Question

The (+) enantiomer of compound A has an optical rotation of 125

∘
. If a pure sample of compound A has an optical rotation of 100
∘
, what is the composition of the sample?

Answer 1

The sample of compound A with an optical rotation of 100° is a mixture of both (+) and (-) enantiomers, more enriched in the (+) enantiomer, given that the (+) enantiomer has a known optical rotation of 125°.

Step-by-step explanation:

The question is asking to determine the composition of a sample of compound A, which exhibits an optical rotation less than the known optical rotation of the pure (+) enantiomer. The optical rotation of a pure enantiomer is a characteristic known as its specific rotation. Given that the pure (+) enantiomer of compound A has an optical rotation of 125° and the mixed sample has an optical rotation of 100°, we can conclude that the sample is not purely one enantiomer, but a mixture of both (+) and (-) enantiomers.

To calculate the exact composition, we would use the percent enantiomeric excess (% ee), which is defined as [(observed rotation) / (specific rotation of pure enantiomer)] × 100%. Since the specific rotation is taken as the maximum rotation when the sample is 100% one enantiomer, the percentage of the (+) enantiomer in the mixture can be determined. However, without knowing the specific rotation of the (-) enantiomer, we cannot precisely calculate the ratio of enantiomers in the sample, only that it is enriched in the (+) form due to the positive optical rotation.

Answer 2

The composition of compound A, which exhibits an optical rotation of 100° when the (+) enantiomer alone has a rotation of 125°, is 80% (+) enantiomer and 20% (-) enantiomer.

Step-by-step explanation:

The student is asking about the optical rotation of a compound and how to determine the composition of a sample given a known optical rotation for the pure enantiomer. The optical rotation of a pure sample of compound A, which is a chiral molecule, is given as 125° for the (+) enantiomer.

If a sample of compound A exhibits an optical rotation of 100°, this suggests that the sample is not purely the (+) enantiomer but contains a mixture of both (+) and (-) enantiomers.

To calculate the specific rotation of the sample, we use the observed rotation of the enantiomerically pure substance and the observed rotation of the sample in question. The enantiomeric excess can be determined by the following relation: percentage of (+) enantiomer = (observed rotation / specific rotation of the pure enantiomer) × 100.

Given the pure enantiomer has a specific rotation of 125° and the sample has an observed rotation of 100°, the composition of the sample in terms of the (+) enantiomer can be calculated as (100° / 125°) × 100%, which equals 80%. This implies that the sample is made up of 80% of the (+) enantiomer and 20% of the (-) enantiomer.