JEE Main 2026 — Jan 22, Shift 1 — Chemistry solution

JEE Main2026Jan 22, Shift 1Chemistry

The correct order of reactivity of $C H_{3} B r$ in methanol with the following nucleophiles is

$F^{-}, I^{-}, C_{2} H_{5} O^{-}$ and $C_{6} H_{5} O^{-}$

Solution

The solvent (methanol) is protic, so nucleophilicity rankings follow polarisability and basicity in a protic medium:

$I^{-}$ is the largest, most polarisable, and the least solvated — the strongest nucleophile here despite being a weak base.

$C_{2} H_{5} O^{-}$ is a strong base with localised negative charge — a good nucleophile.

$C_{6} H_{5} O^{-}$ has its charge delocalised into the aromatic ring, so it’s weaker than $C_{2} H_{5} O^{-}$ .

$F^{-}$ is small and tightly H-bonded by methanol, so it’s the poorest of the four.

Overall: $I^{-} > C_{2} H_{5} O^{-} > C_{6} H_{5} O^{-} > F^{-}$ .

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