Select the appropriate reagent for the synthesis of tert-Butyl ethyl ether
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Ethyl bromide and sodium tert-butoxide.
In Williamson synthesis an alkoxide ion displaces a halide ion from an alkyl halide by an \(S_N2\) reaction. Alkoxide ions are strong bases as well as nucleophiles. With a primary halide they substitute, but with a tertiary halide they remove a β-hydrogen instead, and an alkene is formed by elimination. So for an unsymmetrical ether, the bulky (tertiary) group is taken as the alkoxide and the small group as the primary halide.
tert-Butyl ethyl ether is \( \ce{(CH3)3C-O-C2H5} \). The tert-butyl part must come from sodium tert-butoxide and the ethyl part from ethyl bromide:
\( \displaystyle \ce{(CH3)3C-O^-Na^+ + CH3CH2-Br -> (CH3)3C-O-CH2CH3 + NaBr} \)
So the reagents are ethyl bromide and sodium tert-butoxide.
Why the other options are incorrect
Option (1) Tert-butyl bromide and sodium ethoxide. — Incorrect. The ethoxide ion acts as a base on the tertiary halide, and the main product is 2-methylpropene, not the ether:
\( \displaystyle \ce{(CH3)3C-Br + C2H5O^-Na^+ -> (CH3)2C=CH2 + C2H5OH + NaBr} \)
Option (2) Butyl bromide and sodium tert-butoxide. — Incorrect. These would give butyl tert-butyl ether, \( \ce{(CH3)3C-O-C4H9} \); the target has an ethyl group, not a butyl group.
Option (3) Propyl bromide and sodium tert-propoxide. — Incorrect. There is no “tert-propoxide”, since a three-carbon group has no tertiary carbon, and a propyl group is not part of the target ether.
NCERT (2026-27 edition):
Williamson Synthesis – “… Better results are obtained if the alkyl halide is primary. In case of secondary and tertiary alkyl halides, elimination competes over substitution. If a tertiary alkyl halide is used, an alkene is the only reaction product and no ether is formed. For example, the reaction of CH3ONa with (CH3)3C–Br gives exclusively 2-methylpropene.”