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

Nov 25, 2015

Since for every atom that can exist in more than one configuration, you have R or S ($s {p}^{3}$), or E or Z ($s {p}^{2}$), 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:

$\setminus m a t h b f \left(\text{Total Stereoisomers" = 2^n - "meso structures}\right)$

where $n = \text{number of stereounits}$.

Note that if we had used the traditional definition of a stereocenter instead of a stereounit (i.e. $s {p}^{3}$ 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!