The specific rotation for a pure enantiomer is known to be +139o g-1 mL-1 dm-1. A sample containing both enantiomers is found to have an observed rotation of +0.87o in a one dm tube at a concentration of 0.025 g/mL. What is the optical purity of the sample?



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[a] = a/(c x l)

 


The specific rotation, [a] = +139o

The observed rotation, a = +0.87o

The length of the analysis tube is one dm;  l = 1.0 dm

The concentration of the sample is 0.025 g/mL;  c = 0.025 g/mL

You should calculate the "apparent specific rotation"...



















































Solution:

The specific rotation, [a] = +139o

The observed rotation, a = +0.87o

The length of the analysis tube is one dm;  l = 1.0 dm

The concentration of the sample is 0.025 g/mL;  c = 0.025 g/mL


For this sample, the apparent specific rotation is:




[a]apparent = +0.87o/(0.025 g/mL  x 1.0 dm)


[a] apparent = +34.8o g-1 mL-1 dm-1



 
If the fraction of the (+) enantiomer is x, then (1x) gives the fraction of the (-) enantiomer.  For any mixture of the two, the apparent specific rotation will be given by:


x(+139o) + (1x)(-139o) = [a]apparent


For this mixture:

x(+139o) + (1x)(-139o) = +34.8o

(+139x) - 139 (+139x) = +34.8

278x = 173.8

x = 0.63

Therefore, the mixture contains 63% of the (+) enantiomer and 37% of the (-) enantiomer.
























































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