When 2-methyl butyl bromide is treated with sodium ethoxide in ethanol, what will be the major product?

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{ONa}/{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}$ gives an elimination product.

The given reaction is 

$C{\mathrm{H}}_3{\mathrm{CH}}_2\mathrm{CH}\left({\mathrm{CH}}_3\right){\mathrm{CH}}_2-\mathrm{Br}+{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{ONa} \stackrel{\mathrm{in} \mathrm{ethanol}}{\to } {\mathrm{CH}}_3{\mathrm{CH}}_2 \mathrm{CH} {\mathrm{CH}}_3 = {\mathrm{CH}}_2$

So, the final product will be  2-methyl but-2-ene.