Question

What is the percentage of enantiomeric excess of a mixture containing 12.8 mol (R)-2-bromobutane and 3.2 mol (S)-2-bromobutane?

Answer 1

The percentage of enantiomeric excess for a mixture of 12.8 mol (R)-2-bromobutane and 3.2 mol (S)-2-bromobutane is 60%.

Step-by-step explanation:

The percentage of enantiomeric excess (ee) of a mixture containing two enantiomers can be calculated using the formula: ee = [(major enantiomer) - (minor enantiomer)] / (total of all enantiomers) × 100%. In the given case, we have 12.8 mol of (R)-2-bromobutane and 3.2 mol of (S)-2-bromobutane. So, the ee is calculated as follows:

1. Calculate the excess of one enantiomer, which is 12.8 mol - 3.2 mol = 9.6 mol.
2. Calculate the total amount of enantiomers, which is 12.8 mol + 3.2 mol = 16.0 mol.
3. Plug the values into the equation to find the ee: ee = (9.6 mol / 16.0 mol) × 100% = 60%.

Therefore, the enantiomeric excess of the mixture is 60%.