Which of the following is the correct order of reactivity of the given nucleophiles when treated with \( \text{CH}_3\text{Br} \) in methanol?
\[
\text{F}^-,\ \text{I}^-,\ \text{C}_2\text{H}_5\text{O}^-,\ \text{C}_6\text{H}_5\text{O}^-
\]
Show Hint
In \textbf{polar protic solvents}:
Nucleophilicity increases down the group for halides
Strong solvation drastically reduces nucleophilicity of small ions like \( \text{F}^- \)
Resonance-stabilized anions are weaker nucleophiles
Concept:
The reaction of methyl bromide (\(\text{CH}_3\text{Br}\)) with nucleophiles proceeds via an SN2 mechanism.
Since the solvent is methanol (polar protic), nucleophilicity is strongly influenced by:
Solvation effects Size of the nucleophile Resonance stabilization
In polar protic solvents, nucleophilicity generally follows:
\[
\text{Larger, more polarizable ions}>\text{smaller ions}
\]
Step 1: Compare halide ions.
\(I^-\): Large size, weak solvation, highly polarizable
\(F^-\): Very small size, strongly solvated by methanol
Thus,
\[
\text{I}^- \gg \text{F}^-
\]
Step 2: Compare alkoxide and phenoxide ions.
\(C_2H_5O^-\) (ethoxide):
Negative charge is localized on oxygen → stronger nucleophile.
\(C_6H_5O^-\) (phenoxide):
Negative charge is delocalized into the benzene ring due to resonance → reduced nucleophilicity.
Hence,
\[
\text{C}_2\text{H}_5\text{O}^→ \text{C}_6\text{H}_5\text{O}^-
\]
Step 3: Overall comparison in methanol (polar protic solvent).
\(\text{I}^-\): Least solvated, most polarizable → strongest nucleophile
\(\text{C}_2\text{H}_5\text{O}^-\): Strong base and good nucleophile
\(\text{C}_6\text{H}_5\text{O}^-\): Resonance-stabilized → weaker nucleophile
\(\text{F}^-\): Strongly solvated → weakest nucleophile
Step 4: Final order of nucleophilic reactivity.
\[
\boxed{\text{I}^→ \text{C}_2\text{H}_5\text{O}^→ \text{C}_6\text{H}_5\text{O}^→ \text{F}^-}
\]
This matches Option (1).
