Question

If the enantiomeric excess of a mixture is 77%, what are the % compositions of the major and minor enantiomers?

Answer 1

Concepts and reason

The concept used to solve this problem is based on enantiomeric excess.

The chiral molecules that are mirror images of each other are known as enantiomers and measurement of purity used for chiral molecules is known as enantiomeric excess.

Fundamentals

It helps to calculate the degree to which a solution contains one enantiomer in greater amount than the other. Thus, optical purity is equal to the percentage excess of major enantiomer over the minor enantiomer.

A racemic mixture contains equal amount of two enantiomers. A single completely pure enantiomer has an enantiomeric excess of $100\%$ and a racemic mixture has $0\%$ of enantiomeric excess.

(1)

Out of two enantiomers X and Y, X is in excess. Now the enantiomeric excess of the mixture is $77\%$ , it means that the solution contains an excess of $77\%$ X and remaining $23\%$ of racemic mixture of X and Y.

According to the definition, a racemic mixture contains equal amount of both enantiomers X and Y. Thus, the amount of X and Y is $11.5\%$ each in a racemic mixture of $23\%$ .

The total amount of X the solution is calculated as follows:

$\begin{array}{c}\\{\rm{X = 77\% + 11}}{\rm{.5\% }}\\\\{\rm{ = 88}}{\rm{.5\% }}\\\end{array}$

(2)

Now, total amount of Y in the solution is calculated as follows:

$\begin{array}{c}\\{\rm{Y = 100\% }} - {\rm{88}}{\rm{.5\% }}\\\\{\rm{ = 11}}{\rm{.5\% }}\\\end{array}$

Ans: Part 1

The percentage composition of major enantiomer X is $88.5\%$ .

Part 2

The percentage composition of minor enantiomer Y is $11.5\%$ .