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

Question

Quizplus · Accepted Answer

First, we need to calculate the observed specific rotation ([]obs) of the mixture:

[]obs = Observed rotation / (concentration x length of cell)

= (–2) / (0.2 x 1)

= –10

Next, we can use the following equation to calculate the enantiomeric excess (%ee):

%ee = ([]obs / []max) x 100

where []max is the specific rotation of the pure enantiomer.

[]max = +20

%ee = (–10 / 20) x 100

= –50%

Since the observed rotation is negative, this indicates that there is an excess of the S enantiomer in the mixture. Therefore, the correct answer is C) 50% S.

A Pure Sample of the R Enantiomer of a Compound α\alphaα

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