Enantiomeric excess is defined as

$$ee = \frac{[R]-[S]}{[R]+[S]}$$

I found this problem in an IChO paper:

"When using the enantiomerically pure (BINOL)Al(OiPr) as a catalyst for reduction of αbromoacetophenone, the ee of the product equals 81%. What is the ee of the product if the catalyst ee equals 50%?"

The problem as stated is not that difficult: if given the ee of the product mixture ($ee_{p1}$) when the catalyst is enantiomerically pure ($ee_{c1}=1$) then (after noting that mole fraction of [R] and [S] are simply $f_R=\frac{1}{2}(1+ee)$ and $f_S=\frac{1}{2}(1-ee)$ respectively) we say that $f_{Rp2} = f_{Rc2}f_{Rp1} + f_{Sc2}f_{Sp1}$ from which we can calculate $ee_{p2}$, which after a little algebra reduces to $ee_{p2} = ee_{c2} ee_{p1}$. In the above case, then, we'd get $ee_{p2} = 0.5*0.81$.

It's quickly possible based on this formula ($ee_{p2} = ee_{c2} ee_{p1}$) to show that, if given one value $ee_{p1}$ for the product mixture's enantiomeric excess when the catalyst is of a certain enantiomeric excess $ee_{c1}$, we can calculate the product's predicted enantiomeric excess $ee_{p2}$ for any other catalyst enantiomeric excess we try $ee_{c2}$ through the equation

$$ee_{p2} = ee_{p1} \frac{ee_{c2}}{ee_{c1}}$$

What I am wondering is, is there a way to derive this latter formula from the beginning without going through the step of first considering an enantiomerically pure catalyst case? I would not have thought of that case had the contest writers not specifically mentioned it. But not being able to derive the formula from first principles makes me feel I don't really understand the way the various mole fractions would contribute etc.


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