Statement I: 2-Bromobutane, on reaction with sodium ethoxide in ethanol, gives but-1-ene as the major product.
Statement II: But-1-ene is less stable than but-2-ene.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Statement I is incorrect but Statement II is correct.
Sodium ethoxide is a strong base. With 2-bromobutane it removes a β-hydrogen together with the bromine (dehydrohalogenation). 2-Bromobutane has β-hydrogens on C1 and on C3, so two alkenes can form; the one with more alkyl groups on the double bond is the major product:
\( \displaystyle \ce{CH3CH2CH(Br)CH3 + C2H5ONa} \xrightarrow{\ce{C2H5OH}} \ce{CH3CH=CHCH3 + C2H5OH + NaBr} \)
But-1-ene, \( \ce{CH2=CHCH2CH3} \), forms only as the minor product.
Statement I: 2-Bromobutane, on reaction with sodium ethoxide in ethanol, gives but-1-ene as the major product. — False. By Zaitsev rule the major product is but-2-ene, which has two alkyl groups on the doubly bonded carbons; but-1-ene has only one.
Statement II: But-1-ene is less stable than but-2-ene. — True. An alkene with more alkyl groups on the double bond is more stable, so but-2-ene is more stable than but-1-ene; this is why it is the preferred product.
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.”