Question

The specific rotation of a pure sample of the S form of compound a is +15. A a mixture of the R and S form of "A" shows a specific rotation of +3. Provide the % enantiomeric excess and amount for each isomer in % in this mixture.

Answer 1

Specific rotation of the S form of the compound A = +15

Let the specific rotation of R form of A = - R

Given +15 + (- R) = +3

=> R = 12

Hence specific rotation of R form of A = - R = - 12

Therefore, % of S form of A = (15 /15+13) x100 = 53.57 %

% of R form of A = (13 /15+13) x100 = 46.43 %

Enantiomeric excess is a measure of the optical purity of a chiral organic compound which reflects the degree to which one enantiomer is present in excess in comparison to the other. Here S enantiomer is in excess and can be calculated through the following formulae

enantiomeric excess(ee)

ee = [(% of major enantiomer, S - % of minor enantiomer, R) / % of total enantiomer(R+S)] x 100

= [(53.57 - 46.43) / (53.57 + 46.43)]x100 = 7.14 % (answer)