How do you calculate the enantiomeric excess in the following problem? How would you calculate the percentage of each enantiomer in the mixture?
The specific rotation of Compound X is +15.2 degrees. A mixture of Compound X and its enantiomer are placed in the polarimeter and the observed rotation is -5.1degrees. Which enantiomer is in excess? (+ or -)
The specific rotation of Compound X is +15.2 degrees. A mixture of Compound X and its enantiomer are placed in the polarimeter and the observed rotation is -5.1degrees. Which enantiomer is in excess? (+ or -)
1 Answer
The (-)-X is in excess. The mixture contains 67 % (-)-X and 33 % (+)-X.
Explanation:
The rotations of the two enantiomers cancel each other, so the rotation of the mixture will be that of the excess enantiomer.
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.
The formula for enantiomeric excess is
We can calculate the percent of each enantiomer as described in this Socratic question.
If we have a mixture of (+) and (-) isomers and (-) is in excess,
We have a 34 % enantiomeric excess of (-).
∴
So, the mixture contains 67 % (-)-X and 33 % (+)-X.