Which of the following alkyl halides does NOT undergo β-elimination (dehydrohalogenation) when heated with alcoholic potassium hydroxide?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1-Bromo-2,2-dimethylpropane
In β-elimination, alcoholic KOH removes a hydrogen atom from the β-carbon and the halogen from the α-carbon, giving an alkene. A halide can do this only if it has a hydrogen atom on a β-carbon.
Option (1) 2-Bromopropane — Undergoes β-elimination. Both β-carbons are CH3 groups with hydrogens, so propene is formed:
\( \displaystyle \ce{CH3CHBrCH3 + KOH} \xrightarrow{\text{alcohol}, \Delta} \ce{CH3CH=CH2 + KBr + H2O} \)
Option (2) 1-Bromo-2,2-dimethylpropane — Does not undergo β-elimination. In \( \ce{(CH3)3C-CH2Br} \) the α-carbon is the CH2Br carbon. Its only neighbouring carbon, the β-carbon, is bonded to three methyl groups and carries no hydrogen, so no alkene can form by β-elimination.
Option (3) 2-Bromo-2-methylpropane — Undergoes β-elimination. Its three β-carbons are CH3 groups, so 2-methylpropene is formed:
\( \displaystyle \ce{(CH3)3CBr + KOH} \xrightarrow{\text{alcohol}, \Delta} \ce{(CH3)2C=CH2 + KBr + H2O} \)
Option (4) Bromoethane — Undergoes β-elimination. The CH3 group is the β-carbon, so ethene is formed:
\( \displaystyle \ce{CH3CH2Br + KOH} \xrightarrow{\text{alcohol}, \Delta} \ce{CH2=CH2 + KBr + H2O} \)
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “2. Elimination reactions: When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β-carbon and a halogen atom from the α-carbon atom. As a result, an alkene is formed as a product. Since β-hydrogen atom is involved in elimination, it is often called β-elimination.”