The correct order of reactivity of $$CH_{3} Br$$ in methanol with the following
nucleophiles is

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

Find the correct order of reactivity of $$CH_3Br$$ in methanol with nucleophiles $$F^-$$, $$I^-$$, $$C_2H_5O^-$$, $$C_6H_5O^-$$.

$$CH_3Br$$ (primary substrate) undergoes $$S_N2$$ reaction with nucleophiles.

Methanol is a protic solvent. In protic solvents, nucleophilicity is strongly affected by solvation. Smaller, more charge-dense ions are more heavily solvated and thus less nucleophilic.

- $$I^-$$: Large, polarizable, weakly solvated in protic solvents → strongest nucleophile

- $$C_2H_5O^-$$: Good nucleophile (strong base, alkoxide), but solvated by H-bonding in methanol

- $$C_6H_5O^-$$: Weaker nucleophile than ethoxide because the negative charge is delocalized into the benzene ring

- $$F^-$$: Small, highly solvated in protic solvents → weakest nucleophile

Order: $$I^- > C_2H_5O^- > C_6H_5O^- > F^-$$

The correct answer is Option (1).