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?

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Answer 1 · 2016-03-05T19:51:50

The specific rotation of the pure $R$ isomer is -272°.

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°.