why is the answer b? B) The rate would double C) The rate would triple D) The ra

Question

why is the answer b?

B) The rate would double C) The rate would triple D) The rate would increase 4-fold E) The rate would increase 6-fold A particular sample of compound Z containing a mixture of the R and S enantiomer has a specific rotation of +15.0 degrees. The specific rotation of the pure S enantiomer of compound Z is-50.0 degrees. What is the enantiomeric excess of this sample, and how much of the S enantiomer is present? A) 70% ee, 30% S B) 30% ee, 35% S C) 30% ee, 65% S D) 35% ee, 70% S 0. hybridization of the oxygen atom and the approximate For the molecule shown, what are the C-C-O bond angle, respectively? CH3CH2CCH2CE CH

Explanation / Answer

let fraction of R enantiomer be x
then fraction of S enantiomer will be 1-x
specific rotation of R enantiomer = 50.0
specific rotation of S enantiomer = -50.0

net rotation = fraction of R enantiomer * specific rotation of R enantiomer + fraction of S enantiomer * specific rotation of S enantiomer

15.0 = x * 50.0 + (1-x) * ( -50.0)
15.0 = x * 50.0 - 50.0 + x * 50.0
100.0*x = 65.0
x = 0.65
so,
fraction of R enantiomer = 0.65
fraction of S enantiomer = 0.35
percentage of R enantiomer = 65 %
percentage of S enantiomer = 35 %

%ee = 100 - 2*S
= 100 - 2*35
= 30 %
Answer: B

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