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.

Question

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.

DrBob222 · Answer

%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.