If a pure R isomer has a specific rotation of –142.0°, and a sample contains 77.0% of the R isomer and 23.0% of its enantiomer, what is the observed specific rotation of the mixture?

1 Answer
Ernest Z.
May 16, 2016

The observed specific rotation of the mixture is -76.7 °.

Explanation:

The rotations of the two enantiomers cancel each other, so the rotation of the mixture is that of the excess enantiomer.

The enantiomeric excess (#ee#) of the #R# isomer is

#ee = 77.0 % - 23.0 % = 54.0 %#

∴ The observed rotation is 54.0 % that of the #R# isomer.

#[α]_"obs" = 0.540 × ("-142.0 °") = "-76.7 °"#

Related questions
See all questions in Enantiomers
Impact of this question
9759 views around the world
You can reuse this answer
Creative Commons License