# Enantiomeric excess catalysed

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.