Which of the following reactions will give t-butyl ethyl ether by Williamson synthesis?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → sodium tert-butoxide + ethyl chloride
In Williamson synthesis an alkyl halide reacts with a sodium alkoxide by SN2; the alkoxide ion attacks the carbon carrying the halogen. SN2 works best on a primary alkyl halide. With a tertiary halide the strongly basic alkoxide removes a β-hydrogen instead, and only an alkene forms.
So in t-butyl ethyl ether the tertiary group must come from the alkoxide and the ethyl group from the primary halide:
\( \ce{(CH3)3C-ONa + CH3CH2Cl -> (CH3)3C-O-CH2CH3 + NaCl} \)
Why the other options are incorrect
Option (2) sodium ethoxide + tert-butyl chloride — Incorrect. The halide is tertiary, so elimination wins and 2-methylpropene is formed, not the ether:
\( \ce{(CH3)3C-Cl + CH3CH2ONa -> (CH3)2C=CH2 + CH3CH2OH + NaCl} \)
Option (3) isopropyl bromide + sodium isopropoxide — Incorrect. Both groups are isopropyl, so even where substitution occurs the product is diisopropyl ether, \( \ce{(CH3)2CH-O-CH(CH3)2} \); with a secondary halide, propene is also formed.
Option (4) sodium propoxide + isopropyl bromide — Incorrect. The product would be isopropyl propyl ether, \( \ce{CH3CH2CH2-O-CH(CH3)2} \), along with propene; it has no t-butyl group.
NCERT (2026-27 edition):
Williamson Synthesis – …
