How to Calculate Enantiomeric Excess and Optical Purity?

Optical purity of compound is defined as percentage of major isomer present in the sample of compound.

Key words: Enantiomeric excess, Isomers, Enantiomers, optical purity

Hi Friends, in this article we will discuss about how to calculate optical purity and enantiomeric excess formula for organic compounds by using specific optical rotation values. 

As we know that a chiral compound consists of two types of isomers or enantiomers. If a sample consist of only one isomer, then such sample is called as optically pure. But if the given sample consists of both the isomers in same or different proportions, then the purity of compound is specified by percent optical purity.

Definition of Optical purity

Optical purity of compound is defined as percentage of major isomer present in the sample of compound.

Formula for percent optical purity is;

Figure : Percent optical purity

So, to calculate the percent optical purity we need to add specific rotation values of sample and pure enantiomer in above equation.

Let’s see this with one example;

A sample of tartaric acid shows specific rotation of +8 so what will be its percent optical purity?

To solve this problem, first we need to know optical rotation value of pure isomers.

The R,R-isomer of tartaric acid has specific rotation of -12 ° whereas S,S-isomer has specific rotation of +12°. Since the sample has positive specific rotation value, it means that the sample consists of S,S-isomer in major proportion.

Therefore, according to the formula;

Hence percent optical purity of given sample is 66.6 %. In other words, the sample consists of S,S-isomer of tartaric acid in 66.6 %.

So, amount of R,R-isomer in the sample = 100-66.6 = 33.4 %

Now we will learn about enantiomeric excess.

Definition of Enantiomeric excess

The enantiomeric excess value shows which isomer is present in greater extent. 

Enantiomeric excess is calculated by the formula;

Let’s, understand this with above example; the sample has R,R-isomer (33.4 %) and S,S-isomer (66.6%). So, Enantiomeric excess is calculated as;

Hence the enantiomeric excess of the given sample is 33.2 %

Now take one more example to calculate percent optical purity and enantiomeric excess. 

Suppose a stereoselective reduction of 2-butanone provides chiral 2-butanol. The isolated compound shows specific rotation -10.5. So, what will be its optical purity and enantiomeric excess?

Figure : Stereoselctive reduction of 2-butanone

Here first thing we need to find specific rotation of pure R-isomer and S-isomer. It is reported that (R)-2-butanol has specific rotation of +13.9 ° whereas (S)-2-butanol has specific rotation of −13.9 °.

Since the specific rotation value is negative that means the sample has major S-isomer. Therefore;

Hence it is understood that the isolated compound has S-isomer with 75.5 %

The percentage of R-isomer = 100−75.5 = 24.5 %

Now we will calculate Enantiomeric excess

Therefore, isolated compound from stereoselective reduction has 75.5 % optical purity and 51 % enantiomeric excess.

I hope that this article would been informative for you. Please write in the comment section below if you have any queries regarding this topic. Also, you can suggest me subjects for next articles.  

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