# 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#), o **E** or **Z** (#sp^2#), you have two configurations for each of those atoms.

- Se tu avessi #2# of those atoms, then you have #4# configuration combinations: (
**R**,**R**), (**R**,**S**), (**S**,**R**), (**S**,**S**). - Per qualificarti per il #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 NON.*

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!