Question

The optical rotation of a sample composed of only D-fructose is determined. [α]obs = -0.37°. [α]D is known to be -92.4° for D-fructose. If the cell path length was 0.6 dm and the concentration of D-fructose in the sample was 0.15 g/mL, what is the enantiomeric excess (ee)?

Answer 1

observed rotation = - 0.37^o

specific rotation of pure = - 92.4^o

specific rotation = observed rotaion / l x c

                          = - 0.37 / 0.6 x 0.15

                         = - 4.11 ^o

optical purity = observed specific rotation / specific rotation of pure

                     = - 4.11 / - 92.4

                     = 0.0445

enantiomeric excess %ee = optical purity x 100

                                        = 0.0445 x 100

enantiomeric excess %ee   = 4.45 %