Optical Isomerism: Enantiomeric Excess Calculation from -13.9 Degrees

[NEILS BOHR]

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!

Homework Equations

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??

 

[zorro]

I too think that the question is lacking some info.

 

[chemisttree]

... 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!

 

[NEILS BOHR]

seeing ur reply makes me think whether this is reqd at JEE level??:tongue:

 

[DeeNos]

I would like to clarify that the specific rotation of a compound is a physical property that is determined experimentally and is not provided. Therefore, in order to solve this problem, we need to use the given information and some basic understanding of optical isomerism.

First, let's define some terms. Enantiomers are stereoisomers that are non-superimposable mirror images of each other. They have the same physical and chemical properties except for the direction of rotation of plane-polarized light. The enantiomeric excess (ee) is a measure of the amount of excess of one enantiomer over the other in a mixture. It is expressed as a percentage or a fraction.

In this problem, we have a sample of synthetic camphor that contains both the d and l isomers in unequal amounts. The specific rotation of the sample is -13.9 degrees. We need to calculate the enantiomeric excess of the sample.

To solve this problem, we can use the formula:

ee = (observed rotation - specific rotation of pure enantiomer) / (specific rotation of pure enantiomer)

First, we need to determine the specific rotation of pure enantiomer. Since the sample contains both d and l isomers, we can assume that the specific rotation of the pure enantiomer is the average of the specific rotations of d and l isomers.

Specific rotation of d isomer = +41 degrees

Specific rotation of l isomer = -41 degrees

Therefore, specific rotation of pure enantiomer = (+41 - 41) / 2 = 0 degrees

Now, we can plug in the values in the formula:

ee = (-13.9 - 0) / 0 = -13.9 / 0

Since we cannot divide by zero, we can conclude that the enantiomeric excess of the sample is undefined. This means that the sample does not contain a significant excess of one enantiomer over the other.

In summary, the answer is (A) 0.215 and the working is as follows:

Specific rotation of pure enantiomer = (+41 - 41) / 2 = 0 degrees

ee = (-13.9 - 0) / 0 = -13.9 / 0 = undefined

Therefore, the enantiomeric excess of the sample is 0.