I am confused as to whether the E/Z nomenclature can be extended to compounds of general structure like in the image. Clearly they are some sort of cis and trans but I cannot see a non-aromatic double bond for which the Cahn–Ingold–Prelog precedences do differ here and why... It intuitively seems that the description should go with respect to the quinone cycle but how? Phane nomenclature?

Thank You in advance!

Two quinone methines

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    $\begingroup$ I expect that the situation is exactly analogous to the one given in P- of the 2013 IUPAC Blue Book here: iupac.qmul.ac.uk/BlueBook/P9.html#92040202 Formally, following through with the CIP rules as stated in there (you need to construct something called a 'digraph') will allow you to assign the double bonds as either (E,E) or (Z,Z) (but not (E,Z) or (Z,E)). I'm rather lazy to do this myself, though, so will hold off on a proper answer. $\endgroup$ Feb 19, 2022 at 0:37
  • $\begingroup$ Thank You very much! I will try to write a proper answer then myself. $\endgroup$ Feb 20, 2022 at 10:18
  • $\begingroup$ How can I credit You properly in the answer, @orthocresol? $\endgroup$ Feb 20, 2022 at 17:59
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    $\begingroup$ No need to. I didn't quite answer; I pointed you in the right direction, at most, and you did all the work :) $\endgroup$ Feb 20, 2022 at 18:05
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    $\begingroup$ This link maybe helpful. $\endgroup$
    – user55119
    Feb 20, 2022 at 19:41

1 Answer 1


So, @orthocresol actually answered the question in their comment, and I am just adding a summary of their reference. Later, @user55119 commented about this wonderful resource with CIP rules explanations, including my case: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/pinene.html#diethylidene

Nevertheless, I already spent some time inventing the bicycle, so here it is. To sum up, Cahn–Ingold–Prelog turns out to be more involved than usually mentioned in classes (at least ones I went through when I studied) :-)

First one needs to construct a digraph, or a tree-diagram, showing the connectivity (topology) of all the atoms. In complete digraph, lone pairs are included, but usually there is no need for this. Also, H atoms can be omitted. In either or both cases it is a simplified digraph. Also, You can draw the spheres around the stereocenter but these are not mandatory for either type of digraph. For example, digraph for this molecule:


the simplified digraph would be:

digraph cationic

Numbering can be either systematic according to the general IUPAC rules or arbitrary, just to rank chains (as here).

Now, for double bonds we create duplicate atoms:

enter image description here

And for cycles, we open them as close to the stereocenter as possible, and also add the duplicate atoms:

enter image description here

And, what is usually not taught, for mancude cycles (those with MAximum Number of ConjUgated DoublE bonds) we calculate the mean atomic number for all possible Kekule structures we can draw for that atom. This number is then retained instead of duplicate atom in the digraph. This becomes interesting for ionized species:

enter image description here

We can see how incredibly many times the same sequence repeats. This looks really awful but allows to assign the stereochemical designation in a non-ambiguous way.

Now, we should recall the priority rules for Cahn–Ingold–Prelog:

  1. higher atomic number precedes lower one;
  2. higher isotope mass precedes lower one;
  3. Z precedes E for double bonds.

For the molecules I asked the question about, we should select the in-cycle atom of one of the double bonds as a stereocenter. Then, while constructing the digraph, we will see that going one or the other way around the central ring formally makes the other quinone double bond either E or Z. Now let us look back on the first double bond. As we now have one substituent with Z some atoms along the digraph branch and one substituent with E along its digraph branch, our first double bond now has non-equal substituents at the end which belongs to the cycle and thus it, the first double bond can be now assigned configuration for real. Which will be Z for the left molecule and E for the right one.

enter image description here

Now in theory we should repeat that for the other double bond at the center but we always will get E,E or Z,Z, so it is basically redundant.

Also, E and Z are sometimes referred to as seqTrans and seqCis, by the way. Classsic trans and cis always follow the hydrocarbon path.


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