Question

Consider a solution that contains 66% R isomer and 34% S isomer. If the observed specific rotation of the mixture is -87.039; what is the specific rotation of the pure R isomer?

Answer 1

The specific rotation of the pure `R` isomer is -272°.

Explanation:

The `R` isomer is (-), because it is in excess and the rotation of the mixture is negative.

Its enantiomeric excess (`ee`) is

`% ee = "66 % - 34 % = 32 %"`

Also,

`"% ee" = [α]_text(mixture)/[α]_text(pure sample) × 100 %`

`32 color(red)(cancel(color(black)(%))) = "-87.039°"/[α]_text(pure sample) × 100 color(red)(cancel(color(black)(%))`

`[α]_text(pure sample) = "-87.039° × 100"/32 = "-272°"`

The optical rotation of the pure `R` isomer is -272°.