\[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{ - C}}{{\rm{H}}_{\rm{2}}}{\rm{ - C}}{{\rm{H}}_{\rm{2}}}{\rm{Br + KOH}}\left( {{\rm{alc}}{\rm{.}}} \right) \to {\rm{Product}}\]
The product in the above reaction is:
A. CH₃−CH=CH₂
​B. CH₃−CH₂−CH₃
​C. A and B both
D. None of these

Hint: When a haloalkane is added to an alcoholic potassium hydroxide (KOH) solution, it leads to the formation of an alkene. This reaction is called dehydrohalogenation.

Complete Step by Step Solution:
In this reaction, a halogen atom and a hydrogen atom from the haloalkane get removed in the form of hydrogen halide. It is an elimination reaction.

Alcoholic KOH is the concentrated alcoholic solution of KOH.

The carbon atom which holds the halogen atom is α-carbon. The carbon atom which carries the hydrogen atom is the β-carbon. The hydrogen atom is removed from β-carbon, this reaction is likewise called the β-elimination reaction.
For example - dehydrobromination of bromoethane

Image: Dehydrobromination of bromoethane leading to the formation of ethene.

In the given reaction, the α-carbon loses the bromine atom and the β-carbon loses the hydrogen atom.
The resulting product is an alkene.

Image: Dehydrobromination of bromopropane.

A. Propene is the main product in this reaction. So, A is correct.

B. Propane cannot be the product of this type of elimination reaction. So, B is incorrect.

C. In an elimination reaction, only an alkene is formed as the product. So, C is incorrect.

So, option A is correct.

Note: While attempting the question, one must retain in mind that the product of this type of elimination reaction is an alkene. Alkane is not formed during this type of reaction. The halide is removed from the α-carbon and the hydrogen atom is removed from the carbon atom next to the α-carbon i.e., β-carbon.