The (+) enantiomer of Aromatisse has a specific rotation of +21.5. What is the percent of the (-) enantiomer in a mixture, if the specific rotation of the mixture is -15.05?
The (+) enantiomer of Aromatisse has a specific rotation of +21.5. What is the percent of the (-) enantiomer in a mixture, if the specific rotation of the mixture is -15.05?
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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![**QUESTION 3**
The (+) enantiomer of Aromatisse has a specific rotation of +21.5. What is the percent of the (-) enantiomer in a mixture, if the specific rotation of the mixture is -15.05?
**Explanation:**
To solve this problem, one must understand the concept of specific rotation and the calculation of enantiomeric excess.
**Specific Rotation Calculation:**
Specific rotation is the measure of the degree to which a compound can rotate plane-polarized light. The formula to determine the specific rotation of a mixture of enantiomers is:
\[ \left[ \alpha \right]_{\text{mixture}} = f(+) \left[ \alpha \right]_{(+)} + f(-) \left[ \alpha \right]_{(-)} \]
where \(\left[ \alpha \right]_{\text{mixture}}\) is the specific rotation of the mixture, \(f(+)\) is the fraction of the (+) enantiomer, \(\left[ \alpha \right]_{(+)}\) is the specific rotation of the pure (+) enantiomer, \(f(-)\) is the fraction of the (-) enantiomer, and \(\left[ \alpha \right]_{(-)}\) is the specific rotation of the pure (-) enantiomer.
Since the specific rotation of the pure (-) enantiomer is equal in magnitude but opposite in sign to that of the pure (+) enantiomer,
\[ \left[ \alpha \right]_{(-)} = -\left[ \alpha \right]_{(+)} \]
Therefore,
\[ \left[ \alpha \right]_{\text{mixture}} = f(+) \left[ \alpha \right]_{(+)} - f(-) \left[ \alpha \right]_{(+)} \]
Given:
- \(\left[ \alpha \right]_{\text{mixture}} = -15.05\)
- \(\left[ \alpha \right]_{(+)} = +21.5\)
Applying the enantiomeric excess formula and considering \(f(+) + f(-) = 1\), the calculation can determine the fraction of the (-) enantiomer. Subsequently, convert the fraction to a percentage.
This specific problem could be solved in a thorough explanation to support student comprehension of the enantiomeric excess](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F89d4e4fb-1f7e-4614-824b-fd107892a18d%2F0cec669e-e133-4f6a-b2fd-63173293440c%2F4spb75m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**QUESTION 3**
The (+) enantiomer of Aromatisse has a specific rotation of +21.5. What is the percent of the (-) enantiomer in a mixture, if the specific rotation of the mixture is -15.05?
**Explanation:**
To solve this problem, one must understand the concept of specific rotation and the calculation of enantiomeric excess.
**Specific Rotation Calculation:**
Specific rotation is the measure of the degree to which a compound can rotate plane-polarized light. The formula to determine the specific rotation of a mixture of enantiomers is:
\[ \left[ \alpha \right]_{\text{mixture}} = f(+) \left[ \alpha \right]_{(+)} + f(-) \left[ \alpha \right]_{(-)} \]
where \(\left[ \alpha \right]_{\text{mixture}}\) is the specific rotation of the mixture, \(f(+)\) is the fraction of the (+) enantiomer, \(\left[ \alpha \right]_{(+)}\) is the specific rotation of the pure (+) enantiomer, \(f(-)\) is the fraction of the (-) enantiomer, and \(\left[ \alpha \right]_{(-)}\) is the specific rotation of the pure (-) enantiomer.
Since the specific rotation of the pure (-) enantiomer is equal in magnitude but opposite in sign to that of the pure (+) enantiomer,
\[ \left[ \alpha \right]_{(-)} = -\left[ \alpha \right]_{(+)} \]
Therefore,
\[ \left[ \alpha \right]_{\text{mixture}} = f(+) \left[ \alpha \right]_{(+)} - f(-) \left[ \alpha \right]_{(+)} \]
Given:
- \(\left[ \alpha \right]_{\text{mixture}} = -15.05\)
- \(\left[ \alpha \right]_{(+)} = +21.5\)
Applying the enantiomeric excess formula and considering \(f(+) + f(-) = 1\), the calculation can determine the fraction of the (-) enantiomer. Subsequently, convert the fraction to a percentage.
This specific problem could be solved in a thorough explanation to support student comprehension of the enantiomeric excess
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