What is the role of acetone in the Finkelstein reaction? It has something to do with Le Chatelier's principle but I am not able to understand why it will not proceed without acetone.
$\begingroup$
$\endgroup$
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Finkelstein reaction is a halogen exchange reaction between haloalkane and a salt of a different halogenide. $\ce{R-CH2-Cl + KI -> R-CH2-I + KCl}$. The driving force in this reaction is precipitation of $\ce{KCl}$. It is important to find a solvent that would dissolve $\ce{KI}$ better than $\ce{KCl}$. Then solution will always have $[\ce{I-}] > [\ce{Cl-}]$ (more iodide anions than chloride anions). As a result this equilibrium will be driven to $\ce{R-CH2-I + KCl}$. Acetone is exactly this type of a reagent.
For this case Le Chatelier's principle means that reaction will be driven to the direction in which one of the reagents gets outside of the solution. Since $\ce{KCl}$ keeps precipitating the reaction will be driven to the desired product.
This reaction is important because there are many ways to make $\ce{RCH2Cl}$, but making $\ce{R-CH2-I}$ is difficult. Finkelstein reaction allows to convert available $\ce{R-CH2-Cl}$ to $\ce{R-CH2-I}$.
