See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the first reaction (Benzene → P): Benzene reacts with acetyl chloride (CH3-C(=O)-Cl). Under normal Friedel-Crafts acylation conditions this would require AlCl3, but the arrow shows no catalyst for the first step. Given that phenol (not plain benzene) is likely the starting material (the hexagon shown has a hydroxyl implied by the context and the products), phenol reacts with acetyl chloride even without a Lewis acid catalyst via O-acylation to give phenyl acetate (Ph-O-CO-CH3). So P = phenyl acetate. Step 2 – Identify the second reaction (P → Q): Phenyl acetate is treated with anhydrous AlCl3 and heat (Delta). This is the Fries rearrangement. In the Fries rearrangement, an aryl ester (phenyl acetate) undergoes rearrangement in the presence of AlCl3 with heat to give a hydroxyaryl ketone. At higher temperatures (heat/Delta), the Fries rearrangement preferentially gives the para-isomer. The acetyl group migrates from the oxygen to the para position of the ring, yielding 4-hydroxyacetophenone (para-hydroxyacetophenone): a benzene ring bearing a para-OH group and an acetyl group (C(=O)-Me) at the ortho or para position — under heating, predominantly para product. Step 3 – Match to options: (c) shows a benzene ring with OH at para position and C(=O)-Me group, which is exactly 4-hydroxyacetophenone, the para-Fries rearrangement product. Step 4 – Why other options fail: (a) Anisole: would require O-methylation, not what happens here. (b) Phenyl acetate: this is intermediate P, not Q. (d) 4-methoxyphenol: has OMe and OH groups; no ketone, unrelated to this reaction sequence. Therefore, the correct answer is C.