A solution prepared by mixing 20 mL of a 0.10 M solution of the R enantiomer of a compound and 60 mL of a 0.10 M solution of the S enantiomer was found to have an observed specific rotation of +3.6. W

Question

A solution prepared by mixing 20 mL of a 0.10 M solution of the R enantiomer of a compound and 60 mL of a 0.10 M solution of the S enantiomer was found to have an observed specific rotation of +3.6. What is the specific rotation of each of the enantiomers?

TestHub · Accepted Answer

20 mL x 0.10 M = 2 mmol of R and 60 mL x 0.10 M = 6 mmol of S. 2 mmol of R with 2 mmol of S forms a racemic 4 mmol, leaving 4 mmol of S in excess out of 8 mmol total: enantiomeric excess = 50% S. Since ee = (observed rotation / rotation of pure enantiomer), the pure S enantiomer has [α]S = +3.6 / 0.50 = +7.2, and its mirror image has [α]R = -7.2.

Ans. (A) [α]R = -7.2, [α]S = +7.2.

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