Question

A mixture of enantiomers is tested to be 45% ee (enantiomeric excess) with an observed rotation +37.
There is a larger amount of R enantiomers than S ones in the mixture. What is the observed rotation of
another mixture with 96% of S enantiomers? Give your answer with 1 decimal point.​

Answer 1

Answer:

The mixture has a negative sign of rotation, so (-)-X is in excess. To calculate the enantiomeric excess, you divide the observed specific rotation by the maximum specific rotation of the excess enantiomer