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?

1 Answer
Mar 5, 2016

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°.