(c) (S)-Amlodipine is used to treat high blood pressure and angina. The specific rotation of clinical grade $(S)$-amlodipine is $-25.1^{\circ}$.
$(S)$-amlodipine
The experimentally determined specific rotation of a batch prepared in the development laboratory using an alternative synthesis route is $-19.3^{\circ}$.
(i) Determine the optical purity and enantiomeric excess for this sample.
(ii) Use these values to calculate the percentage of the (+) and (-) enantiomer present in the laboratory sample. Show clearly how you reached your answers.

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate the observed specific rotation of the sample**: The observed specific rotation of the laboratory sample of (S)-amlodipine is given as $-19.3^\circ$. ‖ ‖ 2 ⋮ **Calculate the specific rotation of the pure enantiomer**: The specific rotation of the pure (S)-amlodipine is given as $-25.1^\circ$. ‖ ‖ 3 ⋮ **Determine the optical purity**: Optical purity (also known as enantiomeric excess, ee) is calculated using the formula $ ext{ee} = \frac{ ext{observed specific rotation}}{ ext{specific rotation of pure enantiomer}} imes 100\%$. Substituting the given values, $ ext{ee} = \frac{-19.3^\circ}{-25.1^\circ} imes 100\%$. ‖ ‖ 4 ⋮ **Calculate the enantiomeric excess**: Performing the calculation from step 3, $ ext{ee} = \frac{-19.3^\circ}{-25.1^\circ} imes 100\% \approx 76.9\%$. ‖ ‖ 5 ⋮ **Calculate the percentage of each enantiomer**: The enantiomeric excess gives the excess of one enantiomer over the other. Since the sample is partially optically pure, we assume a 50% baseline for a racemic mixture. The percentage of the (S)-enantiomer is $50\% + \frac{ ext{ee}}{2}$ and the percentage of the (R)-enantiomer is $50\% - \frac{ ext{ee}}{2}$. ‖ ‖ 6 ⋮ **Determine the percentage of the (S)-enantiomer**: Using the ee value from step 4, the percentage of the (S)-enantiomer is $50\% + \frac{76.9\%}{2} \approx 88.45\%$. ‖ ‖ 7 ⋮ **Determine the percentage of the (R)-enantiomer**: Similarly, the percentage of the (R)-enantiomer is $50\% - \frac{76.9\%}{2} \approx 11.55\%$. ‖ ∻1 Answer∻ ⚹ Optical purity (ee) is approximately 76.9%. The laboratory sample contains approximately 88.45% of the (S)-enantiomer and 11.55% of the (R)-enantiomer. ⚹ ∻Key Concept∻ ⚹ Optical purity and enantiomeric excess are measures of the composition of a chiral mixture in terms of its enantiomers. ⚹ ∻Explanation∻ ⚹ The optical purity is calculated by comparing the observed specific rotation of a sample to that of the pure enantiomer. The enantiomeric excess is used to determine the relative percentages of each enantiomer in the sample. ⚹◊What is the formula to calculate enantiomeric excess in a sample of (S)-amlodipine with a specific rotation of -19.3°?⍭ Generate me a similar question◊

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