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?

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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
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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