Why does Finkelstein reaction necessarily involve: R-Cl and NaI (in Acetone)? Why can it not be done with KI? How does potassium instead of Sodium affect the reaction? Why can't R-I react with KCl in the same way?

(I've not included R-I (Iodine) or Fluorine (in the ionic compound) since they are extremes and are very slow or violent in many reactions respectively.)

  • 2
    $\begingroup$ Also, note that the Finkelstein reaction can indeed be done with R-Br! $\endgroup$ – The_Vinz Jun 4 '19 at 16:56

The classic Finkelstein reaction entails the conversion of an alkyl chloride or an alkyl bromide to an alkyl iodide by treatment with a solution of sodium iodide in anhydrouss acetone (Wikipedia). You can find here why this reaction works in this manner.

The reaction is essentially driven toward products by mass action due to the precipitation of the poorly soluble $\ce{NaCl}$ or $\ce{NaBr}$. Solubility of one specific salt (only $\ce{NaI}$) is important to the reaction. I emphasize above that your acetone has to be very dry, otherwise $\ce{NaCl}$ and $\ce{NaBr}$ are also going to be soluble large scale in acetone and reaction won't go to completion, and thus, won't be useful. For example, the solubility of $\ce{NaI}$ in acetone is found to be 29.2% by weight at $\pu{25 ^\circ C}$ but it is increased to 40.5% when 6.1% (by weight) of water added to the solution (Ref.1). Similar effect should be expected for the solubility of
$\ce{NaCl}$ and $\ce{NaBr}$ (Ref.2).

To answer your particular question about $\ce{KI}$, you need to look at the solubility of it in acetone: Unlike $\ce{NaI}$, which has maximum solubility in acetone at $\pu{25.7 ^\circ C}$, maximum solubility of $\ce{KI}$ in acetone is at $\pu{-57.5 ^\circ C}$ (Ref.3):

[....]. In this paper are listed the solubilities of potassium iodide in acetone from $+54.5$ to $\pu{-78.5 ^\circ C}$. While, in general, the solubility curves of sodium and potassium iodides are similar in form, the maximum solubility of potassium iodide is about threefold less than that of sodium iodide, and it occurs at $\pu{-57.5 ^\circ C}$, instead of $\pu{+25.7 ^\circ C}$ as was reported for the sodium salt. The solid phase which is in equilibrium with the solution below $\pu{-57.5 ^\circ C}$ is a solvated form containing five (or six?) molecules of acetone per molecule of potassium iodide.

As stated in above abstract, $\ce{KI}$ molecule is surrounded by 5 to 6 acetone molecule, instead of 3 in $\ce{NaI}$ case. That makes it crowded (steric effect) as well. Its solubility is reported as 1.330% by weight at $\pu{+25 ^\circ C}$ (Ref.3), compared 29.2% of $\ce{NaI}$ under the same conditions. These facts make $\ce{KI}$ unsuitable for used in Finkelstein reaction. Yet, there are several instants of using $\ce{KI}$ in this type of reactions, e.g., see Ref.4.

There are also instances of using substrates other than $\ce{R-Cl}$ or $\ce{R-Br}$, e.g., see Ref. 5.


  1. R. Macy, E. W. Thomas, “The System: Sodium Iodide-Acetone-Water,” J. Am. Chem. Soc. 1926, 48(6), 1547–1550 (https://doi.org/10.1021/ja01417a015).
  2. M. Li, D. Constantinescu, L. Wang, A. Mohs, J. Gmehling, “Solubilities of $\ce{NaCl, KCl, LiCl}$, and $\ce{LiBr}$ in Methanol, Ethanol, Acetone, and Mixed Solvents and Correlation Using the LIQUAC Model,” Ind. Eng. Chem. Res. 2010, 49(10), 4981–4988 (https://doi.org/10.1021/ie100027c).
  3. R. Livingston, R. R. Halverson, “Solubility of Potassium Iodide in Acetone,” J. Phys. Chem. 1946, 50(1), 1–6 (https://doi.org/10.1021/j150445a001).
  4. F. G. Bordwell, W. T. Brannen, “The Effect of the Carbonyl and Related Groups on the Reactivity of Halides in $\mathrm{S_N2}$ Reactions,” J. Am. Chem. Soc. 1964, 86(21), 4645–4650 (https://doi.org/10.1021/ja01075a025).
  5. D. J. Maloney, S. M. Hecht, “A Stereocontrolled Synthesis of $\delta$-trans-Tocotrienoloic Acid,” Org. Lett. 2005, 7(19), 4297–4300 (https://doi.org/10.1021/ol051849t).
| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.