Optical Isomerism: Enantiomeric Excess Calculation from -13.9 Degrees

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SUMMARY

The discussion centers on calculating the enantiomeric excess of a synthetic camphor sample with a specific rotation of -13.9 degrees. Participants emphasize the necessity of knowing the specific rotation of pure R or S camphor to solve the problem accurately. The enantiomeric excess can be determined using the formula: enantiomeric excess = (observed rotation / specific rotation of pure enantiomer) × 100%. The correct answer choices provided include numerical values that suggest varying degrees of enantiomeric excess.

PREREQUISITES
  • Understanding of optical isomerism and enantiomers
  • Knowledge of specific rotation and its calculation
  • Familiarity with the CRC Handbook for chemical data
  • Basic proficiency in solving chemistry problems related to stereochemistry
NEXT STEPS
  • Research the specific rotation values for pure R and S camphor
  • Learn how to calculate enantiomeric excess using optical rotation data
  • Explore the concept of optical isomerism in organic chemistry
  • Study the applications of enantiomeric excess in pharmaceuticals
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Chemistry students, organic chemists, and educators involved in teaching stereochemistry and optical activity concepts.

NEILS BOHR
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Homework Statement


A sample of synthetic camphor contains both the d and l isomers in unequal amounts . The specific rotation of the sample is found to be -13.9 degrees . What is the enantiomeric excess of the sample?

(A) 0.215

(B) 0.314

(C) 0.417

(D) 0.167

pleasez xplain ur working as well!


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The Attempt at a Solution



i think the optical rotation has to be provided . without it i can't think of how to solve it??
 
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I too think that the question is lacking some info.
 
... unless your teacher actually expects you to do a little work and find the specific rotation of pure R or S camphor. Try the CRC Handbook. Even Wiki has it!
 
seeing ur reply makes me think whether this is reqd at JEE level??:-p
 

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