Calculate the enantiomeric excess for the sample of (S)- ibuprofen sample and (R)-ibuprofen sample based on the data in attached image? **Data Section**- **Amount of Racemic Ibuprofen Used**: 3.000 g- **Amount of (S) methylbenzyl amine used**: 0.9 mL = 0.9 g- **Amount of (S) - Ibuprofen Recovered in Step C**: 1.782 g- **Amount of (R) - Ibuprofen Recovered in Step D**: 0.925 g- **Observed Rotation of (S) - Ibuprofen from Step C**: +0.54 degrees- **Observed Rotation of (R) - Ibuprofen from Step D**: +5.5 degrees- **Melting Point of Ibuprofen from Step C**: 113.4–122.8°C- **Melting Point of Ibuprofen from Step D**: 55.8°C–62°C- **Amount of (S) - Ibuprofen Recovered from Step D After Polarimetry**: 0.745 g- **Amount of (R) - Ibuprofen Recovered from Step E After Polarimetry**: 0.065 g

Given, Amount of (S) - ibuprofen recovered from polarimetry = 0.745 g Amount of (R) - ibuprofen recovered from polarimetry = 0.065 g From the given amount of (R) & (S) isomer, it is seen that (S) - ibuprofen is the major product as it᾿s recovered amount is greater than (R) - ibuprofen. Total amount of product recovered from polarimetry = 0.745 + 0.065 = 0.81 g Observed rotation of (S) - ibuprofen = + 0.54 degree Observed rotation of (R) - ibuprofen = + 5.5 degreeTherefore, % of (S) - ibuprofen in mixture = ( 0.745 / 0.81 ) * 100 = 91.98 % % of (R) - ibuprofen in mixture = ( 0.065 / 0.81 ) * 100 = 8.02 % Hence, enantiomeric excess of mixture = Major isomer - Minor isomer = (91.98 - 8.02 ) = 83.96 % Enantiomeric excess = ( observed rotation of mixture / actual rotation of pure enantiomer ) * 100 or, 83.96 = ( observed rotation of mixture / 0.54 ) * 100 ( As S- isomer is major, its optical rotation is used) or, Observed rotation of mixture = ( 83.96 * 0.54 ) / 100 = 0.453 degree Therefore, observed rotation of mixture = 0.453 degree .Enantiomeric excess of (S) - ibuprofen = ( observed rotation of mixture / observed rotation of pure enantiomer ) * 100 = ( 0.453 / 0.54 ) * 100 = 83.89 % Enantiomeric excess of (R) - ibuprofen = ( observed rotation of mixture / observed rotation of pure enantiomer ) * 100 = ( 0.453 / 5.5 ) * 100 = 8.23 % Therefore, enantiomeric excess of (S) - ibuprofen is 83.89 % and 8.23 % , for (R) - ibuprofen.