2-Bromopentane is treated with alcoholic KOH solution. What will be the major product formed in this reaction and what is the type of elimination called? 

Pent-2-ene is major product known as Saytzeff's Product and it is more stable alkene.
The process is known as β− elimination as it involves elimination of β− Hydrogen.

$$\begin{gathered} {\mathrm{CH}}_3-\underset{\underset{\mathrm{Br}}{/}}{\mathrm{C}}\mathrm{H}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_3\stackrel{\mathrm{alc} \mathrm{KOH}}{\to }{\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_3+{\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_2-{\mathrm{CH}}_3 \\ \mathrm{Pent} 1- \mathrm{ene}( \mathrm{minor}) \mathrm{pent}2-\mathrm{ene}( \mathrm{major}) \end{gathered}$$