In the Finkelstein reaction, $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{Br}+\mathrm{NaI}\stackrel{\text{dry acetone}}{\to }\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{I}+\mathrm{NaBr}$ proceeds to completion mainly because

1. A$\mathrm{NaI}$ is insoluble in dry acetone and precipitates
2. B$\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{I}$ is a gas that escapes
3. C$\mathrm{NaBr}$ is insoluble in dry acetone and precipitates, shifting the equilibrium forward
4. D$\mathrm{I}{}^{-}$ is a poorer nucleophile than $\mathrm{Br}{}^{-}$, which favours the product