Williamson's synthesis for preparing dimethyl ether involves
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(S_N2\) mechanism
In Williamson synthesis an alkyl halide reacts with a sodium alkoxide. The alkoxide ion is a strong nucleophile; it attacks the carbon carrying the halogen from the side opposite to the halogen and pushes out the halide ion in a single step. This is an \(S_N2\) reaction.
For dimethyl ether, sodium methoxide reacts with a methyl halide:
\( \displaystyle \ce{CH3Br + CH3O^- Na+ -> CH3-O-CH3 + NaBr} \)
A methyl halide is the best possible substrate for \(S_N2\) attack, because the carbon carries no bulky groups and has no β-hydrogen that could be lost in an elimination.
Why the other options are incorrect
Option (1) \(S_N1\) mechanism — Incorrect. No carbocation is formed; the methyl carbocation would be far too unstable, and the alkoxide displaces the halide in one step.
Option (2) Elimination reaction — Incorrect. Elimination competes only with secondary and tertiary halides, where the alkoxide acts as a base and gives an alkene; a methyl halide has no β-hydrogen to lose.
Option (4) Nucleophilic addition reaction — Incorrect. Addition needs a multiple bond such as C=O; the alkyl halide has none, and the alkoxide replaces the halogen instead of adding.
NCERT (2026-27 edition):
Williamson Synthesis
