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According to my book it is optically active, but no valid reason is given. But what I found is there exists an alternating axis of symmetry, and so my conclusion is the compound would be inactive or has a meso isomer. Am I wrong?

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    $\begingroup$ Axes of symmetry have nothing to do with chirality unless it's an improper axis (which does not sound like what you're referring to). $\endgroup$ – Zhe Jan 5 '18 at 14:21
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First off ...

There is a difference between "chiral" and "optically active". Something that's optically active has to have chiral molecules. However, something can be chiral but if you have equal amounts of the mirror-image isomers, or if molecular motion can make an equal number of isomers like an umbrella-like inversion of a lone pair or an exchange of protons, then you do not see optical activity. All we are looking at here is whether the molecule is chiral, not whether a sample of the stuff is in general optically active.

With that cleared up:

As Weijun says in the comments you have to have a mirror plane, or an improper rotation axis, or else the molecule is chiral. A (proper) rotation axis does nothing in this regard.

Each of the carbon atoms with a methyl group on it is, in fact, a chiral center. The four different groups attached to it are:

No 4. A hydrogen atom

No 3. The methyl group that's attached outside the ring

And...

No 2, No 1. The ring going clockwise and going counterclockwise, as viewed from "above" the C–H bond, are counted as two different attached groups, as if the ring were "cut off" in either direction before rejoining the carbon atom you're looking at.

If you look at the ring attachment that way, you see that the two different directions rank 1 and 2, in some order depending on where the second carbon with the methyl group is, hence my numbering above. So you can figure out an R or S configuration for each chiral center. In the cis isomer you end up with one R and one S (thus mirror-image configurations), and a mirror plane does go between them making a meso rather than a chiral molecule. But the trans isomers give both R or both S, which never allows a mirror plane or any improper rotation axis, and so they are chiral.

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  • $\begingroup$ (Something that's optically active has to have chiral molecules... ) what about allenes, biphenyl...atropisomers? $\endgroup$ – Light Yagami Dec 11 '20 at 5:24
  • $\begingroup$ Chiral molecules in all optically active cases. But there can be chirality even though you don't have a specific carbon atom attached to four different groups. You are offering examples. $\endgroup$ – Oscar Lanzi Dec 11 '20 at 9:41
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Whoa, there! It's unusual to find such imprecise statements in this forum. The necessary and sufficient condition for a molecule to be chiral is that it has no improper rotation in its point group. For example, the point group of 1,3,5,7-tetramethyl-1,3,5,7-cyclooctatetraene is $S_4$ so it is achiral. To see why this is the case, any improper rotation can be written as a proper rotation about some axis followed by a reflection through a plane normal to that axis. For example inversion $i=\sigma_zC_{2z}$ or $S_{4z}=\sigma_zC_{4z}$. To superimpose $\ce{C_{12}H_{16}}$ just apply $C_{4z}=\sigma_z\sigma_zC_{4z}=\sigma_zS_{4z}$. The $S_{4z}$ operation left the molecule invariant because it's in its point group but we can see that it is now a mirror reflection $\sigma_z$ away from a realizable configuration reached via a proper $C_{4z}$ rotation. Also superimposing a molecule on its mirror image means that some proper rotation $R$ and an improper rotation $\sigma$ do the same thing: $R\psi=\sigma\psi$ so multiplying both sides by $\sigma$ we see that $\sigma R\psi=\psi$ so an improper rotation $\sigma R$ is a member of the point group of the molecule.

That is not to say that there can be chiral conformers that can interconvert; the above paragraph is just about geometry. Also, the point group of both trans-1,2-dimethylcyclopropane and trans-1,2-dibromocyclobutane is $C_2$ with no improper rotations so they are both chiral.

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  • $\begingroup$ Missed it the first time I answered, look now at my edit. $\endgroup$ – Oscar Lanzi Dec 11 '20 at 14:29

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