Asymmetric Epoxidation and Kinetic Resolution via the Sharpless Asymmetric Epoxidation

The mnemonic devise on the right predicts the absolute stereochemistry of the epoxide obtained by Sharpless Asymmetric Epoxidation (SAE) on an achiral allylic alcohol. The enantiomer obtained is controlled by the enantiomer of the tartrate ester employed. The (S,S)-tartrate reacts faster with the Si face of the allylic alcohol while the (R,R)-enantiomer reacts faster with the Re face of the allylic alcohol.

This procedure differs markedly from achiral, peracid epoxidation. Both transition states are enantiomeric. A racemic epoxide would ensue. The enantiopure tartrates provide diasteromeric transition states.
Imagine having a chiral, racemic, secondary allylic alcohol shown inside the plane on the right. Assuming that R4 is some simple group such as methyl, ethyl or isopropyl, the (S)-enantiomer is on the left; the (R)-enantiomer on the right. The Re-face of the (S)-enantiomer is less hindered (an H below the plane) while the (R)-enantiomer is more hindered below the plane. Thus, if (R,R)-tartrate is employed, the (S)-enantiomer will epoxide faster than the (R)-enantiomer. To avoid complete oxidation of all the allylic alcohol, approximately 0.55 equivalents of hydroperoxide is employed. By this technique the (S)-enantiomer will be consumed leaving the (R)-enantiomer unreacted. The (S)-derived epoxide and the (R)-allylic alcohol can be separated from one another. What if the (S)-allylic alcohol or the (R)-derived epoxide were required? Simply employ the (S,S)-tartrate!