Which of the following compounds undergo predominantly \(S_N2\) reaction with \(NaOH\) in a polar aprotic solvent?
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) 4-methoxybenzyl chloride, \( \ce{CH3O-C6H4-CH2Cl} \) — Not predominantly SN2. The –\( \ce{OCH3} \) group donates electrons by resonance and strongly stabilises the benzyl carbocation, so SN1 competes strongly.
(B) 4-aminobenzyl chloride, \( \ce{H2N-C6H4-CH2Cl} \) — Not predominantly SN2. The –\( \ce{NH2} \) group is an even stronger electron donor; the carbocation is very stable and SN1 dominates.
(C) benzyl chloride, \( \ce{C6H5-CH2Cl} \) — Not predominantly SN2. Benzyl chloride itself forms a resonance-stabilised carbocation, so it reacts readily by both SN1 and SN2.
(D) 4-nitrobenzyl chloride, \( \ce{O2N-C6H4-CH2Cl} \) — 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 \( \ce{OH-} \).
\( \ce{O2N-C6H4-CH2Cl + OH- -> O2N-C6H4-CH2OH + Cl-} \)