2-Bromopentane is heated with alcoholic KOH. Which is the correct statement for this reaction?
A. Only pent-2-ene is formed
B. Only pent-1-ene is formed
C. Both pent-2-ene and pent-1-ene are formed
D. Pent-2-ene will be the major product as it is more stable
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → C, D only
Alcoholic KOH removes HBr from 2-bromopentane (dehydrohalogenation, β-elimination). 2-Bromopentane has β-hydrogens on two sides of the C–Br carbon, C-1 and C-3, so two alkenes can form:
\( \ce{CH3-CHBr-CH2CH2CH3} \) → \( \ce{CH3-CH=CH-CH2CH3} \) (pent-2-ene, about 81%) + \( \ce{CH2=CH-CH2CH2CH3} \) (pent-1-ene, about 19%)
By Zaitsev’s (Saytzeff’s) rule, the preferred product is the alkene with more alkyl groups on the doubly bonded carbons, i.e. the more substituted, more stable alkene.
A. Only pent-2-ene is formed — Incorrect. Pent-1-ene is also formed, as the minor product.
B. Only pent-1-ene is formed — Incorrect. Pent-1-ene is only the minor product.
C. Both pent-2-ene and pent-1-ene are formed — Correct. Elimination can take a β-hydrogen from either side.
D. Pent-2-ene will be the major product as it is more stable — Correct. It is the disubstituted alkene, favoured by Zaitsev’s rule.
NCERT (2026-27 edition):
Elimination Reactions – “… 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. These form part of a pattern first observed by Russian chemist, Alexander Zaitsev (also pronounced as Saytzeff) who in 1875 formulated a rule which can be summarised as ‘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.”