JEE Chemistry 2026 Question with Solution

Given: $\mathrm{CH_3Br}$ reacts in methanol with nucleophiles $\mathrm{F^-}$, $\mathrm{I^-}$, $\mathrm{C_2H_5O^-}$ and $\mathrm{C_6H_5O^-}$.

Find: The correct order of reactivity.

Step 1: Identify the mechanism.

is a primary alkyl halide and methanol is a polar protic solvent, so the reaction proceeds through the $\mathrm{S_N2}$ mechanism.

Step 2: Effect of solvent on nucleophilicity.

In polar protic solvents, nucleophilicity of anions depends strongly on solvation.

Therefore, among the given halides:

Step 3: Compare alkoxide and phenoxide.

• $\mathrm{C_2H_5O^-}$ is a strong base but is heavily solvated in methanol.
• $\mathrm{C_6H_5O^-}$ is resonance stabilized, which reduces its nucleophilicity.

Hence:

Step 4: Final order.

Combining these comparisons:

Therefore, the correct option is C.

Given: The solvent is methanol, which is polar protic.

Find: Why the nucleophiles follow the observed order.

In a polar protic solvent, smaller anions are more strongly hydrogen bonded and solvated. Because of this, their electron pair is less available for backside attack in an $\mathrm{S_N2}$ reaction.

So although $\mathrm{F^-}$ is a strong base, it is strongly solvated. $\mathrm{I^-}$ is less strongly solvated and therefore behaves as a better nucleophile in methanol.

For oxygen nucleophiles, $\mathrm{C_6H_5O^-}$ has resonance stabilization, so its lone pair is less available than in a simple alkoxide. The solution therefore concludes:

Thus, the correct option is C.