[ktfrancis94]

I have the following multiple choice question that I am confused about:

When optically pure (R)-2-iodooctane is stirred with sodium iodide in acetone, the product will be:
a) optically pure (R)-2-iodooctane
b) optically pure (S)-2-octanol
c) optically pure (S)-2-iodooctane
d) racemic 2-iodooctance

I thought it would be c since the reaction would proceed by Sn2 which gives inversion but it is not the right answer. Any help here?

[discodermolide]

I would also say the c is correct as Sn2 reactions proceed with inversion of stereochemistry. At the worst case senario you will get a racemate. But the answer is inversion of stereochemistry.

[zsinger]

That is very arguable.  Weird question!

[orgopete]

I think the answer is d). This basically a racemization procedure. Here is why. Since the starting material and product are the same they will each react at the same rate. If 50% of the (R) isomer reacted, the product will be racemic. However, the product will also be reacting restoring some of the optical activity. Therefore as the reaction proceeds, the optical activity will fall until it is completely lost.

[rasoooool]

If one mole of sodium iodide c) is the correct answer, but if is plenty d) is correct,Because my friends said.

[orgopete]

If one mole of sodium iodide c) is the correct answer, but if is plenty d) is correct,Because my friends said.

If only a catalytic amount, it is d). Iodide will react with R or S iodide with an equal rate. Molecules don't know they are reactants or products. Just because the reaction says it begins with optically pure R, there is no restriction making them more reactive than the S product.

[Borek]

d is the correct answer regardless of how much iodide is present. It will just take more time to produce the racemate if the amount of iodide is lower.

[clarkstill]

d is the correct answer regardless of how much iodide is present. It will just take more time to produce the racemate if the amount of iodide is lower.

Agreed.  After an amount of time that is reasonably longer than the timescale for nucleophilic addition, ~50% of molecules will have undergone an even number of SN2 reactions, leading to stereochemical retention, and 50% will have done an odd number, leading to inversion.  So overall you would expect a racemic mixture.