A pure sample of the R enantiomer of a compound has a specific rotation, [α], of +20°. A solution containing 0.2 g/mL of a mixture of enantiomers rotates plane polarized light by −2° in a 1 dm polarimeter. What is the enantiomeric excess (%ee) of the mixture?

Question

Quizplus · Accepted Answer

The observed rotation is -2°, indicating the S enantiomer is in excess. The specific rotation of the pure R enantiomer is +20°, so the specific rotation of the pure S enantiomer would be -20°. The enantiomeric excess (%ee) is calculated as (observed rotation/specific rotation of pure enantiomer) x 100 = (-2°/-20°) x 100 = 10%. Since the S enantiomer is in excess, the %ee of S is 10%, meaning the mixture is 55% S and 45% R, resulting in 50% S excess.

A Pure Sample of the R Enantiomer of a Compound