Chapter 5, Problem 5.65P

Interpretation Introduction

(a)

Interpretation: Enantiomeric excess of the given solution is to be calculated.

Concept introduction: Enantiomeric excess is the excess of major enantiomer over the minor enantiomer. Enantiomeric excess is equivalent to the optical purity. Enantiomeric excess is the fraction of observed rotation of the mixture to specific rotation of the pure enantiomer.

Answer to Problem 5.65P

The enantiomeric excess for the solution with $\left[\alpha \right]$ value $-50$ is $30\%$, with $\left[\alpha \right]$ value $-83$ is $50\%$ and with $\left[\alpha \right]$ value $-120$ is $73\%$.

Explanation of Solution

The specific rotation $\left[\alpha \right]$ of pure quinine is $-165$.

$\left[\alpha \right]$ value for the solution is $-50$.

The enantiomeric excess is calculated by the formula,

$\text{ee}=\frac{\left[\text{α}\right]\text{mixture}}{\left[\text{α}\right]\text{pure}\text{enantiomer}}\times 100\%$

Substitute the values of specific rotation $\left[\alpha \right]$ of pure quinine and $\left[\alpha \right]$ value for the solution.

$\begin{array}{rll}\text{ee}& =& \frac{-50}{-165}\times 100\\ & =& 30\%\end{array}$

The enantiomeric excess for the solution is $30\%$.

$\left[\alpha \right]$ value for the solution is $-83$.

The enantiomeric excess is calculated by the formula,

$\text{ee}=\frac{\left[\text{α}\right]\text{mixture}}{\left[\text{α}\right]\text{pure}\text{enantiomer}}\times 100\%$

Substitute the values of specific rotation $\left[\alpha \right]$ of pure quinine and $\left[\alpha \right]$ value for the solution.

$\begin{array}{rll}\text{ee}& =& \frac{-83}{-165}\times 100\\ & =& 50\%\end{array}$

The enantiomeric excess for the solution is $50\%$.

$\left[\alpha \right]$ value for the solution is $-120$.

The enantiomeric excess is calculated by the formula,

$\text{ee}=\frac{\left[\text{α}\right]\text{mixture}}{\left[\text{α}\right]\text{pure}\text{enantiomer}}\times 100\%$

Substitute the values of specific rotation $\left[\alpha \right]$ of pure quinine and $\left[\alpha \right]$ value for the solution.

$\begin{array}{rll}\text{ee}& =& \frac{-120}{-165}\times 100\\ & =& 73\%\end{array}$

The enantiomeric excess for the solution is $73\%$.

Conclusion

The enantiomeric excess for the solution with $\left[\alpha \right]$ value $-50$ is $30\%$, with $\left[\alpha \right]$ value $-83$ is $50\%$ and with $\left[\alpha \right]$ value $-120$ is $73\%$.

Interpretation Introduction

(b)

Interpretation: The percent of each enantiomer in the given solution is to be calculated.

Concept introduction: Enantiomeric excess is the excess of major enantiomer over the minor enantiomer. Enantiomeric excess is equivalent to the optical purity. Enantiomeric excess is the fraction of observed rotation of the mixture to specific rotation of the pure enantiomer.

Answer to Problem 5.65P

For the solution with enantiomeric excess $30\%$ the amount of A and B is $65\%$ and $35\%$, for the solution with enantiomeric excess $50\%$ the amount of A and B is $75\%$ and $25\%$ and $35\%$ and for the solution with enantiomeric excess $73\%$ amount of A and B is $86.5\%$ and $13.5\%$.

Explanation of Solution

For the solution with enantiomeric excess 30%.

Let A be the major enantiomer and B be the minor enantiomer. The major enantiomer is $30\%$ more than the minor enantiomer. Total racemic mixture is $70\%$. Racemic mixture has equal amount of enantiomers. Therefore, the amount of enantiomers A and B in the racemic mixture is $35\%$.

Total amount of A is $30\%+35\%=65\%$.

Hence, the amount of A and B is $65\%$ and $35\%$.

For the solution with enantiomeric excess 50%.

Let A be the major enantiomer and B be the minor enantiomer. The major enantiomer is $50\%$ more than the minor enantiomer. Total racemic mixture is $50\%$. Racemic mixture has equal amount of enantiomers. Therefore, the amount of enantiomers A and B in the racemic mixture is $25\%$.

Total amount of A is $50\%+25\%=75\%$.

Hence, the amount of A and B is $75\%$ and $25\%$.

For the solution with enantiomeric excess 73%.

Let A be the major enantiomer and B be the minor enantiomer. The major enantiomer is $73\%$ more than the minor enantiomer. Total racemic mixture is $27\%$. Racemic mixture has equal amount of enantiomers. Therefore, the amount of enantiomers A and B in the racemic mixture is $13.5\%$.

Total amount of A is $73\%+13.5\%=86.5\%$.

Hence, the amount of A and B is $86.5\%$ and $13.5\%$.

Conclusion

For the solution with enantiomeric excess $30\%$ the amount of A and B is $65\%$ and $35\%$, for the solution with enantiomeric excess $50\%$ the amount of A and B is $75\%$ and $25\%$ and $35\%$ and for the solution with enantiomeric excess $73\%$ amount of A and B is $86.5\%$ and $13.5\%$.

Interpretation Introduction

(c)

Interpretation: The specific rotation $\left[\alpha \right]$ for the enantiomer of quinine is to be predicted.

Concept introduction: Enantiomers are stereoisomers, which are non-superimposable images of each other. They have identical physical and chemical properties in symmetric environment. They rotate the plane-polarized light in equal amounts and in opposite directions.

Answer to Problem 5.65P

The specific rotation $\left[\alpha \right]$ for the enantiomer of quinine is $+165$.

Explanation of Solution

The specific rotation $\left[\alpha \right]$ of pure quinine is $-165$. Hence, the specific rotation $\left[\alpha \right]$ of its enantiomer is $+165$.

Conclusion

Enantiomeric excess of the given solution is $60\%$.

Interpretation Introduction

(d)

Interpretation: Enantiomeric excess of the given solution is to be calculated.

Concept introduction: Enantiomeric excess is the excess of major enantiomer over the minor enantiomer. Enantiomeric excess is equivalent to the optical purity. Enantiomeric excess is the fraction of observed rotation of the mixture to specific rotation of the pure enantiomer.

Answer to Problem 5.65P

Enantiomeric excess of the given solution is $60\%$.

Explanation of Solution

For the solution containing $80\%$ quinine and $20\%$ of its enantiomer.

The enantiomeric excess of the solution is calculated by the formula,

$\text{Enantiomeric}\text{excess}=\%\text{Major}\text{enantiomer}-\%\text{Minor}\text{enantiomer}$

Substitute the values of percentage of major and minor enantiomers in the above equation.

$\begin{array}{rll}\text{Enantiomeric}\text{excess}& =& 80\%-20\%\\ & =& 60\%\end{array}$

Hence, the enantiomeric excess of the solution is $60\%$.

Conclusion

Enantiomeric excess of the given solution is $60\%$.

Interpretation Introduction

(e)

Interpretation: The $\left[\alpha \right]$ value for the solution is to be calculated.

Concept introduction: Enantiomeric excess is the excess of major enantiomer over the minor enantiomer. Enantiomeric excess is equivalent to the optical purity. Enantiomeric excess is the fraction of observed rotation of the mixture to specific rotation of the pure enantiomer.

Answer to Problem 5.65P

The $\left[\alpha \right]$ value for the solution is $-99$.

Explanation of Solution

For the solution containing $80\%$ quinine and $20\%$ of its enantiomer.

The enantiomeric excess of the solution is $60\%$.

The specific rotation $\left[\alpha \right]$ of pure quinine is $-165$.

The enantiomeric excess is calculated by the formula,

$\text{ee}=\frac{\left[\text{α}\right]\text{mixture}}{\left[\text{α}\right]\text{pure}\text{enantiomer}}\times 100\%$

Substitute the values of specific rotation $\left[\alpha \right]$ of pure quinine and enantiomeric excess.

$\begin{array}{rll}\left[\text{α}\right]\text{mixture}& =& \frac{ee\times \left[\text{α}\right]\text{pure}\text{enantiomer}}{100}\\ & =& \frac{60\times \left(-165\right)}{100}\\ & =& -99\end{array}$

Hence, the $\left[\alpha \right]$ value for the solution is $-99$.

Conclusion

The $\left[\alpha \right]$ value for the solution is $-99$.