See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material and first reaction. The starting material is gamma-butyrolactone (GBL), a 5-membered cyclic ester. Treatment with (i) HCl opens the lactone ring to give 4-chlorobutanoic acid (Cl-CH2-CH2-CH2-COOH). Then (ii) SOCl2 converts the carboxylic acid to the acid chloride, giving 4-chlorobutanoyl chloride: Cl-CH2-CH2-CH2-COCl. This is compound (A). Step 2: Friedel-Crafts acylation of benzene. Compound (A) = 4-chlorobutanoyl chloride reacts with benzene (Ph-H) in the presence of AlCl3 (Friedel-Crafts acylation). The acyl chloride end reacts with benzene to give Ph-C(=O)-CH2-CH2-CH2-Cl. However, under these intramolecular Lewis acid conditions, the pendant chloride can undergo intramolecular Friedel-Crafts alkylation or the 4-chloro group cyclizes. More precisely, after the Friedel-Crafts acylation gives Ph-C(=O)-(CH2)3-Cl, the AlCl3 can promote intramolecular cyclization: the terminal chloride undergoes intramolecular Friedel-Crafts alkylation onto the phenyl ring to form a cyclic ketone - specifically alpha-tetralone (3,4-dihydronaphthalen-1(2H)-one). This is compound (B) = alpha-tetralone (a bicyclic compound: benzene ring fused with a cyclohexanone ring). Wait, let me reconsider. With 4-chlorobutanoyl chloride and AlCl3/benzene: Friedel-Crafts acylation gives Ph-CO-CH2CH2CH2Cl (4-chloro-1-phenylbutan-1-one). Then intramolecular Friedel-Crafts alkylation closes the ring to give alpha-tetralone (a 6-membered ring ketone fused to benzene). So B = alpha-tetralone. Step 3: KOH/MeOH treatment of alpha-tetralone. KOH in MeOH is a base-induced elimination (E2 or E1cb) or retro-aldol? For alpha-tetralone with KOH/MeOH, this is an aldol-type base treatment. Actually, KOH/MeOH on alpha-tetralone: the base deprotonates alpha to the carbonyl, but since it's cyclic, elimination would open or modify the ring. Under harsh base conditions, alpha-tetralone can undergo ring opening via retro-Dieckmann or retro-aldol... Actually, reconsidering: KOH/MeOH causes E2 elimination in the ring of alpha-tetralone to give 2H-naphthalen-1(4aH)-one or beta-naphthol... More likely: the beta-keto system in the ring undergoes base-mediated elimination to give Ph-C(=O)-CH=CH-CH3? No. Let me reconsider B. If B = Ph-CO-CH2CH2CH2Cl (no cyclization yet), then KOH/MeOH on Ph-CO-CH2CH2CH2Cl gives intramolecular aldol or elimination? KOH/MeOH on a halo-ketone: the base deprotonates alpha to carbonyl, then intramolecular SN2 displacement of Cl gives a cyclopropyl ketone! Ph-C(=O)-cyclopropyl. This is compound (C) = option (c). So the revised pathway: A = 4-chlorobutanoyl chloride, B = Ph-CO-CH2CH2CH2Cl (Friedel-Crafts acylation product without cyclization, i.e., 4-chloro-1-phenylbutan-1-one), C = Ph-C(=O)-cyclopropyl (phenyl cyclopropyl ketone) via intramolecular base-promoted cyclization (deprotonation alpha to C=O followed by intramolecular SN2 on the terminal chloride forming cyclopropane ring). Why other options fail: (a) Ph-CH2-C(=O)-cyclopropyl: wrong connectivity, CH2 between Ph and carbonyl not consistent. (b) Ph-CH=CH-C(=O)-CH3: chalcone-type, not consistent with this pathway. (d) Ph-C(=O)-CH=CH-CH3: elimination product, not cyclization. Therefore, the correct answer is C.