The specific rotation of (S)-carvone (at 20°C) is +61. A chemist prepared a mixture of (R)-carvone and its enantiomer, and this mixture had an observed rotation of -55°. A) What is the specific rotation of (R)-carvone at 20°C?
B) Calculate the % ee of this mixture.
C) What percentage of the mixture is (S)-carvone?

a) Specific Rotation of (R)-carvone = -61°

b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%

c) The percentage of (S)-carvone in the mixture = 4.9%

Explanation:

a) The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.

Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.

Specific Rotation of (R)-carvone = -(61°) = -61°

b) Enantiometic excess is used to measure the optical purity of an enantiomeric mixture.

The enantiomeric excess is given mathematically as

ee% = (Observed rotation × 100)/(Specific rotation)

Hence, to calculate the enantiomeric excess of (R)-carvone,

Observed rotation of the mixture = -55°

Specific Rotation of (R)-carvone = -61°

ee% = (-55×100)/(-61) = 90.16% = 90.2%

c) An enantiomeric excess of 90.2% for (R)-carvone indicates that it's actual percentage is 90.2% more than the percentage of its enantiomeric partner, (S)-carvone, in the mixture.

Let the percentage of (R)-carvone in the mixture be x

Let the percentage of (S)-carvone in the mixture be y

x + y = 100

x - y = 90.2

2x = 190.2

x = (190.2/2) = 95.1%

y = 100 - x = 100 - 95.1 = 4.9%

Hence, the percentage of (R)-carvone in the mixture = 95.1%

The percentage of (S)-carvone in the mixture = 4.9%

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