Meso Compouds

Remember way back when I said this?

A molecule that contains more than one stereogenic center might be chiral, but it might not. More on this later.

Well, it's later now. This post is on meso compounds. Please, no jokes about the name. Mostly because I am lazy, I will just start with the exact same example as this textbook: 2,3-dibromobutane.
And of course you spotted the tetrahedral stereogenic centers, right? We have a carbon attached to a methyl group, a hydrogen, a bromine, and another, identical carbon. So that's two chiral centers.

I don't think that I've mentioned it so far, but a molecule with two chiral centers can have, at most, four stereoisomers. I suppose that at this point I should introduce wedges and dashes to make three-dimensional interpretations of these stereoisomers, but I won't, so there. Actually, I should have done that a while ago. Fine, I'll get around to it at some point. Moving on...

Anyway, I'll probably just make a video to explain this, because it's even worse than trying to explain what a tetrahedral stereogenic center is using just words. But the short version is that we can arrange the bonds around both centers to have one version of the molecule, take its mirror image and have a pair of enantiomers, then take one of those forms and switch two bonds to have a third stereoisomer that is neither superimposable on either of the previous two molecules nor a mirror image of either of them. But this molecule is superimposable on its mirror image, so it has no enantiomers. It is achiral, even though it has two stereoisomers that are chiral. And that makes it a meso compound.

Also, the term for the relationship between stereoisomers that are not enantiomers is diastereomers.

Tetrahedral Stereogenic Centers in Cyclic Compounds

A carbon atom that is part of a ring can potentially be a tetrahedral stereogenic center. It should be so obvious that I don't need to tell you this, but I'd better do it anyway: since two of the bonds on the potential center must attach in the ring, the two remaining bonds must be to two different substituents. Just to be safe, I shall illustrate this graphically.

So here is an example of a ring carbon that is not a tetrahedral stereogenic center...
And here is one that is...
That's pretty straightforward. Now, don't cry or anything, but that is not quite all there is to it. There is one more detail about this that sort of warrants a separate post. I think even you will find it rather easy, in principle. There is one further requirement in order for this hypothetical carbon atom to serve as a tetrahedral stereogenic center: there must be some difference between the two bonds in the structural sequence of the ring as we trace the path around it. No really, I worded it that way on purpose to dishearten you. It's actually not difficult.

We start with the central carbon atom and move along both ring bonds. Are the atoms that those two bonds attach to the same? And are the atoms that those atoms attach to the same? And so on. Eventually, both paths will converge (halfway across the ring). If both of those paths are identical, then the carbon in question is not a tetrahedral stereogenic center. However, if the paths are different, then the carbon is a tetrahedral stereogenic center.

And that's it! But just to be sure you don't forget about this, which you will anyway, let's demonstrate with an example. My textbook uses this example. Here's a compound that is achiral...
You've been practicing your nomenclature, right? So you already know that this is a skeletal structure for methylcyclohexane with one of the hydrogens drawn in for some reason. The reason is that the carbon we're focusing on is attached to that hydrogen, a methyl group (Me), and twice to the ring. But tracing both paths along the ring, we find that they are identical, arriving at a ring carbon attached to two hydrogens, a ring carbon attached to two hydrogens, and then meeting halfway along the bond between two ring carbons. So what we have is not a tetrahedral stereogenic center and this molecule is achiral.
If you only learned how to name compounds from this blog, you would not yet know that this is 3-methylcyclohexene. Don't worry about that. The important thing here is that, as before, we have a carbon attached to a methyl group, a hydrogen, and twice to a ring. However, this time, as we trace the paths of both ring bonds, going one way takes us to a ring carbon attached to two hydrogens and going the other way takes us to a ring carbon attached to one hydrogen and double-bonded to another ring carbon. The paths are not identical, so we have a tetrahedral stereogenic center, and this molecule is chiral.