See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Na2S is a nucleophilic sulfide reagent (S^2-) that undergoes nucleophilic aromatic substitution (SNAr) with activated aryl halides. The nitro group at the para position strongly activates the ring toward SNAr by stabilizing the Meisenheimer complex intermediate. Step 1: Identify the substrate. The substrate is 4-chloronitrobenzene (1-chloro-4-nitrobenzene), where the nitro group is para to the chlorine. The nitro group is an electron-withdrawing group that activates the carbon bearing the halide toward nucleophilic attack. Step 2: Identify the reagent. Na2S provides the S^2- (sulfide) dianion as the nucleophile. Since 2 equivalents of 4-chloronitrobenzene are used with Na2S, both the nucleophilic attacks can occur: S^2- first displaces Cl from one molecule of 4-chloronitrobenzene to form ArS^- (4-nitrophenyl thiolate), and then the ArS^- acts as nucleophile to displace Cl from a second molecule of 4-chloronitrobenzene. Step 3: Reaction mechanism. - First substitution: S^2- attacks the ipso carbon (bearing Cl) of 4-chloronitrobenzene via SNAr, the nitro group stabilizes the Meisenheimer complex, Cl^- leaves, giving 4-O2N-C6H4-S^-. - Second substitution: The 4-nitrophenyl thiolate anion (ArS^-) acts as nucleophile, attacks the ipso carbon of a second molecule of 4-chloronitrobenzene, again assisted by the para-nitro group, Cl^- leaves, giving 4-O2N-C6H4-S-C6H4-NO2-4, i.e., bis(4-nitrophenyl) sulfide. Step 4: Product identification. The product is 4,4'-dinitrodiphenyl sulfide: two benzene rings each bearing a para-nitro group, connected by a sulfur (thioether) bridge. This corresponds to option (b). Why other options fail: - (a) 4-chlorothiophenol: This would require only one substitution and hydrolysis; the Cl would remain, which is not what Na2S does with 2 equivalents of substrate. - (c) 4-nitrothiophenol: This would be the product of only one SNAr and protonation (ArSH), but with 2 equivalents of ArCl, the thiolate intermediate reacts further. - (d) Unsymmetrical dinitrodiphenyl sulfide: The second aromatic ring shows ortho-NO2, which is not present in the starting material (only para-NO2 is present). Therefore, the correct answer is B.