Question

The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &S carvone is measured as (-) 23°. Which enantiomer is in excess and what is its enantiomeric excess? What are the percentages of (R) - & (S) - carvone in the sample​

Answer 1

Answer:

Explanation:

ee= -23/-61 × 100= 38%

S + R = 100%

S - R = 38%

Solve simultaneously;

S= 69% ( excess)

R= 100-69= 31%

Answer 2

Answer:

See explanation

Explanation:

% optical purity = specific rotation of mixture/specific rotation of pure enantiomer  * 100/1

specific rotation of mixture = 23°

specific rotation of pure enantiomer = 61°

Hence;

% optical purity = 23/61 * 100 = 38 %

More abundant enantiomer = 100% - 38 % = 62%

Hence the pure  (S) carvone is (-) 62° is the more abundant enantiomer.

Enantiomeric excess = 62 - 50/50 * 100 = 24%

Hence

(R) - carvone  =  38 %

(S) - carvone = 62%