Question

If a solution of a pair of enantiomers has 0.60 g of S and 0.20 g of R, and the pure S enantiomer has a specific rotation of +25°, what will the optical rotation of the solution be?

Answer 1

Given, mixture contains 0.60 g of S and 0.20 g of R

First, we will calculate the enantiomeric excess/optical purity as a decimal number:

% for S 0.6 g/ (0.6+0.2)g = 0.6/0.8 = 0.75 for Major(S) isomer

% for R 0.2g/ (0.6+0.2)g = 0.2/0.8 = 0.25 for Minor (R) isomer.

ee favouring the S isomer,

e.e. = %S – %R = 75%-25% = 50% or 0.50 (In decimal)

enantiomeric excess (ee) or Optical Purity = optical rotation of mixture / optical rotation of Pure

0.50 = optical rotation of mixture / 25^o

Optical rotation of solution = 12.5^o