Question

The specific rotation of L-alanine in water (at 25°C) is +2.8. A chemist prepared a mixture of L-alanine and its enantiomer, and 3.50 g of the mixture was dissolved in 10.0 mL of water. This solution was then placed in a sample cell with a pathlength of 10.0 cm and the observed rotation was +0.70. Calculate the % ee of the mixture.

Answer 1

mass of L - alanine = m

mass of D-alanine = 3.50 - m

excess mass of L - alanine = m - (3.50 -m)

concentration of L-alanine = 2m - 3.50 / 10

                                           = (0.2 m - 0.35 ) g/mL

observed rotation = specific rotation x l x C

         0.70 = 2.8 x 1 x (0.2 m - 0.35)

m = 3

% ee = (excess mass of L- alanine / total ) x 100

         = (2 x 3 - 3.5) / 3.50 ) x 100

% ee = 71.4 %