As far as I can tell, there doesn't seem to be a chiral carbon in this compound. C-4 of the cyclohexane ring has two groups with exactly the same connectivity, and the exocyclic double bond can't give rise to optical isomerism. What am I missing?

  • $\begingroup$ I'd say it's an interesting case of a compound with two stereoisomers, that are in the same time enantiomers (optical isomers) and “cis/trans” (E/Z) geometrical isomers. $\endgroup$ – mykhal Nov 29 '18 at 10:11

The strict criterion for a compound to display chirality is that it must not be superimposable upon its mirror image. Let's ignore the chair conformation of the ring for a while, and assume it adopts a planar conformation. You could draw a side-on view of the ring like this:

Planar conformation

Its mirror image would look like this.

Mirror image

This is an example of axial chirality (Wikipedia; IUPAC). The molecule does not possess a chiral centre itself. The chirality in this case therefore does not arise from disposition of groups about a point (the more familiar case); it arises from the disposition of groups about an axis, in this case the $\ce{C=C}$ axis.

Of course, the cyclohexane ring does adopt a chair conformation. That does not affect the fact that the compound is chiral; it just means that the ring isn't entirely flat as depicted.

The two stereoisomers described can be named using Cahn-Ingold-Prelog rules as described by Loong in this answer.


Actually, this is very similar to the case of an allene, which you may or may not be familiar with. Notice how our planar conformation looks almost like this allene


with the planar ring taking the place of the second double bond. Further explanation of axial chirality in allenes can be found in this question.

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  • 3
    $\begingroup$ The analogy to an allene is a good one. An equivalent description of a double bond views it as a two-membered ring (see here for example). In the OP's molecule the two-membered ring has simply been replaced with a 6-membered ring. $\endgroup$ – ron Jan 16 '17 at 18:52
  • $\begingroup$ Whenever we rotate the mirror image, do we rotste it from the plane of the molecule? $\endgroup$ – Scáthach Sep 19 '18 at 10:35

This question has been answered in elegant fashion by @orthocresol. I'd like to expand the discussion somewhat by detailing how digraphs and Cahn-Ingold-Prelog-(Helmchen) rules account for the ZR and ES assignments for 1-ethylidene-4-methylcyclohexane. To determine the configuration at C4, the double bond in ZR-1 is separated into two double bonds as a function of the pathway to C4. Digraph ZR-2 assigns the R-configuration to the chiral center (CH2CH2Z>CH2CH2E>CH3>H).

To assign the geometry of the double bond (ZR-3), the double bond is kept intact and the methyl group is separated into two pathways to the double bond. The red dots terminating the two phantom chains are copies of the sp2 carbon of the double bond. Temporary Ro/So assignments are made with the phantom chains having the second highest priority. The double bond has the Z-configuration because Ro has priority over So and CH3 has higher priority than hydrogen. The ES enantiomer can be analyzed in a similar fashion.
enter image description here

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I would add that from the organic chemistry nomenclature standpoint, the double bond is enantiomorphic. In the current IUPAC Nomenclature of Organic chemistry, special stereodescriptors seqCis/seqTrans instead of Z/E are used in similar cases, here's an example

P- Specification of exo-cyclic double bonds by stereodescriptors other than ‘E’ and ‘Z’.


Example 2:


(1seqCis,3S)-1-(bromomethylidene)-3-propylcyclobutane (PIN)

(Note: I think there is a typo in the latter M/P axial chirality stereodescriptor locant, should be 1, no 3)

So the preferred names of the two enantiomers are1



(1) as provided by ACD ChemSketch 2018.1.1

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