Which of the following compounds undergo predominantly \(S_N2\) reaction with \(NaOH\) in a polar aprotic solvent?
A. 1-(chloromethyl)-4-methoxybenzene
B. 4-(chloromethyl)benzenamine
C. 1-(chloromethyl)benzene
D. 1-(chloromethyl)-4-nitrobenzene
Choose the most appropriate answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → D only
All four compounds are benzylic chlorides with a primary, unhindered \( \ce{CH2Cl} \) carbon. A benzylic halide can also react by SN1, because the benzyl carbocation is stabilised by resonance; the para substituent decides which path wins. A polar aprotic solvent and a strong nucleophile (\( \ce{OH-} \)) favour SN2.
(A) 1-(chloromethyl)-4-methoxybenzene — Not predominantly SN2. The –\( \ce{OCH3} \) group donates electrons by resonance and strongly stabilises the benzyl carbocation, so SN1 competes strongly.
(B) 4-(chloromethyl)benzenamine — Not predominantly SN2. The –\( \ce{NH2} \) group is an even stronger electron donor; the carbocation is very stable and SN1 dominates.
(C) 1-(chloromethyl)benzene — Not predominantly SN2. Benzyl chloride itself forms a resonance-stabilised carbocation, so it reacts readily by both SN1 and SN2.
(D) 1-(chloromethyl)-4-nitrobenzene — Predominantly SN2. The –\( \ce{NO2} \) group withdraws electrons and destabilises the carbocation, so SN1 is suppressed; the electron-poor \( \ce{CH2} \) carbon is attacked from the back by \( \ce{OH-} \).
\( \ce{O2N-C6H4-CH2Cl + OH- -> O2N-C6H4-CH2OH + Cl-} \)
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 02-06 S3 31-05 S1 | – |