A pure sample of the R enantiomer of a compound has a specific rotation, [\9\alpha\)], of -5^$\circ$. A solution containing 0.6 g/mL of a mixture of enantiomers rotates plane polarized light by -3^$\circ$in a 1 dm polarimeter. What is the enantiomeric excess (%ee) of the mixture?

Question

Quizplus · Accepted Answer

The specific rotation of the pure R enantiomer is -5°. The observed rotation is -3°, indicating that the R enantiomer is in excess. The enantiomeric excess (%ee) is calculated as (observed rotation / specific rotation of pure enantiomer) * 100 = (-3 / -5) * 100 = 60%. Since the mixture is 60% R and 40% S, the %ee is 60% - 40% = 20% R. However, the question asks for the percentage of the R enantiomer in the mixture, which is 50% + 20%/2 = 60% R, so the closest answer is 33% R.

A Pure Sample of the R Enantiomer of a Compound ∘\circ∘

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