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A sample of (S)-(+)-lactic acid was found to have an enantiomeric excess of 72%. How much Risomer is present in the sample?

Short Answer

Expert verified

The actual amount of S-isomer present in the sample is = 72% + =86%.

Hence the amount of R-isomer present in the sample is = 100% - 86% = 14%.

Step by step solution

01

What is enantiomer

Pair of enantiomers will exist if the compound has asymmetric centre or a chiral carbon. So the mirror images of compounds don’t superimpose.

Chiral objects are those objects which have no same mirror images. The mirror images don’t superimpose each other. whose mirror images superimpose each other that compounds are called achiral objects.

02

Formula used

The enantiomeric excess , also called the optical purity, tells us how much of an excess of one enantiomer is in the mixture. It can be calculated from the observed specific rotation:

(S)(+)-lactic acid has an enantiomeric excess = 72 %

Let the total % of the enantiomeric mixture is 100%.

The optical activity is 72% means, the enantiomer has 72% of excess S-isomer and 28% racemic mixture.

Thus the actual amount of S-isomer present in the sample is = 72% + = 86%

Hence the amount of R-isomer present in the sample is = 100% - 86% = 14%.

The sample of lactic acid contain 14% of R-isomer.

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