Question

The optical rotation of the S enantiomer of compound M is +40 deg x g^-1 x mL x dm^-1. what percentage of a mixture of the R and S enantiomers is composed of the S enantiomer if a sample with a concentration of 1 g/mL has an observed optical rotation of 10 degrees counterclockwise in a polarimetry cell that is 10cm

Answer 1

The specific rotation of sample = observed rotation / path length x concentration

given observed rotation = -10 deg

concentration = 1g/mL

cell length = 10cm = 1 dm

Thus specific rotation = -10/1x1 = -10

Now the optical purity is calculated as

optical purity = observed specific rotation x100 /specific rotation of pure sample

= -10 x100/-40

= 25%

Thus the mixture of 25% of R- enantiomer and the remaining 75% is racemised. That is 75/2 is S and 75/2 is R- isomer.

Hence the total R- isomer in the mixture is (37.5+25) =62.5%

and the S- enantiomer in the mixture is =75/2 = 37.5 %