How do I calculate the enantiomeric excess of an isomer?
Often a %ee (enantiomeric excess) of a product that is predominantly one isomer will be required. This means the proportion of one isomer that is in excess of another. For an example, a product that is 75% R-isomer and 25% S-isomer has (75-25=50) a 50% excess of the R-isomer. %ee is commonly not directly tested for but can be calculated using the optical rotation. Example; Product 454486 is the R-isomer has an (optical rotation) [a]20D of +53.80. The ee% is lot specific specific. For lot 07028LC has a total purity of 95.1%, with the optical rotation for this lot being +46.40. Using this information the %ee can be calculated for lot 07028LC; ( [a]20D of lot / [a]20D of Product) x 100% = %ee Therefore; (46.4 / 53.8) x 100% = 86.2%ee. The remaining 13.8% is evenly split between the R and S isomer giving 6.9% each. This means that the total %ee of the R-isomer is (86.2+6.9) x 0.951 (decimal of lot purity) = 88.5%.
