Question

Question #4b454

Answer 1

The mixture contains 75 % `S` and 25 % `R`.

Explanation:

The formula for enantiomeric excess is

`ee = (R-S)/(R+S) × 100`,

where `R` and `S` are the percentages of each enantiomer.

We can solve this for `R` and `S` as a function of `ee`.

Remember that `(R+S)=100`, so `S = 100-R`

Let's assume that `R` is in excess.

`ee(R+S) =100(R-S)`

`ee(R+100-R) = 100(R-100+R)`

`100ee = 100(2R-100)`

`ee = 2R-100`

`2R =ee+100`

`R = (ee)/2 +50`

If `S` is in excess, we simply interchange `R` and `S` in the formula.

We have a 50 % enantiomeric excess of `S`.

∴ `S = (ee)/2 +50 = 50/2+50 = 25+50 = 75`

So, the mixture contains 75 % `S` and 25 % `R`.