Explain elimination reactions of alkyl halides.
or
Explain dehydrohalogenation ($\beta$-elimination) of alkyl halides.

(N/A) When a haloalkane with a $\beta$-hydrogen atom is heated with an alcoholic solution of potassium hydroxide,there is an elimination of a hydrogen atom from the $\beta$-carbon and a halogen atom from the $\alpha$-carbon atom. As a result,an alkene is formed as a product.
Since a $\beta$-hydrogen atom is involved in the elimination,it is often called $\beta$-elimination.
If there is a possibility of the formation of more than one alkene due to the availability of more than one $\beta$-hydrogen atom,usually one alkene is formed as the major product.
These form part of a pattern first observed by the Russian chemist,Alexander Zaitsev. He formulated a rule which can be summarised as:
"In dehydrohalogenation reactions,the preferred product is that alkene which has a greater number of alkyl groups attached to the doubly bonded carbon atoms." Thus,$2-$bromopentane gives $pent-2-$ene as the major product.