This tutorial explains how to convert observed optical rotation into the percentage of each enantiomer and the enantiomeric excess (ee). We will use the example of (+)-2-butanol, which has a specific rotation of +13.5°. If a reaction produces 2-butanol with an observed optical rotation of +1.6°, we can calculate the percentage of each enantiomer and the enantiomeric excess.
The specific rotation \([α]\) of a pure enantiomer is a physical constant. For (+)-2-butanol, \([α] = +13.5°\). This means that pure (+)-2-butanol rotates plane-polarized light by +13.5°.
The enantiomeric excess (ee) is a measure of the excess of one enantiomer over the other in a mixture. It is calculated using the formula:
\[ \text{ee} = \left( \frac{\text{Observed Rotation}}{\text{Specific Rotation of Pure Enantiomer}} \right) \times 100\% \]
For our example:
\[ \text{ee} = \left( \frac{+1.6°}{+13.5°} \right) \times 100\% = 11.85\% \]
The enantiomeric excess (ee) can be used to determine the percentage of each enantiomer in the mixture. Let \(x\) be the percentage of the (+)-enantiomer and \(y\) be the percentage of the (-)-enantiomer. Since the total percentage must add up to 100%, we have:
\[ x + y = 100\% \]
The enantiomeric excess is also given by:
\[ \text{ee} = x - y \]
Substituting the value of ee (11.85%) into the equation:
\[ 11.85\% = x - y \]
Now, solve the system of equations:
\[ x + y = 100\% \]
\[ x - y = 11.85\% \]
Adding the two equations:
\[ 2x = 111.85\% \implies x = 55.925\% \]
Substituting \(x\) into the first equation:
\[ 55.925\% + y = 100\% \implies y = 44.075\% \]
For the reaction producing 2-butanol with an observed optical rotation of +1.6°:
By measuring the optical rotation of a mixture and knowing the specific rotation of the pure enantiomer, you can calculate the enantiomeric excess and the percentage of each enantiomer in the mixture. This is a useful technique in chiral chemistry to determine the success of asymmetric synthesis or resolution processes.