See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Nucleophilic Aromatic Substitution (SNAr). The reaction of aryl halides with HO- proceeds via the addition-elimination (Meisenheimer complex) mechanism. The rate-determining step is the addition of the nucleophile to form the Meisenheimer complex, which is stabilized by electron-withdrawing groups (EWG) ortho and para to the leaving group (Br). More EWG groups ortho/para to Br = greater stabilization of the Meisenheimer complex = faster reaction. Analysis of each compound (Br is the site of nucleophilic attack): (I) meta-nitrobromobenzene: One NO2 group at meta position. Meta NO2 does NOT directly stabilize the Meisenheimer complex (negative charge at ortho/para positions relative to Br), so it provides minimal activation. Least reactive among the four. (III) para-nitrobromobenzene: One NO2 group at para position. Para NO2 directly stabilizes the Meisenheimer complex (negative charge delocalized onto the para-NO2 group). More reactive than meta isomer but only one activating NO2. (IV) 2,4-dinitrobromobenzene: Two NO2 groups, one at ortho (position 2) and one at para (position 4) relative to Br. Both are ortho/para to Br, so both directly stabilize the Meisenheimer complex. More reactive than mono-nitro compounds. (II) 2,4,6-trinitrobromobenzene: Three NO2 groups at positions 2, 4, and 6 (both ortho positions and para position) relative to Br. All three NO2 groups are ortho/para to Br, providing maximum stabilization of the Meisenheimer complex. Most reactive. Decreasing order of reactivity: II > IV > III > I Why other options fail: (a) I > II > III > IV: Incorrect; places meta isomer (I) as most reactive, which is wrong. (c) IV > II > III > I: Incorrect; places IV above II, but trinitro compound II has more activating groups than dinitro IV. (d) II > IV > I > III: Incorrect; places meta isomer (I) above para isomer (III), but para-NO2 activates SNAr more effectively than meta-NO2. Therefore, the correct answer is B.