This section is about selected spectroscopies other than NMR: absorption (UV-vis and IR), Raman, fluorimetry, X-ray crystallography, and mass spectroscopy. The first four of these use photons to probe molecules’ identities and characteristics. Such information is widely applicable for understanding how chemical reactions work in industry, in nature, and in our bodies. Mass spectroscopy involves ionizing the molecules of a sample and then measuring the masses of the ions in order to deduce the identity of the sample molecules.

Below is an diagram of representative energy levels describing the electrons of a molecule. Putting the molecule in an excited electronic state involves exciting an electron of the molecule into a higher energy molecular orbital. For most molecules, the energy difference between valence molecular orbitals is in the range of uv-vis (sometimes ir) photons.

Putting the molecule in an excited vibrational state involves giving the nucleii of the molecule energy to vibrate. The nucleii can vibrate only in certain specific ways (called "normal modes-" you’ll learn more about normal modes in physics or p-chem classes), so like electronic states, vibrational states are quantized. The energy difference between vibrational states is usually in the range of ir photons. Electrons do not jump into different orbitals when the molecule moves into an excited vibrational state. That’s why it makes sense to talk about the "ground vibrational state of an excited electronic state" or the "third excited vibrational state of the ground electronic state," etc. Make sure you can find these states on the energy diagram below.

Putting the molecule in an excited rotational state involves causing the molecule to rotate in different ways. The energy difference between rotational states is very small and is probed by photons of microwave energies.

Note that NMR spectroscopy, which also uses photons to probe molecules’ identities and characteristics, does not make use of any of the electronic energy levels shown above. In NMR, we are concerned with the nucleii of the atoms comprising a molecule. We create an energy difference between the spin states of certain atomic nucleii (not electrons!!) via application of a magnetic field. Then we send in radio wave photons. The energies of the photons that get absorbed tell us how big an energy difference was induced by the magnetic field for the different nucleii. That in turn yields information about the chemical environment of those nucleii in the molecule, which gives us clues about how the molecule is put together. So you can see that the NMR energy difference wouldn’t appear on the diagram above even if it were big enough to be noticable, because the spin states we are probing are degenerate until we place the molecule in a magnetic field!!

In an absorption experiment we send in a beam of photons of varying energies and detect which ones get absorbed. The photons that get absorbed correspond to the energy differences between electronic or vibrational states. The epsilon value e in the equation at the detector above is a measure of how many photons of a certain energy get absorbed, or the spectroscopic intensity of the transition (i.e. the height of the peak on the spectrum). Note that the equation at the detector is a statement of what is known as Beer’s Law. Uv-vis and IR absorption spectroscopies are the spectroscopies we will study for which Beer’s Law holds.

UV-vis and IR spectroscopy and their governance by the Beer’s Law equation are big topics on chem 32 exams. Note that absorption measurements should be taken at the absorption maximum. This is because the derivative of the curve in this region is the smallest and therefore a slight discrepancy in wavelength will not have a huge effect on the spectrum.

The spectroscopic intensity of a transition is governed by "selection rules." Sometimes these selection rules tell us that a transition can’t happen at all. For example, an excited vibrational state will be detected by an absorption spectrometer only if the dipole moment of the molecule changes during the course of the vibration. Look at the following picture of the normal mode vibrations of CO₂, a molecule that does not have a dipole moment. The second two vibrations may have peaks in the IR absorption spectrum; the first will not.

Uv-vis and ir absorption spectroscopy is used mainly to figure out what a molecule is. Organic functional groups have characteristic vibrational normal modes, so ir transitions of certain energies can be used as spectroscopic fingerprints of those functional groups. You should be familiar with some characteristic IR frequencies because they allow you to tell what functional groups are in a molecule just from a quick peek at the spectrum.

Compounds that have electronic transitions in the visible region are colored. For example, if a molecule has a transition of an energy corresponding to red-orange light (say around 580 nm or so), it will absorb red-orange light and allow the other colors to pass through. Therefore, such a molecule looks blue-green to our eyes.

Uv-vis light is sent in, putting the molecule in an excited vibrational state of an excited electronic state. Instantaneously, the molecule emits a photon and falls back to an excited vibrational state of the ground electronic state. We know the energy of the photon going in, and we measure the energy of the emmitted ("scattered") photon at the detector. The difference in energy between these two photons tells us the energy difference between the ground vibrational state and the excited vibrational state of the ground electronic state. Refer to the diagram of the energy levels to make sure that you’re keeping straight what these transitions mean.

Notice that the lightsource for this experiment is a laser, which sends out an especially intense beam of photons. The reason we need so very many photons is that few of them will actually undergo the absorption/re-emmission process described in the preceeding paragraph. Of every million photons that we send in only one will be scattered, and of the ones that are scattered only one in 10,000 will have smaller energy than the ones we sent in. So we can only actually use 1 in 10⁹ of the photons that we send in to make the measurement!! This is the reason that we never need to worry about Raman scattering interfering with absorption or fluorescence measurements, which use lightsources far less intense.

Photons emmitted by a molecule are scattered in all directions, so we can place the detector pretty much anywhere. However, we wouldn’t want to put it right behind the sample, because then we would be doing absorption spectroscopy, and a laser lightsource is too intense to be suitable for absorption spectroscopy (we’d fry the detector if we tried a stunt like that!!). In practice, the detector for Raman is usually placed at an angle 90^o to the incoming laserbeam.

Why would we want to spend thousands of dollars on a laser when we can already see vibrational transitions using a much cheaper ir absorption spectrometer? The reason is that the selection rules for photon scattering are different from the selection rules for photon absorption. Recall, that first symmetric stretch of the CO₂ molecule could not be seen using ir absorption because the dipole moment does not change during the course of the motion. That vibrational transition is observed in Raman. Ir absorption and Raman spectroscopies complement each other beautifully: a transition that is forbidden for one is usually allowed for the other. A slightly more formal statement of the Raman selection rule is that the "polarizability of the molecule’s electron cloud" has to change during the course of the vibration. To put it non-mathematically, symmetric stretches are Raman-active; asymmetric stretches are ir-active.

Another use of Raman is that it allows us to assign peaks in the uv-vis absorption spectrum. The reason for this is that vibrational transitions involving atoms in the MO’s of an electronic transition show enhanced intensity when the excitation photon has the energy of the electronic transition. So if we have a molecule with a carbonyl on it and our Raman excitation photons have the energy of the electronic transition between C-O ¼ and ¼*, then the CO stretching vibration (around 1870-1550 cm^-1) will have extra intensity. So if a chemist sees extra Raman intensity around 1870-1550 cm^-1, she might deduce that there is a carbonyl in her molecule and that her excitation frequency corresponds to the electronic transition between the C-O ¼ and ¼* orbitals. If she’s right, there will be a peak on her absorption spectrum right at her excitation frequency, and this can be assigned to the C-O ¼ to ¼* transition.

Make sure you understand the section on Raman before reading this section!! Like Raman, fluorimetry uses light in the uv-vis energy range to put the molecule into an excited vibrational state of an excited electronic state. The difference is that the molecule stays in the excited electronic state for a while (we say that the excited state has "longer lifetime"). Then the molecule loses some of its energy thermally (i.e. without emitting a photon--the energy is lost as heat that goes into the vibrations of surrounding molecules) so that it ends up in the ground vibrational state of the excited electronic state. Then it emits ("scatters") a photon that brings it back to some vibrational state of the ground electronic state. As in Raman, this results in an emitted photon of lower energy than the photon we sent in.

So the only thing that really distinguishes fluorescence from Raman is the aforementioned thermal loss of energy, which occurs when the excited state has a long enough lifetime to decay a little bit before the transition back to the ground state occurs. Look at the following diagram of a few representative fluorescence transitions (and see if you can think of some other examples of transitions that could happen using this technique).

Thus, what we see on a fluorescence spectrum is a broad band centered around an energy slightly lower than the energetic difference between the ground and excited electronic states.

For most metal atoms, the "fluorescence" signal that we observe is really phosphorescence, which is a slightly different phenomenon with similar general principles. The main idea again is that we get the molecule into an excited electronic state, it relaxes thermally (and radiationlessly) a little bit, then falls back down to some vibrational state of the ground electronic state with the emission of a photon. The difference is that the radiationless decay for phosphorescence involves changing the system from a singlet to a triplet by flipping the spin of one of its electrons (so that the excited state is actually paramagnetic!). This is a "forbidden" process, but it is made possible in heavy atoms like metals by a phenomenon called "spin-orbit coupling." When the system falls back down to the ground electronic state, it becomes a singlet again via the spin-orbit coupling. It is actually phosphorescence that is responsible for the red color of rubipy.

For now, what you should understand about fluorimetry is that we put the molecule in an excited state of the excited electronic state, allow it to lose energy thermally without any emission of photon radiation, and then see a lower energy transition back to the ground state on our spectrum. You should also understand why molecules that absorb blue light might fluoresce red light.

X-ray crystallography is a technique that is used to figure out exact bond lengths and geometries in solid crystals. In performing x-ray crystallography, we are not probing any of the energy level splittings diagrammed at the beginning of this section, although knowledge of bond lengths and geometries can sometimes yield insight into those splittings. It is often the case that the molecule we wish to study resists our efforts to get it into crystalline form; in these situations we have to rely solely on the other spectroscopic techniques to deduce its structure.

X-ray crystallography involves sending x-rays into a crystalline sample. The sample can be a polymer, an enzyme, an organic compound, an inorganic compound: basically anything, but it has to be in crystalline form. Most of the x-rays with which we bombard the sample will pass through it unaffected, but a small fraction of them will be reflected. Some will be reflected from the top layer of the sample and others will be reflected by layers underneath. The key is that the x-ray waves will exit the sample in phase where-ever the interlayer spacing d = nl/2sinq. In this expression, n is any integer, so that nl represents an integral number of wavelengths. The angle q is the angle between the incoming x-rays and the surface of the sample. Note that it must be the same for all of the x-rays: the x-rays must come in parallel. The reason this equation works is that the path of the bottom x-ray in the diagram below is exactly dsinq + dsinq = 2dsinq longer than that of the top x-ray. If the 2dsinq is an integral number of wavelengths, then the x-ray will still be in phase with the x-ray that didn’t traverse the extra space.

So the punchline is that we can send in waves of a certain wavelength at a certain angle and then solve the equation for the interlayer spacing d to calculate how far apart the layers are in the crystal. It seems like quite a jump from that to knowing exactly which atoms are where in any given molecule (and it is), but this is the basis of the technique.

Mass spectroscopy involves ionizing the molecules of a sample and then measuring the masses of the ions in order to deduce the identity of the sample molecules. Again, mass spectroscopy does not probe the energy level splittings of the first page; it is a technique for determining the mass of a sample. The sample molecules are ionized by bombarding the sample with electrons. Then we have to find the masses of these ions. They are measured via the radii of the circles they traverse in a magnetic field.

The force felt by a charged particle in a magnetic field is the particle’s charge times the cross product of the particle’s velocity and the magnetic field strength. The cross product of two vectors is defined as the magnitude of the first times the magnitude of the second times the sin of the angle between them, but as you can see from the diagram above the particle’s velocity is at 90^o to the magnetic field. Therefore the sin term is one. This tells us that the magnitude of F = qvB, where F is the force, q is the charge of the particle, v is the particle’s velocity, and B is the magnetic field strength.

The mathematical definition of the cross product also tells us that the force felt by the particle is always at 90 degrees to both the particle’s velocity and the magnetic field. This is what makes the particle turn in a circle: the force it feels is always pointed inward, towards the center of the circle. The magnetic field is providing the centripetal force that keeps the particle moving in a circle. Thus we can equate the physical expression for centripetal force (F = mv²/r) to the force provided by the magnetic field (F = qvB).

qvB = mv²/r

m/q = rB/v

T magnetic field B is set by the experimenter and the particle’s velocity v is known from the potential difference through which the particle was accelerated. Therefore the experimentally measured radius of the particle’s motion can be used to calculate the mass/charge of the particle. Since the charge is usually +1, the q usually doesn’t even show up.

The experimental result of a mass spec experiment is typically a spectrum with the mass/charge of the ionized molecule on the x axis and its relative abundance on the y axis. The sharp peaks that are shown on the spectrum correspond to the masses of the particles in the sample. In the mass spectrum of air, the following peaks are observed:

The most abundant component of air is diatomic nitrogen, mass 28 amu. The second most abundant component is diatomic oxygen, mass 32 amu. There is also a little bit of argon, mass 40 amu and neon, mass 20 amu. The peak at 14 probably represents molecules of nitrogen that were doubly ionized by the electron bombardment so that their charge is +2 and their mass to charge ratio m/q = 28/2 = 14. Likewise, the peak at 16 is doubly ionized oxygen. The peak at 29 amu represents dinitrogen for which one of the nitrogens is the isotope with mass 15 amu. It is much smaller than the 28 amu peak because ¹⁴N is so much more abundant than ¹⁵N. If you know the exact abundance, you can calculate how high one peak should be relative to the peak of the isotope. Conversely, from looking at the mass spectrum, you can deduce the isotopic abundance of certain elements. Practice doing this in the problems for this section. You should commit 1/90, the isotopic abundance of ¹³C/¹²C, to memory since this ratio comes up so often.