Organic Chemistry

490 CHAPTER 13 Mass Spectrometry and Infrared Spectroscopy

PROBLEM 6

Which molecular formula has an exact molecular mass of 86.1096 amu:

or

13.5 Fragmentation at Functional Groups

Characteristic fragmentation patterns are associated with specific functional groups;

these can help identify a substance based on its mass spectrum. The patterns were rec-

ognized after the mass spectra of many compounds containing a particular functional

group were studied. We will look at the fragmentation patterns of alkyl halides, ethers,

alcohols, and ketones as examples.

Alkyl Halides

Let’s look first at the mass spectrum of 1-bromopropane, shown in Figure 13.4. The

relative heights of the M and peaks are about equal, so we can conclude that

the compound contains a bromine atom. Electron bombardment is most likely to

dislodge a lone-pair electron if the molecule has any, because a molecule does not hold

onto its lone-pair electrons as tightly as it holds onto its bonding electrons. Thus,

electron bombardment dislodges one of bromine’s lone-pair electrons.

The weakest bond in the resulting molecular ion is the bond (the

bond dissociation energy is 69 kcal/mol; the bond dissociation energy is 85

kcal/mol; see Table 3.1 on p. 129). The bond breaks heterolytically, with both elec-

trons going to the more electronegative of the atoms that were joined by the bond,

forming a propyl cation and a bromine atom. As a result, the base peak in the mass

spectrum of 1-bromopropane is at or The

propyl cation has the same fragmentation pattern it exhibited when it was formed from

the cleavage of pentane (Figure 13.2).

m>z= 43 [M-79, 1 M+ 22 - 81].

C¬C

C¬Br C¬Br

M+ 2

C 4 H 6 O 2?

C 6 H 14 ,C 4 H 10 N 2 ,

Tutorial:

Fragmentation of alkyl halides

CH 3 CH 2 CH 2 Br

m/z

100

80

60

40

20

10 20 30 40 50 60 70 80 90 100 110 120 130

Relative abundance

15

(^2741)
43
(^122124)
Figure 13.4N
The mass spectrum of
1-bromopropane.

• 
  


• 
  

  m/z = 122
  CH^79
  3 CH 2 CH 2 Br
  −e− +
  m/z = 124
  CH^81
  3 CH 2 CH 2 Br

  
  


• 
  

  m/z = 43
  CH 3 CH 2 CH 2 Br
  1-bromopropane
  CH 3 CH 2 CH 279 Br + CH 3 CH 2 CH 281 Br +^79 +^81 Br
  The mass spectrum of 2-chloropropane is shown in Figure 13.5. We know that the
  compound contains a chlorine atom, because the peak is one-third the height
  of the molecular ion peak. The base peak at results from heterolytic cleav-
  ageof the bond. The peaks at and have a ratio, indi-
  cating that these fragments contain a chlorine atom. They result from homolytic
  cleavageof a bond at the carbon (the carbon bonded to the chlorine). This
  cleavage, known as A cleavage, occurs because the C¬Cl(82 kcal/mol) and C¬C
  C¬C a
  C¬Cl m>z= 63 m>z= 65 3:1
  m>z= 43
  M+ 2