What is the mechanism of the reaction of NaI in acetone with an alkyl halide?

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What is the mechanism of the reaction of NaI in acetone with an alkyl halide?

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Answer 1 · 2016-01-10T16:41:37

This operates by the $S_{N} 2$ $S_(N)2$ mechanism. The iodide ion is a reasonably good nucleophile, and adds as $I^{-}$ $I^-$

Explanation:

$R - C l + N a^{+} I^{-}$ $R-Cl + Na^+I^-$ $\to$ $rarr$ $R - I + N a C l ↓$ $R-I + NaCldarr$

Because sodium iodide is much more soluble than sodium chloride in acetone, when you do this you see a glassy precipitate of sodium chloride in the acetone. I think there is a named organic spot test for this reaction (its name escapes me at the moment, I should not have finished that bottle of wine last night!). If you have time in an organic laboratory it would be worth your while to do this reaction in a small test tube so you can notice the turbidity of the sodium chloride when it precipitates.

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What is the mechanism of the reaction of NaI in acetone with an alkyl halide?

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