High School

The specific rotation of a pure (S) enantiomer is +61.0 degrees. What percentage of a mixture of the two enantiomeric forms is the (S) form if the specific rotation of this mixture is +36.6 degrees?

A. 80.0%
B. 60.0%
C. 40.0%
D. 90.0%

Answer :

Final answer:

The percentage of the (S) form in the mixture is 60.0%.

Explanation:

The specific rotation of a pure (S) enantiomer is +61.0 degrees. To find the percentage of the (S) form in a mixture, we can use the formula: Percentage of (S) form = (Specific Rotation of the Mixture - Specific Rotation of the (R) form) / (Specific Rotation of the (S) form - Specific Rotation of the (R) form) x 100

To find the percentage of the (S) enantiomer in the mixture, you can use the formula: % (S) enantiomer = (Observed rotation / Specific rotation of (S)) * 100.

% (S) enantiomer = (+36.6 / +61.0) * 100 = 60%.

So, the (S) enantiomer makes up 60% of the mixture, while the (R) enantiomer constitutes the remaining 40%.

Plugging in the values given, we get: Percentage of (S) form = (+36.6 - 0) / (+61.0 - 0) x 100 = 60.0%

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