Each haloalkane in List-I is heated with alcoholic potassium hydroxide. Match it with the major alkene formed, given in List-II.
| List-I (Haloalkane) | List-II (Major alkene) |
|---|---|
| (A) 2-Bromo-2-methylbutane | (I) Pent-1-ene |
| (B) 1-Bromo-3-methylbutane | (II) 3-Methylbut-1-ene |
| (C) 2-Bromopentane | (III) Pent-2-ene |
| (D) 1-Bromopentane | (IV) 2-Methylbut-2-ene |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-II, C-III, D-I
On heating with alcoholic KOH, a haloalkane loses a hydrogen atom from a β-carbon and the halogen from the α-carbon, and an alkene is formed (β-elimination). Where more than one alkene is possible, Zaitsev rule applies: the major product is the alkene with more alkyl groups on the doubly bonded carbon atoms.
(A) 2-Bromo-2-methylbutane — This matches (IV) 2-Methylbut-2-ene. Losing H from the CH2 group gives the alkene with three alkyl groups on the double bond, which is preferred over 2-methylbut-1-ene:
\( \displaystyle \ce{(CH3)2CBr-CH2CH3 + KOH} \xrightarrow[\Delta]{\text{alcohol}} \ce{(CH3)2C=CH-CH3 + KBr + H2O} \)
(B) 1-Bromo-3-methylbutane — This matches (II) 3-Methylbut-1-ene. The bromine is on C-1, so the only β-carbon is C-2, and the double bond can form only between C-1 and C-2:
\( \displaystyle \ce{(CH3)2CH-CH2-CH2Br + KOH} \xrightarrow[\Delta]{\text{alcohol}} \ce{(CH3)2CH-CH=CH2 + KBr + H2O} \)
(C) 2-Bromopentane — This matches (III) Pent-2-ene. It has two β-carbons; Zaitsev rule favours the double bond between C-2 and C-3, which carries two alkyl groups:
\( \displaystyle \ce{CH3-CH(Br)-CH2CH2CH3 + KOH} \xrightarrow[\Delta]{\text{alcohol}} \ce{CH3-CH=CH-CH2CH3 + KBr + H2O} \)
(D) 1-Bromopentane — This matches (I) Pent-1-ene. With the bromine at the end of the chain there is only one β-carbon, so the double bond forms at C-1:
\( \displaystyle \ce{CH3CH2CH2-CH2-CH2Br + KOH} \xrightarrow[\Delta]{\text{alcohol}} \ce{CH3CH2CH2-CH=CH2 + KBr + H2O} \)
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “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.”
“…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.”