1. Chemistry 213 Practical Spectroscopy
Dave Berg
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. electromagnetic radiation emr
p. 1
electromagnetic radiation emr
c = λ n c = speed of light = 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. If l = 1m, then n = 3 x 108 / 1 = 3 x 108 Hz = 300 MHz
p. 2
c = ln l = c/n n = 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 nm

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

9. ENERGY of Radiation: E = h ν = h c E = N h ν = N h c λ λ
p. 3
E = h ν = h c E = N h ν = N h c
λ λ
single photon mole of photons
h = Planck’s constant = x Js
N = Avogadro’s number = x 1023 mol-1
For visible light, l = 5 x 10-7 m
E = x Js x 3 x 108 ms-1
5 x 10-7 m
= 4 x Joules
E = 4 x J x 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:

11. l n NMR MRI - UV/vis IR or Raman 60-600 MHz 1 cm 30 GHz 400-4000 cm-1
p. 5
Type of Energy
Region
Process l n
Instrument
Nuclear Magnetism
radiowave
flip the nuclear magnetic spin
0.2 – 2 m
MHz
NMR
MRI
Rotational
microwave
rotation of molecule
1 cm
30 GHz -
Vibrational
infrared
internal vibrations
bond stretch/bends m
cm-1
IR or
Raman
Electronic
Vis/UV
energy change of valence electrons nm
UV/vis
electronic
nuclear
X-ray
g-ray
core electrons
nuclear change
1 Angstrom
0.1 nm
X-ray diff.

12. only frequency n can cause this excitation
p. 5
Absorbing energy causes changes dependent on the wavelength:
e.g. UV or visible light promotes an ELECTRONIC TRANSITION
from the ground state E0 to 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
so for
= 10 mm = 1 x 10-3 cm,
WN = 1000 cm-1 so the unit of cm-1 is a frequency

14. 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. E = Nhc/l = Nhc x WN but remember then c = 3 x 1010 cm/sec
p. 6
SHORT WAVELENGTH LONG WAVELENGTH
HIGH FREQUENCY LOW FREQUENCY
HIGH ENERGY LOW 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. HIGH ENERGY LOW ENERGY C-H stretch C=O stretch C-H bend
p. 6
HIGH ENERGY LOW ENERGY
C-H stretch
C=O stretch
C-H bend
lighter elements
heavier elements
Bond Stretching
Bond Bending

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

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

19. p. 7
CO2
H2O

20. p. 7
H2O
CO2
CH4

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

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

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

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

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

27. 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. Populations The BOLTZMANN Equation DE = Eupper – Elower J mole-1
R = JK-1mole-1
T = temp in Kelvin
NOTE: If E in J per molecule, use k = R/N = x 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)) =
so only lower level is populated
so in IR, n = 1013 to 1014 so NU/NL = 0.20 to 10-7

31. ASSIGNMENT 1 Transition Spectra DE (kJ/mol) NU/NL at 300K rotation
microwave
0.025
0.99
vibration IR 25
5 x 10-5
electronic
UV-vis
250
10-44
What happens when T increases?
ASSIGNMENT 1