`(+)-`Butan`-2-ol` has `[alpha]_(D)^(20) =+13.9^(@)`.a sample of butane`2-ol` contianing both the enantiomers was found to have a specific rotation value of `-3.5^(@)` under similar conditions. The percentages of the `(+)` and `(-)` enantiomers present in the mixture are, respectively,

Options

A. `37.4%` and `62.6%`
B. `35.5%` and `64.5%`
C. `42.2%` and `57.8%`
D. `62.6%` and `37.4%`

Answer

A

Explanation

The specific rotation of pure `(-)-`butane`-2-`ol is `-13.9^(@)`. Therefore, the optical purity `(OP)` of the sample si
`%` optical purity `= (-3.5^(@))/(-13.9^(@)) xx100% = 25%`
This implies that `25%` of the sample consists of `(-)-`enantiomers in excess and `75%` is the racemate. the total `%` of `(-)` is thus `25% +1//2 (75%) = 62.5%`, and the remaining `37.5%` is `(+)`.