Identify the product 'X' in the following reaction:

The correct answer is Option (4) → \((CH_3)_2C=CH_2\) (2-methylpropene)

Sodium methoxide is not only a nucleophile but also a strong base. With a tertiary alkyl halide such as \( \ce{(CH3)3C-Br} \), substitution at the crowded carbon is hindered, so the methoxide removes a β-hydrogen and elimination takes place:

    \( \ce{(CH3)3C-Br} \) + \( \ce{CH3ONa} \) → \( \ce{(CH3)2C=CH2} \) + \( \ce{CH3OH} \) + NaBr

The alkene X is 2-methylpropene, and methanol is the by-product shown in the equation.

Why the other options are incorrect

• Option (1) \((CH_3)_3C-OCH_3\) — Incorrect. This ether would be the Williamson product, but with a tertiary halide no ether is formed; it is made from \( \ce{(CH3)3CONa} \) and \( \ce{CH3Br} \) instead.

• Option (2) \(CH_2=C(OCH_3)CH_3\) — Incorrect. An enol ether cannot form; the reaction neither substitutes nor keeps the \( \ce{OCH3} \) group on the alkene.

• Option (3) \(CH_2=C(OH)CH_3\) — Incorrect. This enol has only three carbons and an OH group; no OH source or loss of carbon is involved.

NCERT (2026-27 edition):
Preparation of Ethers – “… 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. It is because alkoxides are not only nucleophiles but strong bases as well.”

Note: The same/similar question has been asked in different years by NTA as per the details given below: