# How do you calculate how many stereoisomers a compound has?

##### 1 Answer

Since for every atom that can exist in more than one configuration, you have **R** or **S** (**E** or **Z** (

- 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

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.

...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