Question

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?

Answer 1

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

Explanation:

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.

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 = ("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`).