A compound has a specific rotation of -129°. A solution of the compound (o.290g/mL)has an observed rotation of -37° when placed in a polarimeter tube 10cm long. What is the percent of eact enantiomer in the solution?
2007-02-10 16:43:12 · 3 answers · asked by Kitty 1 in Science & Mathematics ➔ Chemistry
Your aving a laugh!
2007-02-10 16:52:01 · answer #1 · answered by Stu pid 5 · 1⤊ 1⤋
The enantiomeric excess (ee) can be calculated from the specific and observed rotations (see reference). The enantiomeric excess tells you the percentage the isomer is in excess. For example, a solution with 20% ee of the R-isomer would contain 60% R-isomer and 40% S-isomer. If there is a 20% excess in R, the remaining 80% would be racemic - 40% R and 40% S. Add 40% and 20% to get 60% total for the R-isomer.
2007-02-10 17:02:27 · answer #2 · answered by jas 2 · 0⤊ 0⤋
Optical rotation!! Blimey, haven't had to deal with that for a long time. Rates along with Lovibond Tintometer Testing, Refractive Indices and Acid Values expressed as mg KOH/ml.
And the good news is I can't remember bugger-all about it. Sorry!
2007-02-10 17:04:33 · answer #3 · answered by Phish 5 · 0⤊ 0⤋