What happens when
2-bromopentane is treated with alc. KOH?
Write the chemical equations in support of your answer.

Explanation

When 2-bromopentane is treated with alcoholic KOH, it undergoes dehydrohalogenation leading to the formation of two or more different products depending upon the number of different types of `beta`-hydrogen available.
The more highly substituted alkene (i.e. having lesser number of hydrogen atoms on the doubly bonded carbon atoms) is the major product of dehydrohalogenation. This is an accordance with Saytzeff rule.
`underset("2-bromopentane (has 2 "beta"-H-atoms)")(overset(H)overset(|)(C)H_(2)-oversetoverset(Br)(|)(C)H-oversetoverset(H)(|)(C)H-CH_(2)-CH_(3))overset("alc.KOH")rarrCH_(3)-underset("pent-2-ene (Major)")(CH=CH-CH_(2))-CH_(3)+CH_(2)=underset("Pent-1-ene (Minor)")(CH-CH_(2)-CH_(2)CH_(3))`