
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
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Hint: You must remember that enantiomers are isomers that are mirror images of each other. Enantiomeric excess gives us the percentage in which one isomer is present in excess in a mixture. Use this to solve the given question.
Complete step by step answer:
Before answering this question, we have to understand what enantiomers are.
Optical isomers are compounds having the same atoms and bonds but in different spatial arrangements and will have non-superimposable mirror images. These non-superimposable mirror images are called enantiomers. We generally found enantiomers of compounds having a chiral centre and are asymmetric.
Now, let us discuss the meaning of enantiomeric excess.
Enantiomeric excess is basically a measurement of purity of a chiral substance. Chiral centre is an atom to which 4 different groups or atoms are bonded so that they can form a non-superimposable mirror image.
Enantiomeric excess is written as ‘ee’ and it shows which sample contains one enantiomer in greater amounts than the other. It is given by the difference of enantiomers present in a mixture.
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%. Therefore, enantiomeric excess will be:
ee = 85% – 15% = 70%.
Therefore, the correct answer is option (C) 70%.
Note: It is important to remember here that optical activity does not depend solely on the chirality of a molecule. If the molecule is dissymmetric and has at least one non superimposable mirror image, it exhibits optical activity and thus possesses enantiomers. It is also important to remember that meso compounds are also optical isomers, but they contain a plane of symmetry. Therefore, all molecules irrespective of their chirality exhibit optical activity when they are not superimposable on their mirror images.
Complete step by step answer:
Before answering this question, we have to understand what enantiomers are.
Optical isomers are compounds having the same atoms and bonds but in different spatial arrangements and will have non-superimposable mirror images. These non-superimposable mirror images are called enantiomers. We generally found enantiomers of compounds having a chiral centre and are asymmetric.
Now, let us discuss the meaning of enantiomeric excess.
Enantiomeric excess is basically a measurement of purity of a chiral substance. Chiral centre is an atom to which 4 different groups or atoms are bonded so that they can form a non-superimposable mirror image.
Enantiomeric excess is written as ‘ee’ and it shows which sample contains one enantiomer in greater amounts than the other. It is given by the difference of enantiomers present in a mixture.
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%. Therefore, enantiomeric excess will be:
ee = 85% – 15% = 70%.
Therefore, the correct answer is option (C) 70%.
Note: It is important to remember here that optical activity does not depend solely on the chirality of a molecule. If the molecule is dissymmetric and has at least one non superimposable mirror image, it exhibits optical activity and thus possesses enantiomers. It is also important to remember that meso compounds are also optical isomers, but they contain a plane of symmetry. Therefore, all molecules irrespective of their chirality exhibit optical activity when they are not superimposable on their mirror images.
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