Question

The correct order of reactivity of CH$_3$Br in methanol with the following nucleophiles is
$ \mathrm{F^- ,\ I^- ,\ C_2H_5O^- \ and\ C_6H_5O^- }$

• A. $I^->C_2H_5O^->F^->C_6H_5O^-$
• B. $I^->C_6H_5O^->F^->C_2H_5O^-$
• C. $I^->F^->C_6H_5O^->C_2H_5O^-$
• D. $I^->C_2H_5O^->C_6H_5O^->F^-$

Hint:

In polar protic solvents, nucleophilicity of anions depends more on solvation than basicity.

Answer 1

The Correct Option is C

CH$_3$Br is a primary alkyl halide, and methanol is a polar protic solvent. Hence, the reaction proceeds via the SN2 mechanism.
Step 1: Effect of solvent on nucleophilicity.
In polar protic solvents, nucleophilicity of anions decreases down the group due to solvation: \[ I^->Br^->Cl^->F^- \] Thus, among halide ions: \[ I^->F^- \] Step 2: Compare alkoxide and phenoxide ions.
- $C_2H_5O^-$ is a strong base but heavily solvated in methanol.
- $C_6H_5O^-$ is resonance stabilized, reducing its nucleophilicity.
Hence: \[ F^->C_6H_5O^->C_2H_5O^- \] Step 3: Final order.
Combining all factors: \[ I^->F^->C_6H_5O^->C_2H_5O^- \] Final Answer: $\boxed{\text{Option (C)}}$