See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Aromatic nucleophilic substitution (ArSN2, also written SNAr) proceeds via the Meisenheimer complex mechanism. The rate-determining step is the addition of the nucleophile to the aromatic ring to form the anionic Meisenheimer complex. Electron-withdrawing groups (EWG), especially nitro groups, at the ortho and para positions relative to the leaving group stabilize the Meisenheimer complex and dramatically increase the reactivity. Reasoning: - Option (a): 1-chloro-4-nitrobenzene has one NO2 group para to Cl. The negative charge in the Meisenheimer complex is delocalized onto one NO2 group (para position). This provides moderate activation. - Option (b): 1-chloro-3-nitrobenzene has one NO2 group meta to Cl. The meta position does NOT effectively stabilize the Meisenheimer complex because the negative charge cannot be delocalized onto the meta NO2 group. This is the least reactive of the nitro-containing options. - Option (c): 1-chloro-2,4-dinitrobenzene has two NO2 groups, one ortho (position 2) and one para (position 4) to the Cl leaving group. Both nitro groups can directly stabilize the Meisenheimer complex through resonance delocalization of the negative charge. Having two EWGs at ortho and para positions provides the maximum stabilization of the transition state/intermediate, making this compound the most reactive toward SNAr. - Option (d): 1-chloro-2-nitrobenzene has one NO2 group ortho to Cl. The ortho NO2 can stabilize the Meisenheimer complex, but only one group is present, so stabilization is less than in option (c). Why other options fail: Options (a) and (d) each have only one nitro group at ortho or para, giving less stabilization than two nitro groups. Option (b) has a meta nitro group which cannot stabilize the Meisenheimer complex by resonance. Option (c) uniquely has two nitro groups both positioned (ortho and para) to maximally stabilize the negative charge in the Meisenheimer complex, making it the most reactive. Therefore, the correct answer is C.