the R-(+) enantiomer of a certain compound has a specific rotation of +52. A mixture of...
the R-(+) enantiomer of a certain compound has a specific rotation of +52. A mixture of this compound with its enantiomer has a has an observed rotation of -26. what is the %ee of this sample and which enantiomer is in excess? what is the ration of R:S in this mixture?
Homework Answers
Suppose the fraction of R enantiomer in the mixture is x. Then the fraction of S enantiomer is (1-x).
So rotation of mixture = -26° = 52x + (-52)(1-x)
Or -26 = 52x - 52 + 52x = 104x - 52
Or 104 x = 26
Hence x = 0.25
So % R enantiomer = 25
And % S enantiomer = 75, enantiomeric excess = 75-25 = 50.
- Ratio of R:S in the mixture = 25:75 = 1 : 3
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