In SN2 reaction, predict the increasing order of reactivity of the following alkyl halides
(A) CH3-Cl
(B) CH3-CH2-Cl
(C) CH3 — CH(Cl) — CH3
(D) CH3 — C(Cl)(CH3) — CH3
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (D), (C), (B), (A)
In an SN2 reaction, the nucleophile attacks the carbon from the back, opposite to the halogen. Bulky groups around that carbon block this approach (steric hindrance). The more alkyl groups the carbon carries, the slower the reaction:
\( \displaystyle \mathrm{CH_3X} > 1^\circ > 2^\circ > 3^\circ \)
Classify the given halides:
(A) \( \ce{CH3Cl} \): methyl
(B) \( \ce{CH3CH2Cl} \): primary
(C) 2-chloropropane: secondary
(D) 2-chloro-2-methylpropane: tertiary
Shortcut: The tertiary halide (D) is the least reactive and \( \ce{CH3Cl} \) (A) the most, which leaves Options (1) and (4). The secondary halide (C) is less reactive than the primary one (B), so the answer is Option (4).
NCERT (2026-27 edition):
Reactions of Haloalkanes – “… Since this reaction requires the approach of the nucleophile to the carbon bearing the leaving group, the presence of bulky substituents on or near the carbon atom have a dramatic inhibiting effect. Of the simple alkyl halides, methyl halides react most rapidly in SN2 reactions because there are only three small hydrogen atoms. Tertiary halides are the least reactive because bulky groups hinder the approaching nucleophiles. Thus the order of reactivity followed is: Primary halide > Secondary halide > Tertiary halide.”