Exercises
Six Exercises, with Worked Solutions
Hints inline; full solutions below, so cover them if you want the problem first.
The problems
1. Reading an ee
A sample is reported as 90% ee. What is the ratio of the two enantiomers? What ratio would 99% ee require?
Hint: Invert ee = (R−S)/(R+S).
2. Amplification
A chiral ligand is 60% ee and the heterochiral complex is completely unreactive. Using the ML₂ result of Part 2 with random pairing, what is the product ee?
Hint: Set g = 0 and use ee_prod = (r²−s²)/(r²+s²).
3. Reading g off an experiment
You run a reaction with a 60% ee ligand and measure 80% ee in the product. What does that tell you about the relative reactivity of the heterochiral complex?
Hint: Solve the ML₂ expression for g. Expect a value between 0 and 1.
4. The limits of the model
Show that with random pairing and g = 0 the ML₂ model can never give ee_prod = 1 for any ee_aux < 1. What would have to change for a partially resolved ligand to outperform a pure one?
Hint: Look at what happens to the denominator as ee_aux → 1.
5. How small is small enough
If N molecules form with no bias, the excess is of order √N, so ee₀ ~ 1/√N. Estimate ee₀ for a micromole. Then use the linearised Frank equation to estimate how many time constants it takes to reach ee ≈ 1.
Hint: Growth is exponential, so the time goes as ln(1/ee₀).
6. Discussion
Does the Soai reaction explain the homochirality of life? State what it does establish, what it does not, and what evidence would be needed to close the gap.
Hint: Separate the amplification mechanism from the original bias. No single right answer.
Worked solutions
1. Reading an ee
From ee = (R−S)/(R+S), R/S = (1+ee)/(1−ee). At 90% that is 19 : 1, or 95 : 5. At 99% it is 199 : 1, or 99.5 : 0.5. Note how punishing the last few per cent are — tenfold less of the unwanted hand for nine percentage points of ee, which is exactly why ligand purification is expensive and why Part 2 matters commercially.
2. Amplification
With x = 0.6: r = 0.8, s = 0.2, so r² = 0.64, s² = 0.04.
\[ ee_{\text{prod}} = \frac{0.64-0.04}{0.64+0.04} = \frac{0.60}{0.68} = 0.882 \]
88.2% product from a 60% ligand. The numerator is just eeaux as the derivation showed; all the gain comes from the denominator being less than one, because the 2rs = 0.32 of the catalyst that is locked in meso complexes has been removed from the reaction.
3. Reading g off an experiment
Solve 0.80 = 0.60/(0.68 + 2g(0.16)) for g. The denominator must be 0.75, so 0.32g = 0.07 and g ≈ 0.22.
The heterochiral complex reacts at about a fifth the rate of the homochiral ones — sluggish but not dead. This is the diagnostic use described in the medicine page: a number characterising an intermediate you never isolated, obtained by weighing out ligand in two different ratios.
4. The limits of the model
With g = 0 the expression is (r²−s²)/(r²+s²). For this to equal 1 you need s² = 0, that is s = 0, that is eeaux = 1. For any impure ligand some SS complex survives and makes the wrong product, so eeprod < 1 strictly.
For a partially resolved ligand to beat the pure one — the “hyperpositive” case — the meso complex would have to be not merely slow but more selective than the homochiral one, or a different species would have to dominate. Neither is possible within this model, which is a fair statement of its boundary.
5. How small is small enough
A micromole is about 6 × 10¹⁷ molecules, so ee0 ∼ 1/√N ≈ 1.3 × 10⁻⁹.
Linearising Frank's equation gives exponential growth, so the time to reach order unity is
\[ t \approx \frac{2}{k'c}\ln\!\left(\frac{1}{ee_0}\right) \approx 20\ \text{time constants} \]
Nine orders of magnitude cost only a factor of about twenty in time. That is the real force of the argument: the starting imbalance barely matters, because the logarithm crushes it.
6. Discussion — guidance, not an answer
What it establishes: that a mechanism exists, in ordinary solution chemistry, by which a symmetric system reaches an asymmetric state from a fluctuation. Before 1995 that was a hypothesis on paper. What it does not: that this mechanism operated on the early Earth, that the relevant molecules were prebiotic, or that the bias was a fluctuation rather than circularly polarised light, mineral surfaces or parity violation. Note the awkward feature that makes the question hard: because amplification works from almost any bias, the end state preserves almost no information about which bias it was. Closing the gap would need a prebiotically plausible autocatalyst, not a better amplifier.