How many different possible alkenes are formed when 2-chlorobutane is treated with ethanolic solution of KOH?

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Solution

The correct option is C 3
Dehydrohalogenation involves removal of (elimination of) a hydrogen halide by abstracting $β - H$ from a alkyl halide in presence of base (alc. KOH) to form alkenes.
Saytzeff product will be major product.

here one of the product is but-2-ene and it can exist as cis and trans form thus two different alkenes will be formed.

Hence total possible alkene that can be formed is three.

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