Lecture 2- IR - Raman Spectroscopy PDF

Infrared and Raman basics Condition for IR vibration. The IR source Population distributions. Isotopic shift The water problem/ sho...

Infrared and Raman basics Condition for IR vibration. The IR source Population distributions. Isotopic shift The water problem/ short pathlengths Complexity of vibrational spectra Raman spectroscopy, the concept Comparison IR/Raman spectra Jaume Torres - - 1 - Infrared and Raman basics Vibrational spectroscopy (infrared) In IR spectroscopy, the resonant frequency corresponds to the oscillation (vibration) of atoms (masses) around a bond (spring). Vibrational Raman spectroscopy 2 Infrared – condition for IR absorption not all chemical bonds absorb infrared radiati , it depends on properties of e bord Vibrational spectroscopy (infrared) Dipole moment is the product of bond distance and charge difference between two atoms (due to their different electronegativities, or ‘electron loving’ properties). If the charge is the same for both atoms, the charge difference is zero, and the dipole moment is zero. distance A Only if the dipole moment changes, as the molecule vibrates,M there will be an absorption of energy in the infrared. For example bonds that have atoms of different electronegativity (C-O, N-O, N-H, C-H) will absorb in the IR. IR active Consequence: molecules like hydrogen gas (H2), nitrogen gas (N2) or oxygen (O2) do not absorb in the IR. IR inactive Some vibrations of molecules may be invisible in the IR, even though they have atoms of different electronegativity (e.g., C-O bonds). This is because of the symmetry of these molecules, e.g., the symmetric stretching of CO2, where the dipole moment does not change with vibration. & distance befor 2 &0 on both sides of molecule , will Gretch & contract symmetrically asymmetric Gretching is visible in IR , co < bending can be observed in IRI is  Infrared radiation is emitted by hot bodies (anything above 0 Kelvin) & (light) 9000 to 11 000 m cold higher temp, higher radiat * shorter wavelength hot http://www.youtube.com/watch?v=7MAEGJqu6NA&feature=related Range of frequencies in infrared spectroscopy. Most useful part for the biologist are the near IR (nIR) and mid IR. At low temperatures (eg, 200 K, or -73 °C), 12,000 cm-1 objects emit very little radiation (see y axis), at IR long wavelengths, ~10 to 30 m. 4,000 cm-1 1000 cm -1 When temperature is increased, the intensity nIR Mid IR Far IR of radiation increases, and the wavelength emitted tends to be shorter. For a human body at 37°C (310 K), all radiation produced is still in the infrared and reaches about 2.2 m (2,200 nm). If an object was much hotter, say 230 °C (~500 K), radiation would increase in intensity and wavelength would reach ~1 m (almost red; the wavelength of red color is ~ 700 nm (0.7 m). This means the object would start to be visible to our eyes, as a red object. Heating the object more would result in other colors in the visible: orange, yellow, and eventually white.  IR absorption requires longer wavelengths (m long) than UV-Vis resonance in 19 In Somm require less energy 1 UV-Vis - Long wavelengths imply low frequency, i.e., low energy. This means that separation between energy levels is small. Examples IR A small separation implies that the E = h populations in the two levels will be 80 similar. NMR Pot smaller Large energy separation means that energy most molecules will be at the lower level. 995 separat ? pop? up & 999,999 - & down almos 1000,000 1000 1000 1000,000,000,000,000,000 ↑ esame some bords alr in excited state even by excited IR by light  The populations distribute among energy levels according to the AE diff in E Boltzmann equation : ~ E L in − N up =e k BT levels Here I change E (T = 10) Here I change T (E = 20) N down L little system here temp Nup ~ very Nup in K when IE - Ndown ↓ E ↑ Ndown ↑ T ↑ N is no of systems most fe KB is the Boltzmann constant, but systems are 0.007 100 1 0.00004 1 cold here when 10 to simplify lets make it equal to an AET 0.082 50 0.135 5 arbitrary constant, for example, 2. 0.135 40 0.367 10 0.223 30 0.606 20 3 0.368 20 0.716 30 Nup 0.606 0.904 10 2 0.778 0.818 40 50 ? 7 0.975 0.5 0.904 100 ~ not w W Ndown Calmost ~ I = almost same art excited of systems to be in both energy Is (lower a lil move)  The ratio between populations up and down depends on E and T If E >>kT, nup/ndown will be very small If E of low transmittance IR spectrum of lactic acid The X axis is not expressed as ‘wavelength’ (nm or Example, 3000 cm-1 is 1 cm/3000 waves = m), but as the inverse of the wavelength in cm (cm-1), 0.000333 cm/wave, which is… 3.33 m/wave. which means ‘the number of wavelengths () in 1 cm’ The wavelength is 3.33 m. The IR spectrum contains much more information than the UV-Visible spectrum: this is a sample that has both protein and lipid 0.8 0.7 Fingerprint region 0.6 0.5 The Mid IR spectrum of a mixture of protein and lipid bands 0.4 Absorbance are 0.3 0.2 broad & tend to overlap 0.1 0.0 -0.1 -0.2 4000 3500 3000 2500 2000 1500 1000 500 Wavenumbers (cm-1) Aromatic amino acids (W, Y, F) The UV-Vis spectrum of a mixture of protein and lipid What IR spectroscopy can detect diff that I diff molecules/samples will have same Ip spectra IR spectra are extremely complex. There are multiple vibrational modes even for simple molecules. Since IR spectra are very complex, they can be used to unequivocally identify a molecule by creating unique fingerprints: diagnostics, forensic science, cell and tissue classification, etc classify molecules Mid IR is the most useful region to identify biological molecules: structural biology, biophysics, diagnostics or tissue classification lipids chinSo carbohydrates amide (1) & (11) proteins nucleic acids 19 What is the spectrum of a protein in the IR? Surely too complicated? Peptide C=O bonds and N-H bonds all absorb at similar frequency There are thousands of modes of vibration in a molecule, but some are much more intense than others. In proteins, the stretching vibration of the N-H bond and the C=O bond are very intense (every peptidic bond has one of each). The most important diagnostic band for proteins is the amide I (it originates mostly from C=O stretching vibration) diff secondary quest , diff warno. used to monitor secondary struct. in proteins The amide I frequency maximum depends on secondary structure One of the reasons the amide I depends on secondary structure is that the frequency of the peptide C=O bond depends on the strength of the hydrogen bond between C=O and N-H: C=O ·····H-N In -helices, this H-bond is weaker than in - structure. amide I In -structure, the strong interaction with hydrogen has the effect of increasing the apparent reduced mass of the bond, therefore amide II the C=O stretching band appears at lower frequency (lower wavenumbers) in -structure than in -helices. -helix -strands  Vibrational spectroscopy (infrared) Vibrational Raman spectroscopy  Vibrational Raman spectroscopy Raman spectroscopy can be of two types: rotational and vibrational. However, Rotational Raman (where the molecule is excited to rotate faster) is only used in the study of gases, where molecules can freely rotate in space. But gases are not of general interest for the biologist. In contrast, in vibrational Raman, like in vibrational (IR) spectroscopy, the chemical bonds are excited to vibrate faster, although the mechanism of excitation is very different to that in IR spectroscopy. Vibrational Raman can be used in gases, liquids or solids. Biological specimens are generally either solids, semi-solids or liquids (tissues, cells, bacteria, blood, etc). Discovery of Raman radiation Sir Chandrasekhara V. Raman was an Indian physicist born in the former Madras Province (India). He was awarded the 1930 Nobel Prize for Physics for his work “On the scattering of light and for the discovery of the Raman effect". He proved that when light traverses a transparent material, some of the deflected light changes its wavelength. This is now known as Raman scattering. When radiation interacts with the sample, light is scattered at all angles. Light is usually detected at 90 degrees from incident light. The Raman effect is very rare, and only involves a very small fraction of the scattered photons, about 1 in 107. Thus, the incident light has to be very powerful, so more light is scattered. The small differences in wavelength can only be observed if the incident light has a single wavelength (monochromatic). Thus, a laser is used, which is both powerful and monochromatic. When Dr Raman discovered this type of radiation, it was well known that samples can scatter (emit) photons with the same wavelength as the photons used for the excitation. This was called Rayleigh radiation. photons Rayleigh What Raman discovered is that a small number of the scattered photons have a different wavelength (and therefore frequency and energy) than the incident photons. The reason these Raman photons have different energy is that some of their energy (a very small fraction) is added to, or removed from, the sample. This small amount of energy is of the order required for a bond jumping between two vibrational levels. & ground state to excited state  How is Raman radiation measured In rotational, vibrational (IR), UV-Visible, NMR… detector Source photons not absorbed sample Source of light (transmitted) are detected In Raman (like in fluorescence), Raman spectra are obtained by detector only photons emitted by the measuring light emitted by the sample are captured… sample. This is very different from absorbance spectra obtained using UV, visible or infrared spectroscopy, where what is …disregarding transmitted measured is light transmitted photons (which is then converted into Laser absorbance). sample Source of light  R WN The difference in energy between the Raman radiation photons emitted (400 and 423 nm) is  = 7 m very small, and therefore it  = 42 THz corresponds to a photon with a very long wavelength (here, 7 m, i.e, infrared waves). Therefore this = 400 nm means that a vibrational transition (between two vibrational levels) = 378 nm has taken place. cannot subtract wavelengths , = 423 nm only subtract war/energy/fret 7 & convert into > convert to wavelength In comparison, the energy of incident photons or emitted photons (which Laser (eg, = 400 nm) 750 THz (750 x 10-12 Hz) are usually from 350-1000 nm, i.e., from UV to nIR) is much higher (and wavelength much their wavelength much shorter). longer than incident (not related to Raman http://www.translatorscafe.com/cafe/EN/units-converter/frequency-wavelength/1-31/hertz-wavelength_in_micrometres/ Origin of Rayleigh, Stokes and anti-Stokes photons a excitation b emission Virtual level b b b a a a Excited level Ground level If the molecule goes back to a If the molecule goes back to the If the molecule goes back to a higher level, emitted photon (b) same level, emitted photon (b) lower level, emitted photon (b) has less energy than (a). has the same energy as (a). has more energy than (a). These These are Stoke photons. These are Rayleigh photons. are anti-Stokes photons. This is the difference between vibrational (IR) spectroscopy and vibrational Raman spectroscopy. Virtual level In a Raman experiment, a E  IR spectroscopy, to excite sample is excited with a high these bonds, the energy used energy UV-Vis-nIR laser for excitation is very small (IR (e.g., 700 nm). The final light with wavelength of a few result for the molecule is the micrometers). same. Their bonds jump between vibrational levels (for which very small energy is required). Vibrational (IR) Vibrational Raman spectroscopy spectroscopy Energy levels of a vibrating bond  The Vibrational Raman spectrum Anti-Stokes photons are Stokes photons are scattered with more energy scattered with less energy than the incident laser light. than the incident laser This is less likely to light. This is more likely happen since most to happen since most vibrating bonds are initially vibrating bonds are initially at their ground state, not at at their ground state. their excited state. more intense The Raman photons (Stokes and anti-Stokes) and the Rayleigh photons, are all captured in a detector. The Rayleigh photons are shown at the center, forming a single ‘band’ that is very intense. The Stokes photons, are shown on one side of this central band whereas the anti-Stokes photons are shown on the other side.  The Vibrational Raman spectrum Each band, representing a ‘mode of vibration’ (e.g., C=O stretching, N-H stretching, etc) in the molecule appears at a different ‘Raman shift’ that will be shown as wavenumbers (cm-1). The same band will also appear, with an identical shift but opposite sign, in the other half of the spectrum. The Raman shift is the difference in energy - or in wavenumbers - between the Rayleigh radiation (central band) and the other bands of the spectrum. The sign of this shift, positive or negative, is arbitrary and not really important. What is really important is that the Stokes half is always the most intense half. The two halves look like mirror images (but Stokes half several orders of magnitude more intense), and therefore only the Stokes half of the spectrum is used. But even the Stokes half has low intensity and is noisy, so to increase the intensity, the sample is excited for a ‘long time’ (seconds or minutes). This results in an increased intensity detected of emitted photons. Note that in contrast to Raman, in IR or in UV-Vis spectroscopy, the duration of the illumination has no effect on the quality of the spectrum. This is because transmittance depends on a ratio between incident vs transmitted photons. This ratio is always the same, whether illumination is short or long. (Much less intense than this, not to scale) Since it is the most The sign of the shift is just a intense half, only the convention; normally Stokes is Stokes part is analyzed shown as negative shift (but in this example is positive)  537um Numeric example 514. 5 - 8392m 19,435 – 18,596 = 839 cm-1 slokes have less energy , ↓ W N. than Rayleigh be ? ↑ wavelength 839  Comparison of IR and Raman spectra Since the jumps involved in vibrational Raman and in IR spectroscopy are the same, this means that the spectra obtained with these two techniques are expected to be similar in shape. However, they are not identical because the mechanism used for absorption is completely different, and depends on very different properties of molecules: In IR, the dipole moment must change as the bond vibrates. In Vibrational Raman, the polarizability (the ability to be polarized) must change as the bond vibrates. - tendency of electron cloud to be distorted forming temp dipoles Therefore, in some cases, some bonds will have a strong Raman absorption and a weaker IR absorption, or vice versa, whereas in other cases, the bands will appear in both spectra. This makes these two techniques complementary. temp dipoles have diff values during vibration (changing polacisability 34  Spectra of water (IR and Raman comparison) In IR spectroscopy, absorbance IR depends on the difference in electro- IR and Raman spectra of negativity between the atoms water look similar in involved, and oxygen and protons (O- Abs shape, but water is a H bond in water) have very different strong IR absorber and it electronegativity, so the dipole interferes with other IR moment is large. bands from biomolecules. But in Raman, emission intensity is stronger when the polarizability of the bonds is higher. The polarizability of Raman water is low, hence water is a very weak scatterer, and therefore its Raman spectrum is also weak. ? Water is a weak scatterer, and it does not interfere with Raman signals from biomolecules.  Vibrational Raman spectra look similar to IR spectra (but not identical). Human hair Absorbance Infrared Wavenumber (cm-1) Intensity emitted Raman shift (cm-1) Water is a weak scatterer  Water is a weak scatterer In Raman, a bond vibration can be detectable even when the atoms involved in the bond are identical (e.g., C=C in The fact that water is a weak scatterer, the benzene ring, or C-C in cyclohexane), whereas makes Raman very useful when looking vibration of these bonds is undetectable in the IR). at aqueous solutions directly. However, the Raman intensity emitted is very weak, and data collection may be longer than when using IR. Identification of components in blood serum. The x-axis is the Raman shift vs Mixture of 95% Rayleigh (or incident) radiation. water/5% cyclohexane. Cyclohexane is a MUCH stronger Raman scatterer than Despite its much lower water concentration, the only peaks we see are from the cyclohexane. Intensity Water spectrum (too small to see) 37 Summary. The source of IR radiation is a hot object. IR absorption is observed when the bond has a changing dipole moment. This means that small molecules where this condition is not met are invisible in the IR. Water is very abundant in biological samples and has a high absorption that obscures other bands. Its contribution can be subtracted after using short pathlengths: transmission cells or ATR. IR spectra have a high complexity that is useful to identify minute variations between samples. The amide I band in proteins is diagnostic of protein conformational changes (secondary structure). Raman vibrational spectroscopy is based on irradiation of samples with lasers of short wavelengths (UV-Vis-nIR) and the emission of photons with slightly longer (Stokes) or shorter (anti-Stokes) wavelengths. Only the Stokes photons are used because they are far more abundant. These emitted Raman photons are also indicative of vibrational jumps in the sample, therefore IR and Raman spectra are similar, although not identical, because the absorption mechanism is different. The two techniques are complementary because bands absent or weak in one technique may be very strong in the other. The fact that water is a poor scatterer means that its Raman spectrum is very weak, and biological samples can be studied by Raman without water interference, although spectra collection is slower than in IR spectroscopy because the emission of Raman photons is a rare event. The end