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ochem e2 PLEASE HELP

Discussion in 'DAT Discussions' started by theedaddy77, Jun 17, 2008.

  1. theedaddy77

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    KKK it is in topscore problem 92 on the first one. IS THIS TRUE When you are looking for anticoplaner are you looking for the opposite on the next carbon. Like if the leaving group is axial u want a hydrogen that is equitoral? ??? i hate this problem and i cant figure it out. thought and help are greatlyt needed taking test less thana week
     
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  3. Orgodox

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    In the E2 elimination reaction, the carbon-hydrogen sigma bond and the carbon-leaving group sigma bond must lie in the same plane. This allows the orbitals to begin to overlap to form the pi bond as the bonds to the hydrogen and the leaving group are broken. There are two possible planar arrangements of these bonds: both on the same side of the C-C bond (syn-coplanar); or on opposite sides of the C-C bond (anti-coplanar).

    ANTI: In this case the bond from the carbon to the leaving group (green) and the bond from the other carbon to the hydrogen (blue) are anti-coplanar. The dihedral angle between these bonds is 180 degrees. This conformation is staggered about the carbon-carbon bond. Because this conformation is more stable than the eclipsed conformation required for syn elimination, anti elimination is preferred in E2 reactions
    [​IMG]


    SYN In this case the bond from the carbon to the leaving group (green) and the bond from the other carbon to the hydrogen (blue) are syn-coplanar. The dihedral angle between these bonds is zero degrees. This conformation is eclipsed about the carbon-carbon bond.
    [​IMG]

    This may sound complicated but its really not. Just like in the Sn2 you want backside attack so too in the E2 you want "back side attack" But here you have to think of the bond that broke as doing the attacking and also realize that it is happening in one concerted step. So for this to be LIKE a backside approach the bond that mush be broke must allow for this and that is best with the anticoplaner ie Anti elimination.
    Hope this helps!
     
  4. theedaddy77

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    that was an amazing explanation but im still not sure with regard to my specific problem. LETS SAY if you have a cyclopentane were on c1 u have a h and methyl and on the c2 u have iodine. Under what steryoconditions under E2 would the bond form between c1 and c2?? I think it is when H and the Iodine are cis to each other. IS that right or should they be trans??
     
  5. Orgodox

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    Yea so I know which problem you are refering to. For this type of reaction to occur on a ring the 2 groups need to be BOTH IN THE AXIAL position. This way it is "backside attack" mentioned before.
    Cis wouldnt work at all because remeber you need backside attack. What we are trying to do is knock off the I with the forming double bond. Think if an analogy of hitting a cue ball in billiards. The cue ball is the forming double bond which goes and hitts another ball (Iodine) and the ball travels forward because of a hard straight on shot in the back. (SEE ATTACHMENT FOR THE PICTURE OF BACKSIDE SN2 sorry couldnt find an E2 one) If you hit the ball partially on the side and partially in the front it wont go away from you in the direction you want and wont have as much force as a strong backside shot. So same thing here the iodine must be knocked hard by the forming double bond which must come from the opposite side. The trans side. But this alone is not enough since we are on a ring. Fr optimal "force" to make the reaction go you need both groups to be axial (straight up and down)

    [​IMG]

    The dash line in the middle is the double bond forming

    You may find one of these annimations a little helpful its not great
    http://www2.trincoll.edu/~tmitzel/chem212fold/movies/E2/e2arrowfinal.swf
    http://research.cm.utexas.edu/resources/iverson-movies/main.htm


    Heres the pool anaolgy for the SN2 its the same ideal for the E2 but nthe double bond acts as the incoming pool ball and must knock the I off
     

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  6. theedaddy77

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    it seems they just have to be on the same plain. Therefore if leaving group is axial then the other must be axial. but if leaving group is equitoral and so is the hydrogen wouldnt it also work? My logic is that if both groups are either axial or equitoral they will be on he same plan with each other? Thank u btw for ur detailed responses they are exception and rarly seen on sdn. I know get that the leaving group must be on the same plan as H but why is it A wont work. It seems the molecule would be more stable if both I and the tert were equitoal which A would allow for... I dont really get the books eplanttion... i think it is this. If both were equitorall (i and the tert) the H that would need to bounce would be equitoral therefore the nucelophile that would attack would also be on the same plane as the H but the nucelophile would be slowed becaause the tert would be axial to the nucleophile? as u can see im throughly confused.
     
    #5 theedaddy77, Jun 18, 2008
    Last edited: Jun 18, 2008
  7. theedaddy77

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    also can u explain "Anti-periplanar geometry" i went to the site but it talks about it but desont explain what this actually looks like in a molecule were there is no freedom of rotation (cyclo structures or double bond)
     
  8. harrygt

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    Theedaddy, E2 will happen if you can find this situation in any of the chair conformations: I is axial [up], and H is axial down or vice versa: I is axial down and H is axial up.
    In order to have periplanar situation [the same as: anti-coplanar], none of them should be in the equatorial position. E2 happens when both substituents are axial and in the oposite directions. [look for this situation in any of the 2 chair conformations]. If there is no such a conformation, E2 can't happen.
    I hope this helps.
     

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