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In a solution containing 68.0% Risomer and 32.0% S isomer, with an observed specific rotation of -75.0°, what is the specific rotation of the pure Risomer?

a. -110.3°
b. -50.0°
c. -75.0°
d. -25.0°

User Mnieber
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1 Answer

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Final answer:

The specific rotation of the pure R isomer is calculated by considering the proportions of R and S enantiomers and their contributions to the observed specific rotation of a mixture.

Step-by-step explanation:

The question relates to the concept of specific rotation and the relationship between the observed rotation of a mixture of enantiomers and the rotation of each pure enantiomer. The observed specific rotation of -75.0° is for a mixture of 68% R isomer and 32% S isomer. Since enantiomers have opposite specific rotations of equal magnitude, we can set up an equation to find the specific rotation of the pure R isomer: observed rotation = (fraction of R isomer) × (specific rotation of R) + (fraction of S isomer) × (specific rotation of S). We know the observed rotation (-75.0°), the fraction of R isomer (0.68), and we know that the S isomer's contribution will be -1 times the R isomer's specific rotation since they are enantiomers. Therefore, we can simplify the equation since the specific rotation of S is the negative of R. We solve for the pure Risomer's specific rotation to find the answer.

User Luke Smith
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