The correct order of reactivity of $\mathrm{CH}_{3} \mathrm{Br}$ in methanol with the following nucleophiles is $\mathrm{F}^{-}, \mathrm{I}^{-}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}$

(B) $\mathrm{I}^{-}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}>\mathrm{F}^{-}$

(A) $\mathrm{I}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}>\mathrm{F}^{-}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$

(C) $\mathrm{I}^{-}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}>\mathrm{F}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}$

(D) $\mathrm{I}^{-}>\mathrm{F}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Haloalkanes and Haloarenes

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The correct order of reactivity of

CH_{3} Br

in methanol with the following nucleophiles is

F^{-}, I^{-}, C_{2} H_{5} O^{-}

and

C_{6} H_{5} O^{-}

(A) $I^{-} > C_{6} H_{5} O^{-} > F^{-} > C_{2} H_{5} O^{-}$
(B) $I^{-} > C_{2} H_{5} O^{-} > C_{6} H_{5} O^{-} > F^{-}$ ✓ Correct
(C) $I^{-} > C_{2} H_{5} O^{-} > F^{-} > C_{6} H_{5} O^{-}$
(D) $I^{-} > F^{-} > C_{6} H_{5} O^{-} > C_{2} H_{5} O^{-}$

Answer: (B)

Solution

The reaction is a bimolecular substitution on methyl bromide, so the more reactive nucleophile wins, and methanol is a protic solvent in which anions are held back by hydrogen bonding. Iodide, being large, highly polarisable and only weakly solvated, is the strongest nucleophile here. Among the oxygen bases, ethoxide keeps its negative charge wholly on oxygen, whereas in phenoxide the charge is delocalised into the ring, so ethoxide is the better nucleophile of the two. Fluoride is small and very tightly solvated in a protic medium and so is the poorest. The order of reactivity is

I^{-} > C_{2} H_{5} O^{-} > C_{6} H_{5} O^{-} > F^{-}

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Haloalkanes and Haloarenes

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