Predict the correct order of reactivity of the following compounds by SN2 reaction. Isobutyliodide, n-Propylchloride, Ethylchloride and Methyl iodide.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Methyliodide > Ethylchloride > n-Propylchloride > Isobutyl iodide
In an SN2 reaction the nucleophile attacks the carbon carrying the halogen from the back side, in one step. Bulky groups on or near that carbon hinder the approach of the nucleophile, so the reactivity depends mainly on steric hindrance: the order is \( \ce{CH3X} \) > primary > secondary > tertiary, and among primary halides a branched chain is slower.
Methyl iodide, CH3I — first: only H atoms on the carbon; the least hindered, the fastest.
Ethyl chloride, CH3CH2Cl — second: primary, one methyl group on the carbon.
n-Propyl chloride, CH3CH2CH2Cl — third: primary, a longer chain.
Isobutyl iodide, (CH3)2CHCH2I — last: primary, but branched at the next carbon; the most crowded, the slowest.
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.”