The final product of 1-chlorobutane and 2-chlorobutane when treated in KOH (alcohol) gives:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Both 1-butene & 2-butene
Heated with alcoholic KOH, an alkyl halide undergoes dehydrohalogenation (β-elimination): the halogen leaves from the α-carbon and a hydrogen from a neighbouring β-carbon, and a double bond forms between them. Where there are two different kinds of β-hydrogen, the more substituted alkene is the major product (Zaitsev rule).
1-Chlorobutane has β-hydrogens on C2 only, so it gives only but-1-ene:
\( \displaystyle \ce{CH3CH2CH2CH2Cl + KOH} \xrightarrow[\Delta]{\text{alcohol}} \ce{CH3CH2CH=CH2 + KCl + H2O} \)
2-Chlorobutane has β-hydrogens on C1 and on C3. Removing one from C3 gives but-2-ene (major, the more substituted alkene); removing one from C1 gives but-1-ene (minor):
\( \displaystyle \ce{CH3CH2CHClCH3} \xrightarrow[\Delta]{\text{alc. KOH}} \ce{CH3CH=CHCH3}\,(\text{major}) + \ce{CH3CH2CH=CH2}\,(\text{minor}) \)
Taking both halides together, the products are but-1-ene and but-2-ene. The name “3-butene” in Option (2) is not a correct name: numbering from the end nearer the double bond makes it but-1-ene.
NCERT (2026-27 edition):
Elimination Reactions – “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. … 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.”