The specific rotation of pure (-)cholesterol is -39o g-1 mL-1 dm-1. What is the apparent specific rotation of a sample containing 90% of the (-)enantiomer and 10% of the (+)enantiomer?



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[a] = a/(c x l)

 


The specific rotation, [a] = -39o

The fraction of the (-)enantiomer is 0.90 (x), and the fraction of the (+)enantiomer is 0.01

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


x(-39o) + (1x)(+39o) = [a]apparent





















































Solution:

The specific rotation, [a] = -39o

The fraction of the (-)enantiomer is 0.90, and the fraction f the (+)enantiomer is 0.01



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(-39o) + (1x)(+39o) = [a]apparent


For this mixture:

[a]apparent = 0.90(-39o) + (0.10)(+39o)

[a]apparent = (-35.1) + 3.9

[a]apparent = -31.2

























































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