Which of the following alkyl halides is preferred for the preparation of an ether by Williamson synthesis?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → n-Propyl bromide
In Williamson synthesis an alkyl halide reacts with a sodium alkoxide by nucleophilic substitution:
\( \displaystyle \ce{R-X + R'-ONa -> R-O-R' + NaX} \)
The alkoxide is also a strong base, so with a secondary or tertiary halide it removes a β-hydrogen and gives an alkene instead. A primary alkyl halide gives the best yield of ether. n-Propyl bromide is primary:
\( \displaystyle \ce{CH3CH2CH2Br + C2H5ONa -> CH3CH2CH2-O-C2H5 + NaBr} \)
Why the other options are incorrect
Option (2) Isopropyl bromide — Incorrect. It is a secondary halide, so elimination competes with substitution and some propene forms.
Option (3) 2-Bromo-2-methylpropane — Incorrect. It is a tertiary halide; the alkoxide gives only 2-methylpropene and no ether: \( \ce{(CH3)3C-Br + CH3ONa -> (CH3)2C=CH2 + CH3OH + NaBr} \).
Option (4) tert-Pentyl bromide — Incorrect. \( \ce{CH3CH2C(CH3)2Br} \) is also tertiary, so it gives an alkene, not an ether.
NCERT (2026-27 edition):
7.6.1 Preparation of Ethers – “Better results are obtained if the alkyl halide is primary. In case of secondary and tertiary alkyl halides, elimination competes over substitution.”