Question

What is enantiomeric excess of mixture containing 25 % of +2-butanol and 75 % of -2-butanol, if specific rotation of +2-butanol is $13.5{}^\circ $.

Answer 1

Hint: Enantiomers are the same molecules that are mirror images, containing a chiral centre. They differ in rotating the plane polarized light towards left and right.

Formula used: Enantiomeric excess, ee = $\dfrac{[d-l]}{d+l}\times 100%$

Complete step by step answer:
We are given a mixture of a compound that contains some amount of enantiomers of the same substance, 2-butanol. These enantiomers are 25 % of +2-butanol and 75 % of -2-butanol. We have to calculate the enantiomeric excess, ee.
As we know, enantiomers are mirror images, and rotate the plane polarized light, PPL. The one that rotates it towards right is termed with (+) or d, while that rotates it left is termed with (-) or l. Here, 25 % of +2-butanol is d and 75 % of -2-butanol is l. so, according to the formula ee = $\dfrac{[d-l]}{d+l}\times 100%$, we have,
d = 25, and l = 75, so,
ee = $\dfrac{[25-75]}{25+75}\times 100%$
ee = $\dfrac{-50}{100}\times 100$
ee = 50%
Hence, the enantiomeric excess ee of this mixture is calculated to be 50 %.

Note: Specific rotation, $\alpha $ is the rotation change of the compound when PPL passes through it. It has observed value and the value of pure mixture, in this case, the value observed of specific rotation, from the given value can be calculated, by the enantiomeric excess, as ${{\alpha }_{obs}}=13.5\times \dfrac{50}{100}=6.75$.
 Enantiomeric excess, ee is also calculated from the formula , ee = $\dfrac{{{\alpha }_{observed}}}{{{\alpha }_{do\min ant}}}\times 100$