1. 12. Structure Determination: Mass Spectrometry and Infrared Spectroscopy

2. Determining the Structure of an Organic Compound
The analysis of the outcome of a reaction requires that we know the full structure of the products as well as the reactants
In the 19th and early 20th centuries, structures were determined by synthesis and chemical degradation that related compounds to each other

3. Determining the Structure of an Organic Compound
Physical methods now permit structures to be determined directly. We will examine:
mass spectrometry (MS)—this chapter
infrared (IR) spectroscopy—this chapter
nuclear magnetic resonance spectroscopy (NMR)—Chapter 13
ultraviolet-visible spectroscopy (VIS)—Chapter 14

4. 12.1 Mass Spectrometry (MS)
Sample vaporized and bombarded by energetic electrons that remove an electron, creating a cation-radical
Bonds in cation radicals begin to break (fragment)

5. Mass Spectrometer

6. The Mass Spectrum
Plot mass of ions (m/z) (x-axis) versus the intensity of the signal (corresponding to the number of ions) (y-axis)
Tallest peak is base peak (100%)
Other peaks listed as the % of that peak
Peak that corresponds to the unfragmented radical cation is parent peak or molecular ion (M+)

7. MS Examples: Methane and Propane
Methane produces a parent peak (m/z = 16) and fragments of 15 and 14 (See Figure 12-2 a)

8. MS Examples: Methane and Propane
The Mass Spectrum of propane is more complex (Figure 12-2 b) since the molecule can break down in several ways

10. 12.2 Interpreting Mass Spectra
Molecular weight from the mass of the molecular ion
Double-focusing instruments provide high-resolution “exact mass”
atomic mass units – distinguishing specific atoms
Example MW “72” is ambiguous: C5H12 and C4H8O but:
C5H amu exact mass C4H8O amu exact mass
Result from fractional mass differences of atoms 16O = , 12C = , 1H =

11. Other Mass Spectral Features
If parent ion not present due to electron bombardment causing breakdown, “softer” methods such as chemical ionization are used
Peaks above the molecular weight appear as a result of naturally occurring heavier isotopes in the sample
(M+1) from 13C that is randomly present

12. 12.3 Interpreting Mass-Spectral Fragmentation Patterns
The way molecular ions break down can produce characteristic fragments that help in identification
Serves as a “fingerprint” for comparison with known materials in analysis (used in forensics)
Positive charge goes to fragments that best can stabilize it

13. 2,2-Dimethylpropane: MM = 72 (C5H12)

14. Mass Spectral Fragmentation of Hexane
Hexane (m/z = 86 for parent) has peaks at m/z = 71, 57, 43, 29

15. Hexane

16. Practice Problem 12.2: methylcyclohexane or ethylcyclopentane?

17. Mass Spectral Cleavage Reactions of Alcohols
Alcohols undergo -cleavage (at the bond next to the C-OH) as well as loss of H-OH to give C=C

18. Mass Spectral Cleavage of Amines
Amines undergo -cleavage, generating radicals

19. Fragmentation of Ketones and Aldehydes
A C-H that is three atoms away leads to an internal transfer of a proton to the C=O, called the McLafferty rearrangement
Carbonyl compounds can also undergo  cleavage

20. Fragmentation of Ketones and Aldehydes

21. 12.5 The Electromagnetic Spectrum

22. Wavelength and Frequency

23. Absorption Spectra
Organic compounds exposed to electromagnetic radiation can absorb photons of specific energies (wavelengths or frequencies)
Changing wavelengths to determine which are absorbed and which are transmitted produces an absorption spectrum
Energy absorbed is distributed internally in a distinct and reproducible way (See Figure 12-11)

24. Infrared Absorption Spectrum of Ethanol

25. 12.6 Infrared Spectroscopy of Organic Molecules
IR region is lower in photon energy than visible light (below red – produces heating as with a heat lamp)
2.5  106 m to 2.5  105 m region used by organic chemists for structural analysis
IR energy in a spectrum is usually measured as wavenumber (cm-1), the inverse of wavelength and proportional to frequency:
Wavenumber (cm-1) = 1/l(cm)
Specific IR absorbed by organic molecule is related to its structure

26. IR region and vicinity

27. Infrared Energy Modes
IR energy absorption corresponds to specific modes, corresponding to combinations of atomic movements, such as bending and stretching of bonds between groups of atoms called “normal modes”
Energy is characteristic of the atoms in the group and their bonding
Corresponds to molecular vibrations

28. Infrared Energy Modes

29. 12.7 Interpreting Infrared Spectra
Most functional groups absorb at about the same energy and intensity independent of the molecule they are in
Characteristic IR absorptions in Table 12.1 can be used to confirm the existence of the presence of a functional group in a molecule
IR spectrum has lower energy region characteristic of molecule as a whole (“fingerprint” region)

33. Regions of the Infrared Spectrum
cm-1 N-H, C-H, O-H (stretching)
N-H, O-H
3000 C-H
cm-1 CºC and C º N (stretching)
cm-1 double bonds (stretching)
C=O
C=C cm-1
Below 1500 cm-1 “fingerprint” region

34. Regions of the Infrared Spectrum

35. Differences in Infrared Absorptions
Molecules vibrate and rotate in normal modes, which are combinations of motions (relates to force constants)
Bond stretching dominates higher energy (frequency) modes

36. Differences in Infrared Absorptions
Light objects connected to heavy objects vibrate fastest (at higher frequencies): C-H, N-H, O-H
For two heavy atoms, stronger bond requires more energy (higher frequency): C º C, C º N > C=C, C=O, C=N > C-C, C-O, C-N, C-halogen

37. 12.8 Infrared Spectra of Hydrocarbons
C-H, C-C, C=C, C º C have characteristic peaks

38. Hexane

39. Alkenes

40. 1-Hexene

41. Alkynes

42. 12.9 Infrared Spectra of Some Common Functional Groups
Spectroscopic behavior of functional groups is discussed in later chapters
Brief summaries presented here

43. IR: Alcohols

44. Amines

45. IR: Aromatic Compounds
Weak C–H stretch at 3030 cm1
Weak absorptions cm1 range
Medium-intensity absorptions 1450 to 1600 cm1

46. Phenylacetylene

47. IR: Carbonyl Compounds
Strong, sharp C=O peak 1670 to 1780 cm1
Exact absorption characteristic of type of carbonyl compound
1730 cm1 in saturated aldehydes
1705 cm1 in aldehydes next to double bond or aromatic ring

48. Practice problem 12.7:

49. Phenylacetaldehyde

50. C=O in Ketones 1715 cm1 in six-membered ring and acyclic ketones
1750 cm1 in 5-membered ring ketones
1690 cm1 in ketones next to a double bond or an aromatic ring

51. C=O in Esters 1735 cm1 in saturated esters
1715 cm1 in esters next to aromatic ring or a double bond

52. Chromatography: Purifying Organic Compounds
Chromatography : a process that separates compounds using adsorption and elution
Mixture is dissolved in a solvent (mobile phase) and placed into a glass column of adsorbent material (stationary phase)
Solvent or mixtures of solvents passed through
Compounds adsorb to different extents and desorb differently in response to appropriate solvent (elution)
Purified sample in solvent is collected from end of column
Can be done in liquid or gas mobile phase

53. Principles of Liquid Chromatography
Stationary phase is alumina (Al2O3) or silica gel (hydrated SiO2)
Solvents of increasing polarity are used to elute more and more strongly adsorbed species
Polar species adsorb most strongly to stationary phase
For examples, alcohols adsorb more strongly than alkenes

54. High-Pressure (or High-Performance) Liquid Chromatography (HPLC)
More efficient and complete separation than ordinary LC
Coated silica microspheres (10-25 µm diameter) in stationary phase
High-pressure pumps force solvent through tightly packed HPLC column
Detector monitors eluting material
Figure 12.17: HPLC analysis of a mixture of 14 pesticides, using acetonitrile/water as the mobile phase

55. HPLC of Pesticide Mixture

56. Prob. 12.39: Cyclohexane or Cyclohexene?

57. Problem 12.48: Unknown hydrocarbon

58. Problem 12.49: Unknown hydrocarbon2

59. Some Useful Websites:
Interpretation of IR spectra (CSU Stanislaus):
IR Spectroscopy Tutorial (CU Boulder):
NIST Chemistry WebBook:
SDBS Data Base: