Question

If the enantiomeric excess of a mixture is 81 %, what are the percent compositions of the major and minor enantiomer?

Answer 1

major enantiomer = 90.5 %

minor enantiomer = 9.5 %

Step-by-step explanation:

Assuming that the major isomer is in excess;

From, ee/2 + 50

ee = enantiomeric excess = 81%

% major =81/2 + 50 = 90.5 %

% minor = 100% - 90.5 %

% minor = 9.5 %