Ibuprofen is a drug used to manage mild pain, fever, and inflammation. it is a chiral drug, and only the s enantiomer, whose specific rotation is 125.0°, is effective. the r enantiomer exhibits no biological activity. if a particular process is capable of producing ibuprofen in 84% enantiomeric excess of the s enantiomer, then what is the specific rotation of that mixture? what is the percentage s enantiomer in that mixture?

Specific Rotation = 105º

S% in mixture = 92%

Explanation:

Specific rotation in organic compounds varies linearly, then the enantiomeric excess can be express as:

Where alpha is the specific rotation of the mixture and alpha pure is the specific rotation of the excess enantiomer.

Then, knowing that the S enantiomer is in excess you can calculate the specific rotation of that mixture:

For calculating the percentage of S you should remember that the enantiometric excess is the percentage of excess of a certain enantiomer, for example in this case:

Since S is in Excess

Also, it is true:

Then you can solve the system of linear equations, finding:

Hope it helps!