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My textbook says that 1,4-dimethylcyclohexane (cis or trans) does not have a chiral centre, since two groups on the tertiary carbon (that are the part of the ring) are identical, and thus the compound cannot be called a meso compound. But it also says that it has two stereogenic centres (C-1 and C-4), since interchange of groups around them creates a geometrical isomer.

The IUPAC definition of chiral center is:

An atom holding a set of ligands in a spatial arrangement which is not superposable on its mirror image. A chirality centre is thus a generalized extension of the concept of the asymmetric carbon atom to central atoms of any element, for example N+abcd, Pabc as well as Cabcd.

This definition of chiral center is also equivalent to (for coordination number 4) saying that if exchange of any 2 ligands on the carbon atom creates a stereoisomer, then that carbon is chiral center. So according to this, 1,4-dimethylcyclohexane has two chiral center and can be called a meso compound.

But my textbook says the opposite. So, my question is, which one is correct?

(My basics regarding chirality and stereochemistry is clear but the confusion in this question arose as I think that my textbook considered that since the two groups on the tertiary carbon seem to be identical, so it is not a chiral center, but I think it is wrong.)

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    $\begingroup$ There are only 2 stereoisomers here, RR and SS are identical, and RS and SR are identical. $\endgroup$ Jul 7 at 3:35
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    $\begingroup$ @NisargBhavsar if you are able to define R and S, then the carbon must be chiral, and that's what my question is about. $\endgroup$
    – user92687
    Jul 10 at 6:43
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    $\begingroup$ No time for an answer, but see pseudo-asymmetric centre. Google, or searching on SE, will throw up many descriptions. $\endgroup$
    – orthocresol
    Jul 10 at 10:09
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    $\begingroup$ pseudo-asymmetric centres are created when 2 of the 4 components of a carbon are same structurally but have different chirality correct me if I am wrong, @orthocresol but I don't think "pseudo-chirality" applies in this particular case. $\endgroup$ Jul 10 at 13:18
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    $\begingroup$ @napstablook That definition is correct and applies precisely here. To go into why would be way beyond a comment, but in short, you need to construct the full digraph which involves "breaking up" the ring: there is a similar example in chemistry.stackexchange.com/a/35511/16683. $\endgroup$
    – orthocresol
    Jul 10 at 13:26
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As it is said in the definition of IUPAC, for a carbon atom to be chiral it should be attached to 4 different groups:

IUPAC definition of carbon chiral center

In case of 1,4-dimethylcyclohexane, you will see that the two groups on the right and on the left are identical. So in this case it's not Cabcd, it's Cabbc. If you alter one group (let's say the one on the right), these groups will no longer be identical, so you will have 4 non-identical groups attached to a carbon atom, and this atom will become a chiral centre.

1,4-dimethylcyclohexane

Meso-compounds are compounds that have chiral centre but they also have a plane of symmetry so that they are still superimposable with their mirror image. Following is the example of a meso-compound from McMurry's book (Ref.1):

An example of a meso compound

So a meso-compound must have a chiral centre by definition, but if a compound has a chiral centre, that doesn't necessarily mean that it's a meso-compound. And the answer to your question is no, 1,4-dimethylcyclohexane doesn't have any chiral centre.


References:

  1. John E. McMurray, In Organic Chemistry, Ninth Edition; Cengage Learning: Boston, MA, 2016, pp. 133 (ISBN-13: 978-1-305-08048-5).
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  • $\begingroup$ Keep in mind that you should provide all details of the references. Look at my editing as an example, which is in ACS format. $\endgroup$ Jul 13 at 18:16
  • $\begingroup$ @MathewMahindaratne thank you $\endgroup$
    – Azamat
    Jul 14 at 17:56