Asked by Jenn
A sample of 2-butanol has a specific rotation of -9.75 degrees. Determine the % ee (enantiomeric excess) and the molecular composition of this sample. The specific rotation of pure (+)-2-butanol is +13.0 degrees.
Answers
Answered by
DrBob222
%ee = %1-%2 = ??
If we call the mole fraction of the A isomer x (with specific rotation of +13.0) and the mole fraction of the B isomer 1-x (with a specific rotation of -13.0), then will it not be true that
13.0*x + [-13.0*(1-x)] = -9.75
Solve for x and 1-x.
Multiply those by 100 to change to mole percent.
Then mole percent A - mole percent B = %excess enantiometer. You take the absolute value of this number so the sign at the end doesn't matter. Check my thinking on this.
If we call the mole fraction of the A isomer x (with specific rotation of +13.0) and the mole fraction of the B isomer 1-x (with a specific rotation of -13.0), then will it not be true that
13.0*x + [-13.0*(1-x)] = -9.75
Solve for x and 1-x.
Multiply those by 100 to change to mole percent.
Then mole percent A - mole percent B = %excess enantiometer. You take the absolute value of this number so the sign at the end doesn't matter. Check my thinking on this.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.