I read this question and it's answer by @orthocresol. Now, he has written that the compound in question is meso not because there is a plane of symmetry but because of the rapid flipping of the cyclohexane ring, which makes sense to me. In reality the ring is not planar which seems to be in 2D. But, if we don't consider that ring is 3D and that it actually isn't planar, and we assume that it is planar, answer still comes out to be same, that is, for an actual meso compound, we are able to find a plane of symmetry in the 2D drawing and for a non meso compound, there is no plane of symmetry in the 2D drawing.

So, my question is why from that view also answer still comes out to be same? Is it a coincidence?

In this question, @Chakravarthy Kalyan writes "stereochemical analyses such as the preceding one can be done with drawings using planar rings", but with no explanation why. If we consider the compound 1-(chloromethyl)-3,5-dimethylcyclohexane, this is a meso compound with no plane of symmetry in the 2D drawing. So isn't it a counter example to his claim? If that is true, then why at all people use the method of plane of symmetry for determining meso compounds in cyclic rings?

  • 1
    $\begingroup$ As Alchimista points out the chloromethyl can rotate, probably with a period far less than a nanosecond and so, in effect, there is a plane of symmetry involving the Cl atom, but not as you happen to have drawn it. On average the Cl atom can be considered to lie on the 'north-south' line of the figure. $\endgroup$
    – porphyrin
    Commented Feb 15, 2021 at 12:39

2 Answers 2



A plane of symmetry in a 2-D projection of a cyclohexane is a sufficient condition. Alternatively, and lengthy, you conduct an analysis involving conformers but not limited to the flipping of the ring, that means you must let the chloromethyl rotate, in your example.

In more details:

The answer to your title question is YES providing that the molecule isn't blocked in one conformer.

The latter situation requires low T, or the presence of more than one bulky substituent. At room temperature and within the context of common practice and exercises we can be sure that the condition is met. I suspect that cases in which the situation can rapidly escalates towards complexity exist, as for heavy substitutions alters the conformational distribution, too.

The second question is basically "How is that/why a plane of symmetry in a 2-D projection of conformers which themselves have no symmetry but are enantiomers does effectively work as a criterion?".

Well, a laconic but solid answer is that there is no other possible outcome.

There must be a formal mathematical way to express this fact; I am thinking of symmetry and projection operations.

Yet, I can't do better than visualise the situation, but this is again following the reasoning and depicting the sketch within the answer you have linked, although mentally.

It is the flipping between enantiomeric conformers that makes the two simultaneously equalise each other, and this necessarily requires the two stereocentres having opposite absolute configuration and being at opposite sides of a symmetry plane of the molecule's projection, otherwise they would forcedly be different isomers and there won't even be this discussion.

You can also think of the structure as having the axial and equatorial substituents in an averaged intermediate position, as for the two limits can't be resolved. Indeed, a 2-D projection is in a way a real representation of the molecule - not brutal alteration is operated. It just says what it's supposed to tell, i.e. how one would see the molecule as watched from above the projection plane. During the flipping, one would see a rapid on-plane oscillation of the ax and eq substituents in an alternate fashion, and this require a plane of symmetry in order for the two to compensate their effects. To make this visualisation clearer, I've sketched a rudimental figure showing the motion of ax / eq substituents and their average position as ×

enter image description here

Considering that the motion of the substituent can be taken as a handle for the entire conformational flipping, one can see the relevance of a symmetry plane of the projection. In such a picture there is no reason to expect optical activity, as if the molecule would be flat.

However, while the picture just above is a nice visual, the optical inactivity originates from the simultaneous presence of both enantiomers, and this should be right the reason for deprecating the old term internal raceme in favour of meso compound.

  • $\begingroup$ So, there is a plane of symmetry in the 2D projection of 1-(chloromethyl)-3,5-dimethylcyclohexane? $\endgroup$
    – V.G
    Commented Feb 13, 2021 at 9:10
  • $\begingroup$ No. I have said that symmetry in one of the conformers is sufficient, not that must be there for every conformer. In other words the conformer you have drawn isn't meso. The compound is. $\endgroup$
    – Alchimista
    Commented Feb 13, 2021 at 9:14
  • $\begingroup$ I see. But you didn't answer why is it alright to consider only the 2-D projection when in reality it isn't 2-D? $\endgroup$
    – V.G
    Commented Feb 13, 2021 at 9:15
  • $\begingroup$ @LightYagami then I don't understand the question or better it seems to me that no other outcome is possible. It is OK only if a plane is there. In your example isn't possible. If you would have drawn a conformer in which Cl point above or under the projection plane you would have conclude that the compound is meso, again without thinking of flipping. $\endgroup$
    – Alchimista
    Commented Feb 13, 2021 at 9:37
  • $\begingroup$ I know that it is me who lack conceptual clarity. My understanding is, that a compound is meso if there exists a plane of symmetry in it. I would have been satisfied if the chair conformer had a plane of symmetry. Ok, so you said that if a particular conformer has a plane of symmetry, then it would be a meso compound. Does any conformer of the chair form also has a plane of symmetry in it...maybe by some rotations... if there is, then I am satisfied. If still my question doesn't make sense, would it be OK to you if I delete this post? $\endgroup$
    – V.G
    Commented Feb 13, 2021 at 14:16

Overall I agree with Alchimista's answer, though I think it might be quite complex to grasp, seeing the comments that followed it.

If it helps, here's my down-to-earth version, which in fact covers chirality in general, not just cyclohexanes.

The 'executive summary' of my answer would be: not only it is correct to use an 'unnatural' or 'high energy' conformation of a molecule to look for its symmetry elements, but it's a fundamental part of the abstraction / rigidification that inevitably accompanies symmetry considerations.

Side note: the chloromethyl group in your example is not 'asymmetric' just because the Cl atom points to one side: this is just a side effect of the limitation of representing 3D objects in 2D. You could just as well write it with the Cl pointing toward you, and you would have 2 H atoms pointing symmetrically to the back of the group.
Symmetry is more fundamentally related to imagining the object in 3D (which I understand is not everyone's strongest suit, but nowadays you can find free 3D simple molecular modelling software by which you can explore these concepts).

Lengthier explanation below.

First, it should be clear that a molecule is usually an extremely dynamic entity, it never stands still, it keeps vibrating, and its bonds and the groups connected by them rotate in space in very complicated ways (just think of all the docking studies).
By such motions, the molecule will assume different conformations, some of which have higher energy and some lower energy.
However, between the extremes of these energy curves there are a multitude of conformations, and the molecule can spend some time in each of them [except of course those that are not physically allowed (typical case: atropisomers)].

I'm stressing this point just to clarify that there is nothing scandalous or outlandish in imagining the molecule in a conformation that is not the lowest energy one or a relative minimum: it is still one possible conformation, we are not breaking or forming any bonds to make it.

Even more fundamentally, the act of looking for 'planes of symmetry', 'axes of symmetry', etc., of a molecule, implies by necessity that we imagine a rigid abstraction of it: we are looking at it frozen in a state or conformation that is definitely unnatural for it, no matter how well we choose such state, even if it's a relative minimum in the conformational energy curve.
So it is totally acceptable to choose any conformation that is 'convenient' for studying the stereochemistry of the molecule.

Why? Very simply, because in any case if a molecule is chiral, you cannot possibly find any (physically allowed) conformation with a plane or centre (or more generally, rotary-reflection axis) of symmetry (i.e. 'symmetrical' in the common sense).

It follows that, on the contrary, finding even just one (physically allowed) symmetrical conformation guarantees that the molecule is achiral (so, meso if some stereoisomers of the same molecule in other configurations are chiral).

If you want to see this in a mathematical sense, you can think that the existence of one symmetrical conformation implies symmetry in the whole set of conformations the molecule can assume.
For each conformation A there is for sure a conformation A' (of the same molecule, of course) that is a mirror image of A.
In a more physical sense, once you have achieved symmetry in one conformation, for each molecular motion that breaks the symmetry, there is a corresponding, opposite motion that cannot be 'distinguished' (energetically) from the former.
On the other hand, if a molecule has no such symmetry in any of its conformations, this is necessarily never the case.

Here's a concrete example:

enter image description here

I'm sure you would have no doubt that this molecule is a meso stereoisomer.

Even if for some electronic reason the most stable conformer of this molecule were this:

enter image description here

which is not symmetrical, it would not make the molecule chiral, would it?

For any N conformations of the molecule you can imagine, there are for sure N corresponding mirror image conformations of it that are still the same molecule.

If you take this one instead:

enter image description here

it is sure that you will not find even a single conformation whose mirror image is still a conformer of the original molecule.

  • $\begingroup$ Thank you (+1), this cleared a lot of confusion, but you have explained "why it can be assumed a molecule is achiral even if it has a plane of symmetry in only one conformation?" , my question was "why a 2D projection of the molecule takes into account the effect of a 3D molecule?" which has now been answered by alchimista. :) $\endgroup$
    – V.G
    Commented Feb 16, 2021 at 8:46

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