Part 3
Soai: Breaking the Symmetry
Kagan's amplification still needs a handed catalyst to start with. This part is about a stronger claim: that a reaction can begin with no meaningful bias at all and end with one hand. The argument is forty lines of chemistry and four lines of calculus, and it predates the experiment by forty-two years.
Derivation: Frank's model, 1953
Step 1 — Three reactions
Charles Frank asked what the minimum ingredients for spontaneous resolution are. He found two: a product that catalyses its own formation, and some process by which the two hands destroy or neutralise one another.
\[ A + R ;\xrightarrow{;k;}; 2R, \qquad A + S ;\xrightarrow{;k;}; 2S, \qquad R + S ;\xrightarrow{;k';}; \text{inert} \]
Note the symmetry is perfect: both autocatalytic steps have the same rate constant, and the antagonism treats the two hands identically. Nothing here prefers R. That is essential — if we smuggled in a bias the result would be worthless.
Step 2 — Write the rate equations
\[ \frac{dR}{dt} = kAR - k'RS, \qquad \frac{dS}{dt} = kAS - k'RS \]
Each hand grows in proportion to how much of it already exists — that is the autocatalysis — and both are consumed together by the cross term.
Step 3 — Change to the symmetry-adapted variables
Part 1 argued that ee is the variable odd under R ↔ S. Take it and the total: c = R+S and ee = (R−S)/c. Differentiating the quotient and substituting:
\[ \frac{d(R-S)}{dt} = kA(R-S), \qquad \frac{dc}{dt} = kAc - 2k'RS \]
The autocatalytic term cancels out of the difference equation entirely. Then
\[ \frac{d(ee)}{dt} = \frac{(R-S)'c - (R-S)c'}{c^{2}} = \frac{2k'RS\,(R-S)}{c^{2}} \]
and using RS = (c² − (R−S)²)/4:
\[ \boxed{;\frac{d(ee)}{dt} = \frac{k'c}{2}\,ee\,\bigl(1 - ee^{2}\bigr);} \]
A logistic equation in the enantiomeric excess. Note it contains only odd powers of ee, exactly as the symmetry argument in Part 1 demanded — a useful check that no bias crept in.
Step 4 — Read off the fixed points
The right-hand side vanishes at ee = 0 and ee = ±1. Near the racemic state, linearising for small ee:
\[ \frac{d(ee)}{dt} \approx \frac{k'c}{2}\,ee \quad\Longrightarrow\quad ee(t) \sim ee_0\,e^{\,k'ct/2} \]
The coefficient is positive. The racemic state is therefore unstable: any deviation grows exponentially. The two pure states are stable. A racemic mixture is not a resting place but a pencil balanced on its point.
What decides which hand wins
Nothing in the chemistry. The equation is odd in ee, so +1 and −1 are equally available; the outcome is fixed entirely by the sign of ee0, the imbalance present when the autocatalysis takes hold.
In a flask that imbalance is a statistical fluctuation. If N molecules form essentially at random, the excess is of order √N, so ee0 ∼ 1/√N — around 10⁻⁹ for a micromole. Tiny, and entirely sufficient. Run the experiment again and you get the other hand about half the time.
The Soai reaction
Frank's model sat as theory for four decades because no one could find a reaction that did it. In 1995 Soai found one: the addition of diisopropylzinc to a pyrimidyl aldehyde, where the alcohol produced is itself the catalyst, and a handed one.
It behaves as the model says it should. Seeded with product of 0.00005% ee— far below what most instruments can measure — successive rounds drive the mixture to above 99%. Run with no deliberate seed at all, it still ends up nearly pure, and the hand it chooses varies from run to run.
It remains, three decades on, essentially the only chemical reaction known to do this cleanly — which is why the prize names the discovery rather than a general method.
Check your understanding
- What happens to Frank's model if you delete the mutual antagonism, k′ = 0?
Answer: d(ee)/dt = 0. Both hands grow exponentially at the same rate and the initial ee is preserved forever. Autocatalysis alone amplifies the amount, not the purity — the antagonism is what does the selecting. - Why can the equation contain no term in ee²?
Answer: it would break the R ↔ S symmetry built into the three reactions. Its presence would mean an algebraic error. - Starting from ee0 = 10⁻⁹, roughly how long until ee ≈ 1, in units of 2/k′c?
Answer: growth is exponential, so the time is about ln(1/ee₀) = ln(10⁹) ≈ 21 time constants. Nine orders of magnitude cost only a factor of about twenty in time.
What this model leaves out
Frank's scheme is a caricature: it holds A constant, uses one rate constant per step, and treats the antagonism as a simple bimolecular sink. The real Soai mechanism involves zinc alkoxide dimers and is still debated in detail. The model also says nothing about why biology settled on the hands it did — it shows only that a symmetric world need not stay symmetric, which is a weaker and much more defensible claim.