Question

The specific rotation of optically pure (2S,3S)-tartaric acid is -12°. The optical rotation of an impure sample containing a mixture of enantiomers of this compound was measured to be -9°. In the boxes below, write the percentages of the specified stereoisomers present in the sample. Show your work below.

Percentage of (2S,3S)-tartaric acid?

Percentage of (2R,3R)-tartaric acid?

What is the specific rotation of pure (2S,3R)-tartaric acid? How do you know?  

Answer 1

The solution to the problem is as given below.

We have -12 degrees specific rotation for pure (2S,3S)-tartaric acid enantiomer.

Given is -9 degrees specific rotation for mixture of enantiomers. The -ve sign of specific rotation for the mixture shows we have excess of (2S,3S)-tartaric acid enantiomer in the given mixture.

Calculate optical purity

The optical purity of a mixture of enantiomers is given by:

% Optical purity of sample = 100 * (specific rotation of sample) / (specific rotation of a pure enantiomer)

Thus, optical purity = 100 x -9/-12 = 75 %

Thus, we have 75 % excess of (2S,3S)-tartaric acid enantiomer in the mixture.

The remaining amount is 100 - 75 = 25 %, which will be 50% (2S,3S)-tartaric acid and 50% (2R,3R)-tartaric acid.

Thus, we have,

Percentage of (2S,3S)-tartaric acid = 75 + 12.5 = 87.5 %

Percentage of (2R,3R)-tartaric acid = 12.5 %

The specific rotation of pure (2R,3R)-tartaric acid will be exactly the same number with opposite sign as that of (2S,3S)-tartaric acid, that is it would be +12 degrees. The two enatiomers of tartaric acid would rotate the plane of polarized light equally in opposite direction.