Question

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?

Answer 1

Specific Rotation = 105º

S% in mixture = 92%

Step-by-step explanation:

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

`e.e. = (\alpha )/(\alpha pure)`

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:

`84=(\alpha )/(125.0)*100`

`\alpha =105 Degrees`

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:

`84 = S-R`

Since S is in Excess

Also, it is true:

`100=S+R`

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

`8=R\\92=S`

Hope it helps!