Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For the given ee values, calculate the percentage of each enantiomer present.

a. 90% ee

b. 99% ee

c. 60% ee

Short Answer

Expert verified
  1. 95% A and 5% B.
  2. 99.5% A and 0.5% B.
  3. 80% A and 20% B.

Step by step solution

01

Enantiomeric excess (ee) 

The enantiomeric excess comes to explain the particular enantiomer’s excess amount in the given mixture.

02

Calculation of the ee

The formula to calculate the ee is as follows:

ee=%ofoneenantiomer-%ofotherenantiomer

03

Explanation

a. From the formula of ee, we can easily calculate the percentage of enantiomers as:

90% ee means 90% excess of A, and 10% racemic mixture of A and B (5% each); therefore, 95% A and 5% B.

Therefore, the percentage of enantiomers is 90 % of A and 5% of B.

b. From the formula of ee, we can easily calculate the percentage of enantiomers as:

99% ee means 99% excess of A, and 1% racemic mixture of A and B (0.5% each); therefore, 99.5% A and 0.5% B.

Therefore, the percentage of enantiomers is 99.5 % of A and 0.5% of B.

c.From the formula of ee, we can easily calculate the percentage of enantiomers as:

60% ee means 60% excess of A, and 40% racemic mixture of A and B (20% each); therefore, 80% A and 20% B.

Therefore, the percentage of enantiomers is 80 % of A and 20% of B.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A molecule is achiral if it has a plane of symmetry in any conformation. Each of the following conformations does not have a plane of symmetry, but rotation around a carbon–carbon bond forms a conformation that does have a plane of symmetry. Draw this conformation for each molecule.

a.

b.

Label each pair of compounds as constitutional isomers, stereoisomers, or not isomers of each other.

The shrub ma huang (Section 5.4 A) contains two biologically active stereoisomers-ephedrine and pseudoephedrine-with two stereogenic centers as shown in the given structure. Ephedrine is one component of a once popular combination drug used by body builders to increase energy and alertness, while pseudoephedrine is a nasal decongestant.

a. Draw the structure of naturally occurring (–)-ephedrine, which has the 1R,2Sconfiguration.

b. Draw the structure of naturally occurring (+)-pseudoephedrine, which has the 1S,2Sconfiguration.

c. How are ephedrine and pseudoephedrine related?

d. Draw all other stereoisomers of (–)-ephedrine and (+)-pseudoephedrine and give the R,Sdesignation for all stereogenic centers.

e. How is each compound drawn in part (d) related to (–)-ephedrine?

The [α] of pure quinine, an antimalarial drug, is –165.

a. Calculate the ee of a solution with the following [α] values: –50, –83, and –120.

b. For each ee, calculate the percent of each enantiomer present.

c. What is [α] for the enantiomer of quinine?

d. If a solution contains 80% quinine and 20% of its enantiomer, what is the ee of the solution?

e. What is [α] for the solution described in part (d)?

Classify each pair of compounds as constitutional isomers or stereoisomers.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free