A pure sample of the R enantiomer of a compound has a specific rotation, [a], of -15$\degree$. A solution containing 0.6 g/mL of a mixture of enantiomers rotates plane polarized light by -3$\degree$ in a 1 dm polarimeter. What is the enantiomeric excess (%ee) of the mixture?

Question

Quizplus · Accepted Answer

The specific rotation of the mixture is -3°, and the pure R enantiomer has a specific rotation of -15°. The enantiomeric excess (%ee) is calculated as (observed rotation / specific rotation of pure enantiomer) * 100 = (-3 / -15) * 100 = 20%. Since the observed rotation is negative, the R enantiomer is in excess. The %ee is 20%, meaning the mixture is 60% R and 40% S, resulting in 33% R excess.

A Pure Sample of the R Enantiomer of a Compound °\degree°

A Pure Sample of the R Enantiomer of a Compound $\degree$