The specific rotation of pure (+)2-octanol is +10o g-1 mL-1 dm-1. A sample containing both enantiomers is found to have an apparent specific rotation of +4.0o g-1 mL-1 dm-1. What is the optical purity of the sample?



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[a] = a/(c x l)

 


The specific rotation, [a] = +10o

The apparent specific rotation is,  [a]apparent = +4o g-1 mL-1 dm-1

You should calculate the "fraction (x) of the (+)enantiomer" using...


x(+10o) + (1x)(-10o) = [a]apparent





















































Solution:

The specific rotation, [a] = +10o

The apparent specific rotation is,  [a]apparent = +4o 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(+10o) + (1x)(-10o) = [a]apparent


For this mixture:

[a]apparent = x(+10o) + (1x)(-10o) = +4

(+10x) - 10 (+10x) = +4

20x = 14

x = 0.7

Therefore, the mixture contains 70% of the (+) enantiomer and 30% of the (-) enantiomer.

























































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