Question

Consider a solution that contains 81.0% R isomer and 19.0% S isomer. If the observed specific rotation of the mixture is –49.0°, what is the specific rotation of the pure R isomer?

Answer 1

Let x = specific rotation of R

   specific rotation S = -x

81x-19x/100    = -49

81x-19x            = -4900

62x                  = -4900

x                     = -79

specific roration of R = x = -79⁰