InterviewSolution
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InterviewSolution
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`(+)-`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,A. `37.4%` and `62.6%`B. `35.5%` and `64.5%`C. `42.2%` and `57.8%`D. `62.6%` and `37.4%` |
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Answer» Correct Answer - A 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 `(+)`. |
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