Hello everyone, I've been attempting to understand a question that is on one of the new spec papers. It's about e-z isomerism in a molecule. To be honest, I really have no idea how you find the answer, as ir seems equally possible that any of them could cause the z-isomerism. Every single double bond has 2 different groups at each end. I'm really confused basically.
Im pretty sure it is C, also what course is this you are doing?
Yeah I forgot to mention in the question, it is C according to the markscheme. Its the new AQA A level chemistry specimen paper. Any ideas how to work it out?
Yeah I forgot to mention in the question, it is C according to the markscheme. Its the new AQA A level chemistry specimen paper. Any ideas how to work it out?
Hello everyone, I've been attempting to understand a question that is on one of the new spec papers. It's about e-z isomerism in a molecule. To be honest, I really have no idea how you find the answer, as ir seems equally possible that any of them could cause the z-isomerism. Every single double bond has 2 different groups at each end. I'm really confused basically.
Why's that? Also, how would you know when the higher priority groups are on the same side just by looking at the skeletal formula?
no idea lol can't wait to find out. Edexcel doesn't go into that much depth aha bare in mind specimen papers are not often representative of the final paper
no idea lol can't wait to find out. Edexcel doesn't go into that much depth aha bare in mind specimen papers are not often representative of the final paper
If yo want to know why, check out the interactive in my post above ...
I'm not really following you, isn't is possible that every one of those double bonds could show stereoisomerism? And once you've established that, how would you know which groups are on which side i.e. whether its an e or z isomer?
I'm not really following you, isn't is possible that every one of those double bonds could show stereoisomerism? And once you've established that, how would you know which groups are on which side i.e. whether its an e or z isomer?
no, I mean the carbon which has a double bond, attached to the carbon which you are assessing, counts as 6, 6
Imagine you have
R-CH=CH-CH=CH-CH3
And you are interested in the configuration of the DB at the right,
Then the carbon to the left with the db counts twice.