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?

Question

Chemistry

Nathen Huffman

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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Answers (1)

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Ellie Bullock · Answer 1

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°