Question

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?

Answer 1

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 (`ee`) of `R` is

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

Another equation for `ee` 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)(%)))`

`"maximum specific rotation" = [α]_"D" = "-40.3 °" × 100/60 = "-67 °"`

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