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If a pure R isomer has a specific rotation of –144.0°, and a sample contains 69.0%...

If a pure R isomer has a specific rotation of –144.0°, and a sample contains 69.0% of the R isomer and 31.0% of its enantiomer, what is the observed specific rotation of the mixture?

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Answer #1

specific rotation of R = -1440

specific rotation of S = 1440

specific rotation of mixture = specific rotation of R + specific rotation of S

                                        = -144*69/100 + 144*31/100

                                       = -99.36 +44.64 = -54.720

specific rotation of mixture = -54.720

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