Enantiomeric excess

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From relative concentrations to ee

A and B are enantiomers, to get an enantiomeric excess (ee) from relative concentration, just do :

ee=|[B]-[A]|   
|| stands for absolue value

example:

we have 70% of A and 30% of B in solution
 so also 40%A + 30%A +30%B 
        [––]   [––––––––]
       excess  racemic part

so the ee in this case is : 40%

From ee to relative concentrations

As enantiomeric says, the ee value is an exces, so do if A is in excess:

%A=100-(100-ee)/2
%B=(100-ee)/2

Best here is an example also:

ee=40%, A is in excess.
%A=100-(100-40)/2=70%
%B=(100-40)/2=30%

From activation energy to ee

Sometimes, it is interesting to access a theoretical value for the enantiomeric excess. 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.

From ee to activation energy

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