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Question

2-Butanol having an observed rotation of +9.72 has the rotation for the pure enantiomer calculated as +13.5
Then choose the correct option(s):

A
Total (-) isomer is 14%
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B
Total (-) isomer is 36%
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C
Racemic mixture is 72%
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D
Racemic mixture is 28%
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Solution

The correct option is D Racemic mixture is 28%
%Optical purity=Observed optical rotationOptical rotation of pure enantiomer×100%

%Optical purity=9.7213.5×100=72%
That means 72% is pure (+) 2-Butanol and 28% is racemic mixture.
Total (+) isomer=(72+14)%=86%
Total (-) isomer=14%



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