Infra-Red Spectroscopy
Luke A. BurkepcIntroduction

Our conceptualization of chemistry is built on theory. Quantum Chemistry is now at the stage where the IR spectrum of organic molecules can be calculated with acceptable accuracy. However, we would still like to have a basic qualitative idea so that we can predict trends in a series of molecules. We start at the very basic models from General Chemistry, such as the idea that the atoms are bonded somehow to each other in molecules. From Rutherford's experiment, we have the idea of the nuclei as tiny spheres and the electrons as somewhere outside, flying around the nuclei and holding the whole thing together. We then assume that the nuclei are vibrating (oscillating) as if they were on springs Another assumption is that the bonds will only vibrate at certain frequencies (thus the latin word quantum, 'how much'), called "Vibrational States".

The frequency of IR light absorbed will be due to the exact difference between these vibrational states. We also know that the higher the bond order (single, aromatic, double, triple bonds, i.e. 1, 1.5, 2, 3) the more energy it takes to vibrate a bond. We can rationalize this by saying that there are more electrons between the nuclei in a double bond than in a single bond and so the 'spring' is stiffer. We see this from the equation relating the energy imparted, E to the frequency of the light absorbed, nu, where the the proportionality constant, h is called Planck's constant:

(eqn. 1a) E = h * n
or
(eqn. 1B) E(oscillator) = h * n

Let me now present you with a paradox. We know that CC single, double, and triple bonds absorb at different frequencies, therefore at different DE. If going from a single to a double bond doubles E1, then it should also double E2 . That means that DE is the same for a single and double bond because E1 and E2 are shifted equally. Thus, the CC single and double bonds should absorb at the same nIR, which contradicts experiment. Let's spot the problem.

We can make a model of molecules which resembles spheres held together by springs. From Hooke's Law we know that the vibrational frequency is directly related to the strength of the spring, k or Hooke's constant, and inversely to the mass of the spheres:

(eqn. 2a) nu(oscillator) = (1/2pi)(k/mass)**1/2
or
(eqn. 2b) E(oscillator)/ h = (1/2pi)(k/mass)**1/2

It is also seen that the energy needed to pull or compress the spring, E is equal to the square of the distance away from the rest position, d times k, Hooke's constant:

(eqn. 3) E = (1/2)k * d**2

Figure 1a.

Figure 1b.

"3N-6", "degrees of freedom", "normal modes"

How many peaks (or troughs) are there in an IR spectrum? Where M= the number of bonds, there are in general: M (to account for bond stretchings) + [M-1] (to account for bond angles) + [M-2] (to account for dihedral angles). I'll be using 'dihedral' angle and 'twist' angle interchangeably throughout this discussion. One can also use the number of atoms (N) times 3 to consider vibrations along the x, y, and z axes; but in this case one must subtract 6 from 3N when there are 4 or more atoms. This formula '3N-6' gives the number of 'degrees of freedom'. We must subtract 3 degrees for the first atom, 2 for the second and 1 for the third. This is because we start off with the first atom at one particular point (usually the origin). Since all the other atoms positions are defined in reference to the first atom, this first atom is not free. Thus 3 degrees are missing (The first atom is not bonded to any previously listed atoms, nor does it make an angle or dihedral angle). Then the second atom stretches along a line from the first atom (usually along the z axis). It can only move along a line in reference to the first atom. Thus 2 degrees are missing (the bond angle and dihedral angle). Then the third atom defines a plane (usually the xz plane). It can only move in a plane in reference to the first two atoms. Thus 1 degree is missing (a dihedral angle). It's only with the fourth and subsequent atoms that the atoms can be placed any where in 3D space, with any possible values of bond length, angle and dihedral angle.

Figure 2.

This rule of '3N-6' is applicable to molecules of 4 or more atoms. What rules are used for triatomic and diatomic molecules? What about linear, triatomic molecules?

I said that there are in general 3N-6 absorbances but there can be more in cases where absorbances are very intense and cause doubling like when the vibration of a piano wire can cause the vibration of another string an octave higher (or lower). Symmetry can also cause absorbances not to appear or to be very weak. We'll explore this latter case next.

Let's now take the case where two bond stretches are very similar or identical such as the two N-H stretches in an amine, -NH2. 'Normal modes' of vibration are stretches, bends, or twists involving similar parts of the molecule. In the case of an amine, the two N-H stretching vibrations are coupled. This means their vibrations are combined. These combinations are exactly like the combinations found for electronic energies. Taking the H2 molecule as an example, two 1s orbitals are combined in a bonding way (no new node) to form the sigma MO and in an antibonding way (one new node) to form the sigma* MO.

Figure 3.

Likewise, there are two combinations of R-N-H and R-N-H' bond angle vibrations. The two remaining normal modes are the N-R (e.g. Cl) stretch and the 'out of plane' [oop] twist of the two N-H bonds (like a bird flapping its wings).

Why should we have such similar energy diagrams for MOs and IR absorbances?

It's because they both involve motion and any motion is quantized, i.e. only certain states are allowed and each state has its own energy. A way to picture the quantization of energy is to observe the motion of a guitar string. Let's keep the string tied at one length so we don't change the notes (pitch, frequency, wavelength). If we pluck the string lightly, the string vibrates up and down in the middle, just like a sine wave from 0 to 180 degrees (or 0 to pi). What happens when we pluck harder? The amplitude is increased and we hear the vibration louder, but the frequency doesn't change. The extra energy has gone into increasing the amplitude. If we keep plucking harder, there will come point when the string will vibrate at a pitch which is one octave higher. (Actually this is accomplished more practically by changing the tension on the string or moving your finger half way down the neck of the guitar). Theoretically, there are octaves or harmonics, but you'll probably snap the string before you attain them. In the macroscopic world, which is described by classical mechanics, any amplitude is permitted in between modes. In the microscopic world of the molecule, which is described by quantum mechanics, only the modes are permitted. Permitted by whom? Experiment; we see only certain absorbances, which come about by changes in energy between certain energy levels. How do we explain (rationalize) the certain energy levels? Theory, we 1) assume the wave-like behavior of matter, 2) solve the Schroedinger equation for the different states described by wavefunctions, and 3) subtract the energy values of the two states obtained from solving the Schroedinger equation. The better our approximations used to solve the equation, the closer our numbers are to experiment. Sometimes, when our approximations are very small, we get better precision than experiment. There have even been cases where the numbers were very different from experiment but the experimentalists went back to their labs. An error was found in procedure, and the new numbers came close to those calculated by Theory.

Figure 4.

It's easy to picture the motion in vibrations but where's the motion in a 1s orbital? The point is that we can't picture it and don't need to picture it in order to calculate physical properties. It's enough that the electron is just moving. But if you insist on a picture, try the H atom: when the electron is described by a 1s orbital, it leads to the electron being found 99% of the time in a smaller sphere than if it was described by a 2s orbital. The motion of an electron (-) will be different when it is closer or further from the nucleus(+). As you can see, there are two different states described by two different wavefunctions (1s and 2s) each leading to two different energy levels. The difference in the energy levels gives you the energy (or frequency, E=h*nu) of the photon coming off when an electron "drops from a 2s to a 1s orbital".  These changes between energy levels (or between states) are called 'electronic transitions'.

The same is true for an electronic transition between the sigma and sigma* levels in the H2 molecule. HOWEVER, THERE'S A BIG DIFFERENCE BETWEEN THE COMBINATION OF STRETCHES AND THE COMBINATION OF ORBITALS. THIS DIFFERENCE IS IN THE OCCUPATION NUMBERS. We say that there is one electron from each 1s orbital, resulting in two electrons "occupying" the sigma MO. We have "two electrons in the sigma MO" because one has spin "up" and the other spin "down". With the vibrations, both H atoms must be vibrating together somehow and there must be two ways -modes- in which they vibrate. These modes are not states like electronic states. Each mode describes the way the atoms vibrate,  NOT HOW FAST they vibrate in that mode. It turns out that each mode has definite energy levels (quantized) and these levels are the vibrational states within each mode. That means that the "N-H symmetric stretch" we see in the IR spectrum at 3500(cm-1) comes from transition of the NH2 vibrating symmetrically at the slowest speed (lowest vibrational state) to the next higher speed (first excited vibrational state). It has nothing to do with a transition between symmetric and asymmetric modes.

Figure 5.

So where do the energy levels come from quantum mechanically? Solving the Schroedinger equation, of course H(Y)=E(Y), but this is very difficult to solve so let's make the approximation that the Hamiltonian operator H is just the operator that describes things vibrating according to Hooke's Law (called harmonic oscillators), i.e. a kinetic energy term and a potential energy term that looks like KE=(1/2)mv**2 and PE = (1/2)kx**2.

Through the marvels of differential calculus, the Schroedinger equation is solved for the harmonic oscillator and the states described by the wavefunctions Y(i) and their energies E(i) are obtained. For the first three states,

the ground vibrational state               Y(0) = ((2a/pi)**0.25) * 1 * exp(-ax**2)
                                                            E(0) = (0+1/2)h*nu

the first excited vibrational state       Y(1) = ((2a/pi)**0.25) * (2*((a**0.5)*x)) * exp(-ax**2)
                                                            E(1) = (1+1/2)h*nu

the second excited vibrational state   Y(2) = ((2a/pi)**0.25) * (4*(a**0.5)*x)**2)-2) * exp(-ax**2)
                                                            E(2) = (2+1/2)h*nu

All the Y(i) have three parts, the same normalization function in front and the same gaussian function on the right. The middle function (in bold, called a Hermite polynomial) differs for each Y(i). The Y(i) look a bit like a sine wave that goes through one-half, one, or one-and-a-half phases. Each higher Y has one more 'bump' which comes from the Hermite polynomial. As for the E(i), each has a quantum number (i+1/2). Remember that nu = (1/2pi)(k/mass)**1/2 from eqn. 2.

The lowest vibrational level Y(0) still has vibrational energy equal to (1/2)h*nu. This means that there is always an amount of vibrational energy (1/2 h nu) in a molecule, even at 0K. The total energy of a molecule is equal to the electronic energy plus the vibrational energy (plus the rotational [microwave] and translational energies).

E(tot) = E(el) + E(vib) + E(rot) +E(trn)

Figure 5.

In  figure 5, I have tried to match the classical parabola for an oscillator with the dissociation curve for the hydrogen molecule. The value of the E(vib) contribution from (1/2)h*nu is approximately 0.015 Ha (atomic units) or nearly 10 kcal/mol. (This is called the 'Zero Point Energy' ZPE because the zero'th quantum level still has (1/2)h*nu worth of energy). The value for the dissociation of H2 is approximately 100 kcal/mol. In the H2 molecule the ZPE represents 10% of the dissociation energy. I have only drawn three vibrational energy levels in figure 5. As you can see, the harmonic oscillator approximation (parabola within the dissociation curve) gets worse at higher levels. For molecules in general, the ZPE is calculated by summing over all its vibrations.

We have seen that one can consider each bond length and angle in a molecule to be a geometric parameter that can be described by a vibrating spring between two spheres or a swinging hinge for angles. In a classical oscillator, each energy level has the spheres vibrating between the two points at each end of the parabola in figure 5. These are the "turning points" at which extension in one direction halts and compression towards the middle begins. These points are where the kinetic energy is zero and all the energy is potential energy. The point in the middle is where all the energy is kinetic. Classical mechanical harmonic oscillators can have any energy level because you can extend them to any length and let go (provided the spring doesn't snap or warp). Quantum mechanical harmonic oscillators (because of their wave-like behavior) can have only certain levels. Also, as can be seen in figure 4, the wavefunctions extend past this barrier asymptotically towards infinity (which can not be seen in figure 4). This means there is a small but measurable probability of finding the vibrating particles past the 'walls' of their classical barriers. [This is the origin of effects such as beta decay in isotopes and tunnelling through energy barriers, which are topics in Physical Chemistry.]

Instructions:

In order to calculate vibrational frequencies and the motion of the nuclei in each vibrational mode of a molecule, you can use Gausview and Gaussian98:
1)if you're on a SUN type: xhost +hpc
2)telnet to hpc
3)type: gv
4)select element button
5)select N atom from periodic chart
6)bring cursor to center and click once. The NH3 molecule will appear.
7)select element button again, then F atom.
8)bring cursor to one of the H atoms and click once, making it a F atom.
9)select Calculate at top, then Gaussian98.
10)On the Gaussian 98 Calculation Setup box,
 a)delete everything in the Link0 Commands box
 b)in 'job type' box select 'opt+freq' in top box (it might read 'energy' originally) [This is different from class in that the program will search for the 'optimized geometry' where all the forces on the nuclei are zero. This is the equilibrium geometry.]
 c)in 'Method' box click choose AM1 from the middle pull-down menu
 d)click on edit, then answer 'Save' when it tells you you must save an input file
 e)type in a "filename.com" into the 'Save Your Gaussian File:' box. Choose an appropriate filename such as nh2f. Then click OK
 f)save and exit the edit window that appears.
 g)select 'cancel' in the little box that appears.
 h) You can then select cancel in the large 'Gaussian Setup' box.
11)In the hpc window, type
g98run  filename [There is no need for the g98x command and the g98run command works for file.dat or file.com}
12)A few seconds to a few minutes later check to see if the gaussian job is completed. This is done typing the command:

tail filename.out

in the hpc window. If the job is done, there will be a line that says "Normal termination".
13)When the job is done, click on open file, then type in filename.out. Don't worry if any of the sticks in the bonds are missing. Just click 'bond' in the build menu and then on the two atoms where a bond is missing. Then activate the single bond button and choose OK.
14)Click on Results, then Vibrations...
15) Move the 'Display Vibrations' box out of the way, and click on start.
16) click on each of the 6 vibrations. In reporting IR spectra, three things are noted: 1) the absorbance (usually reported in reciprocal cm, but also in microns in older publications), 2) the assignment of a name of the vibrational mode (str, bend; sym or asym), 3) the relative absorbance (on a basis of 0 to 1 where the strongest peak is assigned the value 1, or 0 to 100%, or s, m, w standing for strong, medium, or weak, which is a very subjective rating but is the most popular).

Now do the same for the carbonyl X2C=O, where X=H,F,Cl,Br,I and record all frequencies and assign names to their modes. Pay particular attention to the asymmetric C-X stretching mode.

Experiment for October 18th:

See if you can graph a trend for the C-X asymmetric vibrations. Remember that:

(eqn. 2) nu(oscillator) = (k/mass)**1/2

If you make the assumption that the C-X bond is a spring of the same strength for H and each of the halogens, you should obtain a graph (use gnuplot on clam, if you want) where the equation,

(eqn. 3)          nu = const / sqrt(mass).

is a straight line.

The 'mass' quantity expressed in equations 1-3 is only valid for homonuclear diatomic molecules such as N2 of F2. for heteronuclear diatomics (such as CO or NO) and other molecules with the structure of formaldehyde, X2C=O, one must use a quantity called the 'reduced mass' This quantity can account for the case where two parts of the molecule having different masses (m1 and m2) vibrate at different distances (r1 and r2) from the center of mass (M) of the molecule:

m1<----r1---->M<--------r2-------->m2

(eqn. 4)          reduced mass = (m1*m2) / (m1+m2)

You can now replot your data to see whether the reduced mass gives a straight line with less scatter. Use the atomic masses of C=O for m1 and (2 * atomic masses of H or the halogens) for m2. Due to the symmetry, the two C-H vectors can form one vector along the C=O axis and the form of the reduced mass in equation 4 can be used.

H
\
.........C=O
/
H

2H<-r1->M<--r2-->C=O

It was also assumed in equations 1-3 that the C-X bond strengths are the same. From the Handbook of Chemistry and Physics (71st edition), the bond strengths (D) in kcal/mol for the series of cyanides X-CN are H: 124, F: 112, Cl: 101, Br: 88, I: 73. You can now plot the following function:

(eqn. 5) nu = const * sqrt( D / reduced mass).

If there is still non linearity and scatter, try to think of possible causes. There is another set of bond strengths that you could try, Et-X, Et=ethyl. H:100, F:106, Cl:86, Br:69, and I:53. If you are curious, try the symmetric C-X stretches, then the C=O stretch. Do the symmetric bond angle wagging modes fit?

Why I prefer to say 'spheres and springs'

I once built a tiny benzene,
just like a classical machine.
On the forty ninth stroke,
the damned thing broke;
and its spheres were smashed to cream.

anonymous.

P.S. Something more to think about if you're still curious:

Let us suppose the dissociation energy of H2 calculated using molecular orbitals is 100.0 kcal/mol. This value is the same for the deuterium isotope D2 because the mass of the nucleus doesn't affect MO's, only the charge. Yet the rates of dissociation are different. Why?

Because the E(tot) = E(electronic) + E(vib) + E(rot) +E(trn)

Neglecting E(rot) +E(trn) for now (until Phys Chem), the E(vib) for H2 is different than that for D2. Why? (What do the energy levels depend on? What does the oscillation frequency, nu depend on?)

The ZPE for H2 is calculated to be 8.47 kcal/mol, that for D2 5.99 Kcal/mol. Thus the dissociation energy for H2 is (100.0 - 8.47)Kcal/mol and (100.0 - 5.99)kcal/mol for D2.

Although dissociation isn't strictly speaking an activation barrier, we can use Arrhenius's equation as an example:

Eqn. (6)    k(i) = A(i) *exp(- Ea(i)/RT),      where (i) is the species H2 or D2.

Assuming that the pre exponential factor A(i) is negligibly different for H2 and D2, we divide the equation for H2 by that for D2 and get

Eqn. (7)      k(H) / k(D)     =  exp(-Ea(H)/RT) /  exp(-Ea(D)/RT)

                 ln [ k(H)/k(D) ] = {[Ea(D)-Ea(H)] / RT}

                 ln [ k(H)/k(D) ] = {[94.0 - 91.5]Kcal/mol  /  (1.99cal/mol.K * 298K )}

                 ln [ k(H)/k(D) ] = 4.22

                       k(H)/k(D)    =  68.   or H2 dissociates 68 times faster than D2 at 298K

                  but k(H)/k(D)   =  23. at 400K,  9. at 600K,  3. at 1000K, and  2. at  2000K

You might want to check my arithmetic.

The ratio, k(H)/k(D) can be important in organic chemistry for distinguishing between mechanisms. In fact, it is not only stretches leading to bond breaking or H-atom abstraction which are used. The ratio, k(H)/k(D) when applied to (D)H-C-H(D) bond angle bending is used to distinguish between SN1 and SN2  and between E1 and E2 reactions. Another use of k(H)/k(D) is in solvation. Reactions carried out in D2O as a solvent are usually slower than in H2O, especially where there is significant H-bonding. It has also been shown that mammals given a steady diet of D2O have moderate to severe growth problems;
but don't get me started on that...