Question

How would you determine the percentage of each isomer in the mixture?

An attempt at synthesizing a certain optically active compound resulted in a mixture of its enantiomers. The mixture has an observed specific rotation of +11.2'. If it is known that the specific rotation of the R enantiomer is -39.839, determine the percentage of each isomer in the mixture.

R enantiomer = %
S enantiomer= %?

Answer 1

The mixture contains `64 % S` and `36 % R`.

Explanation:

I assume that the specific rotation of the mixture is +11.2 ° and that of the `R` enantiomer is -39.8 °.

The rotations of the two enantiomers cancel each other, so the rotation of the mixture will be that of the excess enantiomer.

The mixture has a positive sign of rotation, so the `S` isomer is in excess.

The formula for enantiomeric excess is

`color(blue)(|bar(ul(color(white)(a/a) ee = "observed specific rotation"/"maximum specific rotation" × 100 %color(white)(a/a)|)))" "`

`ee = ("11.2" color(red)(cancel(color(black)(°))))/("39.8" color(red)(cancel(color(black)(°)))) × 100 % = 28 %`

We can calculate the percent of each enantiomer as described in this Socratic question.

If we have a mixture of (+) and (-) isomers and (+) is in excess,

`% ("+") = (ee)/2 +50 %`

We have a 28 % enantiomeric excess of (+).

∴ `% ("+") = (28 %)/2 +50 % = (14+50) % = 64 %`

So, the mixture contains 64 % (`S`) and 36 % (`R`).