Part 1
Measuring a Mixture
Before anything can be amplified it has to be measured, and the choice of variable turns out to matter more than it looks. This part sets up enantiomeric excess, derives the relationship everyone expected between catalyst and product, and so defines precisely what Kagan found to be false.
Enantiomeric excess
A sample contains [R] of one enantiomer and [S] of the other. The natural measure is the normalised difference:
\[ ee = \frac{[R]-[S]}{[R]+[S]} \]
The name is literal. Think of the mixture as a racemic part plus a surplus: 2[S] molecules pair off into a racemate and [R]−[S] are left over. The ee is the fraction of the sample that is that leftover excess, which is why a 3 : 1 mixture is 50% ee and not 75%.
Why this combination and not another
Because the chemistry is mirror-symmetric, every physical law in this subject must be unchanged when you swap R and S everywhere. That swap sends ee → −ee and leaves the total [R]+[S] alone.
So ee is the odd variable under the symmetry and concentration is the even one. Any correct equation must therefore be odd in ee — which is why both derivations later in this course collapse to equations in ee alone, with only odd powers. That is a symmetry argument, not a convenience.
The linear law: what everyone expected
Suppose a chiral catalyst is used, itself a mixture with excess eeaux. Assume — and this is the assumption that fails — that catalyst molecules act independently of one another.
Then a fraction (1 + eeaux)/2 of them are R, each delivering product with excess +eemax, and (1 − eeaux)/2 are S, each delivering −eemax. Averaging:
\[ ee_{\text{prod}} = ee_{\max}\left[\frac{1+ee_{\text{aux}}}{2} - \frac{1-ee_{\text{aux}}}{2}\right] = ee_{\max}\cdot ee_{\text{aux}} \]
A straight line through the origin. It says you cannot get out more purity than you put in, and it is what every chemist assumed until 1986. It is also a perfectly sound derivation — from a premise that is wrong.
The premise is independence. If two catalyst molecules can associate, or if one metal centre can hold two chiral ligands, then the minority hand is not simply outvoted: it can be sequestered. Part 2 counts what happens then.
Check your understanding
- A sample is 90% ee. What is the ratio of enantiomers?
Answer: ee = (R−S)/(R+S) = 0.9 gives R/S = 19, so 95 : 5. - Why can a reaction between achiral reagents in an achiral solvent never give ee ≠ 0?
Answer: the mirror image of the whole experiment is the same experiment, and it maps ee to −ee. The only value equal to its own negative is zero. - If an equation for ee contained a term in ee², what would be wrong with it?
Answer: it would not change sign under R ↔ S, so it would predict different behaviour for a mixture and its mirror image. Only odd powers are allowed.