Finkelstein reaction explicitely entails on the conversion of an alkyl chloride or an alkyl bromide to an alkyl iodide by treatment with a solution of sodium iodide in acetone. $$\ce{R−X + NaI→[acetone] R−I + NaX↓ }$$ $\ce{(X= Cl, Br; R= alkyl group)}$
What I think I understand
Acetone is a polar aprotic solvent so $\ce{NaBr and NaCl}$ formed are insoluble due to being mostly ionic while $\ce{NaI}$ has much covalent character (which can be thought of by Fajans's rules) (also organic solvents can dissolve covalent compounds) so, $\ce{I-}$ will be in solution while $\ce{Br- or Cl-}$ will not be there due to being precipitated.
So, Le Chatelier's principle will make the reaction to be towards the product. Therefore I also understand how can a weaker nucleophile can substitute a stronger nucleophile.
Nucleophilicity order of halides in polar aprotic solvent: $\ce{F- > Cl- > Br- > I-}$
What I ponder
- Why can't we do it on alkyl fluorides?
- Why only acetone? Will the reaction be same in other polar aprotic solvents too?
Some specific reactions which seems to deny all this
- $$\ce{CH3-CH2-Cl →[NaF/DMF] CH3-CH2-F}$$
- $$\ce{CH3-CH2-Cl →[NaBr/DMSO] No rxn}$$
- $$\ce{CH3-CH2-Br' →[NaBr/DMSO] CH3-CH2-Br}$$ where $\ce{Br'}$ is isotopic bromine
Note-
- Here I am talking about major organic products.
- DMF & DMSO are polar aprotic solvents.
I would like to know the theoretical reasoning for above reactions.