2
$\begingroup$

When I have drawn stereoisomers of substituted cyclohexanes and checked if they are chiral, I typically represent the ring as a planar hexagon and look for symmetry elements. For example, I would use this representation of cis-1,2-dimethylcyclohexane, and I would say that it is achiral due to reflectional symmetry:

enter image description here

enter image description here

However, I was recently reading about how some disubstituted cyclohexanes such as cis-1,2-dimethylcyclohexane are achiral not due to the symmetry in their 2D representation, but due to the fact that they exist as rapidly equilibrating chiral conformers that cancel out each other's optical activities. In this case, both the planar representation and the conformational analysis lead to the same conclusion about the chirality of cis-1,2-dimethylcyclohexane (though for different reasons). Will the planar representation and conformational analysis always lead to the same conclusion about chirality?

I think they will: I think that if a molecule can exist in an achiral conformation (the planar hexagon conformation), and you could rotate bonds to turn the achiral conformation into a chiral conformation (one of the chair conformations), then you must be able to obtain the enantiomer of that chiral conformation (here that enantiomer is the flipped chair) by performing mirrored bond rotations on the original achiral conformation (since the achiral molecule has reflectional/center of inversion/improper rotation symmetry, I think that guarantees that the mirror operations must be possible, though I am not certain). Since enantiomers have the same physical and chemical properties, they have the same thermodynamic stability, so at equilibrium, they will exist in equal amounts (the equilibrium constant for their interconversion is 1), so a sample of the disubstituted cyclohexane must exist as a racemic mixture. Is this thinking correct? Could you extend this argument to other molecules that we represent as achiral, but can exist as chiral conformers (such as butane, monosubstituted cyclopentanes, etc.)?

I also was thinking about whether it was correct to find all of the stereoisomers of the cyclohexanes by analyzing only the planar hexagon conformation. For example, to count the stereoisomers of inositol, I would use the following diagram, moving certain OH groups up or down to find the eight relative configurations, with one being chiral, which gives a total of nine stereoisomers:

enter image description here enter image description here I am wondering whether this always leads to the correct result, despite the majority of the inositol in a sample likely existing in a chair conformation, not a planar conformation. The nine stereoisomers drawn must be possible stereoisomers of inositol, so we know that there are at least nine stereoisomers. This leaves us with the question of whether the nine stereoisomers we have drawn represent all possible stereoisomers of inositol, or whether there are some stereoisomers that can not achieve a planar conformation, and are therefore not included in the nine we drew. I am not sure how to ensure that we have drawn all possible stereoisomers, besides saying that we can generally correspond each 3D structure of a cyclic molecule to a planar representation, and therefore all possible stereoisomers of inositol would have to be represented by one of the 9 planar representations from above.

Is it correct to analyze the planar depiction of substituted cyclohexanes for chirality and stereoisomerism, or do we need to look at the true 3D conformations?

$\endgroup$
3
  • $\begingroup$ I'd say yes. All molecules are somewhat dynamic. None of the (necessarily static) representations we are able to make can possibly be realistic. However, chirality is a geometric concept. So when we study chirality we need to use an 'ideal', rigid representation, where the molecule is 'as symmetric as possible'. That is your 2D depiction. Another way to look at it could be: is the average of all possible conformations symmetric? Take the amine example in the post you linked: the average is a planar species that does not exist, but justifies that the molecule behaves as achiral. $\endgroup$ Nov 8, 2022 at 19:09
  • $\begingroup$ The only way the correct result for chirality using a planar structure might be wrong would be if there were very bulky groups as substituents that prevented easy interconversion of conformers (which otherwise interconvert very rapidly). This is certainly known in other classes of molecule like biphenyls with bulky groups or in hexahelicene where the entire chirality is a product of long distance spatial interactions preventing the possibility of a "flat" structure. $\endgroup$
    – matt_black
    Nov 9, 2022 at 14:55
  • $\begingroup$ @matt_black : indeed, I had thought of biphenyls; hence the wording "average of all possible conformations". Biphenyls with bulky substituents cannot have planar conformations, thus the average will still be asymmetric, and they are chiral (only) for that reason. Still, it's true that conformations that are not possible in standard conditions can become possible when more energy is available, e.g. by heating, so this is indeed a thorny issue. To keep it simple, however, the answer to the OP is yes for the most common cases. $\endgroup$ Nov 9, 2022 at 18:20

1 Answer 1

1
$\begingroup$

Is it correct to analyze the planar depiction of substituted cyclohexanes for chirality and stereoisomerism, or do we need to look at the true 3D conformations?

Yes, you can do that. As the six membered ring transitions from chair to boat to twisted to planar (with the latter having an unrealistically higher energy, so it would not be populated in reality), the configuration around the carbon atoms does not change. So if you construct the mirror image of a substituted cyclohexane, you make no mistake if you first "flatten it out" for easier drawing.

Many of the conformations of substituted cyclohexanes are chiral if you don't allow for conformational change. To take a simpler example than the cyclohexan one, if butane were frozen in the gauche conformation, it would be chiral. Because it is not frozen in that conformation, as we know, it is not chiral.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.