Quoting from Wikipedia,

Hexachlorocyclohexane (HCH), $\ce{C6H6Cl6}$, is any of several polyhalogenated organic compounds consisting of a six-carbon ring with one chlorine and one hydrogen attached to each carbon. This structure has nine stereoisomers (eight diastereomers, one of which has two enantiomers), which differ by the stereochemistry of the individual chlorine substituents on the cyclohexane.

According to Wikipedia, only $\alpha-$hexachlorocyclohexane can exhibit optical isomersim. Here is an image of $\eta-$hexachlorocyclohexane,enter image description here Clearly it doesn't have any plane or centre of symmetries. So why it isn't an optically active compound?

  • $\begingroup$ Make a mirror image of this molecule and try to superimpose the structure you have here. $\endgroup$
    – AChem
    Sep 16, 2022 at 16:41
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    $\begingroup$ There is a mirror plane of symmetry normal to the plane of the ring. $\endgroup$ Sep 16, 2022 at 16:42
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    $\begingroup$ The plane is between the 12 o'clock to 2 o'clock and 6 o'clock and 8 o'clock. $\endgroup$ Sep 16, 2022 at 17:38
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    $\begingroup$ Your compound is related to allo-inositol where OH groups replace Cl. All nine stereoisomers are discussed here. $\endgroup$
    – user55119
    Sep 16, 2022 at 22:19
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    $\begingroup$ @ToddMinehardt that is not a plane of symmetry when considering the 3D conformation $\endgroup$ Sep 17, 2022 at 5:01

2 Answers 2


OP's comment: Clearly it doesn't have any plane or center of symmetries. So why it isn't an optically active compound?

It has a plane of symmetry:


Thus it is not optically active compound (a meso-compound):

Super imposable image of Hexachlorocyclohexane

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    $\begingroup$ If we consider the chair conformer of cyclohexane I don't observe any plane of symmetry. $\endgroup$
    – Infinite
    Sep 17, 2022 at 0:36
  • $\begingroup$ this plane of symmetry is only a plane of symmetry in this representation of cyclohexane—a chair representation of cyclohexane does not have this plane of symmetry and thus it cannot be a meso-compound by this reason $\endgroup$ Sep 17, 2022 at 2:15
  • $\begingroup$ i have left another response that i believe is more accurate—in re Matthew's answer, note that the mirror plane splits two sets of adjacent chlorines: to point in the same direction, one side of the mirror must have the axial orientation, and the other HAS to be equatorial. $\endgroup$ Sep 17, 2022 at 4:47
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    $\begingroup$ i am not saying it is optically active, i’m saying “it has a mirror plane of symmetry” is not the reason it’s achiral $\endgroup$ Sep 17, 2022 at 8:45
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    $\begingroup$ @ToddMinehardt I don't quite get your comment; you can't show that a conformer is optically active by the presence of a symmetry element, only the absence. The conformer doesn't have any improper rotation axes, thus it is chiral. The lack of optical activity in a real sample arises not because of the presence of a symmetry element, but because of conformational interconversion (between the two enantiomeric chair conformers). $\endgroup$ Sep 17, 2022 at 9:44

This molecule does not have any symmetries—however, we can show that its enantiomers are equivalent with some visualization and demonstrate thus that it is not optically active. We'll start by drawing out each enantiomer, like so:

the two enantiomers of eta-hexachlorocyclohexane

One property of substituted cyclohexanes is that it is isomerically equivalent with the isomer where the axial and equatorial substituents are swapped. This conformational isomerism is known as a ring-flip—after a ring-flip, all upward facing carbons face downward and vice versa; every axial substituent becomes equatorial; and every equatorial substituent becomes axial. Let's now perform a ring-flip on the left enantiomer of $\eta$-hexachloro-cyclohexane:

a demonstration of what impact a ring-flip has on eta-hexachlorocyclohexane

Remember: a ring-flip is a conformational isomerism—no exchange of bonds needs to occur for this isomerism to happen. This means that these enantiomers are equivalent to each other and so the molecule is not optically active. They are not enantiomers because they can "freely" (ignoring the kinetics and thermodynamics of the transition states/intermediates) move between each other. There does not need to be a plane of symmetry in this case. In fact, this molecule does not have any standard symmetries—it belongs to the $C_1$ point group, as it lacks a symmetry across any mirror plane, axis of rotation, and inversion center.

You can prove mathematically via cyclic permutations as well that there won't be superimposability between the isomers, but they will be isomeric across a ring-flip. Let's represent a conformation $\eta$-hexachlorocyclohexane according to its sequence of axial ($A$) and equatorial ($E$) chlorines:

$$ AAEEAE $$

Rotation of the molecule about the z-axis can be represented as a cyclic permutation of this sequence. A 60$^\circ$ rotation clockwise would look like this:

$$ AAEEAE \longrightarrow EAAEEA $$

A reflection across a mirror plane looks like reflecting the sequence, like so:

$$ AAEEAE \longrightarrow EAEEAA $$

We do not need to note the orientation (up or down) of each individual chlorine, because we know that relative to each other there will always be four chlorines in one direction, and two going the other. However, rotating the molecule by any axis other than the one previously described 180$^\circ$ reorients the molecule and swaps ups for down and downs for up. This is isomorphic to a binary sign-flip operation, so we will represent all 180$^\circ$ rotations with a sign flip, as such:

$$ (+) AAEEAE \longrightarrow (-) EAEEAA $$

This would change our mirror reflection notation to:

$$ (+) AAEEAE \longrightarrow (+) EAEEAA $$

The question is: is this above transformation possible using only the rotations discussed? This is the mirror reflection; if rotations bring us to the same sequence as a mirror reflection, then our compounds are superimposable. If you play around with this for a while, it isn't possible. Any 180$^\circ$ rotation to get the sequence in the right orientation will change the sign such that they can't be lined up anymore.

However, we can prove that a ring-flip will bring us to the same sequence as the mirror reflection sequence. A ring-flip takes all axial to equatorial and all equatorial to axial, so we can represent a ring-flip like so:

$$ (+) AAEEAE \longrightarrow (+) EEAAEA $$

A ring-flip does not change the direction the chlorines are facing—only their orientation—so the sign is maintained. The following sequence incorporating the ring-flip brings us to the enantiomer sequence:

$$ (+) AAEEAE \longrightarrow (+) EEAAEA \longrightarrow (+) EAEEAA $$

The first operation was the ring-flip, and the second operation was a 120$^\circ$ clockwise rotation. This proves that even though there is no symmetry in this molecule it is conformationally isomeric with its enantiomer via the ring-flip isomerization.

It turns out that if the skeletal representation of a cyclohexane ring has an internal plane of symmetry, that molecule will be achiral—this is a convenient shortcut for assessing the achirality of a compound, but the mirror plane in the skeletal representation should not be mistaken for a mirror plane in the chair conformation.

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    $\begingroup$ @toddminehardt to have a mirror plane of symmetry is not the only reason something can be achiral—that is a shortcut for looking at the flat representation of a molecule and seeing if it is achiral. the molecule itself does not have a true mirror plane, ONLY its skeletal representation does. the reason the literal molecule itself is achiral is because the enantiomers are equivalent through a ring-flip isomerization—achirality comes from conformational equivalence of two enantiomers, which we typically think of as rotating but can also be through changes in internal degrees of freedom $\endgroup$ Sep 17, 2022 at 8:52
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    $\begingroup$ @ToddMinehardt does cis-1,2-dibromocyclohexane have a mirror plane of symmetry? no, one of the bromines is equatorial and the other is axial—it’s literally impossible. but its skeletal representation does. the reason it is achiral is via a ring-flip isomerization. $\endgroup$ Sep 17, 2022 at 8:55
  • $\begingroup$ @ToddMinehardt i also—just to make sure i wasn’t crazy—made the molecule in GaussView and tried to superimpose it with its reflection (couldn’t figure it out after toying for like 10 minutes) and the only point group GaussView would tolerate was C1, even with the highest tolerance for RMSD. so i’m pretty confident it’s not superimposable with its enantiomer without a ring-flip. $\endgroup$ Sep 17, 2022 at 9:02
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    $\begingroup$ cf. my and others' answers on cis-1,2-dimethylcyclohexane $\endgroup$ Sep 17, 2022 at 9:24
  • $\begingroup$ Yes, ring flips are the reasons it isn't optically active. But the fact that the flat configuration of the ring has a mirror plane is a good indicator that this will happen when cyclohexanes are known to interconvert between common configurations easily at room temperature. There is little point in considering the chirality of an individual chair form when the molecule is not frozen and the various forms interconvert rapidly. $\endgroup$
    – matt_black
    Sep 19, 2022 at 16:32

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