Could someone please illustrate how to determine the E/Z configuration of this molecule?

image below

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What I understand so far is that if the two highest priority substituent on each side are on the same side, then it's Z and vice versa.

Now, how can I handle such a structure? I've considered it and I thought it might be $\ce{CH-CH3}$ on one side (up), and $\ce{CH2CH2}$ on the other. However, I don't think it's right.

Bottom line is that I don't understand how to fractionate the ring into two substituents.

  • $\begingroup$ This molecule has not isomers about the double bond. Simple as this. $\endgroup$
    – Alchimista
    Dec 27 '19 at 8:25

The CH3 and C2H5 groups are both smaller than the rest of the cycle. So these two groups can be forgotten and this molecule is Z.

  • 3
    $\begingroup$ While the configuration is "Z", the IUPAC 2013 Blue Book recommends no designation.: "P-91.2.2 Omission of stereodescriptors The omission of stereodescriptors specifying double bonds is recommended in the case of three- through seven-membered unsaturated alicyclic compounds where any double bond has a fixed configuration, 'Z' in the case of hydrocarbons...." $\endgroup$
    – user55119
    Dec 23 '19 at 23:31
  • $\begingroup$ Thank you. So, just to understand this properly, we consider the cycle as a shared branch for the two pi-bond carbons? (each of them is considered to bear the same ring). In other terms, I want to know if I could break down this structure to something like this (a,b)>C=C<(e,d) if (a) is the methyl branch (CH3) and e is the C2H5, would it be correct to say b and d are the same? $\endgroup$
    – Hussein
    Dec 24 '19 at 0:16
  • $\begingroup$ I have added to your post your structure numbered. At C1, C6 ranks higher than the CH2 of the ethyl group because it is attached to C5 (CHH) while the ethyl group terminates with CH3 (HHH). AT C2, C3 (CCH) ranks higher than the methyl group (HHH). Therefore, the ring is unofficially, as I commented above, of the Z-configuration. Read up on Cahn-Ingold-Prelog (CIP) rules. $\endgroup$
    – user55119
    Dec 24 '19 at 0:48
  • $\begingroup$ Thank you for both the clarification, I do have a better understanding now, and the recommendation on the CIP rules; that's exactly what I need. $\endgroup$
    – Hussein
    Dec 24 '19 at 1:12

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