused into the ring system, and aromatic rings (circles in the structure indicate aromatic rings). 2. The product has the aromatic pyrene core bearing a -C(=O)CH3 group, and the original ketone position is now reduced to CH2 (the cyclohexane ring no longer has C=O). 3. To achieve this: First perform Friedel-Crafts acylation using CH3COCl/AlCl3 to introduce the acetyl group on the aromatic ring (Step A). Then perform Clemmensen reduction using Zn(Hg)/HCl to reduce the alicyclic ketone to CH2 (Step B). 4. The order matters for best yield: If Clemmensen reduction is done first (option a), the aromatic acylation still works but there is no issue in principle. However, option (b) does Clemmensen first then acylation - but option (a) does acylation first then Clemmensen. Actually, the key issue is: Clemmensen reduction conditions (Zn(Hg)/HCl) could also reduce an aryl ketone, so it is better to do Clemmensen reduction FIRST (to reduce the alicyclic ketone) and then do Friedel-Crafts acylation. This way the newly introduced acetyl group is not subjected to Clemmensen reduction conditions. 5. Option (b): A = Zn(Hg), HCl (reduces the alicyclic ketone first), B = CH3COCl, AlCl3 (then introduces acetyl group via Friedel-Crafts). This sequence protects the acetyl group from being reduced and gives the best yield. 6. Why other options fail: - Option (a): Acylation first, then Clemmensen - the Clemmensen step would reduce both the alicyclic ketone AND the newly introduced aryl ketone (acetyl group), destroying the desired product. - Option (c): Uses alkyl chloride (CH3CH2Cl) instead of acyl chloride - this would give alkylation, not the acetyl group shown in product. - Option (d): Wolff-Kishner first, then alkylation with CH3CH2Cl - wrong reagents for the required transformation. Therefore, the correct answer is B.