My book states that:
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.
$$\ce{R-X~+~NaI->[\ce{acetone,~reflux}]R-I~+NaX}$$ $(\ce{X=Cl,~Br;~R=alkyl ~group})$
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 question:
Why are $\ce{NaCl}$ and $\ce{NaBr}$ not soluble, whereas $\ce{NaI}$ is soluble in acetone? All we can observe is that, sodium iodide has greater molecular weight than sodium chloride or sodium bromide. But, how do this account for solubility of sodium iodide?
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