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 °"

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