A mixture of 1-chloropropane and 2-chloropropane when treated with alcoholic KOH, it gives

When alkyl halides are treated with alcoholic KOH, an elimination reaction (dehydrohalogenation) occurs. This reaction removes a hydrogen atom and a halogen atom, forming a double bond and resulting in an alkene. This is a β-elimination reaction where the halogen atom is removed from the α-carbon, and the hydrogen atom is removed from the β-carbon.

In this case:

1. 1-Chloropropane (CH₃CH₂CH₂Cl):

When treated with alcoholic KOH, it undergoes elimination to give propene:

CH₃CH₂CH₂Cl → CH₃CH=CH₂ + HCl

2. 2-Chloropropane (CH₃CHClCH₃):

When treated with alcoholic KOH, it also undergoes elimination to give propene:

CH₃CHClCH₃ → CH₃CH=CH₂ + HCl

Final Product:

Both 1-chloropropane and 2-chloropropane produce the same product, which is propene (CH₃CH=CH₂), through dehydrohalogenation with alcoholic KOH.

The reaction gives propene (CH₃CH=CH₂) as the product.