Question

A sample solution of [(+/-)-Co(en)₃]I₃ (with no impurities) is found to have [α]_D = +77^o, whereas a pure enantiomer would give a value of [α]_D = +89^o. Calculate the exact percentage of (+) and (-) isomers present in the sample solution. (note: you must first reason whether the rotation value indicates the total amount of + isomer or the enantiomeric excess).

Answer 1

When a beam of plane polarized monochromatic light falls on chiral chemical compound,there occurs a change in it's orientation depending on concentration of the chiral chemical compound and the pathlength travelled by the light.

Compounds (enantiomeric forms) which rotate light clockwise are said to be dextrorotary (+ isomer), and correspond with positive specific rotation values, while compounds (enantiomeric forms) which rotate light counterclockwise are said to be levorotary (- isomer) and correspond with negative values .

In our case, rotatory value, [α]D = +77^o and [α]D = +89^o indicate presence of a + isomer.

Specific rotation is a physical property of a chiral chemical substance which can be used as reference for identifying the substance. In other words, specific rotation can be used for assessing optical purity of substance like other physical properties of substance such as boiling point can be used to assess purity of substance.

Specific rotation is defined as,

Specific rotation [α]_D = α_obs / c l

where "α_obs" is the experimentally observed rotation, "c" is the concentration in g/ml and "l" is the pathlength of the cell used expressed in dm

When the two enantiomeric forms are present in 50:50 ratio they cancel each other's optical rotation resulting in no rotation. Such a mixture of enantiomers is called racemic mixture.

Therefore, the optical purity of a substance can be calculated by using formula:

% Optical purity of sample = 100 * (specific rotation of sample) / (specific rotation of a pure enantiomer)

Substituting values given in our problem,

% Optical purity of sample = 100 * (77) / (89)

% Optical purity of sample = 86.51

This means sample contains 86.51% of + isomer and

% of - isomer=(100-86.51)=13.48% (The compound does not contain any other impurity)

Enantiomeric excess is excess of + isomer over the - isomer

Therefore, the % enantiomeric excess (%ee) can be calculated by using formula:

ee % = 100 * ( [ R ] - [ S ] ) / ( [ R ] + [ S ] )

ee % = 100 * ( [ 86.51] - [ 13.49 ] ) / ( [ 86.51 ] + [ 13.49 ] )

ee %=73.033