See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the target intermediate T and product V. T is 4-(bromomethyl)benzophenone: a benzene ring with a -CH2Br group and a -C(=O)Ph (benzoyl) group at the para positions. V is 4-(cyanomethyl)benzophenone: same scaffold but -CH2CN instead of -CH2Br. Step 2 - Determine reagent mu (conversion of T to V). The transformation T -> V replaces -CH2Br with -CH2CN. This is a nucleophilic substitution (SN2) at the benzylic position. The CN- nucleophile displaces Br-. To get a nitrile (not an isonitrile), the preferred reagent is KCN (potassium cyanide), because KCN in polar aprotic or protic solvents gives predominantly the nitrile product via carbon attack. HCN is a weak acid and a poor source of CN- for SN2 reactions; KCN provides CN- ion efficiently. Therefore mu = KCN. Step 3 - Determine R and S (synthesis of T from R using S). T is 4-(bromomethyl)benzophenone. The benzoyl group (-C(=O)Ph) on the ring suggests a Friedel-Crafts acylation. If R = benzene, then S must supply the acyl group and the bromomethyl-substituted benzene ring simultaneously. S = 4-(bromomethyl)benzoyl chloride [Cl-C(=O)-(4-CH2Br-phenyl)] with AlCl3 as catalyst would acylate benzene to give T directly via Friedel-Crafts acylation. This matches option (c): R = benzene, S = Cl-C(=O)-(4-CH2Br-phenyl) with AlCl3, mu = KCN. Step 4 - Elimination of other options. (a) R = chlorobenzene: Chlorobenzene as starting material would introduce a chlorine substituent on the ring of the product, which is not present in T. Incorrect. (b) R = benzene, S = Cl-C(=O)-(CH2Br-phenyl) (AlCl3), mu = HCN: The R and S parts are correct, but HCN is not a good reagent for SN2 nitrile synthesis; KCN is preferred. Incorrect. (d) R = benzoyl chloride, S = benzyl bromide (AlCl3): This would not cleanly give T via standard Friedel-Crafts chemistry because benzoyl chloride is not the aromatic substrate and the regiochemistry would be problematic. Incorrect. Therefore, the correct answer is C.