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}=(72)/(100)=0.72`

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

`[\alpha]=\text{Enantiomer excess}* [\alpha]_(Pure)`

`[\alpha]=0.72* -165`

`[\alpha]=-118.8`

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