# E/Z configuration of the central double bond in a highly branched poly-ene

What would be the configuration of the marked double bond in the attached molecule?

• From Cahn Ingold Prelog rules, it can not be assigned E/Z configuration. Nov 6 '16 at 18:38
• The substituents on the right side of the marked double bond are symmetrical. Thus, the marked double bond doesn’t give rise to cis/trans isomerism.
– user7951
Nov 6 '16 at 18:39
• Maybe Cahn-Ingold-Prelog (CIP) rules cannot describe such a system but IUPAC rules can. In IUPAC rules, everything should have a priority and, according to what I obtained on the latest ChemDraw 16, E has priority over Z (just like R has priority over S in atom configuration. See Pseudo-asymmetry). In your example, the central double bond should have geometry Z according to ChemDraw, but I do not have the IUPAC rule for that. I will go on searching. Nov 6 '16 at 19:59

For simplicity, I shall assign numbers to the double bonds so that it is easier to reference them. This numbering is chosen solely after their position in space and does not correspond to the number of carbons in a PIN or otherwise systematic name.

1                 10
> 5         8 <
2    \       /    11
> 7 <
3    /       \    12
> 6         9 <
4                 13


The terminal bonds, 1–4 and 10–13, are easy since they are all disubstituted: (1Z,2Z,3E,4E,10E,11Z,12Z,13E)

1Z                 10E
> 5         8 <
2Z    \       /    11Z
> 7 <
3E    /       \    12Z
> 6         9 <
4E                 13E


Next up are the intermediate double bonds. If we simply follow the atoms, neither side will ever have priority, thus the somewhat less common rule that Z has priority over E takes effect. This allows us to label the double bonds (8Z) and (9Z).

1Z                  10E
> 5         8Z <
2Z    \       /     11Z
> 7 <
3E    /       \     12Z
> 6         9Z <
4E                  13E


However, we cannot determine a stereodescriptor for bonds 5 and 6: both 1 and 2 have the same geometry (Z) as do both 3 and 4 (E). Thus, neither of these double bonds is asymmetric as they only feature three different substituents with the two identical ones being on the same side.

1Z                   10E
> 5Ø         8Z <
2Z     \       /     11Z
> 7 <
3E     /       \     12Z
> 6Ø         9Z <
4E                   13E


Moving on to the final double bond 7, we face the same problem. The substituents 8 and 9 are identical, both feature an E double bond attache to E and Z. Thus again, we are unable to determine a stereodescriptor since the double bond is not asymmetric.

1Z                    10E
> 5Ø          8Z <
2Z     \        /     11Z
> 7Ø <
3E     /        \     12Z
> 6Ø          9Z <
4E                    13E


Note that we can assign priorities to the left-hand side of double bond 7: the bottom half (3,4,6) has a lower priority than the top half (1,2,5) since the latter features Z double bonds where the former features E. But assigning a descriptor to 5, 6 and 7 is impossible due to their symmetric other side.

• So what is the answer to the question? Nov 6 '16 at 21:42
• @ketbra Last sentence.
– Jan
Nov 6 '16 at 21:45