Read the passage carefully and answer the questions.
Replacement of a hydrogen atom in a hydrocarbon by an alkoxy or carboxyl group yields a class of compounds known as ethers. Ethers are classified as symmetrical or unsymmetrical on the basis of groups attached to the oxygen atoms. Diethyl ether, a symmetrical ether, has been widely used as an inhalation anesthetic. Ethers can be prepared by acid catalyzed intermolecular dehydration of alcohols and Williamson's synthesis. Acid catalyzed dehydration of alcohols is not generally preferred as it gives a mixture of elimination and substitution products. In Williamson's synthesis, an alkyl halide is allowed to react with sodium alkoxide. Ethers containing substituted Alkyl groups may also be prepared by this method. The C-O bond in ether is weakly polar and is cleaved under drastic conditions with excess of hydrogen halides. In electrophilic substitution, the alkoxy group deactivates the aromatic ring and directs the incoming group to ortho and para positions.
In Williamson synthesis, the alkoxide ion attacks the alkyl halide via which pathway?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(S_N2\)
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:
\( \displaystyle \ce{R-X + R'-O^- Na+ -> R-O-R' + Na+ X^-} \)
For example, sodium ethoxide and methyl bromide give ethyl methyl ether:
\( \displaystyle \ce{CH3Br + C2H5O^- Na+ -> CH3-O-C2H5 + NaBr} \)
Because the attack is \(S_N2\), the method works best with a primary alkyl halide. With a tertiary halide the strongly basic alkoxide removes a proton instead, and an alkene is formed by elimination.
Why the other options are incorrect
Option (2) \(S_N1\) — Incorrect. An \(S_N1\) reaction goes through a carbocation, which forms only from a tertiary halide, usually in a polar protic solvent. With a tertiary halide the strongly basic alkoxide removes a proton instead, so an alkene is formed by elimination, not an ether by \(S_N1\) substitution.
Option (3) Depends on nature of alkoxide ion — Incorrect. The alkoxide decides only which ether forms; the mechanism stays \(S_N2\).
Option (4) Depends on the nature of Alkyl halide — Incorrect. The kind of halide decides whether substitution or elimination wins, but the ether itself is always formed by the \(S_N2\) route.
Note: It seems that there is a mistake in the original NTA paper, which prints alkoxy or carboxyl group when it should read alkoxy or aryloxy group, and deactivates when it should read activates.
NCERT (2026-27 edition):
Williamson Synthesis
