Optical activity is the ability of a chiral molecule to rotate the plane of plane-polairzed light. It is measured using a polarimeter, which consists of a light source, polarizing lens, sample tube and analyzing lens.

Measuring Optical Activity:

When rotation is quantified using a polarimeter it is known as an observed rotation, because rotation is affected by path length (l, the distance the light travels through a sample) and concentration (c, how much of the sample is present that will rotate the light).  When these effects are eliminated a standard for comparison of all molecules is obtained, the specific rotation, [a].

[a] = 100a / cl      when concentration is expressed as g sample /100ml solution

Enantiomers will rotate the plane of polarization in exactly equal amounts (same magnitude) but in opposite directions.

Dextrorotary designated as (+), clockwise rotation (to the right)
Levorotary designated as (-), anti-clockwise rotation (to the left)

For example, a 50:50 mixture of two enantiomers or a racemic mixture will not rotate plane polarised light and is optically inactive.  A mixture that contains one enantiomer excess, however, will display a net plane of polarisation in the direction characteristic of the enantiomer that is in excess.

Determining Optical Purity:

The optical purity or the enantiomeric excess (ee%) of a sample can be determined as follows:

Optical purity = % enantiomeric excess = % enantiomer₁ - % enantiomer₂
                     = 100 [a]_mixture / [a]_{pure sample}
 

ee%  =  100 ([R]-[S]) / ([R]+[S])

where [R] = concentration of the R-isomer
          [S] = concentration of the S-isomer

Diasteromeric substances can have different rotations both in sign and in magnitude. 

Optical Activity in depth:

Consider that (S)-2-bromobutane has a specific rotation of +23.1^o and (R)-2-bromobutane has a specific rotation of -23.1^o
 

Question:  Determine the optical purity of a racemic mixture.

Answer:  The specific rotation, [a], of the racemate is expected to be 0, since the effect of one enantiomer cancel's the other out, molecule for molecule.

Optical purity, %  = 100 [a]_mixture / [a]_{pure sample}
                            = 100 (0)  /  +23.1^o
                            = 0%

Question:  Determine the enantiomeric excess of the racemic mixture.

Answer:  You would expect [R] = [S] = 50%.

ee%  =  100 ([R]-[S]) / ([R]+[S])
         = 100 (50-50) / (50+50)
         = 0%

Let's consider something a bit harder......

Question: Which isomer is dominant and what is the optical purity of a mixture, of (R)- and (S)-2-bromobutane, whose specific rotation was found to be -9.2^o?

Answer: The negative sign tells indicates that the R enantiomer is the dominant one.

Optical purity, %  = 100 [a]_mixture / [a]_{pure sample}
                            = 100 (-9.2)  /  -23.1^o
                           = 40%  this indicates a 40% excess of R over S!

Question: What is the percent composition of the mixture?

Answer:  The 60% leftover, which is optically inactive, must be equal amounts of both (R)- and (S)-bromobutane.  The excess 40% is all R so there is a total of 70% (R) and 30% (S).

Try a similar problem.

Copyright ©2000 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use and Privacy Policy.
McGraw-Hill Higher Education is one of the many fine businesses of The McGraw-Hill Companies.
For further information about this site contact mhhe_webmaster@mcgraw-hill.com.