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9 July, 07:23

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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  1. 9 July, 09:19
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    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°
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