# How does iodide replace bromide when the latter is more basic?

The following reaction follows the $S_N2$ pathway: $$\ce{CH3CH2CH2-Br + I- (presence of DMSO) -> CH3CH2CH2I + Br-}$$ I found this reaction in my textbook and according to me this reaction should not even take place. $\ce{Br-}$ is a stronger base than $\ce{I-}$ and keeping in mind, stronger the base, lesser is the leaving ability, why will $\ce{I-}$ replace $\ce{Br-}$ in the first place?

As you suggest "$\ce{Br−}$ is a stronger base than $\ce{I−}$ and keeping in mind, stronger the base, lesser is the leavability, $\ce{I−}$ should not replace $\ce{Br−}$ "

I actually agree with the logic and according to me, infact, the reverse reaction, i.e.

$$\ce{CH3CH2CH2I + Br- (presence of DMSO) -> CH3CH2CH2Br + I-}$$

should be much more favourable. Because we already know, that in the aprotic solvents, the nucleophilicity of halides is $\ce{F− > Cl− > Br− > I−}$, so $\ce{I−}$ should be a better leaving group and $\ce{Br−}$ should be a better nucleophile .

This reaction would be similar to Finkelstein reaction. Here instead of using acetone, you are using $\ce{NaI}$ in DMSO.

We usually use $\ce{R-Cl}$ in a non polar solvent in Finkelstein reaction, not only for increasing nucleophilicty of the attacking reagent, but also to precipitate the $\ce{NaCl}$ formed in the reaction.

The increasing order of solubility in non polar solvent for sodium halides is: $\ce{NaF < NaCl < NaBr < NaI}$

The order is similar for potassium halides. This order is due to polarisation. In non polar solvents, $\ce{KI}$ is soluble but $\ce{KCl}$ is not. This drives the reaction forward to completeness.

The difference in solubility in $\ce{NaI}$ and $\ce{NaBr}$ (or the potassium derivatives) is not very appreciable, so the reaction will take place in the forward direction, but will not give a very good percentage yield of product.