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Topic: Question about E2 mechanism.  (Read 2576 times)

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Offline Ahmed Abdullah

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Question about E2 mechanism.
« on: March 31, 2007, 09:00:58 AM »
In the presence of polar, non-protic solvents (i.e., DMSO or DMF), alkyl halides undergo reaction with bases to generate alkenes by an E2 mechanism. In the E2 reaction, the proton removed by the base must be anti- to the leaving group, and they must also be periplanar (the hydrogen, the leaving group and the two carbons must all be in the same plane). In the E2 reaction, both the alkyl halide and the base are present in the rate-limiting transition state, making the reaction bimolecular and concerted.

  Now my question is why hydrogen which is removed must be anti-(trans-) and planar relative to the leaving halogen??

I have found in some books it is about electron coming from backside and forming C-C double bond. HERE what is meant by electron coming from backside.

Offline Dan

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Re: Question about E2 mechanism.
« Reply #1 on: March 31, 2007, 10:24:35 AM »
I think this is due to the overlap required beween the sigma* C-X and sigma C-H orbitals to form the double bond (which is the combination of two coplanar p orbitals). See pic.
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