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After treating the compound with NaI, we should follow SN2 mechanism and the product will follow an optical inversion, giving us (-)2-iodobutane, but the answer is (±)2-iodobutane.

After going through various articles, I am getting reasons, such as there will be an equilibrium created between the reactants and the products, hence we will get both the products (±)2-iodobutane.

And there is the same question where we are treating (R)-2-bromobutane with NaI in acetone, but the answer is (S)-2-iodobutane.

What could be the reason?

I believe we should simply get one product in the question, and there will be no equilibrium be created, as there shouldn't be enough NaI left to form another compound.

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    $\begingroup$ Iodide is attacking and kicking out another iodide, so you cannot ever run out of it. $\endgroup$ – orthocresol Sep 26 '18 at 17:07
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    $\begingroup$ If iodide can react with the (R)-enantiomer, why not the (S)-enantiomer? As orthocresol states, it goes on and on until racemization occurs. This is classic experiment that shows that racemization is twice the rate of SN2 displacement. Imagine you have two (R)-molecules. You only have to invert one to (S) to achieve racemization. At that point, only one of the two (R)'s has reacted. Think about how radioactive iodide would be useful in this regard. $\endgroup$ – user55119 Sep 26 '18 at 19:52
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There is no equilibrium in the second reaction because the sodium bromide formed by the iodide displacement is not soluble in acetone so there is no Br- to displace iodide. In contrast with the first reaction, NaI is soluble in acetone so there is always excess I- around to continue the reaction. This is called the Finkelstein reaction, further information here finkelstein reaction

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  • $\begingroup$ But in the first reaction, could the final major product be only (-)2-iodobutane? $\endgroup$ – kiv Sep 26 '18 at 16:35
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    $\begingroup$ No, because the iodides keep exchanging so you get the mixtures. $\endgroup$ – Waylander Sep 26 '18 at 16:46

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