Question

1. Percent Enantiomerie Excess (ee) is given by the equation if in moless (The difference between Rand S isomers/sum of R & S isomer) - 100. Thus, the sample prepared by Dr. Z. Evans in the PSB 340 Lab contained 12.8 moles of isomer and 3.2 moles of the S isomer. a. Which isomer is in excess in Dr. Evans product? 3.2 moles b. By how much: 12.8 - 3.2 x100 c. What is the percent ce? 2. Enantiomeric Excess and optical activity Introduction: The enantiomeric excess can also be calculated from the observed specific rotation of the sample by the following formula: % ce can also be expressed as (observed ay (a of pure enantiomer ) (100) Let us assume the % ce to be 70%. It then mean the 30% is a racemic mixture and thus (15%R and 15% S). Thus if the %ee is for the "R" isomer, then the % of "R" isomer in the sample is 70% + 15%. Which is 85% "R" isomer and 15% "S" isomer. Now it your turn to solve Dr. P. Sarah's problem. Thus, Dr. P. Sarah recrystalized an acid that was initially contaminated with its diastereomer and observed the optical rotation of the recrystallized product to be + 18.5. However, pure "R" isomer of the acid bought from Ramgbo Innovative Research Concepts has la ID of -23.1 while the "S" enantiomer has [a ]D of +23.1 a. Which isomer is in excess ? b. What is the % ee? c. What is the distribution of the two isomers in the Doctor's recrystallized sample?

Answer 1

Dr 2 EYans a) mdus CR)-mds (s)=12.8-3.29.6ma 6) Thre is a 9.6 md Eaun d H P mdy d Exus imte) Tatel nd 9.6 100 6oY 12.843-2 f.eece) - 6or. drees(R-S) x1001 R+S 2 Dr pSonah Tue Comend optulnt +1S a enantranur in Eaun becar &eubrratahr& ty isenr + 23-1 obsermd purt enauti T.ee X100x T-ders t18.5 x 1002ニ 80y 23) Rnetomr meoy ted ekses c u rte Oftfer 2 0 fg a Salinsr ct hob Sand R Rso equd to 10 - 907, ノ/R =(0% ys' REDMI NOTE 5 PRO