Bromoethane is treated with $\mathrm{KCN}$ in ethanol to give $\mathrm{P}$. $\mathrm{P}$ is reduced with $\mathrm{LiAlH}{}_4$ to give $\mathrm{Q}$, and $\mathrm{Q}$ is treated with a large excess of $\mathrm{CH}{}_3\mathrm{I}$ in the presence of base. The final product is

1. A$\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{N}(\mathrm{CH}{}_3){}_3{}^{+}\mathrm{I}{}^{-}$
2. B$\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{N}(\mathrm{CH}{}_3){}_3{}^{+}\mathrm{I}{}^{-}$
3. C$\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{N}(\mathrm{CH}{}_3){}_2$
4. D$\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{NH}{}_3{}^{+}\mathrm{I}{}^{-}$