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

Question

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= %?

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Answer 1 · 2016-04-25T01:56:46

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

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.

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

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

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 % ( $R$ ).