What is the major product formed when 2-bromopentane is heated with alcoholic KOH?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Pent-2-ene
Heating a haloalkane with alcoholic KOH brings about dehydrohalogenation (β-elimination): H is lost from a β-carbon and Br from the α-carbon, giving an alkene.
In 2-bromopentane, C2 carries Br, and both C1 and C3 are β-carbons carrying hydrogen, so two alkenes are possible:
\( \displaystyle \ce{CH3CH(Br)CH2CH2CH3 + KOH} \xrightarrow[\Delta]{\text{alc.}} \ce{CH3CH=CHCH2CH3 + KBr + H2O} \)
By Saytzeff’s rule, the major product is the alkene with more alkyl groups on the doubly bonded carbons. Pent-2-ene (H from C3) has two alkyl groups on its C=C, while pent-1-ene (H from C1) has only one, so pent-2-ene is the major product.
Why the other options are incorrect
Option (1) Pentan-2-ol — Incorrect. An alcohol is the substitution product, formed with aqueous KOH, not with alcoholic KOH on heating.
Option (3) Pent-1-ene — Incorrect. It forms only as the minor product, because its C=C carries fewer alkyl groups.
Option (4) 2-Ethoxypentane — Incorrect. An ether is a substitution product; heating with alcoholic KOH favours elimination to the alkene.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “If there is possibility of formation of more than one alkene due to the availability of more than one β-hydrogen atoms, usually one alkene is formed as the major product. … ‘in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.’ Thus, 2-bromopentane gives pent-2-ene as the major product.”