Ethyldimethylpropylammonium hydroxide, $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2-\mathrm{N}{}^{+}(\mathrm{CH}{}_3){}_2-\mathrm{CH}{}_2\mathrm{CH}{}_3\mathrm{OH}{}^{-}$, is heated. Two alkenes can be formed; the one formed in larger amount is

In Hofmann elimination the hydrogen lost is the most acidic and least hindered beta hydrogen. The ethyl group has three primary beta hydrogens, while the propyl group has only two secondary ones. The ethyl group is therefore preferentially eliminated, giving ethene as the major alkene.