Question

A sample of 2-butanol has a specific rotation of 3.25. Determine the % ee and molecular composition of this sample. The sepcific rotation of pure (+)-2-butanol is +13.0.

is it (3.25/13.0)x100%= 25%... i'm a little confused on the molecular composition though.

Answer 1

Yes, %ee is 25% and it can be used to check later calculations as shown.
If the + enantiomer is +13.0, then the - enantiomer is -13.0.
Let x = + enantiomer
1-x = - enantiomer.
Then x(13) + (1-x)(-13)=3.25
Solve for x and 1-x.
I get 0.625 for x and
0.375 for 1-x.
Therefore, % composition is 62.5% dextro and 37.5% levo.
As a way of checking the first number,
%ee = enan(1) - enan(2) = 62.5% - 37.5% = 25% which agrees with your first calculation done a different way (and called optical purity). Voila!
A third way of doing the same thing is
%ee = (R-S/R+S)*100 =
[(0.625-0.375)/(0.625+0.375)]*100 =
(0.25/1.0)*100 = 25% where R is one isomer and S is the other.
I hope this helps.