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# Need hekp with e/z isomerism

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Why bother with a post grad course - waste of time? 17-10-2016
1. 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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2. 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
3. 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.
4. (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?
5. (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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