The specific rotation of (R)-(+)-glyceraldehyde is +8.7. If the observed specific rotation of a mixture of (R)-glyceraldehyde and (S)-glyceraldehyde is +2.2, what percent of glyceraldehyde is present as the R enantiomer?

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Answer 1 · 2016-06-10T18:43:01

The mixture contains $"63 %" color(white)(l)R"-glyceraldehyde"$ and $"37 %"color(white)(l) S"-glyceraldehyde"$ .

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 $R$ isomer is in excess.

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

$ee = ("2.2" color(red)(cancel(color(black)(°))))/("8.7" color(red)(cancel(color(black)(°)))) × 100 % = 25.3 %$

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,

$color(blue)(|bar(ul(color(white)(a/a) % ("+") = (ee)/2 +50 %color(white)(a/a)|)))" "$

We have a 25.3 % enantiomeric excess of (+).

∴ $% ("+") = (25.3 %)/2 +50 % = (12.6+50) % = 63 %$

So, the mixture contains 63 % ( $R$ ) and 37 % ( $S$ ).