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Mn Catalyst for Asymmetric Epoxidation

  1. Nov 30, 2004 #1
    For lab we used an Mn Catalyst for epoxidation of trans-B-methyl styrene. One of the problems is that we have to figure out which enantiomer of the trans-B-methylstyrene epoxide is formed by looking at the H NMR. I have never looked at an NMR before to try to figure out which enantiomers were present. Does anyone know what I should look for in order to determine which enantiomers are present or at least a website that explains how to look at NMR for different enantiomers? Also, what is enantiomer excess (ee) and how can it be calculated from the NMR? Is it by integration?
     
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  3. Nov 30, 2004 #2

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    Enantiomeric excess is the excess of one enantiomer over the other. For example, say you had a mixture of enantiomers, 75% R and 25% S. The enantiomeric excess is the amount of chiral material beyond the amount of racemic material. In this case the ee would be 50%, meaning that 50% of the material is a racemic mixture and the other 50% is R only.

    The most common way to do this by NMR is to add a "chiral shift reagent" to the NMR sample. These reagents are usually lanthanide metals with 2 or 3 chiral ligands. When your product complexes to the lanthanide, you will have diastereomeric complexes arising from the two enantiomers. Diasteremers often have significantly different NMR shifts. You can calculate the ee from the integration of a particular "diagnostic" NMR signal.

    I assume that you used the Jacobsen-Mn system. In the original Jacobsen paper they determined ee's by using the chiral shift reagent Eu(hfc)3.

    Reference: J. Am. Chem. Soc. 1991, 113, 7063-7064
     
  4. Nov 30, 2004 #3
    AHHHH thank you very much, you hit the nail right on the head.
     
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