Question

An attempt at synthesizing a certain optically active compound resulted in a mixture of its enantiomers. The mixture had an observed specific rotation of 16.3°. If it is known that the specific rotation of the R enantiomer is –31.9°, determine the percentage of each isomer in the mixture.

Answer 1

observed rotation of mixture = 16.3

specific rotation of pure R - enantiomer = -31.9

optical purity = observed roationx100/ specific rotation of pure sample

= + 16.3 x100 / +31.9

= 51.09 %

Thus 51.09 % is pure S - isomer and the remaining 48.9 % is racemised.

In the racemic mixture ( which is 1:1 mixture of both R and S enantiomers) has half R and half S - isomers.

Thus the total S- isomer in the mixture = 51.09 + 48.9/2

= 75.54 %

and th e % of R - isomer in the sample = 24.46 %