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The following question has multiple correct answers:

Identify the compound which has a stereo center

The book out of which the question is taken shows that a, b and d are correct answers.

According to me c should also be correct, as it shows cis-trans isomerism, both X could go to the front (wedge bond), or one could go to the front (wedge) and one to the back (dash bond). After all geometrical isomerism is also a sterioismersim. I came to this conclusion considering the definition given on Wikipedia:

A stereocenter or stereogenic center is any point in a molecule, though not necessarily an atom, bearing groups, such that an interchanging of any two groups leads to a stereoisomer.

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    $\begingroup$ @OP I have tried to edit the question to highlight the actual conceptual problem you face (Sorry for the confusion caused by this earlier). You can always edit your question to include more context, e.g. if someone asks something in the comments. I did this because there was already a close-vote on it, which I think is not appropriate. (It would be great if you could add a source, e.g. author and title, for this exercise/problem, so that it is easier for other people with the same book to find.) $\endgroup$ – Martin - マーチン Oct 29 '18 at 18:15
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    $\begingroup$ To the close-voters Please read the question carefully before casting a vote. I have de-queued it once, because I think it is not in the least bit a homework-like question. $\endgroup$ – Martin - マーチン Oct 29 '18 at 18:16
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Good point. The (c) compound has two tetrahedral stereogenic carbon atoms, with four “substituents”

  • $\ce{-H}$,
  • $\ce{-X}$,
  • $\ce{-CH=CH-\overset{\text{*}}{C}HX-C=C-\bond{...}}$,
  • and the same one, $\ce{-CH=CH-\overset{\text{*}}{C}HX-C=C-\bond{...}}$ but from the other side, thus with the opposite configuration, or its mirror image.

(‘*’ denotes (the other) stereogenic carbon atom)

They are called pseudo-asymmetric carbon atoms.

A different relative configuration at these two pseudochiral centers leads to these two ‘cis’ and ‘trans’ isomers.

The definition of stereogenic center (for more complete one, see e.g. this answer) is fulfilled.

(Maybe this topic is too early, for this textbook stage, and the author did not notice this complication.)

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Now, the key to understanding stereoisomerism is that we have to look for elements of symmetry of which:

  • Plane of Symmetry(PoS)

  • Centre of Symmetry(i)

  • Alternating Axis of Symmetry(AAoS) are the prominent ones. Again, please keep it in mind that chiral centres may or may not enough to decide the chirality of the entire molecule(examples are meso compounds with two or more chiral centres, or "pseudo chiral" molecules which are chiral in one configuration but not in another) Now, since you are clear on the other three options, I am going to jump straight to the third option.

Now, as you can notice, the molecule does possess a centre of symmetry. What this means is that now, when you will try to make those cis-trans configurations, you will always find a plane to cut through to make identical halves. Also, since they are present on a ring,so there is no scope of "weird" configurations like keeping both the X and H on cis or trans. Hence, it is safe to say that as soon as you have located even one symmetry element, you can call the molecule achiral.

If the molecule turns out to be achiral, so any change at any group of atoms won't give you a steroisomer. So it also won't possess any stereocentres

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  • $\begingroup$ Yes, we have looked upon chirality elements so as to find whether optical isomerism will be shown or not. Well, we jump on the conclusion that the molecule is a chiral. OK. But why there is no scope of geometrical isomerism? $\endgroup$ – Mathomania Oct 29 '18 at 17:01
  • $\begingroup$ @Saifuddin Have you seen that there is a centre of symmetry in part (c)? $\endgroup$ – YUSUF HASAN Oct 29 '18 at 17:03
  • $\begingroup$ Yes the molecule is achiral, but what about geometrical isomerism? Why cis or trans can't be shown on the ring? $\endgroup$ – Mathomania Oct 29 '18 at 17:05
  • $\begingroup$ @ Saifuddin Just try to draw the stereoisomer you are talking about and visualize it properly. You will find that you will always be able to send a plane of symmetry passing through the two opposite substituted carbons and perpendicular to the ring. Whenever you will try to make the trans isomer, the other X and H will pick up a symmetric positioning with it. And as I have already explained in my answer, both the groups cannot be kept on cis as the ring will not have such flexibility due to steric reasons. So ultimately, the molecule will exist is one form only. $\endgroup$ – YUSUF HASAN Oct 29 '18 at 17:11
  • $\begingroup$ @Martin-マーチン Can you explain what really is the difference between chirality and stereocentres? $\endgroup$ – YUSUF HASAN Oct 29 '18 at 17:19

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