Question

The specific rotation of Compound X is -30.5 degrees. A mixture of Compound X and its enantiomer are placed in the polarimeter and the observed rotation is +5.1 degrees. Which enantiomer is in excess? (+ or -) Calculate the enantiomeric excess. Show your work. Calculate the percentage of each enantiomer in the mixture.

Answer 1

Lets assume R enantiomer has positive specific rotation.

let fraction of R enantiomer be x

then fraction of S enantiomer will be 1-x

specific rotation of R enantiomer = 30.5

specific rotation of S enantiomer = -30.5

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

5.1 = x * 30.5 + (1-x) * ( -30.5)

5.1 = x * 30.5 - 30.5 + x * 30.5

61.0*x = 35.6

x = 0.5836

so,

fraction of R enantiomer = 0.5836

fraction of S enantiomer = 0.4164

percentage of R enantiomer = 58.36 %

percentage of S enantiomer = 41.64 %

%ee = 100 - 2*S

= 100 - 2* 41.64

= 16.72 %

Answer: 16.72 %