Specific Rotation and Enantiomeric Excess in Mixtures

What is the specific rotation of a mixture containing 75% of a pure substance and 25% of the (-) isomer?

A) +1.68°
B) 0°
C) +1.26°
D) +0.84°
E) +.042°

Answer:

Specific rotation of the mixture is D. 0.84 °

The specific rotation of a pure substance is given as 1.68°. In this case, we have a mixture containing 75% of this pure substance and 25% of the (-) isomer. We need to determine the specific rotation of this mixture.

First, we calculate the enantiomeric excess, which is the difference between the percentages of the two enantiomers in the mixture. In this case, the enantiomeric excess is 75% - 25% = 50%.

Enantiomeric excess can also be calculated using the formula:

Enantiomeric excess = (Optical rotation of the mixture / Optical rotation of the pure enantiomer) x 100

Given that the optical rotation of the pure enantiomer is 1.68°, we substitute the values into the formula:

50 = (Optical rotation of the mixture / 1.68°) x 100

Optical rotation of the mixture = (1.68° / 100) x 50

Specific rotation of the mixture = 0.84°

Therefore, the specific rotation of the mixture containing 75% of the pure substance and 25% of the (-) isomer is 0.84°.