Omeprazole Resolution (II)

From WikidChem

Omeprazole_resolution_(ii) Chiral Catalysis by Titanium/Diethyltartrate

When synthesizing the drug for commercial purposes, the most efficient use of resources would be to only make one of the two enatiomers, rather than to make both and throw away the unwanted one. A chiral catalyst can hopefully lead to synthesis of only one of the two enatiomers.

The molecule originally has a plane of symmetry vertically through the -OR, -Ti, -O, and -Ti, making it achiral. The Ligand -OR on the Ti gets exchanged for a different group. Then the R groups on the other two ligands undergo a double exchange with the diethyltartarate molecule, which is chiral. This now makes the molecule chiral.

This chiral compound loses the -OR group. Then peroxide (-OOR) attaches where the -OR left. Then the high HOMO of the lone pair on the -SRR' (basically the omeprazole) reacts with the low LUMO of the O-O sigma*. This double bonds the S to the O and removes the -OR from the -OOR group. (The O=SRR' group is the desired omeprazole, but is produced as a racemate (It was thought that a chiral catalyst would favor the formation of one enantiomer over another, but it didn't work - MJM). However, when this reaction is carried out in the presence of the compound iPr2NEt, the omeprazole is produced in 94% e.e., which stands for enantiomer excess. This means that the desired enantiomer (S) makes up 97% of the final product while the other enantiomer (R) makes up 3% of the final product. -SR) The -OR ligand starts the process again. (This is a circular process in which a peroxide is consumed to make each omeprazole. -SR)