SolutionsStrategies in asymmetric synthesis
Exercise 1
An asymmetric reduction gives a 92:8 mixture of the R and S alcohols. What is its enantiomeric excess? What energy difference between the two transition states is needed to obtain 98% ee at 25 °C and at −78 °C?
ee = (92 − 8)/(92 + 8) × 100 = 84%. 98% ee corresponds to a 99:1 ratio; ΔΔG‡ = RT ln 99. At 298 K: 8.314 × 298 × 4.60 ≈ 11.4 kJ/mol; at 195 K: 8.314 × 195 × 4.60 ≈ 7.5 kJ/mol. At lower temperature a smaller difference is enough, which is why many asymmetric reactions are run cold.
Exercise 2
In the kinetic resolution of racemic 1-phenylethanol with CAL-B and vinyl acetate, what is the maximum yield of each enantiomer? What happens to the ee of the remaining alcohol if the reaction is allowed to go beyond 50% conversion? How could all of the racemate be obtained as the R acetate?
At most 50% of each: the R acetate and the S alcohol. Beyond 50% conversion, the enzyme begins to acetylate the S alcohol as well: the remaining alcohol becomes even purer (its ee keeps rising), but the acetate loses ee and the yield of alcohol falls. To convert everything, a catalyst is added that racemizes the alcohol without touching the acetate (for example, a ruthenium complex that dehydrogenates it to acetophenone and hydrogenates it back): this is a dynamic kinetic resolution.