Review of Circular Dichroism Absorption of photons by molecules requires that electrons in the molecule be able to absorb the energy of the photon and also its momentum (energy is conserved and momentum is conserved). Quantum-mechanically, we talk about the transition moment of the molecule, which embodies the displacement of electrons in conjunction with their excitation from one orbital to another. Dichroism is when a molecule is best able to absorb photons with particular polarity (that matches the molecule s own polarizability). Polymers oriented in this way (such as polaroid) are used to generate polarized light in which the electrical field vector oscillates in one direction only. Spectroscopic Methods in Biochemistry, chapter 4 http://www.uni-konstanz.de/fuf/bio/folding/4-cd
Components of a spectropolarimeter Eisenberg and Crothers 1979 Plane-polarized light is created by passing the beam through a material that absorbs all of the light polarized in one direction (here X) and therefore transmits only the Y-polarized light. Alternately a Nicol prism can be used to reflect all light polarized in one direction, transmitting only light polarized perpendicular to that direction. NOTE, most reflection is polarized to some extent (hence the utility of polaroid glasses). Optical Rotation temp [ α ] = λ 100α l C angle in degrees path length (dm), concentration g/100 ml 20 For α-d-glucose [α] = +112.2 D + means rotation to the right as look into the beam. Eisenberg and Crothers 1979
Optical activity is a property of molecules with chiral centres. Tetrahydrocannabinol A sample problem If a sample of bromobutane measured under standard conditions has an observed rotation of 9.2, what is the enantiomeric excess? Examples (S)-2-Bromobutane +23.1 (R)-2-Bromobutane 23.1 D-Fructose 92.4 D-Glucose +52.5 D-Sucrose +66.47 D-Lactose +52.3 Camphor +44.1 Cholesterol 31.5 Paclitaxel 49 Penicillin V +223 (+)-Cavicularin +168.2 Hexol bromocamphorsulphonate 2640 [2] All values are given in units of deg dm 1 cm 3 g 1.
Optical Rotatory Dispersion = Optical rotation vs. λ Due to physics we will not discuss, the lines are dispersive and can be difficult to deconvolute when they overlap. We prefer Circular Dichroism for structural studies of complicated macromolecules. The following exercises allow you to see for yourself how both Optical Rotation and Circular Dichroism arise from differential effects on the two rotationally polarized components that make up plane-polarized light. i.e. one is affected differently than the other. λo is the λmax for absorption. Eisenberg and Crothers 1979 Plane-polarized light can be thought of as the sum of two oppositely circularly polarized light components. (Sum, left-polarized, right-polarized). As an exercise: complete each time-point cartoon by drawing in the blue vector and the sum vector (black) to confirm that the sum of two oppositely rotating components of the same amplitude is plane-polarized.
Optical Rotation is the result of different refractive indices for the two different circularly polarized components of light. This causes one component to rotate more slowly than the other (v=c/n, n = refractive index). As the light passes through the material, one polarization of the light is delayed relative to the other. Once they emerge, they both recover the same speed. Optical Rotation: After passing through a birefringent material, one of the two circularly polarized components has been delayed more than the other. (Sum, left-polarized, right-polarized) As an exercise: draw in the red and blue components after each has progressed another 45 and then draw in the sum at each point.
Circular dichroism Δε = ε L ε R ε is a function of λ therefore so is Δε. Δε can be either negative or positive (vs. ε which is always positive). CD only occurs near an absorption band (i.e. ε 0) It also only occurs for molecules that are optically active. Optical activity: a molecule cannot be superimposed on its mirror image. Opposing enantiomers have oppositely signed CD. 11 Circular Dichroism: After passing through a dichroic material, one of two oppositely circularly polarized light components is weaker than the other (has been absorbed more due to a higher ε). (Sum, left-pol, right-pol). As an exercise: complete each time-point cartoon by drawing in the blue vector and the sum (black) vector to confirm that the sum of two oppositely rotating components of different amplitudes is elliptically polarized but not rotated. (Consider that the light has now emerged from the material and the blue vector does not shrink further.)
Solution to sample problem If a sample of bromobutane measured under standard conditions has an observed rotation of 9.2, what is the enantiomeric excess? Examples (S)-2-Bromobutane +23.1 (R)-2-Bromobutane 23.1 D-Fructose 92.4 D-Glucose +52.5 D-Sucrose +66.47 D-Lactose +52.3 Camphor +44.1 Cholesterol 31.5 Paclitaxel 49 Penicillin V +223 (+)-Cavicularin +168.2 Hexol bromocamphorsulphonate 2640 [2] All values are given in units of deg dm 1 cm 3 g 1. -9.2 is 40% of -23.1 so we have 40% excess of R form. The remainder of the sample is an equal mix of R- and S- because it contributes no net optical rotation. Thus we have an additional 30% R and 30%S. Total R is 70% and total S is 30 %. Enantiomeric excess is 40%. Plane-polarized light can be thought of as the sum of two oppositely circularly polarized light components. (Sum, left-polarized, right-polarized) Answer
Optical Rotation: After passing through a birefringent material, one of the two circularly polarized components has been delayed more than the other. (Sum, left-polarized, right-polarized) Answer (note that the black sum vector always stays in the same plane although its magnitude and sign oscillate). Circular Dichroism: After passing through a dichroic material, one of two oppositely circularly polarized light components is weaker than the other (has been absorbed more due to a higher ε). (Sum, left-pol, right-pol). Answer: Copying all the sum arrows to start at a single point, we see that they map out an ellipse.