Question

мар ur An attempt at synthesizing a certain optically active compound resulted in a mixture of its enantiomers. The mixture had an observed specific rotation of +14.2°. If it is known that the specific rotation of the R enantiomer is - 39.5°, determine the percentage of each isomer in the mixture. Number Renantiomer: IT Number Senantiomer: Previous Give Up & View Solution Check Answer Next Exit - 7 Hint

Answer 1

who specific Rotation of Mixture a +14. 2° specific Rotation of R enantiomer - 39.5° The rotations of the two enantiomers cancel each other. a So the rotation of the mixture will be that of the excess enantiomere. The mixture has tve sign of rotation s isomer is in excess.

enantiomeric exus obserwed specific rotation -X100% maximum specific rotation > 1422 x 100 39.50 = 35.95% ~ 36 7.1 We have 36), on enantiomeric execes of +) o s isomer = le + 50%

We have 36 ) ene enantiomerio execes of + %ť pre up of Sisomente + 50% Vous 7 367 + 50 / olmasa 268 7.68 % (-) on % of R isomer (100-68)% 2 32% w