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Chapter 7: Stereochemistry

Summary | Isomer types | Enantiomers | The Stereogenic center | Optical Activity | Absolute Relative Configuration | Cahn-Ingold-Prelog R/S notation | Fischer projections | Self Assessment | Quiz |

Optical Activity

Chapter 7 : Stereochemistry

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.

schematic diagram of a polarimeter
When light passes through a sample that can rotate plane polarized light, the light appears to dim to the eye because it no longer passes straight through the polarizing filters.  The amount of rotation is quantified as the number of degrees that the analyzing lens must be rotated by so that it appears as if no dimming, of the light has occurred.

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
Specific rotation is a physical property like the boiling point of a sample and can be looked up in reference texts.    Take a look at a problem.

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)
If only one enantiomer is present a sample is considered to be optically pure.  When a sample consists of a mixture of enantiomers, the effect of each enantiomer cancels out, molecule for molecule.

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

Look at some problems like these more in depth.

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 +23.1o and (R)-2-bromobutane has a specific rotation of -23.1o

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)  /  +23.1o
                            = 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 -9.2o?

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)  /  -23.1o
                           = 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.



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