The specific rotation of -Carvone is +61 degrees, while the specific rotation of (R)-Carvone is -61 degrees. If a sample of mixture of the two enantiomers has a specific rotation of -40.58 degrees
Part I. What is the percent enantiomeric excess of the mixture?
a) +24.75
b) -24.75
c) 24.75
d) -66.52
e) 66.52
Part II. What is the percentage of each enantiomer in the above mixture?
a) of and of S
b) of and of
c) of and of
d) racemic mixture of and
e) of and of
The Correct Answer and Explanation is:
Part I: What is the percent enantiomeric excess (ee)?
We are given:
- Specific rotation of pure (S)-Carvone: +61°
- Specific rotation of pure (R)-Carvone: –61°
- Specific rotation of the mixture: –40.58°
The formula for enantiomeric excess (ee) is: ee=(observed rotationrotation of pure enantiomer)×100%ee = \left( \frac{\text{observed rotation}}{\text{rotation of pure enantiomer}} \right) \times 100\%
Since the observed rotation is negative, we take the value for (R)-Carvone as the pure enantiomer reference: ee=(−40.58−61)×100%=66.52%ee = \left( \frac{-40.58}{-61} \right) \times 100\% = 66.52\%
✅ Correct answer for Part I: e) 66.52
Part II: What is the percentage of each enantiomer?
Enantiomeric excess (ee) also means the excess percentage of one enantiomer over the racemic mixture. That is: ee=%major enantiomer−%minor enantiomer\text{ee} = \% \text{major enantiomer} – \% \text{minor enantiomer}
And since the total must be 100%: %major+%minor=100%\% \text{major} + \% \text{minor} = 100\%
Let’s solve this: %major−%minor=66.52%major+%minor=100\% \text{major} – \% \text{minor} = 66.52 \\ \% \text{major} + \% \text{minor} = 100
Add both equations: 2×%major=166.52⇒%major=83.26%%minor=100−83.26=16.74%2 \times \% \text{major} = 166.52 \Rightarrow \% \text{major} = 83.26\% \\ \% \text{minor} = 100 – 83.26 = 16.74\%
Since the mixture has a negative rotation, the (R)-Carvone is in excess.
✅ (R)-Carvone: 83.26%, (S)-Carvone: 16.74%
So the correct answer for Part II is:
e) 83.26% of (R) and 16.74% of (S)
✅ Final Answers:
- Part I: e) 66.52
- Part II: e) 83.26% of (R) and 16.74% of (S)
