Enantiomeric excess

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Sometimes, it is interesting to access a theoretical value for the enantiomeric excess, written ee. This value can be accessed using the Free Energy of activation, assumed that:

• the pre-exponential factors are equivalent for both pathways leading to enantiomer A and B.
• Starting compound is identical for A and B

       exp(-δΔG˜/RT)-1	
ee(%)= ––––––––––––––– *100
       exp(-δΔG˜/RT)+1

where δΔG˜ is the difference of Free Energy of activation separating the TS leading to A and B

δΔG˜=ΔG˜(favoured)-ΔG˜(unfavoured)

This theoretical value doesn't take into account the Gibbs Free Energy of the compounds, the formula given is derived from reaction rate constants. Thus, after equilibration in solution,depending on the temperature, the observed ee might change in respect to this.

Finally, one can get, under same conditions, a Gibbs Free Energy activation difference :

            1+ee
δΔG˜= RT ln ––––
            1-ee

/!\ The value of R is often in cal K−1.mol−1 [1], so if your δΔG˜ value is in kcal/mol, choose R=1.9858775*10^-3 Kcal K−1.mol−1