Question

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 -)

Answer 1

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

`ee = "observed specific rotation"/"maximum specific rotation" × 100 % = ("-5.1" color(red)(cancel(color(black)(°))))/("-15.2" color(red)(cancel(color(black)(°)))) × 100 % = 34 %`

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,

`% ("-") = (ee)/2 +50 %`

We have a 34 % enantiomeric excess of (-).

∴ `% ("-") = (34 %)/2 +50 % = (17+50) % = 67 %`

So, the mixture contains 67 % (-)-X and 33 % (+)-X.