See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Diazo-coupling is an electrophilic aromatic substitution (EAS) reaction where the diazonium ion (ArN2+) acts as the electrophile attacking an activated aromatic ring (phenol/phenoxide). The reactivity of the diazonium ion as an electrophile depends on the electrophilicity of the diazonium nitrogen, which is governed by the electron density on the diazonium group. The more electron-withdrawing the substituent on the diazonium-bearing ring, the more electrophilic (reactive) the diazonium ion. Conversely, electron-donating groups reduce the electrophilicity of the diazonium ion by pushing electron density toward the N2+ group, stabilizing it and making it less reactive. Step 1: Assess each substituent's electronic effect on the diazonium group. - (I) Me2N- (para): Strong electron-donating group (EDG) via resonance. Donates electrons into the ring and toward N2+, greatly reducing electrophilicity. Least reactive. - (III) CH3O- (para): Moderate electron-donating group (EDG) via resonance. Also reduces electrophilicity, but less strongly than Me2N-. Second least reactive. - (IV) CH3- (para): Weak electron-donating group (EDG) via hyperconjugation/induction. Slightly reduces electrophilicity. Third in reactivity. - (II) O2N- (para): Strong electron-withdrawing group (EWG) via resonance. Withdraws electrons from the ring and away from N2+, greatly increasing electrophilicity. Most reactive. Step 2: Arrange in increasing order of reactivity (least to most reactive): I (Me2N, strongest EDG) < III (CH3O, moderate EDG) < IV (CH3, weak EDG) < II (NO2, strong EWG) This gives: I < III < IV < II, which corresponds to option (b). Step 3: Why other options fail: - (a) I < IV < II < III: Places III as most reactive, but NO2 is a stronger EWG than CH3O is EDG in the wrong direction; NO2 should give highest reactivity. - (c) III < I < II < IV: Places Me2N above CH3O and CH4 as most reactive, incorrect since NO2 should dominate. - (d) III < I < IV < II: Places Me2N above CH3O, but Me2N is a stronger donor than CH3O, so Me2N should give lower reactivity than CH3O, making I less reactive than III. Therefore, the correct answer is B.