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INFRARED SPECTROSCOPY THEORY A molecule is not a rigid assemblage of atoms. The atoms, even in the solid state, vibrate about an equilibrium position. Each atom vibrates with a frequency which depends on its mass and the length and strength of any bonds which it has formed. There are two kinds of fundamental vibrations for molecules, stretching and bending (or deformation). Stretching In stretching vibrations, the distance between the atoms increases or decreases, but the atoms remain in the same bond axis. symmetrical stretching asymmetrical stretching Bending In bending vibrations, the position of the atom changes relative to the original bond axis. + + + - scissoring rocking wagging twisting (in-plane bending vibrations) (out-of-plane bending vibrations) + and – signify vibrations above and below the plane of the paper respectively Each of the various stretching and bending vibrations of a bond occur at a certain particular frequency, in the range 1.20 x 1013 to 1.20 x 1014 Hz, which corresponds to a frequency in the infrared region of the electromagnetic spectrum. Therefore, when infrared radiation of that same frequency falls on the molecule, energy is absorbed and the amplitude of that vibration increases. Only those vibrations which cause a change in the dipole moment of the molecule will result in the absorption of infrared radiation. So that the numbers involved are more convenient, absorption is usually quoted at a particular wavenumber (the reciprocal of the wavelength in cm of the absorbed radiation), which has the unit cm-1. Bending vibrations generally require less energy and therefore occur at lower wavenumbers than stretching vibrations. Stretching vibrations occur in the order of bond strengths. Thus, a triple bond absorbs radiation of a higher wavenumber than the equivalent double bond, which in turn absorbs radiation of a higher wavenumber than the equivalent single bond. TOPIC 13.32: STRUCTURE DETERMINATION 1 THE INFRARED SPECTROMETER An infrared spectrometer has two beams of radiation from the same source, one passing through the sample, the other passing through a reference cell. Sample Chart recorder Detector glowing source Monochromatic grating Reference A rotating segmented mirror takes information from the sample and reference beams alternately. If a particular frequency is absorbed by the sample, less radiation is transmitted, and the detector compares the intensity of radiation of each wavelength passing through the sample with the intensity passing through the reference cell. When the spectrum is determined, a calibration line is usually recorded on the paper. For this purpose, the absorption peak of polystyrene at 1603cm -1 is commonly used. Preparing Samples Glass absorbs infrared radiation. Therefore, cells for use in an infrared spectrometer have to be made from some other material, which is transparent to infrared. Sodium chloride and potassium bromide are frequently used. Potassium bromide is preferred, because sodium chloride absorbs radiation below 625cm-1. Gases Gases are placed in a special cell, typically 10cm in length. Liquids Liquids are examined in the form of a thin film, formed by sandwiching a small drop of liquid between sodium chloride discs. Solutions Solutions, usually in solvents such as CCl4 and CHCl3, are placed in cells with a path length between 0.01mm and 0.5mm. Solids A disc can be made by finely grinding the solid with KBr in a mortar. The powder is then placed in a circular die and compressed to give a transparent disc which is used to record the spectrum. Alternatively, a paste can be made from the solid and a long-chain liquid hydrocarbon called Nujol. This mull is then placed between sodium chloride discs. TOPIC 13.32: STRUCTURE DETERMINATION 2 INTERPRETATION OF INFRARED SPECTRA Infrared spectra are valuable aids in the identification of unknown compounds. A complex molecule has a large number of vibrational modes. Some of these vibrations (mainly above 1500cm-1) are associated with individual bonds or functional groups, while others (below 1500cm-1) are considered as vibrations of the whole molecule. The absorption associated with a particular individual bond or functional group always occurs at approximately the same wavenumber, irrespective of the molecule involved. e.g. the carbonyl (C=O) absorption in the following aliphatic ketones: butanone 1718cm-1 cyclohexanone 1715cm-1 Structural changes close to the vibrating bond can cause small shifts. e.g. the carbonyl absorption in the following classes of compounds: aliphatic ketones 1725-1705cm-1 aromatic ketones 1700-1680cm-1 aliphatic aldehydes 1740-1720cm-1 aromatic aldehydes 1715-1695cm-1 esters 1750-1735cm-1 aliphatic carboxylic acids 1725-1700cm-1 acyl chlorides 1795cm-1 The infrared spectrum can conveniently be split into four regions for interpretation. 4000-2500 cm-1 absorption due to stretching vibrations of single bonds to hydrogen e.g. C-H, O-H, N-H 2500-2000 cm-1 absorption due to stretching vibrations of triple bonds e.g. C=C, C=N 2000-1500 cm-1 absorption due to stretching vibrations of double bonds e.g. C=C, C=O 1500-400 cm-1 absorption due to stretching vibrations of some single bonds e.g. O-H, C-O absorption due to various bending vibrations The region 1500-400 cm-1 is not mainly used for recognising individual bond absorptions. This region contains absorption bands which are characteristic of the compounds under test and no other compound. This region is therefore called the fingerprint region. The fingerprint region can be used to confirm the identity of a substance by comparison with the infrared spectrum of an authentic sample. TOPIC 13.32: STRUCTURE DETERMINATION 3 Correlation Table for Infrared Spectroscopy 4600 4200 3800 aliphatic C-H stretch 3400 3000 2600 2200 2000 1800 N-H stretch amines (v) 1600 1400 1200 1000 C-N stretch C-H bend aromatic nitro cpds. aliphatic (v) C-H stretch C C-H (sh) C N stretch aliphatic nitriles C=O stretch (vs) C-H stretch alkenes (sh) C C-H stretch aromatic ring (w-m) C stretch (v) C-Br stretch haloalkanes 2600 C=C stretch Ar-C=C (w-m) Abbreviations used: w = weak m = medium s = strong vs = very strong v = variable br = broad sh = sharp C-N stretch aromatic nitro cpds. N-H bend (v) 2200 2000 1800 1600 1400 Wavenumber / cm-1 TOPIC 13.32: STRUCTURE DETERMINATION C-Cl stretch haloalkanes C=C stretch aromatic (m) C-H stretch (2 bands) aldehydes 3000 C-N bend amines C=C stretch alkenes (v) C-H stretch alkanes (v) 3400 C-H bend C=C-H (m-s) C-C skeletal vibration (w) O-H stretch ROH (sh) 3800 C-H bend C C-H (s) N-O stretch aromatic nitro cpds. H-bonded O-H stretch (br) 4200 400 C-O stretch esters, alcohols, carboxylic acids O-H stretch COOH (w, br) 4600 600 C-O-C asymmetric stretch (vs) free O-H stretch (sh) aromatic C-H stretch (w) 800 4 Fingerprint region 1200 1000 800 600 400 Characteristic Infrared Absorption Ranges Wavenumber range/ cm-1 Functional Group Vibration 3750 – 3200 3500 – 3300 3500 – 3140 3300 – 2500 3095 – 3010 3080 – 3030 2962 – 2853 2900 – 2700 (2 bands) 2260 – 2215 2260 – 2140 1850 – 1800 1790 – 1740 1795 1750 – 1735 1740 – 1720 1730 – 1717 1725 – 1705 1725 – 1700 1700 – 1680 1700 – 1630 1669 – 1645 1650 – 1590 1600, 1500 & 1450 1570 – 1500 1550 – 1330 1485 – 1365 1370 – 1300 1310 – 1100 880 – 700 800 – 600 600 – 500 alcohol & phenol amine amide carboxylic acid alkene arene alkane aldehyde nitrile alkyne carboxylic anhydrides carboxylic anhydrides acyl chloride ester aldehyde aryl ester ketone carboxylic acid aryl carboxylic acid amide alkene amide & amine arene aromatic nitro compound aromatic nitro compound alkane aromatic nitro compound ester, alcohol, carboxylic acid arene chloroalkane bromoalkane O-H stretch N-H stretch N-H stretch O-H stretch C-H stretch C-H stretch C-H stretch C-H stretch C=N stretch C=C stretch C=O stretch C=O stretch C=O stretch C=O stretch C=O stretch C=O stretch C=O stretch C=O stretch C=O stretch C=O stretch C=C stretch N-H bend C=C stretch C-N stretch N-O stretch C-H bend C-N stretch C-O stretch C-H bend C-Cl stretch C-Br stretch TOPIC 13.32: STRUCTURE DETERMINATION 5 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY Nuclear magnetic resonance (NMR) spectroscopy gives information on the environment in which the nuclei of atoms are found in molecules. In many cases, so much information can be obtained from a single spectrum that the structure of the molecule can be determined. THEORY All nuclei with an odd mass number have a property called spin, e.g. 1H, 13C, 15N, 19F, 31P. Since the nuclei are positively charged, this spin is associated with a circulation of electric charge and thus gives rise to a magnetic field. When placed in an external magnetic field, the nuclei tend to align themselves in the magnetic field. For a nucleus of spin ½ only two orientations are allowed: with the external field and against it. nuclear magnetic moments with field against field external magnetic field These two orientations have different energies and hence there are two energy levels populated by the nuclei. Transitions between the two energy levels can occur if radiation of the correct frequency (E = h) is absorbed. This radiation is in the radiofrequency region. The “resonance” absorption is recorded as an NMR spectrum. 1H NMR is the most widely used. Energy against field E with field no field TOPIC 13.32: STRUCTURE DETERMINATION field applied 6 OBTAINING A NMR SPECTRUM 1. The sample is dissolved in a proton-free solvent e.g. CDCl3, CCl4 2. The solution is placed in a glass NMR tube which is then spun at 20 revolutions per second between the poles of a strong electromagnet. Spinning averages out any lack of uniformity in the sample or the glass tube. The magnetic field must be stable and homogeneous to a few parts in 108. 3. The sample is irradiated with radio waves of constant frequency and the absorption measured as the magnetic field is varied. Alternatively, the magnetic field can be kept constant and the range of radiofrequencies scanned. SHIELDING AND CHEMICAL SHIFT When a molecule is placed in a magnetic field, the electrons surrounding the nuclei behave like perfectly conducting shells, and weak electric currents are induced in them, which flow in such a way as to produce a magnetic field which opposes the applied field. Therefore, the nuclei at the centre experience a magnetic field which is fractionally smaller than the applied field. Since the electron distribution around chemically different hydrogen atoms in a molecule is not the same, the nuclei are shielded from the applied field to different extents. Each unique proton gives a unique NMR signal, which gives a map of the C-H framework of a molecule. The place at which a particular proton absorbs is called its chemical shift. It is measured relative to the standard TMS (tetramethylsilane) (CH3)4Si. Chemical shift is given the symbol and is measured in ppm. = signal (Hz) –TMS (Hz) ppm machine frequency (MHz) Defining chemical shift in this way means that it is independent of the applied magnetic field which is used. Resonances at high field Resonances at low field (shielded) (deshielded) 2 integral 10 1 Chemical shift for hydrogen /ppm 0 By definition, the value for TMS is 0. Most organic proton resonances occur on a scale of 0-10 on the higher frequency/lower field side of zero. The greater the value of , the more the proton is deshielded by adjacent electron-withdrawing atoms or groups. TOPIC 13.32: STRUCTURE DETERMINATION 7 The area under each peak in the spectrum is measured by electronic integration, and the heights of the steps on the integration curve indicate the relative number of protons giving rise to each peak. At A-level, you will not be expected to measure the integral, instead you will be told the number of protons in each peak. A high resolution NMR spectrum reveals that many peaks are not singlets (i.e. single lines), but are multiplets. This arises because of spin-spin coupling. SPIN-SPIN COUPLING The magnetic field experienced by a particular proton is influenced by neighbouring NONEQUIVALENT protons. This interaction is known as coupling, and it causes peaks to be split into a number of lines. If a proton couples with n neighbouring non-equivalent protons, the signal of that proton is split into (n+1) lines The separation between peaks is called the spin-spin coupling constant, J, and is measured in Hz. To decide if protons are equivalent, substitute any one by chlorine and count the possible isomers. If only one isomer is possible, the protons are equivalent. Examples: CH3CH3 all protons are equivalent 1 line only CH3CH2X CH3 protons signal split by CH2 protons 3 lines (1:2:1 triplet) CH2 protons signal split by CH3 protons 4 lines (1:3:3:1 quartet) CH3CHX2 CH3 protons signal split by CH proton 2 lines (1:1 doublet) CH proton signal split by CH3 protons 4 lines (1:3:3:1 quartet) In CH3CHX2 CH CH3 The relative intensities of the peaks are 1:3 J J J J TOPIC 13.32: STRUCTURE DETERMINATION 8 The same coupling constant is seen in both signals Protons separated by more than two carbon atoms show little or no coupling. In the example below, Hb and Hc both couple with Ha but do not couple with each other. Hb Hc Ha Ha appears as a quartet, where Jab>Jac Jab Hb and Hc appear as doublets with splitting Jab and Jac respectively -C-C-CHa Jac Jac Benzene Derivatives In benzene derivatives, such as methylbenzene, the five adjacent aromatic protons have similar chemical shifts and undergo mutual coupling. The overlapping sets of peaks give a multiplet at around = 7.3 which is difficult to analyse. The NMR spectrum of propyl ethanoate below shows the integration curve and how it is used to obtain information on the number of protons resonating at each frequency. In this case, the ratio of protons (not necessarily the absolute number of protons) b:a:c:d is 2:3:2:3 H3C.CO.O.CH2.CH2.CH3 a b c d 3 2 ppm protons 0.9 1.7 2.0 4.0 CH3 CH2 CH3.CO CH2O 3 2 8 TOPIC 13.32: STRUCTURE DETERMINATION 7 6 9 5 4 /ppm 3 2 1 0 Table of Proton Chemical Shifts Methyl protons CH3-R CH3-C=C CH3-CN CH3-COOR CH3-COR CH3-Ar CH3-NRCH3-COOAr CH3-COAr CH3-NCOR CH3-NAr CH3-OR 0.7 – 1.6 1.6 – 1.9 2.0 2.0 2.2 2.3 2.3 2.4 2.6 2.9 3.0 3.3 CH3-OCOR CH3-OAr CH3-OCOAr 3.7 3.8 4.0 – 4.2 ppm Methylene protons ppm Methine protons R-CH2-R 1.4 -CH-R 1.5 R-CH2-CN 2.3 -CH-CN 2.7 R-CH2-COR R-CH2-Ar R-CH2-N 2.4 2.3 – 2.7 2.5 -CH-COR -CH-Ar 2.7 3.0 R-CH2-COAr R-CH2-NCOR 2.9 3.2 -CH-COAr -CH-NCOR 3.3 4.0 R-CH2-OR R-CH2-I R-CH2-Br R-CH2-Cl R-CH2-OH Ar-CH2-COR R-CH2-OCOR R-CH2-OAr 3.4 3.2 3.5 3.6 3.6 3.7 4.1 4.3 -CH-OR -CH-I -CH-Br -CH-Cl -CH-OH 3.7 4.3 4.3 4.2 3.9 -CH-OCOR -CH-OAr 4.8 4.5 Ar-CH2-OCOR 4.9 H-C C -CH=C- 2.5–2.7 5.2–5.7 CH2=C- 4.6 - 5.0 ppm Protons in Other Environments R-NH2 * ArNH2 * R-OH * 1.1 – 1.5 3.4 – 4.0 3.0 – 5.2 R2-NH * 0.4 – 1.6 Ar-OH * 4.5 – 7.7 * The absorption positions of these groups are concentration-dependent and are shifted to higher values in more concentrated solutions. Ar-H R-CHO 6.6 – 8.0 9.7 – 9.8 -NH-CO- 5.5 – 8.5 Ar-CHO 9.7 – 10.0 TOPIC 13.32: STRUCTURE DETERMINATION 10 Carbon-13 nmr spectroscopy After hydrogen, the most useful atom providing information is carbon-13. Natural carbon contains about 1% of this isotope so the instruments for its detection need to be sensitive and spectra will take longer to record. Only the chemical shift is important as each spectrum (in proton decoupled 13C nmr) gives only single lines for each chemically equivalent carbon. Environment C - C (alkanes) C - C=O C - Cl or C - Br C - N (amines) C - OH C = C (alkenes) aromatic C’s (benzene rings) C=O (esters, acids, amides) C=O (aldehydes, ketones) Chemical shift / 10 - 35 10 - 35 30 - 70 35 - 65 50 - 65 115 - 140 125 - 150 160 - 185 190 – 220 Carbon-13 nmr has wide applications in the study of natural products, biological molecules and polymers. Let us consider the 13C nmr spectrum of ethanol (CH3CH2OH) The spectrum in the box shows where the 1H nmr spectrum would appear. We can see that there are 2 large peaks in this spectrum at = 19 and = 58ppm. This tells us that there are 2 non-equivalent C atoms in the molecule. Using the chemical shift data above we can see that the peak at 19ppm is due to a C – C whilst the peak at 58ppm is due to a C-OH. The first peak is due to CH3CH2OH whilst the other is for CH3CH2OH. TOPIC 13.32: STRUCTURE DETERMINATION 11 Exercise How many peaks would you expect there to be in the carbon-13 spectrum of… • • • • • • • • • • • • • ethylethanoate • cyclohexane butane …….. CH3CH2CH2CH3 2-methylpropane CH3CH(CH3)CH3 …….. butanal CH3CH2CH2CHO …….. butanone CH3COCH2CH3 …….. pentan-2-one CH3COCH2CH2CH3 …….. pentan-3-one CH3CH2COCH2CH3 …….. CH3COOCH2CH3 …….. C6H12 …….. TOPIC 13.32: STRUCTURE DETERMINATION 12 ? MASS SPECTROMETRY Mass spectrometry is a useful technique for determining the structure of a compound. It is about 1000 times more sensitive than either infrared or NMR spectroscopy and requires only a microgram or less of the sample. Basic features 1. The apparatus is kept under vacuum to prevent ions produced from colliding with air particles. 2. Ionisation: The sample is dissolved in a volatile solvent and a high voltage is applied. Positively charged droplets are released as a fine spray and the solvent evaporates. This leaves individual positively charged ions. 3. Acceleration: The ions are accelerated by attraction towards a negatively charged plate which gives all of the ions the same kinetic energy. Since kinetic energy = ½ mv2, lighter ions move faster than heavier ones. 4. Ion drift: The ions are passed through a hole in the negatively charged plate forming a beam. The ions drift down the ‘flight tube’ towards a detector. They are separated based on their different velocities. The faster (lighter) ions reach the detector first. 5. Detection: The separated streams of ions strike a negatively-charged detector plate, which is contained in the detector and flight times are recorded. The arriving ions generate a current which sends a signal to a computer or a chart recorder. A plot of the intensity of the signal against m/z is drawn out. TOPIC 13.32: STRUCTURE DETERMINATION 13 e.g. stick diagram of the mass spectrum of ethanol. Relative abundance % [CH2OH]+ 100 80 60 parent ion m/e 46 [C2H5O]+ 40 [C2H5]+ [C2H3]+ [C2H5OH]+ 20 [M+1]+ 10 20 30 40 50 m/z Applications of Mass Spectrometry 1. Mass spectrometry can be used for the determination of the isotopic composition of a sample of an element and hence for the measurement of relative atomic mass. 2. A mass spectrum can be used to determine relative molecular masses of compounds since the line with the highest m/z value is usually the molecular ion, M+. There is always a very small peak at (M+1) for organic compounds which is due to the 13C isotope, present with a natural abundance of 1.1%. 3. The high resolution of the double focussing mass spectrometer enables very precise measurements of masses to be made. Since the mass defect varies from element to element, the masses of ions with the same nominal mass show small differences if precise measurements are made. E.g. for a nominal m/z of 28: Ion Precise mass CO+ N2+ CH3N+ C2H4+ 27.994914 28.006148 28.018724 28.031300 Thus it is possible with a high resolution spectrum to determine not only the molecular mass but also the molecular formula. 4. The structure of a molecule can be determined from the fragmentation pattern. Once the molecular ion has been formed, it will usually decompose into smaller fragments. By looking at the major peaks in the mass spectrum, it is possible to decide which fragments have been lost and thence to build up a picture of the original molecule. Occasionally a rearrangement of an ion may take place. TOPIC 13.32: STRUCTURE DETERMINATION 14 Interpretation of the Mass Spectrum The mass of the molecular ion gives the Mr of the compound. Double resolution machines can measure the mass of the molecular ion with precision sufficient to determine the molecular formula of the compound. If the likely structure of the compound is known, it can usually be confirmed by considering the fragmentation pattern. It is not always possible to distinguish between isomers e.g. the different disubstituted benzenes. If the structure is unknown, the mass spectrum can be compared with libraries of mass spectra which are commercially available on computer software. The mass spectra of all compounds having the same Mr (or molecular formula) as the unknown can be called up and compared. A compound containing one atom of chlorine will give two molecular ions which are two units apart. The higher m/z peak will have an intensity one third that of the lower m/z peak. This is because chlorine contains the isotopes 35Cl and 37Cl in the ratio 3:1. A compound containing one atom of bromine will give two molecular ions which are two units apart. Both peaks will be equal in intensity. This is because bromine contains two isotopes 79Br and 81Br in the ratio 1:1. TOPIC 13.32: STRUCTURE DETERMINATION 15 Common Losses from Molecular Ions Ion M-15 M-16 M-17 M-18 M-27 M-28 M-29 M-30 M-31 M-32 M-36 M-41 M-42 Group being lost Possible inference CH3 O NH2 OH H2O HCN CO C2H4 -NO2 -CONH2 alcohol alcohol, aldehyde, ketone nitrile quinone aromatic ethyl ether (ArOC2H5) ethyl ester (RCOOC2H5) propyl ketone (RCOC3H7) aldehyde ethyl ketone (RCOC2H5) aromatic methyl ether (ArOCH3) aromatic nitro compound methyl ester (RCOOCH3) methyl ester (RCOOCH3) primary haloalkane propyl ester (RCOOC3H7) methyl ketone (RCOCH3) aromatic ethanoate (CH3COOAr) butyl ketone (RCOC4H9) aromatic propyl ether (ArOC3H7) carboxylic acid, ester, acid anhydride carboxylic acid ethyl ester (RCOOC2H5) aromatic nitro compound CHO C2H5 OCH2 NO OCH3 CH3OH HCl C3H5 CH2CO C3H6 M-44 M-45 M-46 CO2 COOH C2H5OH NO2 Aromatic compounds often produce fragment ions at m/z 77 (C6H5+) and m/z 91 (C6H5CH2+). TOPIC 13.32: STRUCTURE DETERMINATION 16