When methyl bromide is treated with sodium tert-butoxide, the compound formed is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → tert-Butyl methyl ether
This is the Williamson synthesis: an alkyl halide reacts with a sodium alkoxide, and the alkoxide ion displaces the halide by an SN2 reaction. Best results are obtained with a primary (here methyl) halide, because the bulky alkoxide only has to attack an unhindered carbon:
\( \ce{CH3-Br} \) + \( \ce{(CH3)3C-O^{-}Na^{+}} \) → \( \ce{(CH3)3C-O-CH3} \) + NaBr
The product is tert-butyl methyl ether (2-methoxy-2-methylpropane). (The reverse choice, \( \ce{(CH3)3C-Br} \) with \( \ce{CH3ONa} \), would give only 2-methylpropene by elimination.)
Why the other options are incorrect
Option (1) Diethyl ether — Incorrect. \( \ce{C2H5OC2H5} \) needs two ethyl groups; neither reactant contains an ethyl group.
Option (3) Diethyl bromide — Incorrect. There is no such compound; the bromine leaves as NaBr.
Option (4) Methyl propyl ether — Incorrect. It would need a propoxide (or propyl halide); the tert-butyl group is not converted into a propyl group.
NCERT (2026-27 edition):
Williamson Synthesis – “It is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.”
Williamson Synthesis – “… Better results are obtained if the alkyl halide is primary.”