See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Nucleophilic Aromatic Substitution (SNAr) with aqueous NaOH involves the displacement of a leaving group (such as Cl) from an aromatic ring. The rate of SNAr is governed by the electron-withdrawing groups (EWGs) on the ring, which stabilize the Meisenheimer complex intermediate. More EWGs ortho/para to the leaving group = faster reaction. Step 1: Identify each product. - M: Mono-nitration of chlorobenzene. Cl is an ortho/para director, so nitration gives predominantly ortho- and para-nitrochlorobenzene (mixture, but major products are 2-nitrochlorobenzene and 4-nitrochlorobenzene). M contains one NO2 and one Cl on the ring. - N: Mono-chlorination of nitrobenzene. NO2 is a meta director, so chlorination gives predominantly meta-chloronitrobenzene (3-chloronitrobenzene). N has one NO2 and one Cl, but the NO2 is meta to Cl, so it does NOT activate Cl toward SNAr (EWGs must be ortho/para to leaving group). - P: Mono-nitration of anisole. OCH3 is ortho/para director, so P is predominantly ortho- or para-nitroanisole. P has OCH3 (electron-donating) and NO2; no Cl present, so SNAr with NaOH is not straightforward (OCH3 is not a good leaving group under mild aq. NaOH conditions). - Q: Mono-nitration of 2-nitrochlorobenzene. 2-nitrochlorobenzene already has NO2 ortho to Cl. Adding another NO2: the incoming NO2 group is directed by both substituents. The product Q is 2,4-dinitrochlorobenzene (the second NO2 goes para to Cl and ortho to the existing NO2), or possibly 2,6-dinitrochlorobenzene. The major product is 1-chloro-2,4-dinitrobenzene (Q), where Cl has two NO2 groups at ortho and para positions. Step 2: Assess SNAr reactivity toward aq. NaOH (nucleophile = OH−, leaving group = Cl−). - For SNAr, EWGs ortho and para to the leaving group (Cl) stabilize the transition state. - M (e.g., 2- or 4-nitrochlorobenzene): one NO2 ortho or para to Cl → moderate activation. - N (3-chloronitrobenzene): NO2 is meta to Cl → does NOT activate toward SNAr → very slow. - P (nitroanisole): no Cl leaving group → SNAr with NaOH not applicable in standard sense; OCH3 is not easily displaced. - Q (1-chloro-2,4-dinitrochlorobenzene): TWO NO2 groups at ortho and para to Cl → very strong activation of SNAr → fastest reaction with aq. NaOH. Step 3: Why other options fail. - M has only one EWG activating Cl, so slower than Q. - N has NO2 meta to Cl, which does not assist SNAr; slowest among those with Cl. - P lacks a suitable leaving group for SNAr under aq. NaOH. - Q has the maximum activation (two NO2 groups correctly positioned relative to Cl), making it react fastest. Therefore, the correct answer is D.