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OPTICAL ROTATION
GENERAL SIMPLE MODEL
This effect arises from the ability of a changing magnetic field to induce an electrical dipole, and for a changing electric field to induce a magnetic moment.
H
m c t E c t
G
G
G
G
β and γ are constants that depend on the molecular structure under consideration. To get a qualitative feeling how this takes place, consider the following diagram (from Kauzmann, “ Quantum Chemistry ”).
The wavelength of the light is drawn much too short compared to the size of the molecular helix (just for graphics sake). Remember, from Maxwell’s equations, we
have (^) s
B
E dl n ds t
∫ ∫∫ ∂
G
G G G
v i^ w i , or in differential form:^
B
E
t
∇ × = −
G
G
. This is from the
Faraday induction (see the second lecture). There will be an emf produced in the helix when the helix axis is parallel to the component of the time derivative of the magnetic field. This emf is due to a circulating electric field that will cause the electron to travel along the helix, and this will produce an electric moment in the direction of the helix axis. We are not considering an electric moment perpendicular to the helical axis because we assume a mirror symmetry of the charge distribution along the helix – with the mirror plane parallel to the helical axis. The electric moment will only change in the direction of the helical axis (by symmetry). This induced electric moment is perpendicular to the
incoming electric field of the E&M wave (see figure above). Note that if the helix is right handed, then β is positive; if the helix is left-handed, then β is negative (this is an indication that we will be able to tell the stereochemistry of the molecule under examination!). Each molecule is therefore an oscillating electric dipole (at the same frequency as the incoming light), and will radiate just as any dipole. The wave front of this dipole will combine with the light wave, and they will add as vectors (Remember the way that the fields added when we calculated refraction.). Thus there will be a change in the direction of the resulting electric field of the light wave, and as the light wave passes through the solvent, the direction of the electric field of the light will slowly rotate. That is the plane of polarization of the light wave is rotated. So, any time that an electric dipole is produced by a changing magnetic field, the plane of polarization of the light passing through the sample will rotate.
Now think of the effect on a sample of the oscillating electric field. The oscillating electric field will act on the electrons of the helical molecule and it will force the electrons back and forth along the helical axis, but the electrons will move in a circular (actually helical) path. This circular motion of electrons will make a magnetic dipole along the axis of the helix. The oscillations of the electrons produce an oscillating magnetic dipole, and this will radiate just as an electric dipole, but with the directions of the fields as given in the figure on the next page. This will again rotate the magnetic field as the light beam passes through the sample, because the magnetic field direction of the emitted wave is perpendicular to the magnetic field of the incoming E&M wave.
The above is the qualitative reasoning why a linearly polarized light beam will have the plane of polarization rotated as it passes through the sample.
CALCULATE THE MAGNITUDE OF OPTICAL ROTATION DEPENDING ON
THE VALUES OF β and γ.
In the next section we will calculate β and γ for a simple model. For now, we assume they are just constants, and calculate their values depending on how much the polarization is rotated. If we are in a condensed phase, the electric field acting on the molecules will be affected by the surrounding molecules. This is given by the Lorentz
field: ( ( ))
2 2 0
cos 3 3 external
n n E E E ω t x c
G G G
. For now, we assume only
orientations of the helical axis along the x-axis (see the first figure of this lecture):
2 0
sin 3 x
n m c H H t x c c
G G^ G
and
2 0
sin y 3
n c E E t x c c
G G^ G
The magnetic dipole will produce a radiation field with the electric field coplanar to the incoming magnetic field, so that the electric fields of the radiated fields due to m (^) x and μ y are coplanar. We also see that the phase of the fields of the induced radiation is 90 degrees lagging behind that of the incoming magnetic field (this comes from the time derivative).
We consider the induced oscillating dipole, m = α E 0,int ernal cos( ω( t − x c ))
G G
Remember that the expression of the emitted field of this induced oscillating dipole is:
( ) ( (^ )) ( ) ( (^ ))
2 2 2 2
E ' 0 = − α E 0 ω rc sin θ cos ω t − r / c = − E 0 4 π α r λ sin θ cos ω t − r / c
G G G
And we showed earlier in the derivation for refraction that the electric field of the scattered forward-moving plane wave at position Δx from a front surface of atoms is:
( ) ( (^ ))
2
E ' s = − 4 π N α λ Δ xE 0 sin ω t − x c.
N is the number of molecules per cm^3.
Similarly (but not the same – we do not derive this) we can derive the following for the case that the oscillating dipoles are produced by the time derivatives of (both) E and H:
( ) (^ )^ ( (^ ))
2 2 0
' 4 cos 3 s
n E N x H t x c c
G G
This can be written as:
( ) (^ )^ (^ (^ ))^ ( ) (^ )
2 2 3 2 2 2 0
' 8 cos 8 s 3 3
n n
E π N λ x β γ E ω t x c π N λ x β γ E
G G
Now we have the expressions to calculate the rotational angle of the plane polarized light wave. Remember that the electric field of the radiated field is perpendicular to that of the incoming direction of the electric field oscillation. Therefore:
( ) (^ )
2 tan ' 8 2 2 2 s 3
n
χ χ E E π N λ β γ x
= = + Δ. The angle is rotating positively
clockwise when looking toward the incoming light wave. That corresponds to a positive χ. This is the convention in experimental optical rotation experiments.
Experimentally measuring optical rotation
One defines an optical rotational angle for a certain amount of substance and for a thickness of the solution. We use the conventional notation. For a pure substance
α = [ α ]Aρ , where A is the sample thickness, ρ is the density (gm/cm
(^3) ), α is the
measured rotational angle, and [α] is the specific rotation (/cm/gm). The χ calculated
above is per number of molecules, where N = N A ρ M , where M is the molecular weight
in gms. Also, A is usually given in decimeters (believe it or not!) and therefore A= Δ x 10
where Δ x in cm, and the angle α is in degrees, whereby χ is in radians, so that
α = 180 χ π. So we can write: [ ]
x
. Similarly for a solution at concentration
c (moles per cm^3 ), we have N = N c MA , and this leads to: [ ]
c x
. Substituting
in the expression for χ, we can write:
[ ]
2
A 3
n N M
π^ β^ γ
Often one defines what is called the molecular rotation , which is:
[ ] [ ] ( )
2 2 2
100^ A 3
M n M N
We now have an expression to calculate the specific rotation of a substance, and we see
that it depends, as expected, linearly on ( β + γ).
CALCULATING β AND γ FOR A SIMPLE, BUT REALISTIC, MODEL.
We consider the following model:
Calculating β ( m = − ( β c )∂ H ∂ t
G G
This helical model will have a dipole oriented parallel to the helical axis at all times. We will do this in two steps; first, for a constantly changing field, and then for a sinusoidally changing field. We will calculate β, and just indicate that a similar calculation gives the same expression for χ. Then we will show the averaging over all orientations of the helix to the incoming light wave. We indicate what happens when one goes through an absorption band. Also, we discuss what this all has to do with absorption (circular dichroism) and the Kramers/Kronig relation for dispersion functions in the complex plane.
The electrons will respond to the electric field, and they are confined to travel along the helical axis (the distance they travel is defined as q). Positive q means the electrons have moved clockwise around the helix, looking into the direction of the incoming light wave. The x-coordinate is along the helical axis. The radius of the helix is r, and the distance
between the equivalent points on the helical axis is 2 π s , which is the pitch of the helix
(this defines s, and is done for convenience). Then, when an electron travels a distance q along the helix, the distance traveled on the x-axis, x, can be calculated as:
x = qs r^2 + s^2. Now we just go back to the Newton’s equations of motion, where we
take the position variable to be q. The restoring force is: f (^) r = − kq , and the inertial force
is f (^) i = − m q (^) e . According to the Faraday electromagnetic induction law, the integration of
the potential around a loop (^) s c area
Δ V = ∫ E d =∫∫ B n ds
G G G
i A i
v w is related directly to the change
in the magnetic field dotted into and integrated over the area of the loop. The integration
over the area gives Δ V = ( area c ) H =( π r^2 c ) H
G G
(^). The distance along the curve “c” for
the line integral is 2 π r^2 + s^2 along the helix. Thus we can calculate the electric field
acting along the helix, which will allow us to calculate the force on the electron. Putting
everything together we have
2 ' (^2 2 2 22 )
V r E H
π r s c r s
G G
(^). Because of Lenz’s law,
the electron moves according to a left-handed rule, as shown in the picture. We can then
calculate the force on the electron to be
2 e '^22
er f e E H c r s
G
∓^
Now we can just use Newton’s equations (balancing forces, ignoring dissipative effects – absorption): 2 2 2
e
er m q kq H c r s
G
. This is what we want to solve. me is here the mass of an
electron. The dipole will always have an arrow above.
We consider three different cases.
- H ^ =constant , and at equilibrium : We are looking for the equilibrium maximum distance “q”, so we set q ^ = 0. This is the position of maximum extension. We can solve for q directly:
2 e 2 2 2
er q H kc r s
G
So the induced electrical moment (remembering that the model has two electrons opposite to each other) is
2 2 e^2 eq ˆ 2 2
e r s m ex x H kc r s
G G
So,
2 2 eq 2 2
e r s k r s
This is the maximum electric dipole moment that can be formed when there is a constant rate of change of the magnetic field.
between the normal of the area perpendicular to the helical axis and H
G
(^). So the
expression for the induced moment will be m = −( β avg c H )
G
(^) , where
cos^2
β avg = φ avg β = β. Thus
2 2 2 2 2 2 0
avg e
e r s m r s
β ω ω
Calculating γ
We will not do this calculation, but just state that:
2 2 2 2 2 2 0
avg (^3) e
e r s m r s
γ ω ω
So the total expression for ( β+γ ):
2 2 2 2 2 2 0
(^3) e
e r s m r s
β γ ω ω
Compare this to the off resonance expression for the polarizability:
2 2 2 2 0
(^4) e
e m
; they are very similar expressions, and depend on the frequency in
an identical fashion, which we might expect.
We can put the expression for (β+γ) into our formula for [M] to get:
[ ] [ ] ( )
2 2 2 2 2 2 0 2 2 2 2 2 2 0 2 2 2 (^12 ) 2 2 2 2 0
A A e
M n n e r s M N N m c r s n r s r s
= ×
Note that now we have an expression to compare to molecular dimensions (r and s) and to stereochemistry (rotating right or left – right handed helix or left handed helix).
