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!