# Cis , trans E or Z?? Tricky question😭

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#1
I can't figure out whether this structure is E or Z cis or trans. Looking at the structure & the textbook to help me better it seems like its neither but it has to be either one since this is a multiple choice question.

Thank you!!!
Last edited by Sidd1; 1 year ago
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1 year ago
#2
You should approach this methodically.

1st: check whether EZ is possible: is there a double bond and does each C attached to the double bond have two different groups attached. Check.
2nd: check whether cis/trans is possible: is there one (or both) group on the left C that is common to the right C. Nope. No cis/trans.
3rd: identify priority groups. Find the first point of difference on atomic numbers*.
LHS first: The C with the double bond is attached to 2x C atoms with the same atomic numbers, so we look to see what they're attached to: top = HHH i.e. atomic numbers 1,1,1 bottom = CHH i.e. 12,1,1 comparing the first number 12>1 (we have a winner, stop looking) so the bottom group wins.
RHS: The C with the double bond is attached to 2xC, so next step: top = CCH, i.e. 12,12,1 bottom = CHH, i.e. 12,1,1 comparing 12=12 (carry on) 12>1 (stop) top wins
4th: priority groups are on opposite sides E.

*if you get to the end of the checks and there is still no winner, you consider masses of isotopes or e.g. H vs D.
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#3
(Original post by Pigster)
You should approach this methodically.

1st: check whether EZ is possible: is there a double bond and does each C attached to the double bond have two different groups attached. Check.
2nd: check whether cis/trans is possible: is there one (or both) group on the left C that is common to the right C. Nope. No cis/trans.
3rd: identify priority groups. Find the first point of difference on atomic numbers*.
LHS first: The C with the double bond is attached to 2x C atoms with the same atomic numbers, so we look to see what they're attached to: top = HHH i.e. atomic numbers 1,1,1 bottom = CHH i.e. 12,1,1 comparing the first number 12>1 (we have a winner, stop looking) so the bottom group wins.
RHS: The C with the double bond is attached to 2xC, so next step: top = CCH, i.e. 12,12,1 bottom = CHH, i.e. 12,1,1 comparing 12=12 (carry on) 12>1 (stop) top wins
4th: priority groups are on opposite sides E.

*if you get to the end of the checks and there is still no winner, you consider masses of isotopes or e.g. H vs D.
Thank you so much! What do you mean by "masses of isotopes eg H vs D ?"
Also, the top carbon on the RHS is actually CH(CH3)2 - not CCH. Sorry if that wasn't clear from my diagram!
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1 year ago
#4
(Original post by Sidd1)
Thank you so much! What do you mean by "masses of isotopes eg H vs D ?"
Also, the top carbon on the RHS is actually CH(CH3)2 - not CCH. Sorry if that wasn't clear from my diagram!
Imagine ethene, except swapping an H on both sides with D (deuterium, i.e. H-2). If you just consider atomic numbers, then both sides you have 1=1 and it looks like there is no priority. So, now you consider atomic masses and you get 2>1 on both sides.

The top C on the RHS is attached to H and to 2x CH3. The three atoms that that C is attached to are CCH.
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