There are two different groups bonded to each terminal C in the C=C=C system, and there is restricted rotation about the C=C=C. So why doesn't it show cis-trans isomerism?

The most common answer I get for this is that the set of bonds C-CH3 and C-H on C1 and C3 in the C=C=C are not in the same plane. Why do they have to be in the same plane for the exhibition of cis-trans isomerism?

  • $\begingroup$ Because of the way cis-trans isomerism about a double bond is (sensibly) defined, you need a plane. Note that in all cases, there are two enantiomers. $\endgroup$
    – Alchimista
    Commented Nov 22, 2021 at 10:16
  • 1
    $\begingroup$ The cis-trans isomerism makes sense when the final substituents are in the same plane. In 2,3-penradiene, the final substituants $\ce{H}$ and $\ce{CH3}$ are not in the same plane. They are in planes which are perpendicular to one another. $\endgroup$
    – Maurice
    Commented Nov 22, 2021 at 10:18

2 Answers 2


The (cis)/(trans) nomenclature (or, more universally applicable, (E)/(Z) nomenclature) works fine for isolated and conjugated double bonds. In the case of 2,3-pentadiene, however, where double bonds literally are adjacent to each other, you have a cumulene. As a result, the mutual orientation of two terminal methyl groups is close to perpendicular.

To put things in perspective (literally), build a model with a model kit:*

enter image description here

and apply the nomenclature for axial chirality instead. You assign the methyl groups a higher priority than the hydrogen atoms (CIP rules), then, while looking along atoms (2) and (3), the «rotation» along the sequence 1 -> 2 -> 3 -> 4 is clockwise, thus it is (2P)-penta-2,3-diene (where P (as in plus, to the right, clockwise), and M (as in minus) is used to describe the axial chirality and helical chirality.

The relevant rules in the Blue Book (see resources page) are P-, P-, P-, and P- If you want to continue to use (R) and (S) to assign axial chirality, add a subscript to the label (here, they correspond inversely and hence (2P)-penta-2,3-diene is ($2S_\text{a}$)-penta-2,3-diene).

*In case you do not have access to a model kit, you may paste the following SMILES string (a structure description a [dedicated] computer program understands) CC=[C@@]=CC as input for molcal.org. Don't mind that the angle enclosed by the methyl groups differs from the one you get with a physical model kit (example), focus on the relative orientation.


Favre, H. A., Powel, W. H. Nomenclature of Organic Chemistry. IUPAC Recommendations and Preferred Names 2013. Excerpt prepared by Moss G. P. at https://iupac.qmul.ac.uk/BlueBook/

  • $\begingroup$ Thanks. Actually what exactly is cis-trans isomerism? Is being on the same plane a pre-requisite? $\endgroup$
    – user119212
    Commented Nov 22, 2021 at 10:56
  • $\begingroup$ I get that it's chiral, but don't get why it cannot show cis-trans. $\endgroup$
    – user119212
    Commented Nov 22, 2021 at 10:56
  • $\begingroup$ To show cis/trans, you need two substituents at each end of the double bond. $\endgroup$ Commented Nov 22, 2021 at 11:16
  • $\begingroup$ Eventually, it is convention to use (R) and (S) here to provide an unambiguous description about the atoms' arrangement. cis is about from the same side of a reference plane, and trans about from the other side. For but-2-ene, this reference plane a would be orthogonal to a plane b passing the double bond and one of the two methyl groups, while a would pass the double bond. If you consider wiggling the 2,3-pentadiene of the methyl groups (conformational changes, not constitutional ones), this however could yield a fuzzy description despite no bonds were detached/rebuilt. $\endgroup$
    – Buttonwood
    Commented Nov 22, 2021 at 11:25
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    $\begingroup$ @OscarLanzi Thank you for the reporting the error. A turn to the right to yield a helical P, which here indeed is (in CIP) an $aS$ (or alternatively, $S_a$). $\endgroup$
    – Buttonwood
    Commented Mar 20 at 16:18

Let me speak of geometrical isomerism as such the discussion is more general.

Geometrical isomerism of the cis-trans / E-Z type can happen about a plane containing two atoms each with two different substituents (though in cyclic compounds the latter requisite is merely formal). In fact the two identical substituents or the pair with high priority can be at the same or at the opposite sides of that plane, leading to different compounds.

If such a plane does not exist, or better if there are not such two atoms, there are not geometrical isomers (as properly intended*).

*Other isomerisms lead to different geometrical arrangements of atoms and moieties in molecules, of course. This often confuses newbies. The term geometrical in geometrical isomerism solely refers to the arrangement discussed above.


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