Dextrororatory `alpha-`pinene has a specific rotation `[alpha]_(D)^(20) =+51.3^(@)`. A sample of `alpha-`pinene containing both the enantiomers was found to have a specific rotation value `[alpha]_(D)^(20) =+30.8^(@)`. The percentages of the `(+)` and `(-)` enantiomers present in the sample are, respectively,

Options

A. `60%`and`40%`
B. `70%`and`30%`
C. `80%`and`20%`
D. `20%`and`80%`

Answer

C

Explanation

First calculate optical purity `(OP)` or enantiomeric excess `(ee)`
`%` optical purity `= (alpha_(obs))/([alpha]_(D)) xx 100%`
`=(+30.8^(@))/(+51.3^(@)) xx 100%`
`60%`
When we say that the enantiomeric excess (or optical purity) of the given sample is `60%`, we mean that `60%` of the sample consists of the `(+)-` enantiomer (the excess ) and the other `40%` consists of the racemic form. Since for the `40%` that is racemic, the optical rotations cancel one another out, only the `60%` of the sample that consist of the `(+)-` enantiomer contributes to the observed optical rotation. Therefore, the observed rotation is `60%` of what it would have beem if the mixture had consisted only of the `(+)-` enantiomer. Let's find the actual stereiosomeric composition of the sample. Of teh total mixture, `40%` consists of teh racemic form, which contains equal numbers of the two enantioners. Therefore, half of this `40%`, or `20%` is the `(-)-` enantiomer and `20%` is the `(+)-` enantiomer. The other `60%` of the mixture (the excess) is also the `(+)-` enantiomer. Consequently the mixture is `80% (+)-` enantiomer and `20%(-)-` enantiomer.