How much can we extend the Finkelstein reaction?

Question

Finkelstein reaction explicitly entails the conversion of an alkyl chloride or alkyl bromide to alkyl iodide by treatment with 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, being mostly ionic, are insoluble while $\ce{NaI}$ have covalent character (in accordance with Fazans's rules) & 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.

Le Chatelier's principle will make the reaction to be towards the product. Therefore, I also understand how a weaker Nucleophile can substitute a stronger Nucleophile.

Nucleophilicity order of Halides in Polar Aprotic solvent: $\ce{F- > Cl- > Br- > I-}$

What I ponder

1. Why can't we do it on alkyl fluorides?
2. Why only acetone? Will the reaction be same in other polar aprotic solvents too?

Some specific reactions which seem to deny all this

1. $\ce{CH3-CH2-Cl →[NaF/DMF] CH3-CH2-F}$
2. $\ce{CH3-CH2-Cl →[NaBr/DMSO] No rxn}$
3. $\ce{CH3-CH2-Br\text{*} →[NaBr/DMSO] CH3-CH2-Br}$

where $\ce{Br\text{*}}$ is Isotopic bromine

Note-

1. Here I am talking about major organic products.
2. DMF & DMSO are polar aprotic solvents.

I would like to know the theoretical reasoning for the above reactions.

Answer 1

1. The Finkelstein doesn’t occur on alkyl fluorides because it is an equilibrium $S_N2$ process, and the C–F bond is very strong, so the reaction rate is reduced. You can also think of it as the transition-state of an $S_N2$ reaction requires thermal access to a vibrational mode that bends the system into a trigonal bipyramidal geometry, and the vibrational modes of C–F bonds (including bending modes) are high energy. They need something like a silane to activate them because the Si–F bond is of a competitive strength.

2. Yes, the Finkelstein can work in other solvents, especially with the correct activators. You can even do the Finkelstein on aromatic rings if you add catalytic copper (I) iodide and a diamine to activate it.

3. Acetone is used often because it is cheap, accessible, was used when this reaction was dicovered early in chemistry, and has the best solubility of sodium iodide relative to sodium bromide that is easy to achieve.

Now those reactions.

1. This is perfectly fine as a type of Finkelstein. The fluoride is highly nucleophilic, so it has a high forward rate for $S_N2$ and a slow backwards rate.

2. Sodium bromide is not very soluble in DMSO. Nor is sodium chloride, but it is the bromide ion that kickstarts the $S_N2$ reaction. Can’t start if there’s no ions in solution. Adding a catalytic amount of soluble bromine species where the bromine can dissociate could kickstart the reaction at least a little bit. Even then, chlorine is more nucleophilic, and since their sodium salts have similar solutibilities nucleophilicity will win.

3. NaBr and NaBr* have the same solubility and the same bromide nucleophilicity, meaning the rates of the forward and backward $S_N2$ are the same. If you start with 100% CH$_3$CH$_2$Br*, you will approach a 50-50 mix.

Answer 2

Finkelstein Iodination

The Wikipedia article on Finkelstein reaction mentions a halide exchange equilibrium:

$$ \ce{R-X +X'- <=> R-X' +X-} $$

As you correctly mentioned, in acetone, $\ce{NaCl}$ and $\ce{NaBr}$ are not soluble, which drives the reaction forward in accordance with Le Chatelier's Principle.

$$ \ce{R-X +MX' ->[\text{dry acetone}] R-X' +MX}\\ \ce{MX' $=$ NaI, KI} $$

Swarts Fluorination

Swarts reaction, not an extension of Finkelstein reaction, is the recommended method to substitute $\ce{Cl}$ and $\ce{Br}$ with $\ce{F}$, that is to directly heat chlorides and bromides with fluorides of antimony, silver, mercury, or copper. According to NCERT$^\text{1}$

The synthesis of alkyl fluorides is best accomplished by heating an alkyl chloride/bromide in the presence of a metallic fluoride such as $\ce{AgF}$, $\ce{Hg2F2}$, $\ce{CoF2}$ or $\ce{SbF3}$. The reaction is termed as Swarts reaction.

$$ \ce{R-X + \text{fluoride} ->[\Delta] R-F + X-}\\ \ce{\text{fluoride} $=$ SbF3, AgF, Hg2F2, CoF2} $$

Further Reading

You may like to read the article by Evano, G., Nitelet, A., Thilmany, P., and Dewez, D. F. (2018)$^\text{2}$ for a thorough review. From the same article:

References

1. Chemistry Part II: Textbook for Class XII. Chapter 6: Haloalkanes and Haloarenes. Available as PDF.
2. Evano, G., Nitelet, A., Thilmany, P., and Dewez, D. F. (2018). Metal-Mediated Halogen Exchange in Aryl and Vinyl Halides: A Review. Front. Chem., 6. 10.3389/fchem.2018.00114.