Question

In a mixture, two enantiomers are found to be present in $85\%$ and $15\%$ respectively. The enantiomeric excess $(e, e)$ is

• A. $85\%$
• B. $15\%$
• C. $70\%$
• D. $60\%$

Answer 1

The Correct Option is C

$15 \%$ will form racemic mixture with another $15 \%$. Hence, the enantiomeric excess is $=(85-15)=70 \%$

Answer 2

Before we can answer this question, we must first define enantiomers.
Compounds with the same atoms and bonds but different spatial arrangements will have non-superimposable mirror images. Enantiomers are mirror images that cannot be overlaid on one another. Enantiomers of compounds containing asymmetric chiral centers were often identified.
Let us now look at what enantiomeric excess means.
Enantiomeric excess is a chiral compound purity assessment. The chiral center is an atom that is attached to four distinct groups of atoms in order to generate a non-superimposable mirror image.
Enantiomeric excess is denoted by the letter 'ee,' and it indicates which sample has more of one enantiomer than the other.It is determined by the enantiomer difference in a combination.
A racemic mixture (a mixture of enantiomeric isomers that contains both isomers in equal proportions) has ee of 0%, whereas a pure enantiomer has ee of 100%.
Now, let us see the question given to us.
The two enantiomers are present in 85% and 15%, respectively. Therefore, the enantiomeric excess will be:
ee = 85% – 15% = 70%.
Therefore, the correct answer is option (C) 70%.