What is the correct order of reactivity of the following compounds in an SN2 reaction?
CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl
In an SN2 reaction the nucleophile must reach the carbon that carries the halogen. The more bulky groups there are on that carbon, the harder the approach and the slower the reaction.
\( \ce{CH3Cl} \) (methyl) — first. The carbon carries only three small hydrogen atoms, so the approach is easiest.
\( \ce{CH3CH2Cl} \) (primary) — second. One methyl group on the carbon begins to hinder the nucleophile.
\( \ce{(CH3)2CHCl} \) (secondary) — third. Two methyl groups crowd the carbon further.
\( \ce{(CH3)3CCl} \) (tertiary) — fourth (least reactive). Three methyl groups block the approach of the nucleophile.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes: (a) Substitution nucleophilic bimolecular (SN2) – “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.”
Fig. 6.3: Steric effects in SN2 reaction. The relative rate of SN2 reaction is given in parenthesis