Question

The [α] of pure quinine, an antimalarial drug, is −165. If a solution contains 86% quinine and 14% of its enantiomer (ee = 72%), what is [α] for the solution?

Answer 1

Answer : The [α] for the solution is, -118.8

Explanation :

Enantiomeric excess : It is defined as the difference between the percentage major enantiomer and the percentage minor enantiomer.

Mathematically,

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

Given:

% major enantiomer = 86 %

% minor enantiomer = 14 %

Putting values in above equation, we get:

$\%\text{ Enantiomer excess}=86\%-14\%=72\%$

$\text{ Enantiomer excess}=\frac{72}{100}=0.72$

Now we have to calculate the [α] for the solution.

$[\alpha]=\text{Enantiomer excess}\times [\alpha]_{Pure}$

$[\alpha]=0.72\times -165$

$[\alpha]=-118.8$

Thus, the [α] for the solution is, -118.8