1. Spectroscopy of Organic Compounds
Chemistry 344
Spectroscopy of Organic Compounds

2. Modern Spectroscopic Methods
Revolutionized the study of Organic Chemistry
Determine the exact structure of small to medium size molecules in a few minutes.
Nuclear Magnetic Resonance (NMR) and Infrared Spectroscopy (IR) are particularly powerful techniques which we will focus on and use in this course.

3. Interaction of Light and Matter The Physical Basis of Spectroscopy
Quantum properties of light (photons)
Quantum properties of matter (quantized energy states).
Photons of light act as our “quantum probes” at the molecular level giving us back precise information about the energy levels within molecules.

4. The Electromagnetic Spectrum
Continuous
Covers a wide range of wavelengths of “light” from radio waves to gamma rays.
Wavelengths (l) range from more than ten meters to less than meter.

5. The Electromagnetic Spectrum

6. Relationship Between Wavelength, Frequency and Energy
Speed of light (c) is the same for all wavelengths.
Frequency (n), the number of wavelengths per second, is inversely proportional to wavelength:
n = c/l
Energy of a photon is directly proportional to frequency and inversely proportional to wavelength:
E = hn = hc/l
(where h = Plank’s constant)

7. Wavelength/Spectroscopy Relationships
Spectral Region
Photon Energy
Molecular Energy Changes
UV-Visible
~ 100 kcal/mole
Electronic
Infrared (IR)
~ 10 kcal/mole
Bond vibrations
Radio
< 0.1 kcal/mol
Nuclear Spin states in a magnetic field

8. Spin of Atomic Nuclei Spin 1/2 atoms: mass number is odd.
examples: 1H and 13C.
Spin 1 atoms: mass number is even.
examples: 2H and 14N.
Spin 0 atoms: mass number is even.
examples: 12C and 16O.

9. Magnetic Properties of the Proton Related to Spin

10. Energy States of Protons in a Magnetic Field= l a b s o r e d i g h t A p M n c F H x

11. Nuclear Magnetic Resonance (NMR)
Nuclear – spin ½ nuclei (e.g. protons) behave as tiny bar magnets.
Magnetic – a strong magnetic field causes a small energy difference between + ½
and – ½ spin states.
Resonance – photons of radio waves can match the exact energy difference between the + ½ and – ½ spin states resulting in absorption of photons as the protons change spin states.

12. The NMR Experiment
The sample, dissolved in a suitable NMR solvent (e.g. CDCl3 or CCl4) is placed in the strong magnetic field of the NMR.
The sample is bombarded with a series of radio frequency (Rf) pulses and absorption of the radio waves is monitored.
The data is collected and manipulated on a computer to obtain an NMR spectrum.

13. The NMR Spectrometer

14. The NMR Spectrometer

15. The NMR Spectrum
The vertical axis shows the intensity of Rf absorption.
The horizontal axis shows relative energy at which the absorption occurs (in units of parts per million = ppm)
Tetramethylsilane (TMS) in included as a standard zero point reference (0.00 ppm)
The area under any peak corresponds to the number of hydrogens represented by that peak.

16. The NMR Spectrum

17. Chemical Shift (d)
The chemical shift (d) in units of ppm is defined as:
d = distance from TMS (in hz)
radio frequency (in Mhz)
A standard notation is used to summarize NMR spectral data. For example p-xylene:
d 2.3 (6H, singlet)
d 7.0 (4H, singlet)
Hydrogens in identical chemical environments (equivalent hydrogens) have identical chemical shifts.

18. Shielding – The Reason for Chemical Shift Differences
Circulation of electrons within molecular orbitals results in local magnetic fields that oppose the applied magnetic field.
The greater this “shielding” effect, the greater the applied field needed to achieve resonance, and the further to the right (“upfield”) the NMR signal.

19. Structure Effects on Shielding
Electron donating groups increase the electron density around nearby hydrogen atoms resulting in increased shielding, shifting peaks to the right.
Electron withdrawing groups decrease the electron density around nearby hydrogen atoms resulting in decreased shielding, (deshielding) shifting peaks to the left.

20. Structure Effects on Shielding
The Deshielding effect of an electronegative substituent can be seen in the NMR spectrum of 1-Bromobutane
Br – CH2-CH2-CH2-CH3
d (ppm):
No. of H’s:

21. Some Specific Structural Effects on NMR Chemical Shift
Type of Hydrogen
d (ppm
Alkyl (C – H)
0.8 – 1.7
Alkyl Halide (RCH2X)
3 - 4
Alkene (R2C=CH2)
4 - 6
Aromatic (e.g. benzene)
6 - 8
Carboxylic Acid (RCOOH)

22. Spin-Spin Splitting
Non-equivalent hydrogens will almost always have different chemical shifts.
When non-equivalent hydrogens are on adjacent carbon atoms spin-spin splitting will occur due to the hydrogens on one carbon feeling the magnetic field from hydrogens on the adjacent carbon.
The magnitude of the splitting between two hydrogens (measured in Hz) is the coupling constant, J.

23. Spin-Spin Splitting Origin of the Doublet

24. Spin-Spin Splitting Origin of the Triplet

25. Spin-Spin Splitting Origin of the Quartet

26. Pascal’s Triangle Pattern Relative Peak Height Singlet 1 Doublet 1 : 1
Triplet : 2 : 1
Quartet : 3 : 3 : 1
Quintet 1 :4 : 6 : 4 : 1

27. The n + 1 Rule
If Ha is a set of equivalent hydrogens and Hx is an adjacent set of equivalent hydrogens which are not equivalent to Ha:
The NMR signal of Ha will be split into n+1 peaks by Hx. (where n = # of hydrogens in the Hx set.)
The NMR signal of Hx will be split into n+1 peaks by Ha. (where n = # of hydrogens in the Ha set.)

28. 1H NMR Spectrum of Bromoethane

29. 1H NMR Spectrum of 1-Nitropropane

30. Unknown A (Figure 9.20 Solomons 7th ed.)
Formula: C3H7I
IHD = 0
No IR data provided
1HNMR d: 1.90 (d, 6H), 4.33 (sept., 1H)

32. Unknown B (Figure 9.20 Solomons 7th ed.)
Formula C2H4Cl2
IHD = 0
No IR data given
1HNMR d: (d, 3H), 4.32 (quartet, 1H)

34. Unknown C (Figure 9.20 Solomons 7th ed.)
Formula: C3H6Cl2
IHD = 0
No IR data given
1HNMR d: 2.20 (pent., 2H), 3.62 (t, 4H)

36. Infrared Spectroscopy
Energy of photons in the IR region corresponds to differences in vibrational energy levels within molecules.
Vibrational energy levels are dependent on bond types and bond strengths.
IR is highly useful to determine if certain types of bonds (functional groups) are present in the molecule.

38. IR Spectrum of Ethanol

39. IR Correlation Table

40. Key Functional Groups by Region of the IR Spectrum

41. IR Spectrum of Benzaldehyde

42. IR Spectrum of Cyclohexanone

43. IR Spectrum of Propanoic Acid

44. NMR: Exceptions to the n+1 Rule
The n+1 rule does not apply when a set of equivalent H’s is split by two or more other non-equivalent sets with different coupling constants.
The n+1 rule does not apply to second order spectra in which the chemical shift difference between two sets of H’s is not much larger than the coupling constant.

49. NMR: Some Specific Functional Group Characteristics
O-H and N-H will often show broad peaks with no resolved splitting, and the chemical shift can vary greatly.
Aldehyde C-H is strongly deshielded.
(d = 9-10 ppm) and coupling to alkyl H’s on adjacent carbon is small.
Carboxylic Acid O-H is very strongly deshielded. (d = ppm)

51. NMR: Some Specific Functional Group Characteristics
Ortho splitting on aromatic rings is often resolved, but meta and para splitting is not.
Cis and trans H’s on alkenes usually show strong coupling, but geminal H’s on alkenes show little or no resolved coupling.

58. NMR: Mixtures of Compounds
Unlike most textbook examples, “real world” NMR samples commonly contain mixtures of compounds.
It is very common to see extra signals due to impurities. The impurities can often be identified.
In some cases it is useful (or necessary) to analyze mixtures that contain comparable amounts of two or more compounds.

59. NMR: Mixtures of Compounds
Synthesis of Ethyl Acetate by Fischer Esterification Results in an Equilibrium Mixture:

62. Unknown A (Figure 14.27 Solomons 7th ed.)
Formula = C9H12
IHD = 4
IR shows no medium or strong bands above 1650 cm-1 except C-H stretching bands around 3,000 cm-1
1HNMR d: (d, 6H), 2.90 (sept., 1H), (m, 5H)

64. Unknown B (Figure 14.27 Solomons 7th ed.)
Formula = C8H11N
IHD = 4
IR shows two medium peaks between 3300 and 3500 cm-1 . No other medium or strong bands above 1650 cm-1 except C-H stretching bands around 3,000 cm-1
1HNMR d: 1.4 (d, 3H), 1.7 (s, br, 2H),
4.1(quart., 1H), (m, 5H)

66. Unknown C (Figure 14.27 Solomons 7th ed.)
Formula = C9 H10
IHD = 5
IR shows no medium or strong bands above 1650 cm-1 except C-H stretching bands around 3,000 cm-1
1H NMR d: (pent., 2H),
2.90 (trip., 4H), (m, 4H)

68. Unknown H (Figure 9.48 Solomons 7th ed.)
Formula = C3H4Br2
IHD = 1
No IR data given
1HNMR d: (2H), 5.63 (1H), 6.03 (1H)

69. Unknown Y (Figure 14.34 Solomons 7th ed.)
Formula = C9H12O
IHD = 4
IR shows a strong, broad, absorbance centered at 3400 cm-1
1HNMR d: (t, 3H), 1.75 (m, 2H),
4.38 (s, br, 1H), 4.52 (t, 1H),
(m, 5H)