Question

If a sample of pure R-enantiomer of a molecule has a specific rotation of $( - 40^\circ )$ and an enantiomeric mixture of that molecule has 82% enantiomeric excess. What would be the observed rotation of the sample?

Answer 1

Enantiomers are optical isomers which are chemically identical but differ in the way they rotate plane polarised light. Enantiomers can be either dextro rotatory (d) or levorotatory (l). Enantiomers are pairs of molecules and they are mirror images of each other, but these images cannot be superimposed on each other.

Formula used:
$\% ee = \dfrac{{{\alpha _{obs}}}}{{\left| \alpha \right|}} \times 100\%$
Here, %ee is an enantiomeric excess, $\alpha_{obs}$ denotes the observed specific rotation and $\mid {\alpha} \mid$ is a specific rotation of pure enantiomers which is in excess.

Complete step by step answer:
The optical purity of the mixture is expressed as enantiomeric excess or ee and it is expressed in percentage. Enantiomeric excess tells us which enantiomer is present in excess as compared to that of another enantiomer.
We need to keep this in mind that the given molecule has two enantiomers, R and S.
The enantiomeric mixture contains 82% enantiomeric excess, therefore $\% ee = 82$.
Now, we need to understand which enantiomer is in excess, R or S in the mixture.
Now, to move further we need to first look at the sign of the specific rotation. The specific rotation of a compound is a physical property of the chiral compound which is measured in terms of rotation of plane polarised light by the enantiomer of that compound.
Two enantiomers will always rotate plane polarised light up to the same extent but in different directions. The specific rotation for R enantiomer is given as -40⁰. The sign of specific rotation is negative, and it tells us that R enantiomer is in excess.
To calculate the observed specific rotation of the sample, we need to use the following formula.
$\% ee = \dfrac{{{\alpha _{obs}}}}{{\left| \alpha \right|}} \times 100\%$
Let us put the respective values, we will get
$82\% = \dfrac{{{\alpha _{obs}}}}{{ - 40}} \times 100\%$
On rearranging the equation, we get
$\Rightarrow \dfrac{{82\% }}{{100\% }} = \dfrac{{{\alpha _{obs}}}}{{ - 40}}$
$\Rightarrow 0.82 = \dfrac{{{\alpha _{obs}}}}{{ - 40}}$
$\therefore {\alpha _{obs}} = 0.82 \times ( - 40^\circ ) = - 32.8^\circ$

The observed specific rotation of the sample is - $32.8^\circ$.

Note: An enantiomeric mixture containing 50-50 percent of both R and S enantiomers is optically inactive. This achiral mixture is also known as a racemic mixture. The two enantiomers will rotate the plane polarised light to the equal extent but in the opposite direction, resulting in cancellation of net rotation, making it optically inactive. The specific rotation of the racemic mixture is zero. Students may get confused in understanding which enantiomer is in excess. The sign of specific rotation of the mixture will depict which enantiomer is in excess. The negative sign of the specific rotation will depict that R enantiomer is in excess, but positive sign will tell us that S enantiomer is in excess.