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Size:  511.7 KB can anyone help me stub 5dii) this is the answer Attachment 532253532255 one of the answer is that one of the carbon is attached to two different group but H is not one of them. How can this work though because the question under it says you need to have a H attached to the c in order to form e/z isomerism. '(In order to have cis or trans isomers) each C atom of the CC double bond must have two different substituent groups and one of those groups must be hydrogen.' Can anyone explain this? Thanks.
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    It's asking you to draw the structural formula of hex-2-ene.
    Since isomers have the same molecular formula but different structural formula, it's asking you to draw the possible isomers of hex-2-ene. Hope that helps
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    E-Z requires each carbon with the double bond must be attached to two different things.

    W.........X
    ...\......./
    ...C=C
    .../.......\
    Y.........Z

    W is not the same a Y and X is not the same as Z

    For cis-trans, W = X or Z OR Y = X or Z.
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    (Original post by Pigster)
    E-Z requires each carbon with the double bond must be attached to two different things.

    W.........X
    ...\......./
    ...C=C
    .../.......\
    Y.........Z

    W is not the same a Y and X is not the same as Z

    For cis-trans, W = X or Z OR Y = X or Z.
    Hi, thanks for helping I am not just not sure whether one of the two different groups has to be hydrogen as that's what is needed for cis-trans isomerism. So if I have CH3 and CH2 one one side, this wouldn't count as e/z isomerism because one of them has to be hydrogen right?
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    (Original post by coconut64)
    Hi, thanks for helping I am not just not sure whether one of the two different groups has to be hydrogen as that's what is needed for cis-trans isomerism. So if I have CH3 and CH2 one one side, this wouldn't count as e/z isomerism because one of them has to be hydrogen right?
    That looks like the old (F322) OCR A cis-trans definition, which seemed to imply that there had to be a H at both ends of the double bond.

    Rest assured, there just has to be a group the same at both ends. If that group happens to be a H, then so be it.
 
 
 
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