Identify the suitable pair of reactants for the preparation of ethyl tert-butyl ether
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Ethyl chloride + sodium-2-methylpropan-2-olate
Ethyl tert-butyl ether, \( \ce{C2H5-O-C(CH3)3} \), is made by Williamson’s synthesis, an SN2 attack of an alkoxide on an alkyl halide. For a good yield the halide must be primary and the bulky group must come from the alkoxide:
\( \ce{C2H5Cl + (CH3)3C-ONa -> C2H5-O-C(CH3)3 + NaCl} \)
With the halide on the tertiary side, the alkoxide acts as a base and elimination takes over:
\( \ce{(CH3)3C-Cl + C2H5ONa -> (CH3)2C=CH2 + C2H5OH + NaCl} \)
Why the other options are incorrect
Option (1) Sodium ethanolate + 2-methylpropan-2-chloride — Incorrect. With a tertiary halide the alkoxide acts as a base, and elimination gives 2-methylpropene instead of the ether.
Option (2) Ethyl chloride + 2-methylpropan-2-ol — Incorrect. The neutral alcohol is a poor nucleophile; the alkoxide ion is needed.
Option (4) Ethanol + 2-methylpropan-2-chloride — Incorrect. Again a tertiary halide with a weak nucleophile; it gives mainly the alkene.
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.”