Question

The [α] of pure quinine, an antimalarial drug, is –165.

a. Calculate the ee of a solution with the following [α] values: –50, –83, and –120.

b. For each ee, calculate the percent of each enantiomer present.

c. What is [α] for the enantiomer of quinine?

d. If a solution contains 80% quinine and 20% of its enantiomer, what is the ee of the solution?

e. What is [α] for the solution described in part (d)?

Short answer

a. The value of ee for[α] values, –50, –83, and –120 are 30%, 50%, and 73%, respectively.

b. The values are mentioned below:

• For 30% ee, the amount of quinine and quinine’s enantiomer are 65% and 35%, respectively.
• For 50% ee, the amount of quinine and quinine’s enantiomer are 75% and 25%, respectively.
• For 73% ee, the amount of quinine and quinine’s enantiomer are 86.5% and 13.5%, respectively.

c. The value of [α] for enantiomer of quinine is +165.

d. The ee of the solution is 60%.

e. The value of [α] of the solution in part (d) is -99.

Step-by-step solution

Formula for ee calculation

The formula that is used to calculate ee is given as follows:

$\text{ee=}\frac{\left[\text{α}\right]\text{mixture}}{\left[\text{α}\right]\text{pureenantiomer}}\text{×100\%}$

Explanation for a, b, and c

a. The[α] values are given as –50, –83, and –120. The calculation of ee for each value is shown below:

Substitute the values in the formula.

$$\begin{gathered} \text{ee=}\frac{\text{-50}}{-165}\text{×100\%} \\ =30\% \end{gathered}$$

$$\begin{gathered} \text{ee=}\frac{\text{-83}}{-165}\text{×100\%} \\ =50.3\% \end{gathered}$$

$$\begin{gathered} \text{ee=}\frac{\text{-120}}{-165}\text{×100\%} \\ =72.7\% \end{gathered}$$

Thus, the value of ee for [α] values, –50, –83, and –120, are 30%, 50%, and 73%, respectively.

b. The % ee value represents the excess of one compound, and the remaining of it contains the mixture of 2 compounds equally.

For 30% ee, 30% represents the excess of quinine, and 70% remaining contains a mixture of 2 compounds; quinine and quinine’s enantiomer.

The amount of quinine and quinine’s enantiomer is calculated as follows:

localid="1648533965427" $$\begin{gathered} \text{Amountofquinine=30+35} \\ =65\% \\ \text{Amountofquinineenantiomer=35\%} \end{gathered}$$

For 50% ee, 50% represents the excess of quinine, and 50% remaining contains a mixture of 2 compounds; quinine and quinine’s enantiomer.

The amount of quinine and quinine’s enantiomer is calculated as follows:

localid="1648533974379" $$\begin{gathered} \text{Amountofquinine=50+35} \\ =75\% \\ \text{Amountofquinineenantiomer=25\%} \end{gathered}$$

For 73% ee, 50% represents the excess of quinine, and 27% remaining contains a mixture of 2 compounds; quinine and quinine’s enantiomer.

The amount of quinine and quinine’s enantiomer is calculated as follows:

localid="1648533984709" $$\begin{gathered} \text{Amountofquinine=73+35} \\ =86.5\% \\ \text{Amountofquinineenantiomer=13.5\%} \end{gathered}$$

c. The value of [α] for a pure enantiomer is given as -165. Thus, the enantiomer of quinine is +165.

Explanation for d and e

d. The ee values are given as 80% quinine and 20% of its enantiomer. The ee of the solution is calculated as follows:

$$\begin{gathered} \text{Amountofsolution=80\%-20\%} \\ =60\% \end{gathered}$$

Thus, the amount of solution is 60%.

e. The ee for the solution in part (d) is calculated as follows:

$$\begin{gathered} \text{60\%=}\frac{\left[\text{α}\right]\text{mixture}}{-165}\text{×100\%} \\ \left[\text{α}\right]\text{mixture=}\frac{\mathrm{60\%(-165)}}{100\%} \\ \text{=-99} \end{gathered}$$

Thus, the value of [α] of the solution in part (d) is -99.