Question

In: Chemistry

If a pure R isomer has a specific rotation of -137.0 degrees and a sample contains...

If a pure R isomer has a specific rotation of -137.0 degrees and a sample contains 84.0% of the R siomer and 16.0% of it anantiomer, what is the observed specific rotation of the mixture?

Solutions

Expert Solution

obsereved specific rotation

if R = -137o

S = +1 37

observed specific rotation of the mixture   = R specific rotation + S -specific rotation

                                                                  = 84 /100 * -137   + 16 / 100 * 137 = -115.08 + 21.92

                                                    = -93.16

observed specific rotation of the mixture   = -93.16 o


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