The Meso Trap
Question: Are these molecules enantiomers, diastereomers, or the same?
If you immediately recognized this as a molecule with an internal plane of symmetry (and thus an achiral molecule, incapable of having an enantiomer), congratulations. If not, you just fell into The Meso Trap.
The Meso Trap is a common feature on exams and tests to make sure that you understand the concept of chirality and that you are paying attention.
Just to make things clear, a meso compound is a molecule that has chiral centers but also has an internal plane of symmetry. This renders the molecule achiral: it does not have an enantiomer, and it does not rotate plane polarized light .
It is similar to a phenomenon found in certain two-headed, two-tailed cats.
The Meso Trap usually comes up in conjunction with the testing of some other important skills, notably:
1) identifying types of isomers
2) manipulating various types of chemical diagrams
3) determining (R)/(S) nomenclature for given stereocenters
4) identifying the stereochemistry of various types of reactions
These are all fundamental skills. We are not talking about trivial stuff here.
It can come up in glaringly obvious ways (like in the above example) or in a number of other ways, as in these examples, where the meso compound is not nearly as obvious, and drawn in different projections.
The Meso Trap can also sneak up on you in other ways, like when you’re being asked for the products of a reaction. In the following example, you might be tempted to draw in not only the product of the reaction but also its enantiomer. In this case, however, the product is a meso compound… meaning that there is only one product.
How To Avoid Falling into The Meso Trap
Identifying a meso compound means being able to identify a plane of symmetry in a molecule. It’s important to realize that this can cut either through a bond or through an atom, as in the following examples.
What if the plane of symmetry isn’t obvious? That’s when you really have to demonstrate that you have a few important skills – like the ability to recognize molecules drawn in different projections, how to do bond rotations, and how to recognize (R)/(S) designations.
There are really two basic strategies for identifying a meso compound if it isn’t drawn in an obvious configuration.
The path of greatest resistance, but greatest reward is to learn to master how to perform bond rotations on line-wedge diagrams, Newman projections, Fischer projections, as well as to be able to convert between the three types of drawings as necessary. In the beginning stages, using a model kit to check your work is invaluable. With these skills you can take any structure given to you and be able to rotate the bonds in a way to test whether the two molecules are mirror images.
A slightly easier method to determine whether a given compound is meso is to take advantage of a simple principle: meso compounds have an internal mirror plane. That is to say that not only must 1) every carbon on each side of a mirror plane have the same substituents, but in addition 2) every (R) stereocenter on one side of the mirror plane must be balanced by an (S) stereocenter on the opposite side. So if you can quickly determine (R)/(S), you can also quickly determine whether a given compound is meso without having to do bond rotations. If the stereocenters aren’t opposite, it can’t be meso. Depending on the complexity of the molecule, this can be a much faster way to do things.
The key ingredient in avoiding falling into the Meso Trap is paranoia. When asked if molecules are diastereomers/enantiomers/the same, ask yourself – “is this a symmetric molecule?”. Every time. After doing an addition reaction to a double bond, ask yourself: “is this a symmetric molecule?”. This comes up a lot more often than you’d expect – especially on tests.
It also comes up later in NMR, where recognizing a plane of symmetry tells you how many distinct proton or carbon signals you will see.\
Answers: 1) the same (meso) 2) a), b), c) are all examples of the same (meso) compounds, drawn differently. 3) a gives cis-dimethylcyclohexane (a meso compound), b) gives (R,S)-2-3-dibromobutane, also a meso compound.