Question

3. You’ve gotten your hands on a sample of carvone, but you aren’t sure which enantiomer you have, or how enantiopure it is. You find that it has a specific rotation of [α]D25 = –52.5°. The specific rotation of enantiopure (S)-carvone is +61°. Show all work.

(a) Which enantiomer does your sample contain?

(b) What is the enantiomeric excess of your sample? Show your work.

(c) What percent composition of each enantiomer present in your sample?

Answer 1

Given,

Specific rotation of enantiopure (S)-carvone = +61°

Sample a specific rotation of [α]D25 = –52.5°

1. Which enantiomer does your sample contain?

(S)-Carvone specific rotation is = +61°, therefore, (R)-Carvone specific rotation will be = -61°,

Since our sample has specific rotation = -52.5o, it contains both the enantiomers but the majority of the compound will be, (R)-Carvone.

% Optical purity of sample = 100 * (specific rotation of sample) / (specific rotation of a pure enantiomer)

= 100 x (-52.5 / -61)

= 86% optically pure (R) –Carvone

Therefore, 86% optically pure (R)- Carvone and remaining 14% is a recemic mixture. Hence, The (S)-Carvone contains = 7%

                (b) What is the enantiomeric excess of your sample?

"enantiomeric excess" or ee:

ee % = Excess of Single Enantiomer/entire mixture

                =(86%/100 %)x 100

                = 86% ee of (R)- carvone.

ee % = Excess of Single Enantiomer/entire mixture

                =(7%/100 %)x 100

                = 7% ee of (S)- carvone.

(c) What percent composition of each enantiomer present in your sample?

We have 86% optically pure (R)- Carvone and remaining 14% is recemic mixture. Hence, The (S)-Carvone contains = 7%

(R)- Carvone = 86 +7 = 93%

(S) –Carvone = 7%