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
Alcoholic KOH promotes $β$-elimination (E2 reaction). Product depends on:
- Position of halogen
- Available ẞ-hydrogens
- Saytzeff rule (more substituted alkene is major)
From 1-Chlorobutane
Structure: $CH_3-CH_2-CH_2-CH_2Cl$
Only $β$-hydrogens available are on adjacent carbon → elimination gives:
→1-butene only
From 2-Chlorobutane
Structure: $CH_3-CHCl-CH_2-CH_3$
$β$-hydrogens are present on both sides → elimination gives:
1-butene
2-butene (major, due to Saytzeff rule)
Final Combined Products
Since both starting compounds are treated separately and give elimination:
Overall products formed:
→ 1-butene
→ 2-butene