1. Chapter 16 Infrared Absorption Spectroscopy An IR spectrum contains information about the functional groups in a molecule, and this is used to uniquely identify the compound.

2. Wavenumbers (cm -1 ) are used since they are directly proportional to energy – e.g. convert 2.5 µm to wavenumbers (cm -1 ) 400 nm780 nm2500 nm or 4000 cm -1 50,000 nm or 200 cm -1 UVVisNear IRMid IRFar IR

3. Typical Infrared Spectrum Higher energy vibrations C-H “overtones” C=O C-C

4. Mechanical Model of a Stretching Vibration in a Diatomic Molecule Treats the vibrating bond like a spring with a given “stiffness” -- = frequency of vibration k = force constant (spring “stiffness”) µ = “reduced mass”

5. Quantum Mechanical Treatment of Normal Modes 1.Only certain vibrations are “allowed” 2.The vibrational quantum number v = 0, 1, 2, 3…… 3.Only transitions between adjacent energy levels are possible, i.e.  v =  1 4.The frequency of the photon has to equal the frequency of the vibration (see next slide) 5.The molecule must have a change in “dipole moment” as a result of the vibration Dipole moment = charge difference X separation distance Homonuclear diatomics never have an Infrared spectrum. Why? ++ --

6. + ++ -- ++ -- -- ++ IR Absorption if the vibration results in a change in dipole moment Frequency of absorbed light  frequency of vibration SoSo S1S1  v =  1

7. Common Types of Vibrations (“Normal Modes”) Higher energy modes Lower energy modes

8. Example – predict the normal modes of CO 2 The number of normal modes = 3N-6 For linear molecules it’s 3N-5

9. Which normal modes are Infrared active?

10. Example 16-1, p.436 Calculate the approximate wavenumber and wavelength of the fundamental absorption peak due to the stretching vibration of a C=O group (k = 1.0 x 10 3 N/m)

11. Force Constants increase with Bond Strength Bond Type* Force Constant, k (N/m) Wave- number (cm -1 ) Bond Energy (kJ/mol) C-C5 x 10 2 800-1200347 C=C10 x 10 2 1600620 CΞCCΞC15 x 10 2 2100812 Bond Type* Reduced mass (kg) Wave- number (cm -1 ) Bond Energy (kJ/mol) C-H1.55 x 10 -27 3000414 C-N1.07 x 10 -26 1000-1350276 @ C-O1.14 x 10 -26 1000-1300351 @ * all bonds are stretches @ consider these as being approx. equal (a). (b) (a).as  then frequency  (b) When masses approx. equal, then see peak at same wavenumbers

12. IR Sources 1.Nichrome – Ni/Cr alloy, resistively heats up and emits IR, 1100 K 2.Globar – SiC rod, 1500 K 3.Nernst Glower – electrically heated rare earth-oxide ceramic, 2000 K Output from a Nernst Glower cm -1

13. Optics – IR spectra are measured in the mid-IR, so have to use halides such as NaCl and KBr Never use aqueous samples, or water to clean salt plates 400 nm780 nm2500 nm or 4000 cm -1 50,000 nm or 200 cm -1 UVVisNear IRMid IRFar IR

14. Sample Handling – a few drops of “neat” sample between “salt plates”

15. Solutions

16. Detectors for the Mid-IR – 200-4000cm -1 1.Pyroelectric – crystal of DTGS (Deuterated Triglyceine Sulfate) Nice link: http://www.doitpoms.ac.uk/tlplib/pyroelectricity/printall.php?question=2&type=1 1.DTGS maintains polarization when heated to just below the Curie Point 2.Above the Curie Point, the permanent polarization of the DTGS crystal disappears. 3.The closer to the Curie Point, the more responsive the detector 4.Much faster response so can take a spectrum much more quickly (FTIR)

17. Detectors for the Mid-IR – 200-4000cm -1 2.Photoconduction - Mercury-Cadmium-Tellurium (MCT) http://www.newport.com Liquid N 2 @ 77K 1.Semiconductor-based 2.Cooled to 77 K using liquid N 2 3.Much faster response so can take a spectrum much more quickly (FTIR) http://www.boselec.com/pdf/Laser_Focus_World.PDF

18. Definition of the Fourier Transform (  ) - The Inverse Fourier Transform - The functions f(t) and  (  ) are called "transform pairs" f(t)   (  )

19. time (t)  frequency (s -1, Hz)

20. e.g. FTIR distance (cm)  wavenumber (cm -1 ) background interferogram P/P o = IR spectrum background spectrum = P o sample spectrum = P sample interferogram

21. e.g. FT-NMR time (s)  ppm (Hz) Free Induction Decay (FID) Signal (time) NMR Spectrum (ppm, Hz)

22. Fourier Transforms "invert dimensionality" – signal domainFT domainInstrument time (t)frequency (s -1, Hz)oscilloscope mirrorwavenumber (cm -1 )FTIR distance (cm) free induction ppm (from Hertz)FT-NMR decay (t) Fast Fourier Transform (FFT) on computers - Cooley-Tukey algorithm very efficient on a PC

23. movable mirror → D fixed mirror 50/50 beamsplitter IR source mirror distance, x (cm) → ¼ interferogram   + + + -- - /42 /43 /44 /4

24. If polychromatic radiation enters the interferometer, then each wavelength produces a separate interferogram. The output from the interferometer will therefore be a superposition of all wavelengths - 1  1 mirror distance (+x, cm) → 2  2 3  3 ←mirror distance (-x, cm)

25. signal at detector S(x) this term contains the IR spectrum Taking the Fourier Transform of S(x) results in the IR spectrum, A( ) -

27. Depends on the distance the mirror moves in the Michelson Interferometer – the further it moves, the greater the resolution. minimum detectable difference in wavenumber x = distance mirror moves e.g. What length of mirror drive in a Michelson Interferometer is required to separate 20.34 and 20.35  m? "retardation“ = 2x

28. Fourier Transform Infrared (FTIR) Spectrometers The Multiplex (Fellgett's) Advantage - entire spectrum obtained virtually instantaneously; results in a higher S/N because during the time a scanning instrument is slowly obtaining the spectrum, a multiplex instrument such as an FTIR can acquire 100’s pf spectra. Averaging these extra spectra causes the increased S/N (see next slide) and allows quantitative IR. S/N increases as where N = number of "resolution elements" 4000 cm -1 400 cm -1 e.g. scan range 400 - 4000 cm -1 at a resolution of 2 cm -1 2 cm -1 "resolution elements" so the S/N increases by IF the FTIR can obtain the entire spectrum for the same amount of time a slower, scanning instrument requires to acquire only one resolution element.

29. Signal-to-Noise Ratio (S/N)

30. Single-Beam – run background first (see next slide) and store in memory (P o ) Higher energy throughput (Jacquinot’s Advantage) and no stray light problems

31. Typical IR background spectrum (P o )

32. inexpensive (~ $25K) range = 350 – 7800 cm -1 (29 to 1.3  m) max resolution = 4 cm -1 scan time = as fast as 1 sec detector = DTGS expensive (~ $100K) range = 10 – 50,000 cm -1 (1000  m to 200 nm) max resolution < 0.01 cm -1 scan time = can be minutes at high-res detector(s) = DTGS, MCT

33. Advantages over Scanning Instruments 1.fast scan times 2.higher S/N because of signal averaging 3.higher resolution 4.because there are no slits, there’s a higher energy throughput (Jacquinot’s Advantage); means larger signals for the same concentration (therefore lower LODs) 5.no stray light problems Disadvantages 1.higher cost than scanning instruments 2.difficult to align the interferometer (automation helps) 3.IR optics water soluble (beamsplitter made of KBr)

34. Chapter 17 Applications of IR Spectrometry

35. 1. Qualitative Analysis

36. "fingerprint region""group frequency region"

42. 2. Quantitative Analysis 1.lower source powers than in UV-Vis (in FTIR Jacquinot's Advantage partially offsets) 2.detectors suffer from thermal noise (i.e. larger backgrounds or  blank ) 3.salt plates have a very short path length (A =  bc) Signal averaging with FTIR's has improved the S/N enough to allow more sensitive quantitative work in the IR. Much less sensitive (i.e. higher LODs) in the IR compared to the UV-Vis because of -

43. 3. Diffuse Reflectance Accessory

44. Typically used for powdered samples, i.e. forensic drug analyses

45. Kubelka-Munk Units = converts the reflectance spectrum to the equivalent of an absorbance spectrum. R’  = sample refectivity/KBr reflectivity f(R’  ) = (1 - R’  )2/2 R’  = k/S k = 2.303  C  = the molar absorptivity C is the sample concentration S = "scattering coefficient" For a sample to follow a linear relationship between f(R’  ) and concentration, the following criteria must be met: 1.the sample must be diluted in a non-absorbing matrix such as KBr or KCl. 2.the scattering coefficient S must remain constant over the entire spectrum. 3.there must be no specular, or regular reflectance off the surface of the sample.

46. 4. Attenuated Total Internal Reflectance (ATR) Accessory Total Internal Reflection and the "evanescent wave" - Samples - conventional (solutions, liquids, etc) as well as powders, pastes, suspensions, colloids

47. 5. Infrared Microscopy