Specific rotation for pure tartaric acid is $-12 ^\circ$. A mixture of d-l isomers is prepared which has a specific rotation of $+6^\circ$. What is the percentage of d-tartaric acid in a mixture of $d$ and $l$ isomers?

The pure tartaric acid must have been $l$-tartaric acid.

My attempt:

$$\frac{l-d}{l+d}=\frac{+6}{-12}=\frac{1}{-2}$$ So $d=3l$. Which implies $$\frac{d}{d+l}=\frac{3}{3+1}=0.75 $$

However the answer given in my textbook is $66.67 \%$. Most probably I didn't do any calculation error. So where did I go wrong?


1 Answer 1


Your calculation is correct.

You can check your result since you know that pure (+)-tartaric acid has a specific rotation of $+12^\circ$ and that pure (−)-tartaric acid has a specific rotation of $-12^\circ$. Therefore, a mixture of $75\ \%$ (+)-tartaric acid and $25\ \%$ (−)-tartaric acid has a specific rotation of $$0.75\times(+12^\circ)+0.25\times(-12^\circ)=6^\circ$$

A similar exercise can be found in Hart, Hadad, Craine, Hart Organic Chemistry – A Short Course: Google Books

The corresponding solution can be found here: Google Books

  • $\begingroup$ And that results in $50~\%~\mathrm{e.e.}$ $\endgroup$
    – Jan
    Sep 14, 2016 at 16:27

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