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A question on enantiomeric excess

  1. Jan 13, 2004 #1
    Why does a racemic mixture of 98:2 enantiomers consist of one enantiomer in 96% over the other? I'm still not clear about this after reading textbooks. And how about the other one? does it have a negative percentage value of enantiomeric excess?
  2. jcsd
  3. Jan 14, 2004 #2
    ee = %(+) - %(-) assuming the (+) enantiomer is predominant.

    also, there is no excess of the (-) enantiomer because there is only an excess of the (+) enantiomer (see the assumption above).

    just think of it as if the lesser enantiomer shades out an equal percentage of the predominant enantiomer. so if you have 98%(+) and 2%(-), 2%(-) will shade out 2%(+) and you're left with 96%(+) pure enantiomer. i don't know if that makes sense, but i hope it helps :smile:.
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