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.

1 Answer
Ernest Z.
Jan 31, 2016

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 %#

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