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?

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Answer 1 · 2016-02-01T01:41:20

The specific rotation of the $R$ isomer is -67 °.

$R$ 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.