Which of the following reaction mechanism is involved in Williamson synthesis for the preparation of symmetrical and unsymmetrical ethers?
R–X + R′–O–Na → R–O–R′ + NaX
If R = CH3
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(S_N2\)
In Williamson synthesis an alkyl halide reacts with a sodium alkoxide. The alkoxide ion R′\( \ce{O^{-}} \) is a strong nucleophile and attacks the carbon carrying the halogen from the back side, pushing out \( \ce{X^{-}} \) in a single step; this is an SN2 reaction. With R = \( \ce{CH3} \) the alkyl halide is a methyl halide, which is unhindered and ideal for SN2 attack:
\( \ce{CH3-X + R'-O^-Na+ -> CH3-O-R' + NaX} \)
Why the other options are incorrect
Option (1) SN1 — Incorrect. A methyl halide cannot form a stable carbocation, so an SN1 path is not possible.
Option (3) SN1 and SN2 both — Incorrect. The reaction does not go partly through a carbocation; only the SN2 route operates.
Option (4) Addition reaction — Incorrect. One group (X) is replaced by another (OR′); nothing adds across a multiple bond, so it is a substitution.
NCERT (2026-27 edition):
Williamson Synthesis
