Calculations with Enantiomeric Percentages: Videos & Practice Problems

Optical activity is a fascinating concept in chemistry, particularly when discussing enantiomers, which are molecules that are mirror images of each other. Understanding how to calculate enantiomeric percentages is crucial for analyzing mixtures of these compounds. When given the specific rotation of a pure enantiomer and the observed rotation of a mixture, you can determine the enantiomeric excess (ee) and the percentages of each enantiomer present.

The relationship between observed rotation, specific rotation, and enantiomeric excess can be expressed with the equation:

$$ \text{ee} = \frac{\text{observed rotation}}{\text{specific rotation}} $$

This equation allows you to calculate the enantiomeric excess directly. Additionally, it is important to remember that the sum of the percentages of the higher and lower enantiomers must equal 100%:

$$ \text{Percentage of higher enantiomer} + \text{Percentage of lower enantiomer} = 100\% $$

To find the percentage of the higher enantiomer from the enantiomeric excess, you can use the following derived equation:

$$ \text{Percentage of higher enantiomer} = \frac{100 + \text{ee}}{2} $$

Conversely, the percentage of the lower enantiomer can be calculated as:

$$ \text{Percentage of lower enantiomer} = 100\% - \text{Percentage of higher enantiomer} $$

For example, if the specific rotation of pure S-epinephrine is +50° and the observed rotation of a mixture is +25°, you would first calculate the enantiomeric excess:

$$ \text{ee} = \frac{25}{50} = 0.5 \text{ or } 50\% $$

Next, using the enantiomeric excess, you can find the percentage of the higher enantiomer:

$$ \text{Percentage of higher enantiomer} = \frac{100 + 50}{2} = 75\% $$

Then, the percentage of the lower enantiomer would be:

$$ \text{Percentage of lower enantiomer} = 100\% - 75\% = 25\% $$

Finally, when sketching the mixture in a polarimeter tube, you would represent the higher enantiomer (S) occupying 75% of the tube and the lower enantiomer (R) occupying 25%. This visual representation helps to understand why the observed rotation is only half of the specific rotation, as the presence of both enantiomers affects the overall optical activity of the solution.

In this problem, we explore the concept of enantiomeric excess, which is a measure of the purity of a chiral compound in terms of its enantiomers. The relationship between specific rotation, observed rotation, and enantiomeric excess can be expressed through the following equation:

Let ee represent enantiomeric excess, [α] the specific rotation, and [α]_obs the observed rotation. The formula to calculate enantiomeric excess is:

$$ ee = \frac{[α]_{obs}}{[α]_{specific}} $$

In this case, the observed rotation is 25 degrees, and the specific rotation is 50 degrees. Plugging these values into the equation gives:

$$ ee = \frac{25}{50} = 0.5 $$

To express this as a percentage, multiply by 100, resulting in an enantiomeric excess of 50%.

Next, to determine the percentage of each enantiomer, we use the following formula:

$$ \text{Percentage of the higher enantiomer} = 100 - \frac{ee}{2} $$

Substituting the calculated enantiomeric excess:

$$ \text{Percentage of the higher enantiomer} = 100 - \frac{50}{2} = 100 - 25 = 75\% $$

This indicates that one enantiomer is present at 75%. To find the percentage of the other enantiomer, simply subtract from 100:

$$ \text{Percentage of the lower enantiomer} = 100 - 75 = 25\% $$

Now, to identify which enantiomer corresponds to these percentages, we examine the sign of the specific rotation. If the specific rotation is positive, it indicates that the enantiomer with the higher percentage (75%) is the one that contributes to the positive rotation. In this case, the S enantiomer is responsible for the positive rotation, confirming that:

- S enantiomer: 75%

- R enantiomer: 25%

Finally, to verify the calculations, subtract the percentage of the lower enantiomer from the higher enantiomer:

$$ 75 - 25 = 50\% $$

This confirms the enantiomeric excess calculated earlier. Understanding these relationships is crucial for analyzing chiral compounds in organic chemistry.