How do you calculate how many stereoisomers a compound has?

Since for every atom that can exist in more than one configuration, you have R or S (`sp^3`), or E or Z (`sp^2`), you have two configurations for each of those atoms.

• If you had `2` of those atoms, then you have `4` configuration combinations: (R,R), (R,S), (S,R), (S,S).
• For `3` of those atoms, you have: (R,R,R), (R,R,S), (R,S,R), (S,R,R), (R,S,S), (S,R,S), (S,S,R), (S,S,S), which is `8`.

Let us call an atom or group of atoms that can exist in more than one configuration a stereounit.

That means for `n` stereounits, you have `2^n` stereoisomers possible.

However, note that if there are any meso compounds (i.e. if the molecule has a chance of having a plane of symmetry dividing two identical halves that each contain asymmetric centers), then we must account for them because the symmetry reduces the number of different compounds.

An example of a meso compound vs. a regular chiral compound...

Thus, we revise the formula to give:

`\mathbf("Total Stereoisomers" = 2^n - "meso structures")`

where `n = "number of stereounits"`.

Note that if we had used the traditional definition of a stereocenter instead of a stereounit (i.e. `sp^3` carbons only), this compound screws things up:

...it has two stereocenters, but three atoms which can be in more than one configuration.

Hence, by the stereocenter definition, it has 4 structures, when in fact it DOESN'T.

Two configurations for the left carbon, two configurations for the right carbon, and two configurations for the middle carbon, meaning `2^3 = 8` stereoisomers. Try drawing them out in your spare time!