Question

If a pure R isomer has a specific rotation of –151.0°, and a sample contains 72.0% of the R isomer and 28.0% of its enantiomer, what is the observed specific rotation of the mixture?

Answer 1

specific rotation of R = –151°

specific rotation of S = + 151°

percent R = 72 % = 0.72

percent S = 28% = 0.28

observed specific rotation of the mixture = 0.72 x (-151) + 0.28 x (+151)

                                                               = - 66.44 oC