See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Nucleophilic aromatic substitution (SNAr) proceeds via a Meisenheimer complex intermediate. The rate-determining step involves attack of the nucleophile on the aromatic ring to form this negatively charged sigma complex. The reaction is facilitated by electron-withdrawing groups (EWGs) at the ortho and/or para positions relative to the leaving group, because they stabilize the negative charge of the Meisenheimer complex through resonance delocalization. Reasoning: In SNAr reactions, the key factor is the stability of the anionic Meisenheimer complex intermediate. Electron-withdrawing groups that are conjugated with the ring (especially at ortho/para positions to the leaving group) stabilize this intermediate and thus increase the reaction rate. Analysis of each option (all have Cl as the leaving group, with the substituent para to Cl): (a) 1-chloro-4-nitrobenzene: The nitro group (NO2) at the para position is a powerful electron-withdrawing group via resonance. It directly stabilizes the negative charge in the Meisenheimer complex by delocalizing the negative charge onto the electronegative oxygen atoms of the nitro group. This provides maximum stabilization of the intermediate. (b) 1,4-dichlorobenzene: The second Cl at para position is weakly electron-withdrawing by induction but electron-donating by resonance. It provides very little stabilization of the Meisenheimer complex. (c) 1-chloro-4-methylbenzene: The methyl group (CH3) is an electron-donating group (EDG) by hyperconjugation and induction. It destabilizes the Meisenheimer complex and thus retards SNAr. (d) 1-chloro-4-methoxybenzene: The methoxy group (OCH3) is a strong electron-donating group by resonance (+M effect), which destabilizes the Meisenheimer complex and strongly retards SNAr. Ranking of SNAr rate: (a) >> (b) >> (c) ≈ last, with (d) also very slow due to OCH3 being strongly electron-donating by resonance. Why other options fail: - (b) The para-Cl provides minimal EWG stabilization, far less than NO2. - (c) CH3 is EDG, disfavoring SNAr. - (d) OCH3 is a strong EDG by resonance, strongly disfavoring SNAr. Only in option (a) does the para-NO2 group powerfully stabilize the Meisenheimer complex through resonance, making it the fastest for nucleophilic aromatic substitution. Therefore, the correct answer is A.