Question

If a pure R isomer has a specific rotation of –139.0°, and a sample contains 83.0% of the R isomer and 17.0% of its enantiomer, what is the observed specific rotation of the mixture?

Answer 1

racemicmixture = equal% of R,S

                              = 17 + 17 = 34%

so that , enantiomeric excess = 100 - 34 = 66%

   enantiomeric excess = observed specific rotation / pure R-form specific rotation *100

               66 = x / -139 *100

   x = observed specific rotation = -91.74 °