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Copy edited (e.g. ref. <https://en.wikipedia.org/wiki/SN2_reaction>, http://www.chemspider.com/Chemical-Structure.552796.html, and <http://www.chemspider.com/Chemical-Structure.21427360.html>).
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edit approved Sep 29, 2018 at 12:27
Peter Mortensen
edit approved Sep 29, 2018 at 12:27
Peter Mortensen
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what What happens when (+) 2-iodobutane is treated with NaI in acetone?

After treating the compound with NaI, we should follow Sn2SN2 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 reasonreasons, 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 athe same question where we are treating (R)2-bromobutane(R)-2-bromobutane with NaI in acetone, but the answer is (S)2-iodobutane(S)-2-iodobutane.

What could be the reason? I

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.

what happens when (+) 2-iodobutane is treated with NaI in acetone?

After treating the compound with NaI, we should follow Sn2 mechanism and the product will follow optical inversion giving us (-)2-iodobutane but the answer is (±)2-iodobutane.

After going through various articles, I am getting reason 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 a 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.

What happens when (+) 2-iodobutane is treated with NaI in acetone?

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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kiv
kiv
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what happens when (+) 2-iodobutane is treated with NaI in acetone?

After treating the compound with NaI, we should follow Sn2 mechanism and the product will follow optical inversion giving us (-)2-iodobutane but the answer is (±)2-iodobutane.

After going through various articles, I am getting reason 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 a 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.