1. Mass Spectrometry, Infrared Spectroscopy, and Ultraviolet/Visible
Organic Chemistry
6th Edition
Paula Yurkanis Bruice
Chapter 13
Mass Spectrometry, Infrared Spectroscopy,
and Ultraviolet/Visible
Spectroscopy

2. Spectrally Identifiable Functional Groups

3. The Mass Spectrometer
A mass spectrum records only positively charged fragments, either cations or radical cations
m/z = mass-to-charge ratio of the fragment

4. Information obtained from a mass spectrum:
The molecular ion (M): measured to the nearest whole number or up to four decimal places (high-resolution mass spectrometry).
Isotope peaks (M + 1, M + 2 etc.). M …
Typically M and the isotope peaks are the highest masses in the spectrum
M + 1
Exception: a compound whose molecular ion completely fragments
The high-resolution mass of the molecular ion provides the molecular formula directly.
The whole-number mass of the molecular ion and the relative intensities of M + 1, M + 2, etc., can also provide the molecular formula.

5. Fragment masses and intensities together provide structural information.
The base peak has the greatest intensity in the spectrum.
Intense peaks correspond to relatively stable cationic and/or relatively stable radical species lost.
The fragments lost also provide structural information. …
M-29
Base peak
M-15
For example:
Fragment m/z 57 resulted from the loss of methyl (m/z = 15) from the molecular ion.
Given its intensity, m/z 57 must be the sec-butyl carbocation (not the primary butyl carbocation). M

6. The Mass Spectrum of Pentane
Note weak m/z = 57 peak, primary butyl carbocation

7. The base peak of m/z 43 in the mass spectrum of pentane indicates the preference for C-2 to C-3 fragmentation:
All fragments originate from the molecular ion
The mass of the radical species lost in a fragmentation is the difference between the m/z values of the fragment ion and the molecular ion

8. The Mass Spectrum of Isopentane
Note strong m/z = 57 peak, secondary butyl carbocation

9. 2-Methylbutane is more likely than pentane to lose a
methyl radical because a secondary carbocation can be formed:

10. Two-Fragment Loss from the Molecular Ion
What are the structures of m/z 42 and 41?
These ions arise from loss of the ethyl radical and either hydrogen atom or H2 from the pentane molecular ion:
Note: All fragments originate from the molecular ion.
Exception: Tandem Mass spectrometry where fragments of fragments are observed.

11. Isotopes in Mass Spectrometry
M + 1 peak: a contribution from 2H or 13C.
M + 2 peak: a contribution from 18O or from two heavy
isotopes (2H or 13C) in the same molecule.
A large M + 2 peak suggests a compound containing
either chlorine or bromine: a Cl if M + 2 is one-third the intensity of M; a Br if M + 2 is the same intensity as M.
To calculate the molecular masses of molecular ions
and fragments, the atom mass of a single isotope of an
atom must be used.

12. Fragmentation Patterns of Alkyl Halides
79Br
81Br

13. The Mass Spectrum of
2-Chloropropane
35Cl
37Cl
35Cl
37Cl

14. a-Cleavage results from the homolytic cleavage of a
C—C bond at the a carbon:

15. a-Cleavage occurs because the C—Cl and C—C bonds
have similar strengths, and the species that is formed is a relatively stable cation:
a-Cleavage is less likely to occur in alkyl bromide
because C—C bond is stronger than C—Br bond

16. Fragmentation Patterns of Ethers

17. A C—O bond is cleaved heterolytically, with the electrons
going to the more electronegative atom:

18. A C—C bond is cleaved homolytically at an a-position
because it leads to a relatively stable cation:

19. Fragmentation Patterns of Alcohols
Because they fragment, molecular ions obtained from alcohols usually are not observed

20. Like alkyl halides and ethers, alcohols undergo
a-cleavage:
In alcohols, loss of water results in a fragmentation peak
at m/z = M-18:

21. Common Fragmentation Behavior in Alkyl Halides,
Ethers, and Alcohols
1. A bond between carbon and a more electronegative
atom breaks heterolytically
2. A bond between carbon and an atom of similar
electronegativity breaks homolytically
3. The bonds most likely to break are the weakest bonds
and those that lead to formation of the most stable
cation

22. Fragmentation Patterns of Ketones
An intense molecular ion peak:

23. McLafferty rearrangement may occur:

24. Spectroscopy and the Electromagnetic Spectrum
Spectroscopy is the study of the interaction of matter and
electromagnetic radiation

25. Electromagnetic radiation has wave-like properties
High frequencies and short wavelengths are
associated with high energy

26. Vibrational Transitions Observed in IR Spectroscopy
Functional groups stretch at different frequencies, and IR spectroscopy is used to identify functional groups

27. Infrared transitions require a bond dipole to occur:
The more polar the bond, the more intense the absorptions:
The intensity of an absorption band also depends on the number of bonds responsible for the absorption

28. Influence of symmetry on IR activity of the alkene stretch:
1-butene — infrared active
2,3-dimethyl-2-butene — infrared inactive
2,3-dimethyl-2-heptene — infrared active, but very weak absorption band

29. The Vibrating Bond as a Quantized Harmonic Oscillator
Quantum levels for a stretching vibration:
Ball-and-spring model:
Fundamental transition: o  1
Overtone: o  2
Overtones are twice the frequency of the fundamental transition and are always weak

30. The approximate wavenumber of an absorption can be
calculated from Hooke’s law:
= wavenumber
c = speed of light
K = force constant
M1 and M2 = masses of atoms

31. Hooke’s law predicts that lighter atoms will vibrate at a higher frequency than heavy atoms:
C—H ~3000 cm–1
C—D ~2200 cm–1
C—O ~1100 cm–1
C—Cl ~700 cm–1
Increasing the s character of a bond (higher K value) increases the stretching frequency: sp
sp2
sp3

32. Note the influence of mass and s character on stretching frequency:

33. An Infrared Spectrum The functional group region (4000–1400 cm–1)
The fingerprint
region (1400–600 cm–1)
High energy
Low energy
The functional group, or diagnostic region, is used to determine the functional group present
The fingerprint region is used for structure elucidation by spectral comparison

34. Functional group regions: Both compounds are alcohols
Fingerprint regions: Compounds are different alcohols

35. The exact position of the absorption band depends on
electron delocalization, the electronic effect of
neighboring substituents, and hydrogen bonding:

36. Carbonyl overtone
Esters have a carbonyl and a C—O stretch Ketones have only a carbonyl stretch

37. Putting an atom other than carbon next to the carbonyl
group causes the position of the carbonyl absorption
band to shift:
The predominant effect of the nitrogen of an amide is
electron donation by resonance
The predominant effect of the oxygen of an ester is
inductive electron withdrawal

39. The position of a C—O absorption varies because of resonance release in acids and esters:
~1050 cm–1
~1050 cm–1
~1250 cm–1
~1250 cm–1 and 1050 cm–1

40. Acids are readily distinguished from alcohols
Higher-frequency
C─O stretch
Broad
OH stretch
C═O stretch

41. The position and the breadth of the O—H absorption
band depend on the concentration of the solution
It is easier to stretch an O—H bond if it is hydrogen
bonded

42. The strength of a
C—H bond depends on the hybridization
of the carbon

43. Examine the absorption bands in the vicinity of 3000 cm–1

44. Benzene ring: Sharp absorption bands at ~1600 cm–1 and 1500–1430 cm–1.
Benzene in-plane and out-of-plane C—H bends
Benzene ring:
Sharp absorption bands at ~1600 cm–1 and 1500–1430 cm–1.
Overtones at 1700–1900 cm–1 for the in-plane and out-of-plane benzene C—H bends.
The benzene overtones in the diagnostic region are readily recognized.

45. Stretch of C—H Bond in an Aldehyde
The stretch of the C—H bond of an aldehyde shows one
absorption band at ~2820 cm–1 and another one at ~2720 cm–1

46. Identifying a functional group by the bending vibrations:
Primary amine: two N—H stretches at 3350 cm–1.
Amine: N—H bend.
“Isopropyl split” at 1380 cm–1 indicates the presence of an isopropyl group.

47. Analyzing Infrared Spectra
The position, intensity, and shape of an absorption band
are helpful in identifying functional groups
The absence of absorption bands can be useful in
identifying a compound in IR spectroscopy
Bonds in molecules lacking dipole moments will not be
detected

48. wavenumber (cm–1) assignment
3075
2950
1650 and 890
absence of bands
1500–1430 and 720
sp2 CH
sp3 CH
a terminal alkene with two substituents
has less than four adjacent CH2 groups

49. wavenumber (cm–1) assignment
3050
2810 and 2730
1600 and 1460
1700
sp2 CH
an aldehyde
benzene ring
a partial single-bond
character carbonyl

50. wavenumber (cm–1) assignment
3300
2950
2100
OH group
sp3 CH
alkyne

51. wavenumber (cm–1) assignment
3300
2950
1660
1560
N—H
sp3 CH
amide carbonyl
N—H Bend

52. wavenumber (cm–1) assignment
>3000
<3000
1605 and 1500
1720
1380
sp2 CH
sp3 CH
a benzene ring
a ketone carbonyl
a methyl group

53. Ultraviolet and Visible Spectroscopy
Spectroscopy is the study of the interaction between
matter and electromagnetic radiation
UV/Vis spectroscopy provides information about
compounds with conjugated double bonds

54. UV and Vis light cause only two kinds of electronic transition:
Symmetry: allowed transition
Forbidden transition: lone pair orthogonal to  system
Only organic compounds with p electrons can produce UV/Vis spectra.
A visible spectrum is obtained if visible light is absorbed.
A UV spectrum is obtained if UV light is absorbed.

55. Only compounds with p electrons can produce UV/Vis spectra
A chromophore is the part of a molecule that absorbs UV or visible light
Only compounds with p electrons can produce UV/Vis spectra
Allowed
Forbidden

56.  = <100 M–1cm–1, Forbidden
The Beer–Lambert Law
 = ~10,000 M–1cm–1, Allowed
A = e c l
 = <100 M–1cm–1, Forbidden
A = log(I0/I)
c = concentration of substance in solution
l = length of the cell in cm
e = molar absorptivity, a measure of the probability of the transition
The molar absorptivity of a compound is a constant that is characteristic of the compound at a particular wavelength

57. Effect of Conjugation on lmax
The lmax and  values increase as the number of conjugated double bonds increases

60. If a compound has enough conjugated double bonds, it
will absorb visible light (lmax >400 nm), and the
compound will be colored

61. An auxochrome is a substituent in a chromophore that
alters the lmax and the intensity of the absorption:

62. The Visible Spectrum and Color

63. Uses of UV/Vis Spectroscopy
Measure the rates of a reaction
Determine the pKa of a compound
Estimate the nucleotide composition of DNA