A question on enantiomeric excess

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The discussion clarifies the concept of enantiomeric excess (ee) in a racemic mixture of 98:2 enantiomers, specifically focusing on the predominance of the (+) enantiomer. The calculation for enantiomeric excess is defined as ee = %(+) - %(-), resulting in an enantiomeric excess of 96% for the (+) enantiomer. The lesser (-) enantiomer does not contribute to excess but instead reduces the percentage of the predominant enantiomer, leading to a clearer understanding of how enantiomers interact in a mixture.

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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?
 
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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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