Among the following halogen compounds, which would undergo an SN2 reaction faster?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → iodobutane
In an SN2 reaction, the nucleophile attacks the carbon from the side opposite the halogen while the C–X bond breaks, all in one step. The rate depends on two things:
Steric hindrance at the carbon: \( \ce{CH3X} \) > 1° > 2° > 3°.
Ease of breaking the C–X bond (leaving group): the weaker the bond and the larger, more stable the halide ion formed, the faster the reaction. The C–X bond enthalpy falls from C–F to C–I:
\( \displaystyle \text{C-F }(452) > \text{C-Cl }(351) > \text{C-Br }(293) > \text{C-I }(234)\ \mathrm{kJ\,mol^{-1}} \)
\( \displaystyle \text{Reactivity: R-I} > \text{R-Br} > \text{R-Cl} > \text{R-F} \)
All four compounds have the same butyl group, so steric hindrance is the same and only the halogen decides the rate. The C–I bond is the weakest and \( \ce{I^{-}} \) is the best leaving group, so iodobutane reacts fastest.
Why the other options are incorrect
Option (2) chlorobutane — Incorrect. The C–Cl bond is stronger than C–Br and C–I, so it reacts more slowly than both.
Option (3) bromobutane — Incorrect. It reacts faster than chlorobutane but slower than iodobutane, because C–Br is stronger than C–I.
Option (4) fluorobutane — Incorrect. The C–F bond is the strongest, so it is the least reactive of the four.
NCERT (2026-27 edition):
Reactions of Haloalkanes – “… For a given alkyl group, the reactivity of the halide, R-X, follows the same order in both the mechanisms R–I > R–Br > R–Cl >> R–F.”