alleryThe reactivity of haloalkanes toward nucleophilic substitution generally follows the order
A$\mathrm{R{-}F} > \mathrm{R{-}Cl} > \mathrm{R{-}Br} > \mathrm{R{-}I}$
Ball four are equally reactive
C$\mathrm{R{-}Cl} > \mathrm{R{-}I} > \mathrm{R{-}Br} > \mathrm{R{-}F}$
D$\mathrm{R{-}I} > \mathrm{R{-}Br} > \mathrm{R{-}Cl} > \mathrm{R{-}F}$
Answer & Solution
Correct answer: D. $\mathrm{R{-}I} > \mathrm{R{-}Br} > \mathrm{R{-}Cl} > \mathrm{R{-}F}$
1. Your syllabus: nucleophilic substitution requires the C-X bond to BREAK as a new bond forms.
2. The C-X BOND STRENGTH decreases down the halogen group: C-F > C-Cl > C-Br > C-I.
3. WEAKER bonds break MORE EASILY, so reactivity is INVERSELY related to bond strength.
4. Result: R-I > R-Br > R-Cl > R-F (iodides most reactive).
5. Additional factor: C-I has the LARGEST polarisability — its electron cloud distorts most easily under nucleophilic attack, further favouring its reactivity.
6. Option A reverses the trend (it would be the bond-strength order, not reactivity). Option C breaks the trend midway. Option D ignores bond strength.
_Source: syllabus Chemistry Part 2, Ch 6, §6.5 (Reactivity vs C-X bond strength), p. 14._
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