See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reagents: Naphthalene reacts with oxalyl chloride (ClCO-COCl) in the presence of AlCl3, a Friedel-Crafts acylation catalyst. Step 2 - Friedel-Crafts diacylation to form [X]: Oxalyl chloride has two acid chloride groups. With AlCl3, it undergoes double Friedel-Crafts acylation on naphthalene. Naphthalene is most reactive at the 1 and 8 positions (peri positions). The two COCl groups acylate both peri positions (C1 and C8) of naphthalene, forming a diketone where the two carbonyl groups bridge the 1,8-positions. This gives naphthalene-1,8-diyl diketone, i.e., a cyclic 1,8-naphthalenediyl-bridged diketone: specifically acenaphthenequinone (acenaphthylene-1,2-dione) — a five-membered ring containing two carbonyl groups fused at the 1,8 positions of naphthalene. So [X] = acenaphthylenedione (acenaphthenequinone). Step 3 - Clemmensen-type reduction of [X] to [Y]: Treatment with Na-Hg (sodium amalgam) and HCl in excess is a reducing condition (similar to Clemmensen reduction) that reduces C=O groups to CH2. Both ketone groups in the five-membered ring of acenaphthenequinone are reduced to CH2, converting the two C=O groups into two CH2 groups. This gives acenaphthylene reduced to acenaphthene — a naphthalene with a five-membered saturated ring (cyclopentane ring) fused at the 1,8 peri positions. Step 4 - Identify the product: Acenaphthene is naphthalene with a -CH2-CH2- bridge across the 1,8 positions, forming a tricyclic system with a five-membered carbocycle fused to naphthalene at the peri positions. This matches option (c), which shows naphthalene with a five-membered saturated ring fused across the top (1,8-positions). Step 5 - Why other options fail: - (a) shows a four-membered ring fused at 2,3-position: incorrect regiochemistry and ring size. - (b) shows a four-membered ring at an angular position: incorrect ring size and regiochemistry. - (d) shows a five-membered ring at peri positions but with a double bond inside the five-membered ring (unsaturated), which would be acenaphthylene, not the fully reduced product from excess Na-Hg/HCl. Therefore, the correct answer is C.