The specific rotation of L-dopa in water (at 15°C) is -39.5. A chemist prepared a mixture of L-dopa and its enantiomer, and this mixture had a specific rotation of -23. Calculate the % ee of this mixture. %ee =
The specific rotation of L-dopa in water (at 15°C) is -39.5. A chemist prepared a mixture of L-dopa and its enantiomer, and this mixture had a specific rotation of -23. Calculate the % ee of this mixture. %ee =
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The specific rotation of L-dopa in water (at 15°C) is -39.5. A chemist prepared a mixture of L-dopa and its enantiomer, and this mixture had a specific rotation of -23. Calculate the % ee of this mixture.
% ee = [___________] %
**eTextbook and Media**
*Explanation:*
This task involves calculating the enantiomeric excess (% ee) of a mixture containing L-dopa and its enantiomer. The specific rotation values provided are:
- Specific rotation of pure L-dopa: -39.5
- Specific rotation of the mixture: -23
The enantiomeric excess (% ee) can be calculated using the formula:
\[ % ee = \left( \frac{\text{observed specific rotation}}{\text{specific rotation of pure enantiomer}} \right) \times 100 \]
In this case:
\[ % ee = \left( \frac{-23}{-39.5} \right) \times 100 \]
\[ % ee = 0.5823 \times 100 \]
\[ % ee = 58.23 \% \]
So, the enantiomeric excess (% ee) of this mixture is 58.23%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3cb00ede-af09-4713-a061-065b07b08661%2F7e18da20-dfe0-4f16-a357-5102de1cd164%2Fqe47zlb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Current Attempt in Progress**
The specific rotation of L-dopa in water (at 15°C) is -39.5. A chemist prepared a mixture of L-dopa and its enantiomer, and this mixture had a specific rotation of -23. Calculate the % ee of this mixture.
% ee = [___________] %
**eTextbook and Media**
*Explanation:*
This task involves calculating the enantiomeric excess (% ee) of a mixture containing L-dopa and its enantiomer. The specific rotation values provided are:
- Specific rotation of pure L-dopa: -39.5
- Specific rotation of the mixture: -23
The enantiomeric excess (% ee) can be calculated using the formula:
\[ % ee = \left( \frac{\text{observed specific rotation}}{\text{specific rotation of pure enantiomer}} \right) \times 100 \]
In this case:
\[ % ee = \left( \frac{-23}{-39.5} \right) \times 100 \]
\[ % ee = 0.5823 \times 100 \]
\[ % ee = 58.23 \% \]
So, the enantiomeric excess (% ee) of this mixture is 58.23%.
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