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My book says:
Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone. $$\ce{R-X + NaI ->[\ce{acetone,~reflux}] R-I + NaX}\\ (\ce{X}=\ce{Cl,Br};\ \ce{R}=\text{alkyl group})$$ NOTE: Refluxing is a process of heating a liquid in a flask provided with a condenser. The vapours of the liquid which rise up get condensed back into the flask.
Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chatelier's principle. The reaction gives the best results with primary halides.

MY QUERY

  • Can we prepare alkyl chloride or alkyl bromide by Finkelstein reaction?
    As we can see, first line reads that Finkelstein reaction is particularly useful for preparing iodoalkanes. The lines followed by it signify that $\ce{NaCl}$ and $\ce{NaBr}$ are less soluble in acetone. If we want to prepare alkyl chloride or alkyl bromide, we will be needed to dissolve $\ce{NaCl}$ or $\ce{NaBr}$ respectively in acetone, which is not so possible. Is it true that, we can't prepare alkyl chloride or alkyl bromide by Finkelstein reaction because of the above reasons?
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The thing is that you have to start with some compound $\ce{R-X}$ where $\ce{X}$ is a good leaving group. The quality of the leaving group is often thermodynamically defined, and in the Finkelstein reaction that you described, the energy motor for the whole thing is the insolubility of the salt that is produced.

I don't see what favourable starting material you could use for this in order to prepare other haloalkanes, especially since there are other ways for doing that.

That being said, it would in principle be possible to do $$\ce{R-X + NaY -> R-Y + NaX v}$$ as long as the solubility of $\ce{NaY}$ is larger than the solubility of $\ce{NaX}$.

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