Ethers are class of organic compounds that contain ether group – an oxygen atom connected to two alkyl groups or aryl groups. They have the general formula R – O – R′, where R and R′ represents the alkyl or aryl groups. Ether, like water have a tetrahedral geometry i.e., oxygen is sp3 hybridised. The C – O – C bond angle in ethers is slightly greater than the tetrahedral angle due to repulsive interactions between the two bulky groups when they are attached to oxygen.
Which of the following cannot be made by using Williamson Synthesis?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Di-tert-Butyl ether
In the Williamson synthesis, an alkyl halide reacts with a sodium alkoxide (or phenoxide) by an SN2 attack of the alkoxide ion:
\( \displaystyle \ce{R-X + R'-ONa -> R-O-R' + NaX} \)
For di-tert-butyl ether both groups attached to oxygen are tertiary, so either way of making it needs a tertiary halide, which gives only 2-methylpropene:
\( \displaystyle \ce{(CH3)3C-Br + (CH3)3C-ONa -> (CH3)2C=CH2 + (CH3)3COH + NaBr} \)
Option (1) Methoxybenzene — Can be made by Williamson synthesis. From sodium phenoxide and methyl iodide:
\( \ce{C6H5ONa + CH3I -> C6H5OCH3 + NaI} \)
Option (2) Benzyl p-nitro phenyl ether — Can be made by Williamson synthesis. From sodium p-nitrophenoxide and benzyl chloride, a reactive primary halide.
Option (3) tert-Butyl methyl ether — Can be made by Williamson synthesis. With the tertiary group on the alkoxide and a methyl halide:
\( \ce{(CH3)3C-ONa + CH3Br -> (CH3)3C-O-CH3 + NaBr} \)
Option (4) Di-tert-Butyl ether — Cannot be made by Williamson synthesis. Both groups are tertiary, so a tertiary halide is needed either way, and it undergoes elimination to 2-methylpropene instead of substitution.
NCERT (2026-27 edition):
Preparation of Ethers – …
