Question

Given the following information what is the enantiomeric excess (%ee) of the mixture?

A pure sample of the R enantiomer of a compound has a specific rotation of +20 degrees. A solution containing 0.2 g/ml of a mixture of enantiomers rotates plane polarized light by -2 degrees in a 1 dm polarimeter.

Answer 1

The enantiomeric excess of `S` is 50 %.

Explanation:

If `R` has a specific rotation of +20°, `S` has a specific rotation of -20°

The negative sign of rotation for the mixture tells us that the `S` isomer is in excess.

The formula for observed optical rotation is

`α_"obs" = [α]_Dcl`

where `c` is the concentration in grams per millilitre, `l` is the length of the tube in decimetres, and `[α_"D"]` is the specific rotation.

For the solution, the specific rotation `[α]_"D"` =

`(α_"obs")/(cl) = ( -2 °)/(0.2×1) = -10 °`

The enantiomeric excess of `S` is

`ee = "observed specific rotation"/"maximum specific rotation" × 100 % = ("-10" color(red)(cancel(color(black)(°))))/("-20" color(red)(cancel(color(black)(°)))) × 100 % = 50 %`