2-Methyl propene can be prepared by the reaction of
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (CH3)3C–Br + CH3ONa
Sodium methoxide is a strong base as well as a nucleophile. With a tertiary alkyl halide, substitution is blocked by the three methyl groups, and the methoxide removes a β-hydrogen instead; elimination gives 2-methylpropene and no ether is formed:
\( \ce{(CH3)3C-Br + CH3ONa -> CH2=C(CH3)2 + CH3OH + NaBr} \)
Why the other options are incorrect
Option (2) (CH3)3C–ONa + CH3Br — Incorrect. The methyl halide undergoes SN2 attack by the tert-butoxide, giving the ether 2-methoxy-2-methylpropane, not an alkene.
Option (3) (CH3)3C–ONa + CH3–CH2–Br — Incorrect. The primary halide gives the ether 2-ethoxy-2-methylpropane by SN2.
Option (4) (CH3)3C–ONa + CH3CHBrCH3 — Incorrect. Elimination from the secondary halide 2-bromopropane gives propene, not 2-methylpropene.
NCERT (2026-27 edition):
Williamson Synthesis – …
