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?

1 Answer
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 (#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.