《材料测试技术及方法》课程教学资源（书籍文献）分子光谱之红外光谱简明教程 INFRARED SPECTROSCOPY

14782_02_Ch2_p015-104.pp2.gxd1/25/0810:27Page15⊕ INFRARED SPECTROSCOPY ovalent bonds whether heeomieswacenghsbereuthoeasocindwhvkbeeLwhidhrneemm m), at wavelengths shorter tha )houhmore thnically coforavelengthn thinfrard ctrum is the micro on infrared spectra.Figure 2. shows that the ortional to the fre the rrhth nergysrcoo heywherePlanck's the latte frequencies have very low energies,only enough to cause nuclear or electronic spin transitions within ⊕ organic molecules.Nuclear magnetic resonance,which occurs in the radiofrequency part of the radiation in the vibrational of spectrum in terms of a unit called a wavenumber ()rather than wavelength (y or um). high high← Energy -RAY ULTRAVIOLET Ultraviolet Visible 80nm← →8e← short- Wavelength(A)- ong

INFRARED SPECTROSCOPY Almost any compound having covalent bonds, whether organic or inorganic, absorbs various frequencies of electromagnetic radiation in the infrared region of the electromagnetic spectrum. This region lies at wavelengths longer than those associated with visible light, which range from approximately 400 to 800 nm (1 nm = 10−9 m), but lies at wavelengths shorter than those associated with microwaves, which are longer than 1 mm. For chemical purposes, we are interested in the vibrational portion of the infrared region. It includes radiation with wavelengths (l) between 2.5 mm and 25 mm (1mm = 10−6 m). Although the more technically correct unit for wavelength in the infrared region of the spectrum is the micrometer (mm), you will often see the micron (m) used on infrared spectra. Figure 2.1 illustrates the relationship of the infrared region to others included in the electromagnetic spectrum. Figure 2.1 shows that the wavelength l is inversely proportional to the frequency n and is governed by the relationship n = c/l, where c = speed of light. Also observe that the energy is directly proportional to the frequency: E = hn, where h = Planck’s constant. From the latter equation, you can see qualitatively that the highest energy radiation corresponds to the X-ray region of the spectrum, where the energy may be great enough to break bonds in molecules. At the other end of the electromagnetic spectrum, radio￾frequencies have very low energies, only enough to cause nuclear or electronic spin transitions within molecules—that is, nuclear magnetic resonance (NMR) or electron spin resonance (ESR), respectively. Table 2.1 summarizes the regions of the spectrum and the types of energy transitions observed there. Several of these regions, including the infrared, give vital information about the structures of organic molecules. Nuclear magnetic resonance, which occurs in the radiofrequency part of the spectrum, is discussed in Chapters 3, 4, 5, 6, and 10, whereas ultraviolet and visible spectroscopy are described in Chapter 7. Most chemists refer to the radiation in the vibrational infrared region of the electromagnetic spectrum in terms of a unit called a wavenumber (n ), rather than wavelength (m or mm). 15 CHAPTER 2 FIGURE 2.1 A portion of the electromagnetic spectrum showing the relationship of the vibrational infrared to other types of radiation. 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 15

p8.0.a0104m2.ga/257091027gag16⊕ Infrared Spectroscop TABLE 2.1 Region of Spectrum Energy Transitions X-ravs Bond breaking Infrared Micro Rotational Wavenumbers are expressed as reciprocal centimeters (cm-)and are easily com outed by taking the reciprocal of the wavelength expressed in centimeters.Convert a wavenumbertoa frequency vby multiplying it by the speed of light (expressed in centimeters per second). 下(cm-=em 1 v()=c=C(cm人ec 入(cm) The main from about 4000 to 400 cm This range coresponds to wave in olde Cmrybyusingthcfolomingrelhationships cmr-dxiaow adm=d×10o0 INTRODUCTION TO INFRARED SPECTROSCOPY 2.1 THE INFRARED ABSORPTION PROCESS of energy absorption,molecul are excited to The absorption of infrared radiation corresponds toenergy changes on theorder/mole R ion in this energy range c onds to the ran frequencies of infrared radiation that match the natural vibrational frequencies of the molecule in question are absorbed,and the energy absorbed serves to increase the amplitude of the vibrationa ons or the bonds in the mo that not all b onds in a mo of the

Wavenumbers are expressed as reciprocal centimeters (cm−1 ) and are easily computed by taking the reciprocal of the wavelength expressed in centimeters. Convert a wavenumber n to a frequency n by multiplying it by the speed of light (expressed in centimeters per second). n (cm−1 ) = l ( 1 cm) n (Hz) = n c = c ( l cm (c / m se ) c) The main reason chemists prefer to use wavenumbers as units is that they are directly proportional to energy (a higher wavenumber corresponds to a higher energy). Thus, in terms of wavenumbers, the vibrational infrared extends from about 4000 to 400 cm−1 . This range corresponds to wave￾lengths of 2.5 to 25 mm. We will use wavenumber units exclusively in this textbook. You may en￾counter wavelength values in older literature. Convert wavelengths (m or mm) to wavenumbers (cm−1 ) by using the following relationships: 16 Infrared Spectroscopy INTRODUCTION TO INFRARED SPECTROSCOPY 2.1 THE INFRARED ABSORPTION PROCESS As with other types of energy absorption, molecules are excited to a higher energy state when they absorb infrared radiation. The absorption of infrared radiation is, like other absorption processes, a quantized process. A molecule absorbs only selected frequencies (energies) of infrared radiation. The absorption of infrared radiation corresponds to energy changes on the order of 8 to 40 kJ/mole. Radiation in this energy range corresponds to the range encompassing the stretching and bending vibrational frequencies of the bonds in most covalent molecules. In the absorption process, those frequencies of infrared radiation that match the natural vibrational frequencies of the molecule in question are absorbed, and the energy absorbed serves to increase the amplitude of the vibrational motions of the bonds in the molecule. Note, however, that not all bonds in a molecule are capable of absorbing infrared energy, even if the frequency of the radiation exactly matches that of the bond motion. Only those bonds that have a dipole moment that changes as a function of time are capable TABLE 2.1 TYPES OF ENERGY TRANSITIONS IN EACH REGION OF THE ELECTROMAGNETIC SPECTRUM Region of Spectrum Energy Transitions X-rays Bond breaking Ultraviolet/visible Electronic Infrared Vibrational Microwave Rotational Radiofrequencies Nuclear spin (nuclear magnetic resonance) Electronic spin (electron spin resonance) cm−1 = (m 1 m) × 10,000 and mm = (cm 1 −1 ) × 10,000 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 16

14782_02_Ch2_p015-104.Pp2.gxd1/25/0810:27age7⊕ 2.2 Uses of the Infrared Spectrum 17 of absorbing infrared radiation.Symmetric bonds.such as those of H,or Cl.do not absorb infrared radiation.A bond must present an electrical dipole that is changing at the same frequency as the in- coming radiation for c y tobe trans For the purposes of an organic chemist,the bonds most likely to be affected by this restraint are those of symmetric or pseudosymmetric alkenes (C=C)and alkynes (C=C). CH3-CH2 C-C一CH CH: CH;-C=C-CH; CH:-CH2-C=C-CH; Symmetric Pseudosymmetric 2.2 USES OF THE INFRARED SPECTRUM pattem.or infrared spec rumAlthough som of the frequencies aborbed in thetwo cses might behe same. tieretmoteuleswilt ca Thus,th paring the infrared spectra of two substances thought to be identical.you can establish whether they are.in fact,identical.If their infrared spectra coincide peak for peak (absorption for absorption).in two substances will be ic oa mocu The ons of each ty C.C=C.C=N,and so on)are regularly fo und only in certain small portions of the vibrational infra Iregion.A sm of bel is range.000主 150 cm-is almo tion in the 171510 cmis normally due to the presence of (carbonyl group)in the molecule.The your 4000 2500 2000 1650 1550 650 0-H C-H VERY C=0 C-N C-CI FEW C=N BANDS N-H c-c X=C=Y (C,0.N,S N-O N-O 2.5 5.5 6.1 6.5 15.d WAVELENGTH ( gions where variou

2.2 Uses of the Infrared Spectrum 17 2.2 USES OF THE INFRARED SPECTRUM Since every type of bond has a different natural frequency of vibration, and since two of the same type of bond in two different compounds are in two slightly different environments, no two mole￾cules of different structure have exactly the same infrared absorption pattern, or infrared spec￾trum. Although some of the frequencies absorbed in the two cases might be the same, in no case of two different molecules will their infrared spectra (the patterns of absorption) be identical. Thus, the infrared spectrum can be used for molecules much as a fingerprint can be used for humans. By com￾paring the infrared spectra of two substances thought to be identical, you can establish whether they are, in fact, identical. If their infrared spectra coincide peak for peak (absorption for absorption), in most cases the two substances will be identical. A second and more important use of the infrared spectrum is to determine structural information about a molecule. The absorptions of each type of bond (NIH, CIH, OIH, CIX, CJO, CIO, CIC, CJC, CKC, CKN, and so on) are regularly found only in certain small portions of the vibrational infra￾red region. A small range of absorption can be defined for each type of bond. Outside this range, absorp￾tions are normally due to some other type of bond. For instance, any absorption in the range 3000 ± 150 cm−1 is almost always due to the presence of a CIH bond in the molecule; an absorption in the range 1715 ± 100 cm−1 is normally due to the presence of a CJO bond (carbonyl group) in the molecule. The same type of range applies to each type of bond. Figure 2.2 illustrates schematically how these are spread out over the vibrational infrared. Try to fix this general scheme in your mind for future convenience. of absorbing infrared radiation. Symmetric bonds, such as those of H2 or Cl2, do not absorb infrared radiation. A bond must present an electrical dipole that is changing at the same frequency as the in￾coming radiation for energy to be transferred. The changing electrical dipole of the bond can then couple with the sinusoidally changing electromagnetic field of the incoming radiation. Thus, a sym￾metric bond that has identical or nearly identical groups on each end will not absorb in the infrared. For the purposes of an organic chemist, the bonds most likely to be affected by this restraint are those of symmetric or pseudosymmetric alkenes (CJC) and alkynes (CKC). CH3 CH3 CH3 CH3 C C CH3 CH3 CH3 CH2 CH3 C C CH3 C C CH3 CH3 CH2 C C CH3 Symmetric Pseudosymmetric FIGURE 2.2 The approximate regions where various common types of bonds absorb (stretching vibrations only; bending, twisting, and other types of bond vibrations have been omitted for clarity). 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 17

Infrared Spectroscop 2.3 THE MODES OF STRETCHING AND BENDING The simplest type.or mode,ofbi i olecule hat ar infared active those. that give ris ng mo C-H Stretching Bending aciesntr sye iher fr iatio rockingwagging,and twisting are commonly used in the literature to describe the origins of p of three or more atoms.at least two of which are identical.there are two modes of stretchin symmetric and asymmetric.Examples of such groupings are-CH3.-CH2-(see p.19). cn i cm modes.A similar phenomenon occurs in the amino group.wherea primary amine (NH)usually has two absorptions in the N-H stretch region,while a secondary amine (RNH)has only one absorp- on peak.Amide stretcn peaks about 1550 cm H H -2853cm 1-.1250cm IN-PLANE OUT-OF-PLANE STRETCHING VIBRATIONS BENDING VIBRATIONS

18 Infrared Spectroscopy 2.3 THE MODES OF STRETCHING AND BENDING The simplest types, or modes, of vibrational motion in a molecule that are infrared active—those, that give rise to absorptions—are the stretching and bending modes. However, other, more complex types of stretching and bending are also active. The following illustra￾tions of the normal modes of vibration for a methylene group introduce several terms. In general, asymmetric stretching vibrations occur at higher frequencies than symmetric stretching vibrations; also, stretching vibrations occur at higher frequencies than bending vibrations. The terms scissoring, rocking, wagging, and twisting are commonly used in the literature to describe the origins of infrared bands. In any group of three or more atoms, at least two of which are identical, there are two modes of stretching: symmetric and asymmetric. Examples of such groupings are ICH3, ICH2I (see p. 19), INO2, INH2, and anhydrides. The methyl group gives rise to a symmetric stretching vibration at about 2872 cm−1 and an asymmetric stretch at about 2962 cm−1 . The anhydride functional group gives two absorptions in the CJO region because of the asymmetric and symmetric stretching modes. A similar phenomenon occurs in the amino group, where a primary amine (NH2) usually has two absorptions in the NIH stretch region, while a secondary amine (R2NH) has only one absorp￾tion peak. Amides exhibit similar bands. There are two strong NJO stretch peaks for a nitro group, with the symmetric stretch appearing at about 1350 cm−1 and the asymmetric stretch appearing at about 1550 cm−1 . C H Stretching H O C Bending 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 18

14782_02_Ch2_p015-104.Pp2.gxd1/25/0810:27Page9⊕ 2.3The Modes of Stretching and Bending 19 Symmetric Stretc Methyl -2872cm-l -2962cnm- 只A -1760cm 1800cm N 3300cm 3400cm =o -1350 cm- 1550 cm The vibrations we have been discussing are called fundamental absorptions.They arise from lowest-energy stat sult from excitation from the ground state to higher energy states.which correspond to integral the undenal()or em yovo kind of erates overtones.If you pluck a string mpanying peak of lower intensity at 1000cm an overtone. combination band.This band is the sum of the two interacting bands( +)Not all possible combinations occur.The rules that govern which combinations are allowed are beyond the scope of our d on here ed frequency in this case results from the difference between the two interacting hands (y =,- One can calculate overtone,combination,and difference bands by directly manipulating fre- n a n Fermi resonance.gain. ance is often observed in carbonyl compounds cquencies of the whole molecule are not infrared active.they often couple sharp in the infrared spectrum is rotational coupling.which may lead to a considerable amount of unresolved fine structure

2.3 The Modes of Stretching and Bending 19 The vibrations we have been discussing are called fundamental absorptions. They arise from excitation from the ground state to the lowest-energy excited state. Usually, the spectrum is compli￾cated because of the presence of weak overtone, combination, and difference bands. Overtones re￾sult from excitation from the ground state to higher energy states, which correspond to integral multiples of the frequency of the fundamental (n). For example, you might observe weak overtone bands at 2n , 3n , . . . . Any kind of physical vibration generates overtones. If you pluck a string on a cello, the string vibrates with a fundamental frequency. However, less-intense vibrations are also set up at several overtone frequencies. An absorption in the infrared at 500 cm−1 may well have an ac￾companying peak of lower intensity at 1000 cm−1 —an overtone. When two vibrational frequencies (n 1 and n 2) in a molecule couple to give rise to a vibration of a new frequency within the molecule, and when such a vibration is infrared active, it is called a combination band. This band is the sum of the two interacting bands (n comb = n 1 + n 2). Not all possible combinations occur. The rules that govern which combinations are allowed are beyond the scope of our discussion here. Difference bands are similar to combination bands. The observed frequency in this case results from the difference between the two interacting bands (ndiff = n 1 − n 2). One can calculate overtone, combination, and difference bands by directly manipulating fre￾quencies in wavenumbers via multiplication, addition, and subtraction, respectively. When a funda￾mental vibration couples with an overtone or combination band, the coupled vibration is called Fermi resonance. Again, only certain combinations are allowed. Fermi resonance is often observed in carbonyl compounds. Although rotational frequencies of the whole molecule are not infrared active, they often couple with the stretching and bending vibrations in the molecule to give additional fine structure to these absorptions, thus further complicating the spectrum. One of the reasons a band is broad rather than sharp in the infrared spectrum is rotational coupling, which may lead to a considerable amount of unresolved fine structure. C C C O O O H H H N H H N O O Methyl Anhydride Amino Nitro ~2872 cm–1 ~1760 cm–1 ~3300 cm–1 ~1350 cm–1 C C C O O O H H H N H H N O O ~2962 cm–1 ~1800 cm–1 ~3400 cm–1 ~1550 cm–1 Symmetric Stretch Asymmetric Stretch 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 19

20 Infrared Spectroscop 2.4 BOND PROPERTIES AND ABSORPTION TRENDS Let us now consider how ond to a simple hetero A diatomic molecule can be considered as two vibrating masses connected by a spring.The bond or,whenod ibraey of yb Eose hvoe isgiven by the eqution which is derived from Hooke's Law for vibrating springs.The reduced mass of the system is given by triple bonds are three tim Tshoud immediately.Oe is that strong bonds haveaforceco stant K and vibrate at higher frequencies than weaker bonds.The second is that bonds betweer and have higher frequencies of vibration (higher wavenumbers): C=c C=C C-C 2150cm11650cm1 1200cm cm As C-H C-C C-0 C-CI C-Br C-I 3000cm- 1200cm 1100cm 750cm-1 600cm 500cm

20 Infrared Spectroscopy 2.4 BOND PROPERTIES AND ABSORPTION TRENDS Let us now consider how bond strength and the masses of the bonded atoms affect the infrared absorption frequency. For the sake of simplicity, we will restrict the discussion to a simple hetero￾nuclear diatomic molecule (two different atoms) and its stretching vibration. A diatomic molecule can be considered as two vibrating masses connected by a spring. The bond distance continually changes, but an equilibrium or average bond distance can be defined. Whenever the spring is stretched or compressed beyond this equilibrium distance, the potential en￾ergy of the system increases. As for any harmonic oscillator, when a bond vibrates, its energy of vibration is continually and periodically changing from kinetic to potential energy and back again. The total amount of energy is proportional to the frequency of the vibration, Eosc ∝ hnosc which for a harmonic oscillator is determined by the force constant K of the spring, or its stiffness, and the masses (m1 and m2) of the two bonded atoms. The natural frequency of vibration of a bond is given by the equation n = 2p 1 c  which is derived from Hooke’s Law for vibrating springs. The reduced mass m of the system is given by m = m m 1 1 + m m 2 2 K is a constant that varies from one bond to another. As a first approximation, the force constants for triple bonds are three times those of single bonds, whereas the force constants for double bonds are twice those of single bonds. Two things should be noticeable immediately. One is that stronger bonds have a larger force con￾stant K and vibrate at higher frequencies than weaker bonds. The second is that bonds between atoms of higher masses (larger reduced mass, m) vibrate at lower frequencies than bonds between lighter atoms. In general, triple bonds are stronger than double or single bonds between the same two atoms and have higher frequencies of vibration (higher wavenumbers): ←⎯⎯⎯⎯⎯ Increasing K The CIH stretch occurs at about 3000 cm−1 . As the atom bonded to carbon increases in mass, the reduced mass (m) increases, and the frequency of vibration decreases (wavenumbers get smaller): ⎯⎯⎯⎯⎯→ Increasing m CII 500 cm−1 CIBr 600 cm−1 CICl 750 cm−1 CIO 1100 cm−1 CIC 1200 cm−1 CIH 3000 cm−1 CIC 1200 cm−1 CJC 1650 cm−1 CKC 2150 cm−1 K m 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 20

14782_02_Ch2_p015-104.Pp2.gxd1/25/0810:27age21⊕ 2.4 Bond Properties and Absorption Trends 21 Bending motions occur at lower energy (lower frequency)than the typical stretching motions be- cause of the lower value for the bending force constant K. C-H stretching C-H bending -3000cm1 -1340cm1 Hybridization affects the force constant K.also.Bonds are stronger in the orderspsppand the observed frequencies of C-H vibration illustrate this nicely. sp- sp =C-H -C-H -C-H 3300cm-3100cm-2900cm- Resonance also affects the strength and length of a bond and hence its force constant K.Thus has its C a ketone that is conjugatec with a( 94 dncy.near 167 cm That is because res ×-天 has the effect of reducing the force connt .and the absorpion The Hooke's Law expression given earlier may be transformed into a very useful equation as follows: 下=frequency in cm e=velocity of light=3x 1040cm/sec K=force constant in dynes/cm masses of atoms in grams. or MM> M1+M26.02×10 masses of atoms in am Rem obtain the =7.76×10区

2.4 Bond Properties and Absorption Trends 21 Bending motions occur at lower energy (lower frequency) than the typical stretching motions be￾cause of the lower value for the bending force constant K. Hybridization affects the force constant K, also. Bonds are stronger in the order sp > sp2 > sp3 , and the observed frequencies of CIH vibration illustrate this nicely. Resonance also affects the strength and length of a bond and hence its force constant K. Thus, whereas a normal ketone has its CJO stretching vibration at 1715 cm−1 , a ketone that is conjugated with a CJC double bond absorbs at a lower frequency, near 1675 to 1680 cm−1 . That is because res￾onance lengthens the CJO bond distance and gives it more single-bond character: Resonance has the effect of reducing the force constant K, and the absorption moves to a lower frequency. The Hooke’s Law expression given earlier may be transformed into a very useful equation as follows: n = 2p 1 c  n = frequency in cm−1 c = velocity of light = 3 × 1010 cm/sec K = force constant in dynes/cm m = m m 1 1 + m m 2 2 , masses of atoms in grams, or , masses of atoms in amu Removing Avogadro’s number (6.02 × 1023) from the denominator of the reduced mass expression (m) by taking its square root, we obtain the expression n = 7.76 2p × c 1011  K m M1M2 (M1 + M2)(6.02 × 1023) K m O C C C O C C C + – • • • • • • • • • • sp3 ICIH 2900 cm−1 sp2 JCIH 3100 cm−1 sp KCIH 3300 cm−1 CIH bending ∼1340 cm−1 CIH stretching ∼3000 cm−1 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 21

C=C bond: Eeng K=10x 10 dynes/cm p4 =1682 cm(calculated) =1650 cm(experimental) C-H bond: 下=42因 K5x1dynes/em 4-2-a2 4 =3032 cm(calculated) =3000 cm-(experimental) C-Dbond: K=5x 10'dynes/cm 4 228 cm(calculated) =2206 cm(experimental)

TABLE 2.2 CALCULATION OF STRETCHING FREQUENCIES FOR DIFFERENT TYPES OF BONDS CJC bond: n = 4.12 K = 10 × 105 dynes/cm m = M M C C + M M C C = ( 1 1 2 2) + (1 1 2 2 ) = 6 n = 4.12= 1682 cm−1 (calculated) n = 1650 cm−1 (experimental) CIH bond: n = 4.12 K = 5 × 105 dynes/cm m = M M C C + M M H H = ( 1 1 2 2) + (1 1 ) = 0.923 n = 4.12= 3032 cm−1 (calculated) n = 3000 cm−1 (experimental) CID bond: n = 4.12 K = 5 × 105 dynes/cm m = M M C C + M M D D = ( 1 1 2 2) + (2 2 ) = 1.71 n = 4.12= 2228 cm−1 (calculated) n = 2206 cm−1 (experimental) 5 × 105 1.71 K m 5 × 105 0.923 K m 10 × 10 5 6 K m 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 22

14782_02_Ch2_p015-104.Pp2.gxd1/25/0810:27Page2⊕ 2.5 The Infrared Spectrometer 23 A new expression is obtained by inserting the actual values ofand. a-4n月 H= MM2 M+M where M and M2 are atomic weights K=force constant in dynes/cm(1 dyne=) This equation may be used to calculate the approximate position of a band in the infrared spectrum 然for sin6 ble ndip心oa5.10an15X1enem Notice that excellent agreer ment in ate in Nevertheless.good ulirive values are obtained by such calculations. 2.5 THE INFRARED SPECTROMETER ctrum for a ntheanbroydispevandorera() ompounds in the common range of infrared spectrometers provide the infrared spectrum much more rapidly than the dispersive instruments. A.Dispersive Infrared Spectrometers odivides it intotwo parallel beams of cqual-intensity radiation.The sar ple is placed in one beam,and the other beam is used as a reference.The beams then pass into the monochromator, into a contin is spe of infrar The mon nately to a diffraction grating (a prism in older inst hy orvelnh of rdition reaching the thed the ratio between the intensities of the reference and sample beams.In whedetetordeevvhrchiefqpsEeeehcabso detector is amplified.the recorder draws the resulting spectrum of the sample on a chart.It is

2.5 The Infrared Spectrometer 23 A. Dispersive Infrared Spectrometers Figure 2.3a schematically illustrates the components of a simple dispersive infrared spectrome￾ter. The instrument produces a beam of infrared radiation from a hot wire and, by means of mir￾rors, divides it into two parallel beams of equal-intensity radiation. The sample is placed in one beam, and the other beam is used as a reference. The beams then pass into the monochromator, which disperses each into a continuous spectrum of frequencies of infrared light. The mono￾chromator consists of a rapidly rotating sector (beam chopper) that passes the two beams alter￾nately to a diffraction grating (a prism in older instruments). The slowly rotating diffraction grating varies the frequency or wavelength of radiation reaching the thermocouple detector. The detector senses the ratio between the intensities of the reference and sample beams. In this way, the detector determines which frequencies have been absorbed by the sample and which frequencies are unaffected by the light passing through the sample. After the signal from the detector is amplified, the recorder draws the resulting spectrum of the sample on a chart. It is important to realize that the spectrum is recorded as the frequency of infrared radiation changes by rotation of the diffraction grating. Dispersive instruments are said to record a spectrum in the frequency domain. The instrument that determines the absorption spectrum for a compound is called an infrared spectrometer or, more precisely, a spectrophotometer. Two types of infrared spectrometers are in common use in the organic laboratory: dispersive and Fourier transform (FT) instruments. Both of these types of instruments provide spectra of compounds in the common range of 4000 to 400 cm−1 . Although the two provide nearly identical spectra for a given compound, FT infrared spectrometers provide the infrared spectrum much more rapidly than the dispersive instruments. A new expression is obtained by inserting the actual values of p and c: This equation may be used to calculate the approximate position of a band in the infrared spectrum by assuming that K for single, double, and triple bonds is 5, 10, and 15 × 105 dynes/cm, respec￾tively. Table 2.2 gives a few examples. Notice that excellent agreement is obtained with the experi￾mental values given in the table. However, experimental and calculated values may vary considerably owing to resonance, hybridization, and other effects that operate in organic molecules. Nevertheless, good qualitative values are obtained by such calculations. n (cm−1 ) = 4.12 K m  m = M M 1 1 + M M 2 2 , where M1 and M2 are atomic weights K = force constant in dynes/cm (1 dyne = 1.020 × 10−3 g) 2.5 THE INFRARED SPECTROMETER 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 23

Infrared Spectroscop oppe DISPERSIVE IR Mior Computer Printer FIGURE.3 Schematic diagrams of (a)dispersive and (b)Fourier transform infrared spectrophotometers of the intensities of the two beams.and percent where /is the intensity of the sample beam,and is the intensity of the reference beam.In man parts of the spectrum.the transmitt nce is n arly 100 ent to radi 5n0 rb it). n th The chemist often obtains the spectrum of a com ound by dissolving it in a solvent (Section 2.6) The solution is then placed in the sample beam,while pure solvent is placed in the reference bear he spectrum of the solvent from tha from the spectrum of the sample (they are presentn both beams) feature is thereason most spectrom sare double-beam bs in b d

24 Infrared Spectroscopy Note that it is customary to plot frequency (wavenumber, cm−1 ) versus light transmitted, not light absorbed. This is recorded as percent transmittance (%T) because the detector records the ratio of the intensities of the two beams, and percent transmittance = I I s r × 100 where Is is the intensity of the sample beam, and Ir is the intensity of the reference beam. In many parts of the spectrum, the transmittance is nearly 100%, meaning that the sample is nearly transpar￾ent to radiation of that frequency (does not absorb it). Maximum absorption is thus represented by a minimum on the chart. Even so, the absorption is traditionally called a peak. The chemist often obtains the spectrum of a compound by dissolving it in a solvent (Section 2.6). The solution is then placed in the sample beam, while pure solvent is placed in the reference beam in an identical cell. The instrument automatically “subtracts” the spectrum of the solvent from that of the sample. The instrument also cancels out the effects of the infrared-active atmospheric gases, carbon dioxide and water vapor, from the spectrum of the sample (they are present in both beams). This convenience feature is the reason most dispersive infrared spectrometers are double-beam (sample + reference) instruments that measure intensity ratios; since the solvent absorbs in both beams, it is in both terms of the ratio Is / Ir and cancels out. If a pure liquid is analyzed (no solvent), Mirror a b Mirror Mirror Reference Cell Beam Chopper Sample Cell Infrared energy source Mirror Diffraction grating Amplifier Recorder Interferogram: the signal the computer receives. FT Transform Detector Beam splitter Infrared source Fixed Mirror Moving Mirror FT-IR Printer Mirror Mirror Detector Slit Slit DISPERSIVE IR Computer Sample Cell Motor FIGURE 2.3 Schematic diagrams of (a) dispersive and (b) Fourier transform infrared spectrophotometers. 14782_02_Ch2_p015-104.pp2.qxd 1/25/08 10:27 AM Page 24