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Question about E2 mechanism.

  1. Mar 31, 2007 #1
    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.:grumpy:

    Your sensible response will broad my understanding. Thnx in anticipation.
     
    Last edited: Mar 31, 2007
  2. jcsd
  3. Apr 2, 2007 #2

    chemisttree

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    The base withdraws the proton in the E2 mechanism. This causes the carbon to begin to rehybridize in the rate determining step from sp3 to sp2. For there to be simultaneous loss of halogen (as X+) they must be interacting by the newly rehybridizing carbon's nascent p orbital. This is the same p orbital that will eventually form the double bond. You can see from a geometrical analysis that the departing proton, nascent p-p orbital formation and halogen leaving group must be coplanar. Aligning the two p "dumbell" shaped orbitals at right angles precludes any interaction, they must be aligned head to head. The only way for this to happen is for the halogen to be either trans or eclipsed cis to the incoming base. For steric reasons the cis orientation is disfavored.
     
  4. Apr 5, 2007 #3
    Isn't something about orientation of molecular orbital?, 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)?
     
  5. Apr 5, 2007 #4

    chemisttree

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    The leaving group, halogen, will produce a p orbital for overlap with the adjacent carbon's newly forming p orbital. For there to be simultaneity, these two orbitals must be aligned in a coplanar fashion.
     
  6. Apr 5, 2007 #5
    Another way to think about it is that the leaving group will run off and leave a carbocation at the other end, which is then promptly taken up by the electron coming in 'the back side'- imagine people in two rooms, someone left the front room and another person replaces that person from the back room. This means there is an empty 'space' for a new person to come into the back room - this electron is the new person! That's an easy way to think about it at any rate- you can go from there! The overlapping orbitals will have an empty space which is filled up by that electron....
     
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