The specific rotation of optically pure adrenaline is $-53$. A mixture of (R)- and (S)-adrenaline was found to have a specific rotation of $-45$. Calculate the % ee of the mixture (enter numeric value only, without % sign, rounded to the nearest whole percent).

This should be $\%~\text{ee}= \text{specific rotation sample}/\text{specific rotation enantiomer}$

But $100\times (-53/{-45})=117.78~\%$

The fraction the other way is $84.9~\%$ but doesn't stick to the formula.


1 Answer 1


Enantiomeric excess (ee) is defined as:

$$\text{ee} = \frac{\alpha(\text{sample})}{\alpha(\text{pure enantiomer})}\tag{1}$$

Here, we are given the specific rotation of our sample ($-45^\circ$) and that of a pure enantiomer ($-53^\circ$). To calculate the ee, all you need to do is to plug these two values into equation $(1)$ and work out how to turn a fractional number into a value in $\%$.

The answer is $85~\%~\text{ee}$.

  • $\begingroup$ "Optically pure" means pure enantiomer? $\endgroup$
    – user41987
    Mar 18, 2017 at 21:23
  • $\begingroup$ @user41987 Yes. $\endgroup$
    – Jan
    Mar 18, 2017 at 21:24

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