Question

Consider a solution that contains 74.0% R isomer and 26.0% S isomer. If the observed specific rotation of the mixture is –73.0°, what is the specific rotation of the pure R isomer?

Answer 1

percentage of S enantiomer = 26 %

percentage of R enantiomer = 74 %

let specific rotation of S enantiomer be x

then specific rotation of R enantiomer will be - x

fraction of S enantiomer = 0.26

fraction of R enantiomer = 0.74

net rotation = fraction of S enantiomer * specific rotation of S enantiomer + fraction of R enantiomer * specific rotation of R enantiomer

-73.0 = 0.26 * x + 0.74 * (-x)

-73.0 = -0.48 * x

x = 152.0833

So,

specific rotation of S enantiomer = 152

specific rotation of R enantiomer = -152

Answer: -152