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 heteronuclear 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 energy 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 constant 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 because 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 resonance 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 spectrometer. The instrument produces a beam of infrared radiation from a hot wire and, by means of mirrors, 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 monochromator consists of a rapidly rotating sector (beam chopper) that passes the two beams alternately 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, respectively. Table 2.2 gives a few examples. Notice that excellent agreement is obtained with the experimental 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 transparent 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