Part 2
Kagan: Getting Out More Than You Put In
Part 1 derived the linear law from one assumption: that catalyst molecules act independently. Drop it, and everything changes. The whole of Kagan's result follows from the observation that two ligands on one metal can pair up in three ways, not two.
Derivation: the ML₂ model
Step 1 — Count the complexes
Let the ligand pool have excess x = eeaux, so the fractions of the two hands are
\[ r = \frac{1+x}{2}, \qquad s = \frac{1-x}{2}, \qquad r+s = 1,\quad r-s = x \]
If two ligands bind a metal independently, the three species appear in the proportions you would get by drawing twice from that pool:
\[ [\mathrm{ML}_R\mathrm{L}_R] \propto r^{2}, \qquad [\mathrm{ML}_S\mathrm{L}_S] \propto s^{2}, \qquad [\mathrm{ML}_R\mathrm{L}_S] \propto 2rs \]
The third is the heterochiral or meso complex, and it is the whole story. It contains one ligand of each hand, so by symmetry it is achiral: whatever product it makes must be racemic.
Step 2 — Let it react at its own rate
Set the reactivity of the homochiral complexes to 1 and let the heterochiral one react at relative rate g. The two homochiral species deliver ±eemax; the heterochiral one delivers 0. Weighting each by how much product it makes:
\[ ee_{\text{prod}} = \frac{ee_{\max}\,r^{2} - ee_{\max}\,s^{2} + 0\cdot g\,(2rs)}{r^{2} + s^{2} + g\,(2rs)} \]
Step 3 — Simplify
The numerator factorises using r+s = 1: r² − s² = (r+s)(r−s) = x. So, remarkably, the top is just the linear law:
\[ ee_{\text{prod}} = \frac{ee_{\max}\cdot ee_{\text{aux}}}{\,r^{2}+s^{2}+2g\,rs\,} \]
Everything non-linear lives in the denominator. Writing β = 2rs/(r²+s²) for the ratio of heterochiral to homochiral complex, and dividing through, this is exactly the form Kagan published:
\[ \boxed{;ee_{\text{prod}} = ee_{\max}\cdot ee_{\text{aux}}\cdot\frac{1+\beta}{1+g\beta};} \]
Reading the answer
The correction factor (1+β)/(1+gβ) decides everything, and it does so through one comparison:
| If | the meso complex is | then |
|---|---|---|
| g < 1 | slower than the homochiral ones | positive NLE — amplification |
| g = 1 | just as fast | exactly the linear law |
| g > 1 | faster | negative NLE — worse than linear |
The mechanism in words: the minority hand cannot help pairing with the majority hand, because there is so much more of it about. So the minority is disproportionately locked into meso complexes. If those complexes are sluggish, the minority has been quietly removed from the reaction, and the majority runs the chemistry almost alone.
The extreme case is g = 0, a completely inactive meso complex. Then eeprod = eemaxx/(r²+s²), and a ligand at 50% ee gives a product at 80%; at 80% ee it gives 97.6%.
Why this was worth a prize
Practically: enantiopure ligands are expensive and partially resolved ones are cheap. A positive non-linear effect means a process can run on a 70% ligand and still deliver 95% product, which changes the economics of a synthesis.
Scientifically, it is a diagnostic. Observing a non-linear effect is evidence that the catalyst is not acting as isolated molecules — that aggregates or multi-ligand species are involved. Measuring the curvature gives you g, and hence information about a reactive intermediate you may never be able to isolate.
What this model leaves out
It assumes exactly two ligands, random binding, a single rate constant for each species, and that the meso complex gives strictly racemic product. Real systems break all four: ligand binding has its own equilibrium constant K, which makes β depend on concentration; aggregates larger than ML₂ occur; and there is a documented “hyperpositive” regime where a partially resolved ligand beats the pure one, which this model cannot produce at all.