A solution prepared by mixing 10 mL of a 0.10 M solution of the R enantiomer of a

Chapter 4, Problem 25

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A solution prepared by mixing 10 mL of a 0.10 M solution of the R enantiomer of a compound and 30 mL of a 0.10 M solution of the S enantiomer was found to have an observed specific rotation of +4.8 . What is the specific rotation of each of the enantiomers? ( Hint: mL * M = millimole, abbreviated as mmol) Solution One mmol (10 mL * 0.10 M) of the R enantiomer is mixed with 3 mmol (30 mL * 0.10 M) of the S enantiomer; 1 mmol of the R enantiomer plus 1 mmol of the S enantiomer will form 2 mmol of a racemic mixture, so there will be 2 mmol of S enantiomer left over. Because 2 out of 4 mmol is excess S enantiomer, the solution has a 50% enantiomeric excess. Knowing the enantiomeric excess and the observed specific rotation allows us to calculate the specific rotation. enantiomeric excess = observed specificrotation specific rotation of the pure enantiomer * 100% 50% = +4.8 x * 100% 50 100 = +4.8 x 1 2 = +4.8 x x = 21+4.82 x = 9.6 The S enantiomer has a specific rotation of +9.6 , so the R enantiomer has a specific rotation of -9.6 .