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Spectroscopy

Contents

• Spectrophotometry.
• Fluorimetry.
• Infrared spectrometry.
• Mass spectrometry.
• Nuclear magnetic resonance.
• X-ray crystallography.

Spectrophotometry

Spectrophotometry is a way of characterising a compound by the way it absorbs and emits light. The absorbance of a compound is determined by how its electrons interact with light.

Light can be considered a wave with wavelength λ, and also as a stream of particles (photons) with a certain energy, thanks to quantum physics. By Planck's equation:

E = L h c ⁄ λ

A short wavelength, λ, of light implies a stream of photons with high energy, E. L, h and c are Avogadro's number (6.022 × 10²³ mol⁻¹), Planck's constant (6.626 × 10⁻³⁴ J s) and the speed of light (299792458 m s⁻¹) respectively.

When light hits a compound, the compound will absorb certain wavelengths of light. The wavelengths it absorbs are related to the differences in the energy of electron orbitals in the compound. In the E₀ 'ground state', the electrons are stacked up in the atomic or molecular orbitals as we saw in the notes on atomic structure, for example, 1s² 2s² 2p² 3s⁰. If the molecule absorbs a photon of light of the correct wavelength (colour), an electron will be promoted from a low energy orbital to a higher energy one. This is the 'excited state', E₁, e.g. 1s² 2s² 2p¹ 3s¹. Photons of the correct energy are those whose wavelength provides the energy difference between E₁ and E₂, i.e. where E₁ − E₀ = L h c ⁄ λ. Note that the colour a compound is is the colour that is not absorbed: a red compound is one that absorbs blue and green light.

Absorbance of shorter wavelength, more energetic light, can promote electrons to even higher excited energy levels.

There are two questions to answer:

1. Why are most spectra continuous? We would expect discrete absorbance lines in the spectrum if only a few specific transitions were possible.
2. Why is there any absorbance at all? Once the energy is absorbed, we would expect the electrons to drop back down to the ground state, emitting a photon of exactly the same wavelength as was first absorbed. There should be not net absorbance of light.

The spectrum is continuous because the molecules are thermally agitated, and they distort one another's energy levels slightly into a variety of 'vibrational substates', as you can see below. The large number of electron transitions that this makes possible all have slightly different energies, which add up to make the spectrum look continuous. Moreover, these interactions between the molecules serve to transfer energy away from the excited molecule in the form of heat, dropping the electrons back to the ground state without emitting any visible photons. The loss of energy as heat means that a bright blue solution of copper sulfate will slowly warm up when irradiated with visible light, as the energy that is absorbed and which gives it its colour, is converted into the random thermal motion of the molecules in the solution.

The spectrophotometer is a machine for measuring the absorbance of solutions. At base, it is just a lamp which shines light at a monochromator, which allows you to pick out a single wavelength, a cuvette holder, in which we place a small volume of the solution we want to measure the absorbance of, and a detector, which sees how much light of the wavelength we chose actually makes it through the solution.

An important relationship for spectrophotometers is the Beer-Lambert Law:

A = ln( I₀ ⁄ I ) = ε C l

I = I₀e ^{− ε C l}

Where A is the absorbance (which is defined as the log of the ratio of the light intensity we shine at the solution divided by the intensity we get out of the other side). The Beer-Lambert law states that absorbance is linear upon concentration (at low concentrations).

• ε is the molar absorption, a.k.a. the extinction coefficient
• C is the concentration of the chemical in the cuvette (in molar, M).
• l is the path length (cm) of the cuvette, which is almost always 1 cm.
• I is the intensity (energy per unit area) getting through the solution.
• I₀ is the light intensity applied to the solution.

The simplest and most important form to remember is that A = εCl.

If we know l, A and ε, we can determine the concentration of the chemical in question. For example, we can use the absorbance of the important biochemical NADH at 340 nm to monitor the rate of an enzyme reaction.

An assay of the enzyme lactate dehydrogenase can be achieved by monitoring the rate of loss of absorbance at 340 nm as the following reaction is catalysed:

lactate + NADH → pyruvate + NAD⁺

The spectrophotometry we have been discussing so far is principally UV/visible spectrophotometry: visible and UV light have energies similar to those of the electron transitions in most compounds. Most such absorbances are due to double and triple bonds, and the hybridised d-orbitals in transition metal complexes. Hence, UV/visible spectrophotometry can be used to look at the multiple bonds, aromaticity and metal ion complexes found in biochemicals. One particularly common use is to monitor the concentration of proteins: the aromatic amino acids tryptophan and tyrosine absorb stringly at 280 nm (DNA and RNA absorb strongly at 260 nm), hence we can use this to non-destructively assay for protein in biochemical samples.

We can also look at other sorts of absorbances in our quest to characterise and monitor compounds. Infrared (heat) absorbance is usually due to the vibrations, bending and stretching of bonds, and the IR 'fingerprint' of a compound can be very distinctive. Interactions with x-rays and radio waves are the basis of x-ray crystallography and NMR, which we will cover shortly.

Fluorimetry

Fluorescence is the production of light at higher wavelength that that absorbed. A fluorescent compound (a fluor) first absorbs some energy at a low wavelength, and electrons are promoted to a high energy excited state. Then it loses a little as heat, and the electrons return to the lowest energy excited state. Finally, the remaining energy (difference between lowest energy excited state and the ground state) is lost as fluorescent light.

Many fluors have conjugated double bonds (single-double-single-double-etc in a ring), such as rhodamine and fluorescein (below).

Fluorescence occurs in all directions, hence a fluorimeter, whilst otherwise similar to a spectrophotometer, measures the fluorescent light at right angles to the incident light (rather than colinear with it as is the case with specs).

An important ratio in fluorimetry is the quantum yield, i.e., the proportion of short wavelength light that is re-radiated as fluorescent light, rather than as heat. i.e. it's the 'efficiency' of the fluorescence. The quantum yield of fluorescence,

Φ_f = photons of light fluoresced ⁄ photons of light absorbed.

Fluorescence can be derived from the Beer-Lambert Law by considering that the light absorbed must be lost either as fluorescence or as heat (for 'typical' absorbance, almost all the energy is lost as heat). The amount that is fluoresced is proportional to the quantum yield: if the yield is high, most of the absorbed light appears as fluorescence, if it is low, most of the energy is lost as heat. The equation for the intensity of fluorescence (F) is therefore:

F = Φ_f I₀ (1-e^{−ε C l})

If you do some maths with that, and chuck away negligible terms, then for a given substance, pathlength and incident light intensity (I0) and at low concentrations, fluorescent intensity is:

F= Φ_f I₀ ε C l = kC

The upshot of all this is that fluorescence is directly proportional to concentration, in the same way that absorbance is. However, because F and I are at different wavelengths, their intensities cannot be directly compared. Consequently you must use reference solutions of known concentrations to calibrate your instrument, and determine k.

There are several phenomena related to fluorescence. A fluorescent compound produces e.g. visible light when irradiated with (higher energy) UV light, but only as long as the UV lamp is on. A phosphorescent compound on the other hand, has a lowest-energy first excited state that is metastable, that is, it only very slowly decays back to the ground state. As a result, phosphors will glow with visible light for seconds to days after the the UV light is turned off. The stability is due to a quantum phenomenon, the stability of triplet electron states. Fluorescent compounds lack this excited metastable state, having only a much shorter lived singlet excited state.

Bioluminescence is another form of fluorescence: it is the production of fluorescent light by living organisms. In most cases, the luminescence is not photofluorescence, but chemifluorescence: in the former case, the excited state that decays to produce the fluorescent light is produced by irradiation with short wavelength light, in the latter, it is produced chemically by a luciferin. A luciferin is activated by ATP, and then oxidised by a luciferase enzyme and oxygen to give an excited product in the first excited state. The excited state then decays, releasing light.

The fluorescence system of the crystal jellyfish (Aequoria victoria) is very useful in molecular biology. It is a two stage system:

• aequorin (luciferin) + Ca²⁺ → blue light.
• blue light + green fluorescent protein (GFP) → green light.

Aequorin is useful in studying nerves and fertilisation, where Ca²⁺ is important, but GFP can be used anywhere where you can irradiate with blue light. GFP can be used:

• As a non-invasive tag.
• Engineer GFP gene into organism of interest.
• Track microtubules, viruses, gene expression, etc. by tagging the protein with GFP.

Modern DNA sequencing relies on fluorescence. The old way of sequencing was:

• Take DNA, split into four samples and mix with dNTPs.
• Add a hint of one ddNTP to each tube.
• ddNTP stops synthesis by polymerase.
• Run four separate electrophoreses.
• Lengths correspond to position of complementary base.

Fluorescent DNA sequencing is much simpler: it only needs one reaction, not four. Each ddNTP is tagged with a different coloured fluor, so we can run all four reactions in the same pot, and rely on the fact that each band in the electrophoresis will be a specific colour. In fact, a column electrophoresis is used: the PCR product is loaded onto the column, and the fluorescence of the column eluate is monitored: imagine the coloured bands in the picture below slowly dropping off the end of the blue rectangle.

ELISA is another technique using fluorescence. It stands for enzyme-linked immunosorbent assay. An antibody to a protein antigen of interest is produced and tagged with an enzyme that breaks down a substrate to produce a fluorescent product. A solution containing an unknown amount of the protein antigen is added (and binded) to a well, and the antibody-enzyme is added. Excess antibody is washed away, and then the substrate is added. The rate of production of the fluorescent product is related to the amount of antigen present in the well.

Infrared spectrometry

Infrared (IR) spectroscopy looks at the effect that IR (heat) has on the bonds in a molecule. It is quite useful to model the bonds between two atoms in a molecule as if they were springs connecting balls together. When we heat a molecule with IR radiation, we can excite the vibrational states of the bonds, and cause them to bend (deform) and stretch. Different bonds have different characteristics: it is easier to bend a spring than it is to stretch one, so absorbances due to bonds bending generally appear at lower energies than absorbances due to bonds stretching. Similarly, short, strong bonds tend to be more difficult to bend or stretch than long weak ones, so absorbances due to C=O bonds stretching are of higher energy than those due to C-O bonds stretching. From these general considerations and empirical data, we can compile long lists of characteristic absorbances for different bonds:

• 3200 cm⁻¹ broad, strong O-H stretch (alcohols)
• 3000 cm⁻¹ broad, medium O-H stretch (carboxylic acids)
• 1200 cm⁻¹ strong, O-H deformation
• 2800 cm⁻¹ strong, C-H stretch
• 1400 cm⁻¹ variable, C-H deformation
• 1700 cm⁻¹ strong, C=O stretch
• 1200 cm⁻¹ strong, C-O stretch

For historical reasons, the units used are cm⁻¹ (the wavenumber, which is the reciprocal of the wavelength of IR radiation absorbed in cm). High wavenumbers imply high energy. The shape and intensity of the absorption can be quite characteristic: O-H groups in alcohols absorb very strongly, whereas in carboxylic acids, the absorbance is rather weaker.

The spectrum above is that of ethyl ethanoate: you can see a variety of characteristic features of an ester: C=O groups, two pieces of evidence for C-H groups, evidence for the existence of C-O and a lack of evidence for O-H. We can also see that the C=O stretch is of higher energy than the C-O stretch, and that C-H deformations are less energetic than C-H stretches, as predicted by our simple model.

Of course, spectra of organic compounds can get much more complex than this, but the principles of analysis are still similar.

Mass spectrometry

Mass spectrometry (MS), often used in combination with gas chromatography (GC-MS) is a very powerful and simple technique for determining the structure of molecules. The basic idea is that sample molecules are bombarded with electrons from an electron gun to strip yet other electrons out of them, leaving them with a positive charge. These positive ions are unstable, and will fragment somewhat. All these positive ion fragments are then accelerated through a high voltage electric field and then swung round using a magnetic field produced by a large electromagnet. Since large fragments are more difficult to divert, they require a stronger magnetic field to move them. Hence, by varying the magnetic field, we can scan across the fragments and determine their masses.

The fragments in the diagram above vary in mass a > b > c. (in fact, it's again the charge density, or (inversely) the m ⁄ z ratio, that's important: the ratio of the charge on the fragment to its mass, however, most of the fragments will have single positive charge). Mass spectra are suprisingly easy to interpret. Here is the MS of the degradation product of IPBC, PBC:

The compound PBC has the structure below, and you can see (by adding up the atomic masses of the atoms in the fragments indicated) where some of the mass spectral peaks come from. The parent ion (i.e. PBC⁺) is the peak at 155, and the base peak (the biggest one) at an m ⁄ z of 112 is PBC less CH₂CH₂CH₃.

Nuclear magnetic resonance

NMR is a another sort of spectrometry, but looking at the absorbance of radio waves rather than visible light. Radio waves are very low energy, so what transitions in an atom have such tiny energies? In fact, it's the transitions that certain sorts of nucleus in a strong magnetic field can have that are responsible for the absorbance and emission of radio waves.

Nuclei are small, spinning balls of positive charge. A moving charge such as this generates a magnetic field. Hence nuclei (with a specific amount of spin) are magnetic, and possess a magnetic moment. If an external magnetic field (B) is applied, this moment precesses, i.e. the direction of the magnetic moment itself rotates, much like the way a spinning penny does when it begins to stop spinning. The frequency of the precession is called the Larmor frequency (ω). The Larmor frequency is given by:

ω = γ B

Where γ is the magnetogyric ratio, characteristic of a particular nucleus. Nuclei that have half integer spin, such as ¹H (proton-NMR) and ¹³C (carbon-13-NMR) act somewhat like tiny bar magnets: if these bar magnets are oriented parallel to the external field, they are at a lower energy state than those oriented antiparallel to the field. The difference in the energies (∆E) is dependent on the strength of the external field (B), hence NMR machines, which require high sensitivity, require a large difference in the energy levels, hence a large magnetic field, hence large and expensive electromagnets.

∆E = h ω

If we irradiate a collection of nuclei held in this strong magnetic field with radio waves, a certain radio frequency (the Larmor frequency, as noted above) will be absorbed: this corresponds to the difference in energy associated with flipping from the parallel to the antiparallel state. This absorbance of radio waves can be measured and the frequencies absorbed tell you something about the environment in which the interacting nuclei reside.

In a hydrogen atom, the single electron rotating around the nucleus will interact with the applied magnetic field in a similar way to the nucleus. In this case, it will tend to oppose the applied field, hence reducing the effective external field that the nucleus experiences. This means that the frequency of the radio waves needed to flip the nucleus will be increased slightly, because the shielding of the nucleus by the electron will decrease the effective magnetic field that the nucleus experiences, and hence decrease the energy difference between the two spin states. The difference between the 'expected' and observed frequency depends on the strength of the external field, hence the chemical shift, which is independent of B, was defined:

Chemical shift = δ (ppm) = 10⁶ ( ω - ω_ref ) ⁄ ω_ref

The chemical shift expresses the shielding (or deshielding) experienced by a nucleus relative to a reference compound's resonant Larmor frequency. This is usually chosen to be tetramethylsilane, (CH₃)₄Si, or TMS. If the proton is close to an area of high electron density (e.g. close to an electronegative atom such as oxygen), the deshielding effect will be increased more significantly than if the proton is close to a more electropositive atom. A typical NMR spectrum looks something like the one below, which is that of ethanol, CH₃-CH₂-OH.

In this spectrum, we can see several things: firstly, there are three clusters of peaks (note the small peak at a shift of 0, this is the TMS reference peak). These correspond to the protons in the OH group, the methylene CH₂ group and the methyl CH₃ group, going left to right. The area under these peaks corresponds to the number of protons in the same environment. Taking the left hand singlet peak (due to -OH₍₁₎), as due to one proton, we can see the area under the methylene quartet group is approximately twice this, indicating there are two protons in this environment (-CH₂-), and that the methyl protons have three times this area, as expected (CH₃-).

The actual order in which the peaks occur (left to right) is due to the previously mentioned deshielding effects of electrons: the peaks (from left to right, in order of increasing deshielding) follow the order hydroxy (highly shielded by the very nearby electrons of the oxygen), methylene (less shielded because they are further from the oxygen atom), methyl (very far from the oxygen atom), just as they do in the ethanol molecule itself. Note that more complex compounds' spectra are much less neat to interpret that that of ethanol!

But why do we get the splitting inside the main chemical shift groups? This is due to spin-spin coupling. The proximity of the methyl protons to the methylene protons means that there are several ways in which the nuclei can orient with respect to one another. The methyl protons can arrange themselves in four ways w.r.t. the external field and the methylene protons in three ways. The four ways the methyl protons can arrange themselves means that the methylene group can experience four slightly different external fields, each shielded slightly by the methyl protons, and hence the methylene group is split into a quartet. The methylene protons can likewise split the methyl group into a triplet. You might expect a similar effect from the O-H group protons, but in general, hydroxyl protons do not to others, because the proton is too far away from the others, and the coupling too weak to show up on a low resolution NMR spectrum. The amount of splitting is determined by the 'N − 1' rule: if a peak is split in three, this indicates that there are 3 − 1 = 2 protons on a nearby atom, if it is split into four, this indicates there are 4 − 1 = 3 protons on a nearby atom.

X-ray crystallography

X-ray crystallography is the most difficult of the techniques so far described to grasp, as it is intensely mathematical. What we would really like to do with X-rays is to take pictures of atoms, like we can take pictures of bigger things with visible light. The reason we must use X-rays is that to 'see' something with light, the light must have a wavelength similar to the size of the thing we are looking at. X-rays fit the bill here (visible light is far too big to see atoms), but they unfortunately cannot be focussed by any current technique. Hence we must rely on making diffraction patterns from beautifully ordered piles of molecules (crystals), rather than looking at the molecules directly. Crystallisation of the compound of interets is the primary problem in X-ray crystallography: making perfect crystals is rather fraught. Once we have a nice crystal of the compound (e.g. a protein), we illuminate it with X-rays and collect the diffracted X-rays on a photographic film, to form a diffraction pattern.

The diffraction patterns can be mathematically transformed back into electron density maps of the molecules in the crystals by Fourier transform techniques we won't go into here. A further problem of X-ray diffraction is that it relies on the size of the atoms, and hydrogen is too small to show up, hence when we 'join the dots' in our electron density map, we get something akin to a line-bond notation of the compound we have 'photographed'.

Test yourself

1. Why do spectrophotometers measure transmitted light parallel to the incident ray, whereas fluorimeters measure fluoresced light perpendicular to the incident ray?
2. How many kJ mol⁻¹ are absorbed by the following compounds at their λ_max, given that the speed of light is 3 × 10⁸ m s⁻¹, Planck's constant is 6.62 × 10⁻³⁴ J s, and Avogadro's constant is 6.02 × 10²³ mol⁻¹.

   Compound	λmax (nm)
   β-Carotene	466
   Lycopene	481
   Vitamin-A	324

3. Calculate the absorbance of a solution of NADH at 50 M given the pathlength is 1 cm and the molar absorption coefficient is 6.22 × 10³ L mol⁻¹ cm⁻¹.
4. Open this IR spectrum of an unknown compound. Which group of compounds does it belong to?
5. Why must the cuvettes (sample holders) used in infrared spectroscopy be made from sodium chloride, rather than plastic?
6. Why is GFP so widely used as a marker gene?
7. Interpret the following mass spectrum. It is the same compound as the IR spectrum above.

   

8. How many (major) peaks would you expect in the NMR spectrum of propan-2-ol?

Answers