The dehydrohalogenation of neopentyl bromide with alcoholic $\mathrm{KOH}$ mainly gives

${\mathrm{CH}}_3-\underset{{\mathrm{CH}}_3}{\overset{{\mathrm{CH}}_3}{\underset{\mathrm{I}}{\overset{\mathrm{I}}{\mathrm{C}}}}}-{\mathrm{CH}}_2-\mathrm{Br}+\underset{\left(\mathrm{alc}\right)}{\overset{ }{\mathrm{KOH}}}\to$ ${\mathrm{CH}}_3-\underset{ }{\overset{{\mathrm{CH}}_3}{\underset{ }{\overset{\mathrm{I}}{\mathrm{C}}}}}=\mathrm{CH}-{\mathrm{CH}}_3+\mathrm{KBr}+{\mathrm{H}}_2\mathrm{O}$

In this reaction $1°$carbonium ion is formed which rearranges to form $3°$carbonium ion from which base obstruct proton. Hence 2-methyl-2-butene is formed as a main product.