See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Esterification: 1-naphthaleneacetic acid reacts with EtOH/H+ to give ethyl 1-naphthaleneacetate (the carboxylic acid becomes an ethyl ester). Step 2 - Alpha-methylation: Treatment with NaH (strong base, deprotonates alpha carbon) followed by MeI (methylating agent) installs a methyl group alpha to the ester carbonyl. The product is ethyl alpha-methyl-1-naphthaleneacetate, i.e., ethyl 2-(naphthalen-1-yl)propanoate. Step 3 - Hydrolysis: NaOH/H2O saponifies the ester, then H+ acidifies to give the free carboxylic acid: 2-(naphthalen-1-yl)propanoic acid (alpha-methyl-1-naphthaleneacetic acid). Step 4 - Friedel-Crafts acylation: SOCl2 converts the carboxylic acid to the corresponding acid chloride: 2-(naphthalen-1-yl)propanoyl chloride. AlCl3 then promotes intramolecular Friedel-Crafts acylation. The acylium ion formed attacks the naphthalene ring intramolecularly. Because the side chain is at the 1-position of naphthalene, the electrophilic attack occurs at the peri position (C-8) of the naphthalene, forming a five-membered ring (the acyl chain has 2 carbons: CH(Me)-C(=O)+, closing onto C-8). This gives a cyclopentenone fused across the 1,8-peri positions of naphthalene with a methyl substituent on the ring - a 2-methyl-2H-cyclopenta[def]naphthalenone or equivalently a methylated acenaphthenone-type compound. The product is option (c): a five-membered cyclopentenone ring fused at the peri positions of naphthalene bearing a methyl group on the carbon adjacent to the carbonyl. Option (a) fails because it shows a six-membered ring fusion pattern inconsistent with the two-carbon tether. Option (b) fails because no carbonyl is present (reduction never occurred). Option (d) fails because only one carbonyl is introduced; a 1,2-diketone would require two separate acylation steps. Therefore, the correct answer is C.