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The correct order of reactivity of CH3Br in methanol with nucleophiles
Q. The correct order of reactivity of CH3Br in methanol with the following nucleophiles is

F−, I−, C2H5O− and C6H5O−
A. I− > C2H5O− > C6H5O− > F−
B. I− > F− > C6H5O− > C2H5O−
C. I− > C6H5O− > F− > C2H5O−
D. I− > C2H5O− > F− > C6H5O−
Correct Answer: I− > C2H5O− > C6H5O− > F−

Explanation

CH3Br is a methyl halide, so it undergoes substitution reaction via the SN2 mechanism.

For an SN2 reaction, the rate is directly proportional to the nucleophilicity of the attacking nucleophile.

The solvent used is methanol, which is a polar protic solvent. Polar protic solvents strongly solvate smaller anions through hydrogen bonding.

Hence, smaller nucleophiles become less reactive. The nucleophilicity order of halide ions in methanol is:

I− > Br− > Cl− > F−

Now comparing alkoxide ions:

C2H5O− is more nucleophilic than C6H5O−.

In phenoxide ion, the negative charge is delocalized due to resonance with the benzene ring, which reduces its nucleophilicity.

Ethoxide ion has localized negative charge on oxygen, hence it reacts faster.

Therefore, the correct order of reactivity is:

I− > C2H5O− > C6H5O− > F−

This concept is extremely important for JEE Main 2026, JEE Advanced and IIT JEE.

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