# Consider a solution that contains 80.0% R isomer and 20.0% S isomer. If the observed specific rotation of the mixture is –43.0°, what is the specific rotation of the pure R isomer?

Feb 1, 2016

#### Answer:

The specific rotation of the $R$ isomer is -67 °.

#### Explanation:

$R$ is in excess, so it must have a negative specific rotation; $S$ has an equal positive specific rotation.

The formula for the enantiomeric excess ($e e$) of $R$ is

ee = %R - %S = 80 % - 20 % = 60 %

Another equation for $e e$ is

ee = "observed specific rotation"/"maximum specific rotation" × 100 %

Inserting numbers, we get

60 color(red)(cancel(color(black)(%))) = "-40.3 °"/"maximum specific rotation" × 100 color(red)(cancel(color(black)(%)))

$\text{maximum specific rotation" = [α]_"D" = "-40.3 °" × 100/60 = "-67 °}$

This is the value of [α]_"D" for the $R$ isomer.