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

Answer 1

Answer: (B), (D)

Total (-) isomer is 14%

$\%\text{Optical purity}=\frac{\text{Observed optical rotation}}{\text{Optical rotation of pure enantiomer}}\times 100\%$

$\%\text{Optical purity}=\frac{9.72}{13.5}\times 100=72\%$
That means 72% is pure (+) 2-Butanol and 28% is racemic mixture.
$\text{Total (+) isomer}=(72+14)\%=86\%$
$\text{Total (-) isomer}=14\%$