Question

An attempt at synthesizing a certain optically active compound resulted in a mixture of its enantiomers. The mixture had an observed specific rotation of 15.8°. If it is known that the specific rotation of the R enantiomer is –33.7°, determine the percentage of each isomer in the mixture.

Answer 1

let fraction of S enantiomer be x

then fraction of R enantiomer will be 1-x

specific rotation of S enantiomer = 33.7

specific rotation of R enantiomer = -33.7

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

15.8 = x * 33.7 + (1-x) * ( -33.7)

15.8 = x * 33.7 - 33.7 + x * 33.7

67.4*x = 49.5

x = 0.7344

so,

fraction of S enantiomer = 0.7344

fraction of R enantiomer = 0.2656

percentage of S enantiomer = 73.44 %

percentage of R enantiomer = 26.56 %

percentage of S enantiomer = 73.44 %

percentage of R enantiomer = 26.56 %