what is the %ee of a mixture of enantiomer that contains 27% of the “R”enantiomer?

Question

Question

science chemistry

Answer 1

Answer #1

Since R isomer is 27%, it means S isomer is 63%

Now, since we see S isomer is in excess

Therefore by using the formula %(S) = (ee/2)+50

Which gives, ee= (63-50)x2 = 26%

Hence the percentage excess of the enantiomer is 26 %

Answer 2

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