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reganjones89
30.03.2020 •
Chemistry
If a pure R isomer has a specific rotation of –126.0°, and a sample contains 64.0% of the R isomer and 36.0% of its enantiomer, what is the observed specific rotation of the mixture?
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Ответ:
The observed specific rotation of the mixture = 68.04°
Explanation:
Given data
Specific rotation of the R isomer = - 126°
The bet rotation of the mixture is the rotation of excess enantiomer.
The R isomer = 64 %
Enantiomer = 36 %
The Enantiomer excess of the R isomer is given by
ee = 64 - 36 = 28 %
The observed specific rotation of the mixture is the rotation of this Enantiomer excess.
Therefore the rotation of this excess enantiomer
α = 0.54 × ( -126° )
α = 68.04°
Thus the observed specific rotation of the mixture = 68.04°
Ответ:
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