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#.

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#).

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Answer 2

Ellie Adams

To determine the percentage of each isomer in a mixture, you would use a technique such as gas chromatography (GC). This method separates the isomers based on their different retention times. The area under each peak in the chromatogram corresponds to the amount of that isomer present. By comparing the area of each peak to the total area of all peaks, you can calculate the percentage of each isomer in the mixture.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.