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Experiment 2 instruction

Infrared Spectroscopy Experiment 1. Objectives 1.1. To learn the basic principles of vibrational spectroscopy 1.2. To determine the relationship between molecular structural features and absorptions in the IR spectrum

2. Background 2.1 Introduction to Infrared Spectroscopy：When a beam of electromagnetic radiation of intensity Io is passed through a substance, it can be either absorbed or transmitted, depending upon its frequency, and the structure of the molecule it encounters. Electromagnetic radiation is energy; thus when a molecule absorbs radiation it gains energy as it undergoes a quantum transition from one energy state (Einitial) to another (Efinal). The frequency of the absorbed radiation is related to the energy of the transition by Planck's law: Efinal - Einitial = ΔE = hν = hc/λ. If a transition exists that is related to the frequency of the incident radiation by Planck's constant, the radiation can be absorbed. If the frequency does not satisfy the Planck expression, then the radiation will be transmitted. A plot of the frequency of the incident radiation vs. some measure of the percent radiation absorbed by the sample is the absorption spectrum of the compound. The type of absorption spectroscopy depends upon the frequency range of the electromagnetic radiation absorbed. Microwave spectroscopy involves a transition from one molecular rotational energy level to another. Rotational energy level spacings correspond to radiation from the microwave portion of the electromagnetic spectrum. Vibrational spectroscopy (or infrared spectroscopy) measures transitions from one molecular vibrational energy level to another, and requires radiation from the infrared portion of the electromagnetic spectrum. Ultraviolet-visible spectroscopy (also called electronic absorption spectroscopy) involves transitions among electron energy levels in the molecule, which require radiation from the UV-visible portion of the electromagnetic spectrum. Such transitions alter the configuration of the valence electrons in the molecule. Molecules may undergo several types of motion. First, the entire molecule may move through space in some direction and with a particular velocity. This is called translational motion and with it we associate the translational kinetic energy of the 1 / 11

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molecule, 1/2mv2 (v = velocity of the center of mass of the molecule). The velocity of translation may be resolved into components along the three axes of a Cartesian coordinate system, so that we may write 1/2mv2 = 1/2mvx2 + 1/2mvy2 + 1/2mvz2, where vx is the x-component of velocity, etc., and m is the mass of the molecule. This equation tells us that the total translational KE of the molecule consists of three parts, each of which represents the kinetic energy along one of the reference directions. Since any translation of the molecule may be viewed as the vector sum of its motions along the three axes, the kinetic energy may always be broken up into the sum of three contributions, one arising from motion along each axis. We say that the molecule has 3 translational degrees of freedom, one corresponding to each Cartesian axis. Second, the molecule may rotate about some internal axis. Again, any such axis may be resolved into components along the x, y, and z axes of a Cartesian coordinate system, so any rotation of the molecule may be resolved into three perpendicular components. The rotational kinetic energy of the molecule can then be written KErot = 1/2Ixωx2 + 1/2Iyωy2 + 1/2Izωz2

where Ix, Iy, and Iz are the moments of inertia and ωx, ωy, and ωz are the angular velocities about the x, y, and z axes respectively. We see that there are 3 degrees of freedom associated with rotational motion, one corresponding to each Cartesian axis. An exception to this statement arises for a linear molecule, for which one of the three axes is normally taken as the molecular axis. Consider a diatomic molecule shown with its bond axis coincident with the z axis of a coordinate system. Because the moment of inertia about this axis is zero, the molecule has no rotational energy about the z axis. For linear molecules there are only two rotational degrees of freedom, rather than 3. Finally, the molecule may vibrate. A diatomic molecule, for example, vibrates by repeated stretching and contraction of the bond joining the two atoms. Such a molecule has, in addition to 3 translational and 2 rotational degrees of freedom, one vibrational degree of freedom. For a polyatomic molecule, we may deduce the number of vibrational degrees of freedom (sometimes called vibrational modes) by subtracting the number of translational (3) and rotational (2 or 3) degrees of freedom from the total number of degrees of freedom possessed by the molecule. The latter is 3N, where N = the number of atoms in the molecule. (Each atom may independently move in any of three directions and therefore has 3 degrees of freedom available to it, for a total of 3N for the molecule.) The number of vibrational degrees of freedom is therefore given by 3N-6 for a non-linear polyatomic molecule; and by 3N-5 for a linear polyatomic molecule. According to this rule, the water molecule, which is non-linear with N = 3, has 3 vibrational degrees of freedom and should undergo three independent types of vibration. The arrows show the directions in which the atoms move during the vibration. 2 / 11

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Each molecular vibrational motion occurs with a frequency characteristic of the molecule and of the particular vibration. The energy of a vibration is measured by its amplitude (the distance moved by the atoms during the vibration), so the higher the vibrational energy, the larger the amplitude of the motion. According to quantum mechanics, only certain vibrational energies are allowed to the molecule (this is also true of rotational and translational energies), so only certain amplitudes are allowed. Associated with each vibrational motion of the molecule is a series of energy levels (or vibrational energy states). The molecule may go from one energy level to a higher one by absorption of a quantum of electromagnetic radiation, such that Efinal-Einitial = h. In undergoing the transition, the molecule gains vibrational energy, and the amplitude of the vibration increases. The frequency of light required to cause a transition for a particular vibration is equal to the frequency of the vibration, so we may measure the vibrational frequencies by measuring the frequencies of light absorbed by the molecule. Since most vibrational motions in molecules occur at frequencies of about 1014sec-1, light of wavelength λ = c/ν = 3 x 1010 cm/sec/1014sec-1 is required to cause transitions. Light of this wavelength lies in the infrared region of the spectrum. IR spectroscopy, then, deals with transitions between vibrational energy levels in molecules. An IR spectrum is displayed as a plot of the energy of the infrared radiation (usually expressed in wavenumbers) versus the percent of light transmitted by the compound. The spectrum of the molecule appears as a series of broad absorption bands of variable intensity, each providing structural information. Each absorption band in the spectrum corresponds to a vibrational transition within the molecule, and gives a measure of the frequency at which the vibration occurs. For water, with three vibrational degrees of freedom, there are three sets of energy levels within which transitions may occur. The spacing between energy levels depends upon the particular vibration considered. Each spacing requires a photon of different energy to cause the transition, so we expect photons of three different energies to be absorbed by H2O. These are found at 3500 cm-1, 1650 cm-1, and 600-300 cm-1, respectively. The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipoles exactly cancel. However, when CO2 undergoes a bending vibration, its dipole 3 / 11

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moment changes from zero to some non-zero value. This vibration produces a change in dipole moment and is IR active. All three vibrational motions of water cause a change in dipole moment. Hence all three vibrations are IR active and we expect each to produce an absorption band in the IR region. We have already seen that this expectation is borne out. The requirement that a vibration cause a change in the dipole moment of the molecule in order to absorb radiation can be understood simply. Exchange of energy between electromagnetic radiation and matter can occur only if the radiation and matter can interact (or couple) in some way. Electromagnetic radiation consists of perpendicular electric and magnetic fields that oscillate sinuosoidally at the frequency of the radiation. Oscillation of the electric field is equivalent to an oscillating dipole moment. A molecule may interact with the radiation by interacting with the oscillating electric field, but this is possible only if the molecule also possesses an oscillating electric field. Furthermore, the frequencies of the oscillations must be the same. This is called the requirement of frequency matching, and is universal in spectroscopy. If a vibrational motion of a molecule is to absorb IR radiation, the motion must generate an oscillating electric field. This is equivalent to saying that the vibration must produce a change in dipole moment. The dipole moment oscillates at the frequency of the vibration, so radiation of this frequency may be absorbed. In addition to the number of vibrations expected for a molecule, we can make statements about where in the spectrum absorptions due to certain types of vibrations are expected to occur. For a diatomic molecule, A-B, the wavenumber (i.e., the reciprocal of the wavelength) of the infrared radiation absorbed by the molecular vibration is given by 1/2 (1) ν = 1/2π (kAB/μAB) where μAB = MAMB/(MA + MB)

kAB is the force constant for the bond between A and B, and is a measure of the bond strength. MA and MB are the masses of the atoms. We conclude from this equation that the heavier the atoms involved in the bond, the lower the absorption frequency, given a constant bond strength. 2.2 Applications of IR Spectroscopy to Organic Molecules. Organic functional groups (atom groups bonded in particular ways) differ both in the strengths of the bond(s) and in the masses of the atoms involved. For instance, the O-H and C=O functional groups each contain atoms of different masses connected by bonds of different strengths. According to equation (1), we therefore expect the O-H and C=O groups to absorb IR radiation at different positions in the spectrum. The presence of a strong, broad band between 3200 and 3400 cm-1 indicates the presence of an O-H group in the molecule, while the presence of a strong band around 1700 cm-1 confirms the presence of a C=O group.

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For organic molecules, the infrared spectrum can be divided into three regions. Absorptions between 4000 and 1300 cm-1 are primarily due to specific functional groups and bond types. Those between 1300 and 909 cm-1, the fingerprint region, are primarily due to more complex vibrational motions; and those between 909 and 650 cm-1 are usually associated with the presence of benzene rings in the molecule. Some particularly important regions are indicated in Table 1. 2.3 Applications of IR Spectroscopy to Inorganic Molecules. Many so-called inorganic compounds are in reality largely organic, and for these we look for the same functional group bands in the IR as we do for purely organic compounds. However, the infrared spectra of relatively simple, purely inorganic compounds containing only a few atoms--specifically, inorganic salts containing polyatomic (complex) ions--are quite distinctive and can be used to rapidly identify the ions. Consider a simple inorganic salt, such as KNO2. On the basis of the empirical formula, we might naively expect there to be a total of 3(4)-6 = 6 normal modes of vibration associated with this material. However, this assumes that KNO2 is covalent. In fact, KNO2 consists of an ionic lattice of K+ and NO2- ions arranged in an infinite and very regular array. The crystal consists of essentially isolated K+ ions and NO2- ions. Thus we are able to consider the vibrational modes of the cation and anion independently of one another. In this case, since the potassium ions are monatomic, they have no vibrations (3(1)-3 = 0), so we need only consider the nitrite anions. The VSEPR (Valence Shell Electron Pair Repulsion) Theory predicts a bent structure for the nitrite ion. We thus anticipate three normal vibrational modes for NO2-, corresponding to the diagrams drawn earlier for H2O, and they should all be infrared active. Indeed, three bands are observed in the IR spectrum of KNO2: the symmetric stretch at 1335 cm-1, the asymmetric stretch at 1250 cm-1, and the bending vibration at 830 cm-1 (bending vibrations occur in general at lower frequencies than stretching vibrations). The frequencies of these vibrations are about the same regardless of counter ion, substantiating the independence of the anion and cation in the crystal. (This independence is only an approximation, but we will not worry about the complicating factors now.) We can, therefore, diagnose the presence of nitrite ion in a salt from the infrared spectrum of the material. This diagnostic application can be implemented only when the spectrum of the material is relatively uncomplicated, but despite this restriction it is an enormously useful application. Let us turn now to the somewhat more complex case of NaNO3. Here we anticipate 3(4)-6 = 6 normal vibrational modes. The infrared spectrum, however, exhibits only three fundamental bands, at 831, 1405, and 692 cm-1. Why? There is no doubt that there are 6 normal vibrational modes. The formula is always valid. In the case of NO3-, however, the symmetric stretch is not IR active because it does not cause a change in the dipole moment of the ion, and so cannot give rise to absorption of IR radiation. This eliminates one of the anticipated bands from the IR spectrum. Among the remaining 5, there are two sets of doubly degenerate vibrations--i.e., two instances in which 2 vibrations occur with exactly the same frequency. Thus although 5 vibrations absorb IR radiation, they are manifested in only three spectral bands. 5 / 11

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However, these absorptions can be used diagnostically just as for nitrite. In similar fashion, other relatively simple anionic (and cationic) species can be identified via their IR spectra. Spectral data for some of the more common polyatomic ions are given in Table 2. In this experiment, you and your colleagues will determine the infrared spectra of a large number of compounds in order to compile a data base from which generalizations relating spectral features to structure may be drawn.

3. Equipment and Materials • • •

IR spectrometer IR cells Pure Substances, liquids: o acetaldehyde, CH3CHO o acetone, CH3C(=O)CH3 o acetonitrile, CH3CN o aniline, C6H5NH2 o benzaldehyde, C6H5CHO o benzene, C6H6 o benzonitrile, C6H5CN o t-butanol, (CH3)3COH o carbon tetrachloride, CCl4 o chloroform, CHCl3 o chloropentane, C5H11Cl o deuterated chloroform, CDCl3 o diallylamine, NH(CH2=CHCH2)2 o dichloromethane, CH2Cl2 o DMSO, (CH3)2S=O o 1,4-dioxane, C4H8O2 o dipropylamine, NH(C3H7)2 o ethanol, C2H5OH o ethyl acetate, CH3C(=O)OCH2CH3 o heptane, C7H16 o hexane, C6H14 o isopropanol, CH3CH(OH)CH3 o methanol, CH3OH o methyl acetate, CH3C(=O)OCH3 o methylethyl ketone, CH3CH2C(=O)CH3 o octane, C8H18 o 1-pentanol, C5H11OH o 2-pentanone, CH3C(=O)(CH2)2CH3 o phenol, C6H5OH 6 / 11

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piperidine, C5H10NH propionaldehyde, CH3CH2CHO o salicylaldehyde, C6H4(OH)(CHO) o styrene, C6H5CH=CH2 o 1,1,2,2-tetrachloroethane, CHCl2CHCl2 o tetrahydrofuran, (CH2)4O o triethylamine, N(C2H5)3 o vinyl chloride, CH2=CHCl Pure substances, solids: o 2-aminoethanethiol hydrochloride, HSCH2CH2NH3Cl o ascorbic acid, C6H8O6 o benzil, (C6H5C(=O))2 o benzoic acid, C6H5COOH o benzophenone, (C6H5)2C=O o citric acid, HOOC-CH2-C(OH)(COOH)-CH2COOH o Cu(Im)4(NO3)2 o p-dichlorobenzene, C6H4Cl2 o imidazole, C3H4N2 o 2-methylimidazole, C4H6N2 o 2-nitrophenol, O2NC6H4OH o 2-nitrobenzaldehyde, O2NC6H4CHO o Ni(dtc)2 o salenH2 o thiourea, NH2C(=S)NH2 o urea, NH2C(=O)NH2 o o

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4. Safety Safety glasses must be worn at all times in the laboratory. Organic liquids are hazardous and many are flammable. If you should spill an organic liquid, tell your instructor immediately so that the liquid may be cleaned up. DISPOSE OF ORGANIC COMPOUNDS ONLY IN APPROPRIATE WASTE BOTTLES, LOCATED IN THE HOODS.

5. Experimental Record all data in your lab notebook. The instructor will assign you several liquid and solid substances to analyze. Based on the formulas of your substances, develop Lewis structures, and translate these into line drawings. Based on the Lewis and line structures for each substance, try to predict where you will see absorption bands in the IR spectrum. Focus on the absorption bands in the region between 4000 and 1300 7 / 11

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cm-1, which may often be associated with vibrational motions of specific structural features. Where necessary, discuss your predictions with the instructor. This predictive aspect of the experiment must be completed before you proceed to obtain actual spectra. Establish a clean work space. You may organize your work space as shown here. Sign out an IR cell from the instructor. Ask the instructor to show you the proper use of the cell before you proceed. Here are some rules for handling and using the cells, and for preparing samples. Study and learn these rules before proceeding. 5.1 Handling IR Cells; Liquid Sample Preparation 1. The IR cell should never be placed directly on a lab bench. Lab benches are DIRTY. As is always recommended, place 3 2-foot strips of clean paper towel on your bench space and place the cell either on that or on a large Kimwipe. 2. The salt crystals (transparent discs) between which the sample is placed must NEVER be exposed to water. They will be dissolved, hence destroyed by, water. 3. The salt crystals must NEVER be handled directly with the fingers. Please wear CLEAN surgical gloves when handling the crystals, or handle them with a Kimwipe. The crystals must be handled by the EDGES only--never touch the polished faces of the discs. 4. Your IR cell may involve threaded parts. UNDER NO CIRCUMSTANCES ARE YOU TO CROSS-THREAD THESE PARTS. THIS WILL RUIN THE THREADS. If the threads are not properly aligned, the parts will not turn together smoothly. DO NOT FORCE THE TURN. Back off, realign the parts correctly, and gently turn the male piece to tighten. 5. Do not use excessive pressure when assembling the IR cell. This will fracture the salt crystals. 6. To load the IR cell with a liquid sample, follow these steps: ① Wear CLEAN gloves. ② Remove the IR cell from the storage desiccator. ③ Carefully disassemble the crystal holder by unscrewing the plastic (male) piece, as shown by the instructor. Place the pieces on a clean paper towel strip or Kimwipe (NOT ON THE BENCH). ④ CAREFULLY remove the crystals by inverting the metal part of the crystal holder. Catch the crystals in a clean Kimwipe. Remember, the crystals are table salt! DO NOT EXPOSE THEM TO WATER OR BARE FINGERS. There is a teflon spacer under the pair of crystals in the crystal holder. BE SURE NOT TO LOSE THIS. ⑤ If necessary, replace the teflon spacer in the metal holder. ⑥ Place 1 of the salt crystals into the metal holder. 8 / 11

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⑦ Using a CLEAN Pasteur pipet, place 1 drop of the liquid to be studied in the center of the crystal in the metal holder. ⑧ Place the other crystal squarely on top of the first crystal, so that the liquid spreads into a film. ⑨ Reassemble the cell by screwing in the plastic holder piece. BEWARE OF CROSS THREADING. 7. Place the cell mount in the IR spectrometer. 8. Place the cell on the cell mount. 9. Obtain the IR spectrum of the sample. 10. Remove the cell from the mount, and the mount from the instrument. 11. Dissassemble the crystal holder. 12. Remove the crystals. 13. Clean the wet crystal surfaces by BLOTTING (NOT rubbing) the liquid from the faces of the crystals using a Kimwipe. Use acetone on the Kimwipe if necessary. Allow the crystals to air dry. 14. Place the Pasteur pipet upside down in the used-pipet beaker. 15. Proceed with the next sample. 5.2 Solid Sample Preparation The first part of the procedure is the same as above (Steps 1-5). Steps 6 and following are modified as follows: 6. To load the IR cell with a solid sample, follow these steps. ① Remove the IR cell from the storage desiccator. ② Place the IR cell on a clean paper towel strip or on a Kimwipe (NOT on the BENCH!). ③ In a 1-dram vial, prepare a very small volume (no more than 0.5 mL) of a somewhat concentrated solution of your solid sample in a volatile solvent (chloroform and dichloromethane are good solvents for most of the solids; others may require methanol or ethanol). Make sure all solid dissolves. (NOTE: if none of these solvents appears to dissolve your solid, it may be necessary for you to obtain the spectrum as a KBr pellet. See the instructor.) To prepare the solution, o Use a clean spatula to transfer a small amount of solid to a clean 1-dram vial. Wipe the spatula with a Kimwipe, return it to its place. o Add 0.5-1 mL solvent to the vial using a CLEAN Pasteur pipet. o Cap the vial, dissolve the solid. ④ Using the same Pasteur pipet, place 1-2 drops of the solution in the center of one of the NaCl plates. Place the pipet in the used pipet beaker. 9 / 11

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⑤ Wait several minutes for the solvent to evaporate from the crystal. To hasten solvent evaporation, wave your hand back and forth over the crystal. DO NOT blow on the crystal--saliva has an IR spectrum. ⑥ Place the salt crystal into the metal holder with the sample on top. The second crystal is not required. ⑦ Reassemble the cell by screwing in the plastic holder piece. BEWARE OF CROSS THREADING. 7. Place your IR cell in the IR spectrometer. 8. Obtain the spectrum of the sample. 9. Remove the cell from the instrument. 10. Dissassemble the crystal holder. 11. Remove the crystal. 12. Clean the crystal surface by BLOTTING (NOT rubbing) the sample from the face of the crystal using a Kimwipe wetted with acetone. Allow the crystal to air dry. 13. Proceed with the next sample. Obtain instruction in the use of the FTIR spectrometer. Then obtain the IR spectra of your samples. Print the spectra. 5.3 Clean-up. When you have finished all of your work: • • • • • •

Clean Labkit glassware by recommended procedures, shake off excess water, and return to the Labkit. Clean Pasteur pipets, graduated pipets, and vials by recommended procedures, shake off excess water, and place in the drying oven. Return CLEAN IR cells to the instructor. Return samples to the front bench where you got them. Clean and return all borrowed equipment to the instructor before leaving lab. Clean up your work area before leaving lab.

6. Focus Questions Enter the observed absorption energies under the appropriate functional group in the table. Can you make any general conclusions from your data regarding the relationship between structure (bond order, atom masses, nature of adjacent atoms, etc.) and the position/intensity of an absorption in the IR? 1. Which region of the IR spectrum is most useful for identifying specfific functional groups in molecules? 2. Which region of the IR spectrum is most useful for unambiguously establishing the identity of a substance? 10 / 11

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3. A carbon-oxygen double bond vibrates with an energy of about 1720 cm-1. With what energy would you expect a carbon-sulfur double bond to vibrate? 4. Calculate the reduced mass, m, for an O-H bond; for an N-H bond; for a C-H bond. What is primarily responsible for the differences in frequency of vibration of these 3 bond types? 5. For each molecule, indicate where you would expect to see distinctive vibrations in the 4000-1400 cm-1 region of the infrared spectrum: (CH3)2NH (CH3)2CH(OH) H2N-CH(CH3)-COOH

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