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    I understand that there's a pi bond surrounding the double bond in an alkene, but does it not rotate well due to electron-electron repulsion above and below the bond?
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    (Original post by Magenta96)
    I understand that there's a pi bond surrounding the double bond in an alkene, but does it not rotate well due to electron-electron repulsion above and below the bond?
    The pi bond cannot rotate or the electron orbitals will no longer overlap, such that the bond is no longer present.
    http://4.bp.blogspot.com/-P_gTwSEVvGY/TdMneuFez4I/AAAAAAAAAEY/587N1eYTHKM/s1600/800px-Pi-Bond.svg.png
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    (Original post by Magenta96)
    I understand that there's a pi bond surrounding the double bond in an alkene, but does it not rotate well due to electron-electron repulsion above and below the bond?
    Unlike a sigma bond, the strength of the bond is highly dependent upon the angle at which you rotate it, going to zero at 90 degrees
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    Ignore the pi bond for a moment and consider the p orbital. The p orbital is dumbbell shaped with the point of origin going through the nucleus of the atom. Each half of the dumbbell is conventionally marked + and - (these are just signs and have nothing to do with charges at all - think of a sine wave above and below the X axis and you get the idea: above is +ve; below -ve)

    Each phase of the p orbital can overlap with another p orbital's phase that has the same sign ie: + to + and - to -. These form (for want of a better word) sausage shapes that lie both above and below the sigma bond forming a pi bond. It is this dual structure of the pi bond that prevents rotation ...
 
 
 
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