Predict the decreasing order of rate of reaction of the following compounds which is progressing via SN2 pathway ?
A. CH3CH2CH2I
B. CH3CH2CH2Cl
C. CH3CH2CH2Br
D. CH3CH2CH2F
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A, C, B, D
An SN2 reaction takes place in one step: the nucleophile attacks the carbon while the halide ion leaves. When the alkyl group is the same, as in these four propyl halides, the rate depends on how easily the C–X bond breaks and the halide leaves. Two related factors decide it:
C–X bond strength: it falls from C–F to C–I (bond enthalpies about 452, 351, 293 and 234 kJ mol−1 for \( \ce{CH3-F} \), –Cl, –Br and –I). A weaker bond breaks faster.
Leaving group ability: \( \ce{I^{-}} \) > \( \ce{Br^{-}} \) > \( \ce{Cl^{-}} \) > \( \ce{F^{-}} \). The larger, weaker-base halide ion is more stable on its own.
Both factors favour the same order:
\( \displaystyle \mathrm{R{-}I > R{-}Br > R{-}Cl > R{-}F} \)
\( \displaystyle \mathrm{CH_3CH_2CH_2I > CH_3CH_2CH_2Br > CH_3CH_2CH_2Cl > CH_3CH_2CH_2F} \)
Matching the given compounds:
(A) \( \ce{CH3CH2CH2I} \): 1-iodopropane
(C) \( \ce{CH3CH2CH2Br} \): 1-bromopropane
(B) \( \ce{CH3CH2CH2Cl} \): 1-chloropropane
(D) \( \ce{CH3CH2CH2F} \): 1-fluoropropane
Shortcut: The iodide (A) reacts fastest and the fluoride (D) slowest. Only Option (2) starts with A and ends with D.
NCERT (2026-27 edition):
Reactions of Haloalkanes – …
