According to Alexander Zaitsev the most preferred product for dehydrohalogenation of 2-Bromopentane will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Pent-2-ene
Dehydrohalogenation of 2-bromopentane can remove a β-hydrogen from C-1 or from C-3:
from C-1: \( \ce{CH2=CH-CH2CH2CH3} \) (pent-1-ene, one alkyl group on the double bond)
from C-3: \( \ce{CH3-CH=CH-CH2CH3} \) (pent-2-ene, two alkyl groups on the double bond)
By Zaitsev’s rule, the preferred product is the alkene with the greater number of alkyl groups on the doubly bonded carbon atoms, so pent-2-ene is the major product.
Why the other options are incorrect
Option (1) Pent-1-ene — Incorrect. It is formed, but only as the minor product because its double bond carries fewer alkyl groups.
Option (3) Pent-3-ene — Incorrect. This is not a correct name; numbering from the nearer end turns it into pent-2-ene.
Option (4) Bromopentane — Incorrect. It is the starting compound, not an elimination product.
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.”