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#).