Question

What is the enantiomeric excess of a mixture whose(+) enantiomer has a specific rotation of 10 degrees and the mixture of (+) and (-)has a specific rotation of 5 degrees?

Answer 1

Answer – We are given (+) enantiomer specific rotation = 10^o

the mixture of (+) and (-) specific rotation = 5^o

we know the formula for calculating the enantiomeric excess as follow –

enantiomeric excess = observed specific rotation / maximum specific rotation x 100 %

                                  = 5^o / 10 ^o x 100 %

                                  = 50 %

So, the enantiomeric excess of a mixture whose(+) enantiomer has a specific rotation of 10 degrees is 50 %.