If a mixture contains 75% of one compound and 25% of its enantiomer, what is the e.e. of the mixture
1 answer:
Answer: The enantiomeric excess ee is 50%
Explanation:
Given that;
compound A = 75%, B = 25%
ee = ?
so
ee = [( A - B) / ( A + B)] × 100
= [( 75 - 25) / ( 75 + 25)] × 100
= [50 / 100] × 100
= 0.5 × 100
= 50%
Therefore enantiomeric excess ee is 50%
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