I have just been pondering upon the question above. I have found this example and I am sure they are enantiomers, but do they also count as geometrical isomers?
4
-
$\begingroup$ As far as I know, you have to call them enantiomers and not geometrical isomers since the right side is symmetrical. However, you can designate it as E or Z.. see this. $\endgroup$– Nikhil AnandCommented Jun 2, 2020 at 10:30
-
3$\begingroup$ Please do not use geometric isomerism, it is obsolete and the usage strongly discouraged. Obsolete synonym for cis-trans isomerism. $\endgroup$– Martin - マーチン ♦Commented Jun 2, 2020 at 10:55
-
$\begingroup$ So can i say them cis-trans Isomers or geometrical isomers clearly? $\endgroup$– Harsh KataraCommented Jun 2, 2020 at 12:37
-
$\begingroup$ This issue has been discussed: chemistry.stackexchange.com/questions/58630/… $\endgroup$– user55119Commented Jun 2, 2020 at 15:23
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
4
The enantiomers 1a and 1b can have their double bonds defined by CIP rules 1a and 5. Rule 1a dictates that CH3>H while Rule 5 has R>S. Thus, 1a has an E-double bond and 1b is of the Z-configuration as exemplified by the red bonds. This issue has been addressed previously on this site.
Addendum: While the enantiomers 1a and 1b (vide supra) have their double bonds labeled in upper case E/Z (ChemDraw), JSmol takes into account the enantiomorphic nature of the double bonds by using lower case e/z (vide infra). See the answer of @orthocresol.
-
$\begingroup$ That doesn't really address the question of whether these enantiomers are cis-trans isomers, too. $\endgroup$– Martin - マーチン ♦Commented Jun 3, 2020 at 10:40
-
$\begingroup$ True enough but cis/trans for double bonds doesn't work here and this nomenclature has been supplanted by E/Z. Old-timers like myself may still refer to 1,2-substituted double bonds as cis/trans. $\endgroup$ Commented Jun 3, 2020 at 12:24
-
1$\begingroup$ In general, I don't think there is anything wrong with cis/trans, especially when it is unambiguous. It is called cis-trans isomerism after all, and that is also pretty unlikely to change. But I believe the question wasn't so much about how to name the different compounds unambiguously, but rather whether enantiomers can show cis-trans isomerism or not. Hence my initial comment. I still like your breakdown though. $\endgroup$– Martin - マーチン ♦Commented Jun 3, 2020 at 12:34
-
$\begingroup$ There is an advantage to having a descriptor for the double bond. By using the IUPAC name preceded by the three descriptors, the correct enantiomer may be drawn. Without employing e/z, one would have to describe the structure, say over the phone. The 2013 Blue Book is now 7 years old. By the next revision, e/z will likely be approved. $\endgroup$ Commented Jun 23, 2020 at 20:26
$\begingroup$
$\endgroup$
3
Technically, this double bond should not be labelled as (E)- or (Z)-.
It is what is known as an enantiomorphic double bond, for which the proper stereodescriptors are seqCis and seqTrans. This is described in P-92.1.1 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book).
Notice that for your compound, we have $(\mathrm a, \mathrm b) = (\ce{CH3}, \ce{H})$ and the two enantiomorphic ligands are the two sides of the cyclopentane ring.
The rules for this are briefly quoted in a previous answer:
P-92.1.3.5 Sequence Rule 5
An atom or group with descriptor ‘R’, ‘M’, and ‘seqCis’ has priority over its enantiomorph ‘S’, ‘P’ or ‘seqTrans’.
As for whether it can be a "geometrical isomer", note first that the term geometrical isomerism has been deprecated in favour of cis–trans isomerism (IUPAC Compendium of Chemical Terminology, 2nd ed. ("Gold Book"), DOI: 10.1351/goldbook.G02620:
Obsolete synonym for cis-trans isomerism. (Usage strongly discouraged).
Also notice that the Blue Book (also P-92.1.1) describes cis/trans isomers as being in a separate category from the enantiomorphic double bond:
I think it is reasonable to conclude that the given compound cannot be considered as an example of geometrical isomerism, although the cynic in me says that this is mostly a matter of semantics.
answered Jun 3, 2020 at 4:37
-
$\begingroup$ After reading your view, I created the structures using JSMol and the double bonds are labeled in lower case e/z, presumably to reflect your point. However, ChemDraw uses the upper case E/Z designation. $\endgroup$ Commented Jun 3, 2020 at 14:50
-
$\begingroup$ @user55119 in general, many software packages do not fully implement the IUPAC recommendations (after all the book is ~1500 pages long (!!)). ChemDraw, as good as it is for drawing structures, is one of them. To me, not using E/Z makes sense, because E/Z is kind of in the domain of diastereomers (for example, we wouldn't use E/Z for allene enantiomers). That said, I think the whole thing is arguably more of a technicality than a real conceptual issue. Everybody including OP seems to be agreeing on the actual concepts. $\endgroup$ Commented Jun 3, 2020 at 16:32
-
$\begingroup$ You and others may be interested in these corrections to the Blue Book 2013. I haven't checked to see if they have been implemented. qmul.ac.uk/sbcs/iupac/bibliog/BBerrors.html . Here is paper about revised CIP rules by Bob Hanson (JSmol). pubs.acs.org/action/showCitFormats?doi=10.1021/acs.jcim.8b00324 $\endgroup$ Commented Jun 3, 2020 at 19:02
$\begingroup$
$\endgroup$
2
Edited version: I originally wrongfully assumed these two structures are not geometrical isomers of the double bond, because it is a trisubstituted double bond with at least two identical groups (although they have two different spatial arrangements as R- and S-designations). As a result, I conclude that the two compounds cannot be distinguished by (E) and (Z) prefixes. Therefore, I thought, it is safe to say they are just stereoisomers (according to the old rules when I was an undergraduate). However, rules have been changed significantly since then and according to Revised Cahn-Ingold-Prelog Rules - IUPAC 2013, (R)-configuration has higher priority than (S)-configuration in this situation. Therefore, actually (E/Z)-configurations exist in these molecules. Accordingly, the top structure has (E)-configuration while the bottom structure has (Z)-configuration (see the diagram below).
About stereoisomerism: Since two stereocenters are having R- and S-designations with identical groups (which are mirror images of each other if double bond is not there), we must check for the symmetric elements such as plane of symmetry. Because of one side of double bond is not symmetric (attached by $\ce{CH3}$ and $\ce{H}$ groups), the molecule is not symmetric as depicted in the diagram:
Keep in mind that, in any case, if $\ce{CH3}$ group is replaced by $\ce{H}$ or vice versa, the molecule becomes meso-isomer, which is previously considered as not optically active. However, as Zhe pointed out correctly, revised Cahn-Ingold-Prelog Rules give rise to pseudo-chiral centers in meso-compounds. Nonetheless, the molecule in hand does not have a plane of symmetry, and hence have non-superimposable mirror images as shown in the image. As a results, the two structures given are enantiomers, one with (E)-configuration while the other with (Z)-configuration (see the diagram above).
answered Jun 2, 2020 at 16:47
-
4$\begingroup$ An (R) stereocenter and an (S) stereocenter are not identical for the purposes for CIP. That gives rise to pseudo-chiral centers in meso compounds. $\endgroup$– ZheCommented Jun 2, 2020 at 18:26
-
3