In: Chemistry
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?
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