July 01, 2022, 07:09:15 PM
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Topic: Alkene reaction stereochemistry  (Read 298 times)

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Offline sharbeldam

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Alkene reaction stereochemistry
« on: May 06, 2022, 10:57:25 AM »
So if we look at this reaction it's anti, I'm trying to find a simple rule for students, I do have set of rules that state (Cis alkene + anti reaction leads to two enantiomers ), but the problem with this question is that the alkene isn't cis or trans.

The carbon on the left isn't chiral so it doesnt really matter if Br is inside or outside the page, but how would i know the stereochemistry of the left carbon? would the Br be inside or outside? I know it should be opposite to the other Br, but the other Br can be drawn linear, or would i just have two enantiomers?

Thanks in advance.
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Offline kriggy

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Re: Alkene reaction stereochemistry
« Reply #1 on: May 09, 2022, 06:56:15 AM »
I would not go for simple rules becaues there are always exception but rather teach them the mechanism so they can deduce it themselves.

In this case you get two products because the bromonium can be formed from the fron or back and then the Br- attacks from the other side. If Im not mistaken, those two compounds are enantiomers since there will be only one chiral center

Offline OrganicH2O

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Re: Alkene reaction stereochemistry
« Reply #2 on: May 15, 2022, 01:03:39 PM »
I think in cases like this it is helpful to emphasize that the stereochemistry of the products will be identical regardless of any of the hundreds of mechanisms we could propose. A single chiral center is forming from a flat/achiral carbon. The two products are enantiomers of each other. Enantiomers have the same chemical and physical properties. It is a fundamental rule for all chemical reactions like this that the product mixture will be racemic (1:1 ratio of enantiomers).

The importance of this is tricky: some professors will require that the racemic mixture be indicated in some way. Others will consider this to be trivial, and accept an answer with the C-Br bonds written as lines, with no indication of the racemic mixture forming.

The second issue: Do we bother to draw it as an anti addition? It's kind of silly to draw the product of an anti-addition, because there is no experimental evidence in the product that an anti-addition has occurred. At this point in the class, some students have failed to gain the ability to quickly recognize chiral centers. In these cases, I suggest that the product be drawn as an anti-addition, because they are likely to lose points on other questions where the anti-addition is actually observable.
I have a Master's in organic chemistry and I am exposed to a LOT of different introductory organic chem classes in the course of my work, ranging from very basic to Harvard. I am here to refine my knowledge and consult with other organic chemistry nerds.

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