How do we recognize the internal plane of symmetry in meso compounds? I read online that the substituted groups attached to a stereocenter could be rotated as such, to obtain an internal plane of symmetry:

Rotation of substituted groups attached to a stereocenter

Why would rotating the bonds in a molecule not affect its optical activity?

Also, exactly how does the presence of an internal plane of symmetry render a molecule optically inactive, as seen in meso isomers?


I see two issues here:

  1. Rotational conformations do not change whether or not the molecule is meso. For meso compounds, the presence of the symmetry element means that for each conformation that is chiral, we are guaranteed to have the enatiomeric conformation with equal probability. Therefore, the net optical rotation is zero.

  2. Fischer projections are no doubt a usual way to represent chiral molecules, but in general, we use Fischer projections most heavily in the case of carbohydrates. The reason is because carbohydrates have shared homology where we care about the specific placement of hydroxyl groups. In general, they are not a great visualization tool. Wedge and dashes are much better. In this case, for a simple two carbon system, where the two carbons have identical connectivity, you could just as well have assigned Cahn-Ingold-Prelog absolute stereochemistry to each carbon to see that the molecule is meso.


I think you are understanding it a bit wrong. What I understand from the picture is that if we have a optically active molecule and we have to find its meso stereoisomer then we rotate groups of a chiral atom untill internal plane of symmetry is achieved. This is right but will work in a very few cases.

You are right that rotating groups will change its optical activity that is why it became optically inactive from being optically active initially.

There is no such exact way of judging whether molecule has plane of symmetry or not but there are some points you should note: 1. Planar molecule always has a plane of symmetry through the plane of the molecule hence always optically inactive. 2. If there are odd nos of chiral atoms then there will never be a plane of symmetry.(not considering pseudo chiral) 3. Always try cutting the molecules through atoms in a line which are different, collinear to the centre of molecule and check if that is a plane of symmetry or not. 4. You should also check centre of symmetry if it exists in the molecule.

In meso compounds there is internal racemisation which means that overall plane polarised light does not rotate when interacts with molecule. This is because one half of the molecule acts as dextro rotatory and other acts as laevo rotatory as the molecule has plane of symmetry. Hence one part rotates the light but other part rotates it in other direction with equal magnitude due to symmetry. Hence meso compounds are optically inactive.

  • $\begingroup$ Thank you for your answer. You mentioned that "This is right but will work in a very few cases." Given that sigma bonds are free to rotate at any one time, will such a molecule (as in my original question) be optically active, or not? Or, will the molecule be sometimes optically active and sometimes optically inactive? $\endgroup$ – Jonathan Smith Mar 15 '17 at 1:26

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