Question

1. Would a 50:50 solution of R and S enantiomers of the same compound have any effect on plane polarized light shining through it? Why or why not?

2. Based on your answer to the last question, how could you determine the enantiopurity of a given concentration of a sample if you know its specific rotation [α] using a polarimeter?

3. Give an example of a drug which can be sold as a racemic (R and S) mixture without any harmful effects.

Answer 1

1. There would be no net effect on plane-polarized light because of a 50:50 solution of R and S enantiomers of the same compound because racemic mixtures are optically inactive. This is because both R and S enantiomers would cancel each other's effects on the plane-polarized light. Let's assume, R enantiomer rotates the plane-polarized light in an anticlockwise direction (hypothetically) by an angle of 180^o. Now, the other half of the mixture, S enantiomer would rotate the same plane-polarized light in the clockwise direction by an angle of 180^o. So, after passing through the solution, there would be no net change in the light. Hence, the mixture would seem optically inactive.

2. The enantiopurity of a sample can be determined by calculating the enantiomeric excess of the sample by using the following formula:

where,

3. Examples of drugs which can be sold as a racemic (R and S) mixture are Ketamine, Ibuprofen, etc.