With $\mathrm{NaOEt}$ in ethanol, the order of increasing proportion of elimination (alkene) for $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{Br}$, $(\mathrm{CH}{}_3){}_2\mathrm{CHBr}$ and $(\mathrm{CH}{}_3){}_3\mathrm{CBr}$ is

Increasing substitution at the alpha carbon blocks SN2 attack and, at the same time, gives more beta hydrogens and more stable alkenes. Primary bromoethane gives mostly ether, the secondary halide a large fraction of alkene, and the tertiary halide essentially only alkene.