Stereochemistry: Specific Rotation and Related Calculations

Discussion: With one exception, the chemical and physical properties of enantiomers are identical in the absence of other chiral materials.  The exception is the rotation of plane polarized light.    Because plane polarized light is effected differently by enantiomers, it can be a useful laboratory tool to quantify the amount of an enantiomer or enantiomers present in a sample.  Recall that equal amounts of mutual enantiomers rotate plane polarized light to an equal extent but in opposite directions.  Rotation of plane polarized light occurs at the molecular level, so the extent of the rotation is proportional to the amount of optically active material that is present as expressed in the following equation.

[α]lT = α/lc

Where:

• [α]lT = specific rotation in degrees. (The correct units are deg cm2 g-1, but are usually just given as degrees).  l is the wavelength of light used for the observation (usually 589 nm, the D line of a sodium lamp unless otherwise specified.  This wavelength is responsible for the orange-yellow color of the common sodium vapor street light). T is the temperature in oC.  This value is characteristic for a given compound, just like the melting point.  Concentration and solvent data is included if relevant.  Example: (c = 10, CH3OH) after the rotation means the specific rotation was determined at a concentration of 10 g  in 100 ml-1 in methanol.
• α = observed rotation in degrees.
• l = cell path length in decimeters. (1 decimeter = 1 dm = 10 cm. A standard polarimeter tube is 1.00 dm in length.)
• c = concentration in g ml-1 for a pure liquid compound (i.e., the liquid's density), or g 100 ml-1 for a solution.

• Armed with these facts, and some basic concepts from general chemistry, we can perform just about any sort of simple calculation involving optical activity and enantiomeric composition of a sample.

Example 1: A sample of pure (S)-2-butanol was placed in a 10.0 cm polarimeter tube.  Using the D line of a sodium lamp, the observed rotation at 20oC was α = +104o.  The density of this compound is 0.805 g ml-1.  What is the specific rotation of (S)-2-butanol?

Solution: Plugging the numbers into [α]lT = α/lc we get: [α]lT = (+104o) / (1.00 dm) (0.805 g ml-1) = +129o.  Thus we would write [α]D20 = +129o (neat). ("Neat" refers to a liquid that has not been diluted.)

Example 2: Calculate the observed rotation of a solution of 0.5245g of (S)-1-amino-1-phenylethane diluted to a volume of 10.0 mL with methanol at 20oC, using the D line of a sodium lamp and a 1.00 dm tube.  Specific rotation of this material: [α]D23 = -30.0o.

Solution: Solving the specific rotation equation for observed rotation, we get α = [α]lTlc.  The sample size is 0.5245 g, but this has been diluted to 10.0 mL, so the sample concentration is 5.245 g in 100 ml.  Plugging in the numbers, we get α = (-30.0o) (1.00 dm) (5.245 g in 100 ml).  Solving, we find α = -157o.

Exercises:

a. Calculate the specific rotation of (2R,3R)-tartaric acid based on the following observation: A 0.856 g sample of the pure acid was diluted to 10.0 mL with water and placed in a 1.00 dm polarimeter tube. The observed rotation using the 589 nm line of a sodium lamp at 20.0 oC was α = +1.06o.

b. What is the expected observed rotation of a 1.0 x 10-4 M methanol solution of the potent anticancer drug paclitaxel (also called Taxol)? [α]D20 = -49o (c = 1, CH3OH). Paclitaxel has a molecular weight of 853.93 g mole-1.

c. A certain compound has a specific rotation of -43.2o (c = 5, toluene). What is the observed rotation of a sample of 1.24 g of the enantiomer of this compound when diluted to a concentration of 1.00 g ml-1 in the same solvent?

d. In his classic studies of stereochemistry and optical activity in organic compounds, Pasteur measured the optical activity of many solutions. For the naturally occurring enantiomer of tartaric acid, [α]D20 = +12.4o (c = 20, H2O). What can be concluded about the ratio of tartaric acid enantiomers present in the solution if the observed rotation is (i) α = -6.0o, or (ii) α =  0o?