See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Nucleophilic Aromatic Substitution (SNAr). The reaction of aryl halides with OH- ions proceeds via SNAr mechanism. The rate-determining step is the addition of the nucleophile to form a Meisenheimer complex. Electron-withdrawing groups (EWG) at ortho and para positions stabilize the Meisenheimer complex and increase reactivity. The number and positions of nitro groups relative to the leaving group (Br) determine reactivity. Step 1: Identify the structures and count activating nitro groups at ortho/para positions relative to Br. - Compound I: m-nitrobromobenzene — one NO2 group at meta position. Meta NO2 provides inductive withdrawal but does NOT directly stabilize the Meisenheimer complex through resonance. Least activating arrangement. - Compound II: 2,4,6-trinitrobromobenzene — three NO2 groups, two at ortho (2,6) and one at para (4) relative to Br. All three are at ortho/para positions; maximum resonance stabilization of Meisenheimer complex. Most reactive. - Compound III: p-nitrobromobenzene — one NO2 group at para position. Para NO2 directly stabilizes the Meisenheimer complex through resonance. More reactive than meta. - Compound IV: 2,4-dinitrobromobenzene — two NO2 groups, one at ortho (2) and one at para (4) relative to Br. Both are at ortho/para positions; strong stabilization. More reactive than mono-substituted but less than trinitro. Step 2: Rank reactivity. - II (3 NO2 at o/p) > IV (2 NO2 at o/p) > III (1 NO2 at p) > I (1 NO2 at m) - This gives the order: II > IV > III > I Step 3: Match with options. - Option (b): II > IV > III > I — matches exactly. Why other options fail: - (a) I > II > III > IV: Completely wrong; more nitro groups at o/p increase reactivity, not decrease it. - (c) IV > II > III > I: Wrong; II has more o/p nitro groups than IV and should be more reactive. - (d) II > IV > I > III: Wrong; III (p-NO2) is more reactive than I (m-NO2) because para position directly stabilizes Meisenheimer complex via resonance. Therefore, the correct answer is B.