See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Nucleophilic Aromatic Substitution (SNAr) reactivity is governed by the ability of electron-withdrawing groups (EWGs) to stabilize the Meisenheimer complex intermediate. The more EWGs present at ortho and para positions relative to the leaving group (Br), the greater the stabilization of the negative charge in the transition state, and the faster the reaction. Step 1: Identify the structures. - Compound I: 1-bromo-3-nitrobenzene. The NO2 is at meta to Br. Meta-NO2 does not effectively stabilize the Meisenheimer complex (negative charge cannot be delocalized onto meta-NO2). Only one NO2, and it is meta — least activating. - Compound II: 1-bromo-2,4,6-trinitrobenzene (picryl bromide). Three NO2 groups at ortho (×2) and para positions relative to Br. Maximum stabilization of Meisenheimer complex. Highest reactivity. - Compound III: 1-bromo-4-nitrobenzene. One NO2 at para to Br. Para-NO2 effectively stabilizes the Meisenheimer complex. One activating group. - Compound IV: 1-bromo-2,4-dinitrobenzene. Two NO2 groups, one ortho and one para to Br. Both stabilize the Meisenheimer complex effectively. More activating than one NO2 (III), less than three NO2 (II). Step 2: Rank by reactivity. - II (three NO2 at ortho/para) > IV (two NO2 at ortho/para) > III (one NO2 at para) > I (one NO2 at meta, ineffective stabilization) - Order: II > IV > III > I Step 3: Evaluate options. - (a) I > II > III > IV — incorrect, reverses the trend. - (b) II > IV > III > I — matches our analysis exactly. - (c) IV > II > III > I — incorrect, II should be highest. - (d) II > IV > I > III — incorrect, I (meta-NO2) should be less reactive than III (para-NO2). Why (d) fails: In option (d), I is placed above III. Compound I has NO2 at meta (cannot delocalize negative charge in Meisenheimer complex), while III has NO2 at para (can delocalize). So III > I, not I > III. Therefore, the correct answer is B.