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arlik [135]
3 years ago
5

The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &S carvone is measured a

s (-) 23°. Which enantiomer is in excess and what is its enantiomeric excess? What are the percentages of (R) - & (S) - carvone in the sample​
Chemistry
2 answers:
jasenka [17]3 years ago
8 0

Answer:

Explanation:

ee= -23/-61 × 100= 38%

S + R = 100%

S - R = 38%

Solve simultaneously;

S= 69% ( excess)

R= 100-69= 31%

shusha [124]3 years ago
7 0

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%

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Answer:

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Explanation:

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