Question

11. (S)-isobutylphenylpropionic acid was synthesized utilizing an enzymatic reduction. This reaction however yielded both the R and the S version of this compound. When the species were analyzed they gave an optical rotation of +12.5°. The concentration and path length used were 0.50g/mL. and 1.Odm. Given that in order for this compound to be pure the literature rotation was +36 what is the enantiomeric excess? And how much of the compound is in solution? 1. 69% and 84.5% 2. 46% and 73% 378% and 89% 4. 55% and 77.5%

Answer 1

Optical rotation (of (S)- isobutylpenylpropionoic acid) = +12.5° ....(given)

Concentration = 0.5 g/ml ....(given)

Path length = 1.0 dm ......(given)

Specific rotation of mixture = ?

Specific rotation of mixture = optical rotation in degree/ (concentration in g/ ml × path length in dm)

So, specific rotation of mixture = 12.5°/(0.5 g/ml × 1.0 dm)

= +25°

Hence now, enantiomeric excess is calculated by formula-

Enantiomeric excess = ( specific rotation of mixture/ specific rotation of pure ) × 100

Specific rotation of pure = +36° ....(given)

So, Enantiomeric excess (of (S)- isobutylpenylpropionoic acid) =( +25°/+36°) × 100

= +69.4% or = +69% (approximately)

Racemic mixture is calculated by formula-

Racemic mixture = 100 - Enantiomeric excess

So, Racemic mixture = 100 - 69 = 31%

Racemic mixture states that both (R) & (S) isobutylpenylpropionoic acid are present in equal quantity.

Hence, 31/2 = 15.5

i.e. 15.5% of (S)- isobutylpenylpropionoic acid

& 15.5% of (R)- isobutylpenylpropionoic acid.

So, now the (S)- compound in the solution will be = 69 + 15.5 = 84.5%.

So, correct option is (1)- 69% and 84.5%.