Chemistry 213 Practical Spectroscopy Dave Berg djberg@uvic Elliott 314

1. Chemistry 213 Practical Spectroscopy Dave Berg djberg@uvic.ca Elliott 314 A course in determining structure by spectroscopic methods

2. SPECTROSCOPY Different types of spectroscopy afford different information about molecules and their structure: INFRARED – what types of functional groups? NMR – what types of H’s, C’s, P’s, F’s......? - how many, what are they connected to? MASS SPEC – what is molecular weight and formula? UV-VIS – how many double bonds?

3. For example: 1H NMR can tell us what kinds of environments each proton is in:

4. Same technique can also be used in medicine: MRI (NMR of water 1H signals)

5. p. 1 Spectrum Intensity of emr emr = electromagnetic radiation

6. p. 1 electromagnetic radiation emr c =λnc = speed of light = 3.00 x 108 ms-1 l = wavelength in meters ( 1 nm = 10-9 m = 10 Å) ν = frequency in s-1 ( 1 Hz = 1 cycle / second = 1 s-1)

7. p. 2 c = lnl = c/nn = c/l If l = 1m, then n = 3 x 108 / 1 = 3 x 108 Hz = 300 MHz If l = 10-8m, then n = 3 x 1016 Hz UV Visible light: from about 4 to 7 x 10-7 m OR 400-700 nm

8. p. 3 ENERGY of Radiation: E = h ν = h c λ where E = energy of a single photon (in Joules ) h = Planck’s constant = 6.626 x 10-34Js E = N h ν = N h c l where E = energy of a mole of photons (in Joules per mole) N = Avogadro’s number = 6.022 x 1023

9. ENERGY of Radiation: p. 3 E = h ν = h cE = N h ν = N h c λ λ single photon mole of photons h = Planck’s constant = 6.626 x 10-34 Js N = Avogadro’s number = 6.022 x 1023 mol-1 For visible light, l = 5 x 10-7 m E = 6.626 x 10-34 Js x 3 x 108 ms-1 5 x 10-7 m = 4 x 10-19 Joules E = 4 x 10-19 J x 6.022 x 1023 = 2.4 x 105 J/mol = 240 kJ/mol For uv light, l = 200nm = 2 x 10-7 m E = 599 kJ/mole

10. Common terminology varies with region of the spectrum: p. 4

11. p. 5

12. p. 5 Absorbing energy causes changes dependent on the wavelength: e.g. UV or visible light promotes an ELECTRONIC TRANSITION from the ground state E0to an excited state E1 only frequency n can cause this excitation - Planck – energy is quantized

13. p. 6 INFRARED l = 10 mm = 10 x 10-6 m so n = c/l = (3 x 108)/(1 x 10-5) = 3 x 1013 Hz Kind of awkward numbers, so we use a convenient stand in for the frequency instead: WAVENUMBER = 1/l where l is in cm = 10,000/l where l is in m • = 10 mm = 1 x 10-3 cm, WN = 1000 cm-1so the unit of cm-1is a frequency so for

14. p. 6 Typical IR spectrum: l runs from 2.5 to 15 mm WN runs from 4000 to 600 cm-1 Note: scale is non-linear

15. p. 6 SHORT WAVELENGTH LONG WAVELENGTH HIGH FREQUENCY LOW FREQUENCY HIGH ENERGYLOW ENERGY E = Nhc/l = Nhc x WN but remember then c = 3 x 1010 cm/sec 36 kJ/mol (or 9 kcal/mol) 12 kJ/mol (or 3 kcal/mol)

16. p. 6 HIGH ENERGYLOW ENERGY C-Hstretch C=Ostretch C-Hbend lighter elements heavier elements Bond Stretching Bond Bending

17. p. 6 For IR peaks to be strong (be seen) the bond dipole must change during the vibration + - + - bond stretch charges move apart, dipole C===O C=O

18. p. 7 GREENHOUSE GASES so N2, O2, Ar do not give IR spectra (no dipole) O H H - + +

19. p. 7 CO2 H2O

20. p. 7 H2O CO2 CH4

21. p. 8 Fine structure = ROTATIONS can only be seen clearly in simple molecules

22. p. 9 QUANTIZATION and SELECTION RULES Molecules are limited to specific energy levels E3 Excited States E2 E1 Ground State E0 True whether electronic, vibrational, rotational, spin

23. Molecular Energy Levels: electronic (S), vibrational (v) and rotational (J) p. 9 E N E R G Y

24. p. 9 A Transition only n causes absorption Need to know: What jumps are possible? What levels are populated?

25. p. 10 1] No selection rules, i.e. all jumps possible In this example only the ground state is populated Absorption of light frequency n2 will cause jump from Eo to E2

26. p. 10 2] The IR CASE Only ground state populated Selection rule = jump of + or – 1 allowed, Dn = ± 1 Only one line, frequency n1

27. p. 10 3] The Microwave (Rotational Spectra) case Most energy levels populated Selection rule is Dn = ± 1 Relative positions depend upon energy level spacings and in microwave these spacings are not all the same

28. p. 11 Populations The BOLTZMANN Equation DE = Eupper – Elower J mole-1 R = 8.314 JK-1mole-1 T = temp in Kelvin NOTE: If E in J per molecule, use k = R/N = 1.383 x 10-23 JK-1 What happens if DE is really small OR really large?

29. p. 11 Rotational (microwave) spectra, DE small, l = 0.1 m DE = Nhc/l = (6.022 x 1023) x (6.626 x 10-34) x (3 x 108) 0.1 = 1.2 J mol-1 NU ----- NL = e -(1.2/(8.3 x 300)) = ~1 so both levels almost equally populated so in IR, vibrational lines not very sharp because lots of similar energy rotational levels, which give lots of similar energy lines (refer to p. 7 and slide 19)

30. p. 12 Vibrational (IR) spectra, DE larger, 1/l = 1700 cm-1 DE = Nhc/l = (6.022 x 1023)x(6.626 x 10-34)x(3 x 1010)x1700 = 20,350 J mol-1 NU ----- NL = e -(20,350/(8.3 x 300)) = 0.00028 so only lower level is populated so in IR, n = 1013 to 1014 so NU/NL = 0.20 to 10-7

31. p. 12 What happens when T increases? ASSIGNMENT 1