The final product of 1-chlorobutane and 2-chlorobutane when treated in KOH (alcohol) gives:

Both 1-butene & 2-butene

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