Identify the major product (P) formed on reaction of

The correct answer is Option (3) → \((CH_3)_2C=CH_2\)

A haloalkane with a β-hydrogen atom reacts with KOH in two possible ways. In aqueous solution, \( \ce{OH^{-}} \) acts as a nucleophile and gives an alcohol (substitution). When the haloalkane is heated with an alcoholic (ethanolic) solution of KOH, \( \ce{OH^{-}} \) acts as a base: it removes a hydrogen from the β-carbon and the halogen leaves from the α-carbon. This is β-elimination (dehydrohalogenation), and an alkene is formed. Tertiary halides, heat and the alcoholic solvent all favour elimination.

In 2-bromo-2-methylpropane (tert-butyl bromide) the carbon carrying Br has three \( \ce{CH3} \) groups, each with β-hydrogens. Loss of HBr gives 2-methylpropene:

    \( \displaystyle \ce{(CH3)3CBr + KOH} \xrightarrow[\text{heat}]{\text{ethanol}} \ce{(CH3)2C=CH2 + KBr + H2O} \)

So the major product P is 2-methylpropene, (CH₃)₂C=CH₂.

Why the other options are incorrect

• Option (1) \((CH_3)_3COH\) — Incorrect. Tert-butyl alcohol is the substitution product, formed with aqueous KOH:

      \( \displaystyle \ce{(CH3)3CBr + KOH(aq) -> (CH3)3COH + KBr} \)

• Option (2) \((CH_3)_2C=O\) — Incorrect. Propanone has one carbon atom fewer than the substrate and needs an oxidation; KOH cannot do this.

• Option (4) \(CH_3CH_2OCH_2CH_3\) — Incorrect. Ethoxyethane contains no tert-butyl group at all; an ether from this substrate would be \( \ce{(CH3)3COC2H5} \).

NCERT (2026-27 edition):
Elimination Reactions