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558 CHAPTER 12 • INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY
place to mount the sample in the infrared beam, and the optics and electronics necessary to
measure the intensity of light absorbed or transmitted as a function of wavelength or wavenum-
ber. Modern IR spectrometers, called Fourier-transform infrared spectrometers (FTIR spec-
trometers), can provide an IR spectrum in a few seconds. (See Further Exploration 12.2.) The
IR spectra in this text were obtained with an FTIR spectrometer.
The sample containers (“sample cells”) used in IR spectroscopy must be made of an in-
frared-transparent material. Because glass absorbs infrared radiation, it cannot be used. The
conventional material used for sample cells is sodium chloride. The IR spectrum of an undi-
luted (“neat”) liquid can be obtained by compressing the liquid between two optically pol-
ished salt plates. If the sample is a solid, the finely ground solid can be dispersed (“mulled”)
in a mineral oil and the dispersion compressed between salt plates. Alternatively, a solid can
be co-fused (melted) with KBr, another IR-transparent material, to form a clear pellet. Simple
presses are available to prepare KBr pellets. IR spectra can also be taken in solution cells,
which consist of sodium chloride plates in appropriate holders equipped with syringe fittings
for injecting the solution.
When mineral-oil dispersions or solvents are used, we have to be aware of the regions in
which the oil or the solvents themselves absorb IR radiation, because these absorptions inter-
fere with those of the sample. A number of solvents are commonly used; chloroform (CHCl 3 ),
its isotopic analog (CDCl 3 ), and methylene chloride (CH 2 Cl 2 ) are among them. As a few stu-
dents learn the hard way, solvents that dissolve sodium chloride, such as water and alcohols,
cannot be used.
12.6 INTRODUCTION TO MASS SPECTROMETRY
In contrast to other spectroscopic techniques, mass spectrometry does not involve the
absorption of electromagnetic radiation, but operates on a completely different principle. As
the name implies, mass spectrometry is used to determine molecular masses, and it is the most
important technique used for this purpose. It also has some use in determining molecular
structure.
A. Electron-Impact Mass Spectra
The instrument used to obtain a mass spectrum is called a mass spectrometer. In one type of
instrument, an electron-impact mass spectrometer , a compound is vaporized in a vacuum
and bombarded with an electron beam of high energy—typically, 70 eV (electron-volts) (more
than 6700 kJ mol
_ 1 ). Because this energy is much greater than the bond energies of chemical
bonds, some fairly drastic things happen when a molecule is subjected to such conditions. One
thing that happens is that an electron is ejected from the molecule. For example, if methane is
treated in this manner, it loses an electron from one of the CLH bonds.
(12.15)
The product of this reaction is sometimes abbreviated as follows:
The symbol (^8)
| means that the molecule is both a radical (a species with an unpaired electron)
and a cation—a radical cation. The species CH 48
| is called the methane radical cation.
C
| H H abbreviated as
H
H
| CH 48
H CH e
_
H
H
C
| H H 2 e
_
H
H
Further Exploration 12. FTIR Spectroscopy
12.6 INTRODUCTION TO MASS SPECTROMETRY 559
Following its formation, the methane radical cation decomposes in a series of reactions
called fragmentation reactions. In a fragmentation reaction, a radical cation literally comes
apart. The ionic product of the fragmentation (whether it is a cation or a radical cation) is
called a fragment ion. For example, in one fragmentation reaction, it loses a hydrogen atom
(the radical) to generate the methyl cation, a carbocation.
Notice that the hydrogen atom carries the unpaired electron, and the methyl cation carries the
charge; consequently, the methyl cation is the fragment ion in this case. The process can be
represented with the free-radical (fishhook) arrow formalism as follows:
Alternatively, the unpaired electron may remain associated with the carbon atom; in this case,
the products of the fragmentation are a methyl radical and a proton.
In this case the proton is the fragment ion. Further decomposition reactions give fragments of
progressively smaller mass. (Show how these occur by using the fishhook notation.)
(12.19a)
(12.19b)
(12.19c)
The ions formed in Eqs. 12.16 and 12.19a–c are very unstable species. They are not the sorts of
species that would be involved as reactive intermediates in a solution reaction. Recall from Sec. 9.6,
for example, that methyl and primary carbocations are never formed as intermediates in SN1 reac-
tions. These ions are formed in the mass spectrometer only because of the immense energy imparted
to the methane molecules by the bombarding electron beam.
Thus, methane undergoes fragmentation in the mass spectrometer to give several positively
charged fragment ions of differing mass: CH 48
| ,
| CH 3 , 8
| CH 2 ,
| CH, C 8
| , and H
|
. In the mass
spectrometer, the fragment ions are separated according to their mass-to-charge ratio, m Ü z
( m = mass, z = the charge of the fragment). Because most ions formed in the electron-impact
mass spectrometer have unit charge, the m Ü z value can generally be taken as the mass of the
ion. A mass spectrum is a graph of the relative amount of each ion (called the relative abun-
dance ) as a function of the ionic mass (or m Ü z ). When the ions are produced by electron im-
pact, the mass spectrum is called an EI mass spectrum. The EI mass spectrum of methane is
shown in Fig. 12.14 on p. 560. Note that only ions are detected by the mass spectrometer—
neutral molecules and radicals do not appear as peaks in the mass spectrum. The mass spec-
trum of methane shows peaks at m Ü z = 16, 15, 14, 13, 12, and 1, corresponding to the various
| CH 8
| C H 8
mass = 12
| CH
| 8 CH 2 H 8
mass = 13
| CH 3
| 8 CH 2 H 8
mass = 14
(12.18)
| CH 4 8 8 CH 3 H
|
mass = 16 mass = 1 methyl radical
| C 8
H
H
H L H H
| C
H
H
H L
| CH 4 8
| CH 3 H
mass = 16 methyl cation mass = 15
12.6 INTRODUCTION TO MASS SPECTROMETRY 561
Possible sources of the m Ü z = 17 peak for methane are
13 CH 4 and
12 CDH 3. Each isotopic
compound contributes a peak with a relative abundance in proportion to its amount. In turn,
the amount of each isotopic compound is directly related to the natural abundance of the iso-
tope involved. The abundance of
13 CH 4 methane relative to that of
12 CH 4 methane is then given
by the following equation:
relative abundance = ! " (12.20a)
= (number of carbons) X
= (number of carbons) X !!
= (number of carbons) X 0.0111 (12.20b)
Because methane has only one carbon, the m Ü z = 17 (M + 1) peak due to
13 CH 4 is about 1.1%
of the m Ü z = 16, or M, peak. A similar calculation can be made for deuterium.
relative abundance = (number of hydrogens) X ! " (12.21)
= (4) X !!
Thus, the CDH 3 naturally present in methane contributes 0.06% to the isotopic peak. Because
the contribution of deuterium is so small,
13 C is the major isotopic contributor to the M + 1
peak. (We’ll ignore contributions of
2 H in subsequent calculations of M + 1 peak intensities.)
In a compound containing more than one carbon, the M + 1 peak is larger than 1.1% of the
M peak because there is a 1.1% probability that each carbon in the molecule will be present
as
13 C. For example, cyclohexane has six carbons, and the abundance of its M + 1 ion relative
to that of its molecular ion should be 6(1.1) = 6.6%. In the mass spectrum of cyclohexane,
the molecular ion has a relative abundance of about 70%; that of the M + 1 ion is calculated
to be (0.066)(70%) = 4.6%, which corresponds closely to the value observed. (With careful
measurement, it is possible to use these isotopic peaks to estimate the number of carbons in an
natural abundance of
2 H !!! natural abundance of
1 H
natural abundance of
13 C !!! natural abundance of
12 C
abundance of
13 C peak !!! abundance of
12 C peak
100
80
60
40
20
0
relative abundance
0 10 20 30 40 50 60 70 80 90 100 110 120 130
142
85
71
57
base peak 43
molecular ion (M)
140
CH 3 (CH 2 ) 8 CH 3
decane
mass-to-charge ratio m/z
Figure 12.15 The EI mass spectrum of decane.
562 CHAPTER 12 • INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY
unknown compound; see Problem 12.41 on p. 577.) Not only the molecular ion peak, but also
every other peak in the mass spectrum has isotopic peaks.
Several elements of importance in organic chemistry have isotopes with significant natural
abundances. Table 12.3 shows that silicon has significant M + 1 and M + 2 contributions; sul-
fur has an M + 2 contribution; and the halogens chlorine and bromine have very important M
- 2 contributions. In fact, the naturally occurring form of the element bromine consists of
about equal amounts of
79 Br and
81 Br. The mixture of isotopes leaves a characteristic trail in the
mass spectrum that can be used to diagnose the presence of the element.
Consider, for example, the EI mass spectrum of bromomethane, shown in Fig. 12.16. The
peaks at m Ü z = 94 and 96 result from the contributions of the two bromine isotopes to the
molecular ion. They are in the relative abundance ratio 100 : 98 = 1.02, which is in good
agreement with the ratio of the relative natural abundances of the bromine isotopes
(Table 12.2). This double molecular ion is a dead giveaway for a compound containing a sin-
gle bromine. Notice that along with each major isotopic peak is a smaller isotopic peak one
mass unit higher. These peaks are due to the isotope
13 C present naturally in bromomethane.
For example, the m Ü z = 95 peak corresponds to bromomethane containing only
79 Br and one 13 C. The m Ü z = 97 peak arises from methyl bromide that contains only
81 Br and one
13 C.
Although isotopes such as
13 C and
18 O are normally present in small amounts in organic
compounds, it is possible to synthesize compounds that are selectively enriched with these and
Exact Masses and Isotopic Abundances of Several Isotopes Important in Mass Spectrometry
Element Isotope Exact mass Abundance, %
hydrogen 1 H 1.007825 99. (^2) H* 2.0140 0.
carbon 12 C 12.0000 98. 13 C 13.00335 1.
nitrogen 14 N 14.00307 99. (^15) N 15.00011 0.
Oxygen
16 O 15.99491 99. (^17) O 16.99913 0. (^18) O 17.99916 0.
fluorine 19 F 18.99840 100.
silicon 28 Si 27.97693 92. 29 Si 28.97649 4. (^30) Si 29.97377 3.
phosphorus 31 P 30.97376 100.
sulfur
32 S 31.97207 95. (^33) S 32.97146 0. (^34) S 33.96787 4.
chlorine 35 Cl 34.96885 75. (^37) Cl 36.96590 24.
bromine
79 Br 78.91834 50. (^81) Br 80.91629 49.
iodine 127 I 126.90447 100.
*^2 H is commonly known as deuterium, abbreviated D.
TABLE 12.
564 CHAPTER 12 • INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY
Next, the molecule splits at the site of electron ejection to give a carbocation with m Ü z = 57
and a radical with mass = 85. In this fragmentation, only the cation is detected.
Notice that there is also a peak in Fig. 12.15 at m Ü z = 85. This does not arise from the radi-
cal, but rather from fragmentation of the same bond in the opposite manner to give the carbo-
cation with m Ü z = 85 and the radical with mass = 57.
A fragmentation of type 2 is illustrated by the mass spectra of many primary alcohols. For
example, in the mass spectrum of 1-heptanol (molecular mass = 116), the molecular ion is
formed by electron ejection from one of the oxygen unshared pairs. (Because unshared elec-
trons are not held in bonds, they are ejected more easily than bonding electrons.)
Loss of the neutral molecule water from this molecular ion gives another radical cation of
mass = 98. The radical cation is detected in the mass spectrum; the neutral molecule water
is not.
If a molecule contains only C, H, O, and halogen, its even-electron fragment ions have odd
mass and its odd-electron fragment ions have even mass. You can verify this with the exam-
ples in Eqs. 12.22–12.26. Thus, from the mass of the fragment ion—odd or even—you imme-
diately know something about its structure and its origin.
(12.26)
CH 3 (CH 2 ) 4 CH 3 (CH 2 ) 4
|
L CHL
CH 2 L L
L
CH CH 2
CH 3 (CH 2 ) 4 L CH L CH 2 L
H
OH
|
|
H OH
H
odd-electron ion, m / z = 98 (detected by the mass spectrometer)
+ OH
..
..
a neutral molecule (not detected by the mass spectrometer)
molecular ion of 1-heptanol
CH 3 (CH 2 ) 4 CH 2 CH 2 (12.25)
OH
.. ..
CH 3 (CH 2 ) 4 CH 2 CH 2
OH
..
.
molecular ion of 1-heptanol m/z = 116
CH 3 CH 2 CH 2 CH 2^.^ CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 (12.24)
.
molecular ion of decane a cation m/z = 85 (detected by the mass spectrometer)
a radical (not detected by the
mass spectrometer)
CH 3 CH 2 CH 2 CH 2^. CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 CH 3 CH 2 CH 2 CH 2 (12.23)
molecular ion of decane (a radical cation, m/z = 142)
+ CH 2 CH 2 CH 2 CH 2 CH 2 CH 3
.
a cation m/z = 57 (detected by the mass spectrometer)
a radical (not detected by the mass spectrometer)
CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 (12.22)
- e – CH 3 CH 2 CH 2 CH 2^. CH 2 CH 2 CH 2 CH 2 CH 2 CH 3
decane molecular ion of decane (a radical cation, m/z = 142)
12.6 INTRODUCTION TO MASS SPECTROMETRY 565
As Fig. 12.15 illustrates, the peaks in a mass spectrum are typically not the same height. What
controls the relative abundances of ions in a mass spectrum? Typically, the most stable ions ap-
pear in greatest abundance. If an ion is relatively stable, it decomposes slowly and appears as a
relatively large peak. If an ion is relatively unstable, it decomposes rapidly and appears as a rel-
atively small peak—or perhaps not at all. The principles of carbocation stability that you already
know can help you to understand why certain fragment ions in a mass spectrum are prominent
and others are not. This idea is illustrated in Study Problem 12.4.
Study Problem 12.
The base peak in the mass spectrum of 2,2,5,5-tetramethylhexane (molecular mass = 142) is at
m Ü z = 57, which corresponds to a composition C 4 H 9. (a) Suggest a structure for the fragment that
accounts for this peak. (b) Offer a reason that this fragment is so abundant. (c) Give a mechanism
that shows the formation of this fragment.
Solution The first step is to draw the structure of 2,2,5,5-tetramethylhexane:
(CH 3 ) 3 CLCH 2 CH 2 LC(CH 3 ) 3.
(a) A fragment with the composition C 4 H 9 could be a tert -butyl cation formed by splitting the
compound at the bond to either of the tert -butyl groups:
(b) The most abundant peaks in the mass spectrum result from the most stable cationic fragments.
Because a tert -butyl cation is a relatively stable carbocation (it is tertiary), it is formed in rela-
tively high abundance.
(c) To form this cation, one electron is ejected from the CLC bond, and the compound fragments
so that the unpaired electron remains on the methylene carbon (see Eqs. 12.23–12.24):
Fragmentation might have occurred at the same bond so that the unpaired electron remains as-
sociated with the tert -butyl group and a primary carbocation with m Ü z = 85 is formed. (In other
words, a more stable free radical and a less stable carbocation would be formed.) There is no peak
at m Ü z = 85. That this mode of fragmentation is not observed demonstrates that carbocation sta-
bility is more important than free-radical stability in determining fragmentation patterns.
(12.27)
C
CH 3
CH 3
H 3 C L^ L^ L "
" CH 2
CH 3
CH 3
L "
" CH 2 3 C CH 3 C
CH 3
CH 3
H 3 C L^ L^ L "
" CH 2
CH 3
CH 3
L "
" CH 2 C CH 3
electron ejection (^) | 8
C
CH 3
CH 3
H 3 C L L L "
" CH 2
CH 3
m / z = 57
L "
" CH 2 C
CH 3
CH 3
8 |
fragmentation
molecular mass = 142 m/z = 142; molecular ion
C 4 H 9
C
CH 3
CH 3
H 3 C L^ L^ L "
" CH 2
CH 3
CH 3
L "
" CH 2 L^ C CH 3
12.6 INTRODUCTION TO MASS SPECTROMETRY 567
The m Ü z = 57 peak is formed by a process called inductive cleavage , which is nothing
more than the radical-cation version of an SN1-like dissociation:
(12.31)
molecular ion of di- sec -butyl ether ( m/z = 130)
CHCH 2 CH 3
CH 3
O .. CHCH 2 CH 3
CH 3
O ..
. ..
CH 3
CH 3 CH 2 CH
CH 3
CH 3 CH 2 CH.
inductive cleavage
CH 3 CH CHCH 3 (12.30)
CHCH 3
CHCH 3
CHCH 3
O ..
b -elimination
CH 3 CH OH .. +
m/z = 101
m/z = 45 H
100
80
60
40
20
0
relative abundance
0 10 20 30 40 50 60 70 80 90 100 110 120 130
57
101
base peak 45
almost no molecular ion at 130
140
mass-to-charge ratio m/z
100
80
60
40
20
0
relative abundance
0 10 20 30 40 50 60 70 80 90 100 110 120 130
73
75
101
115
131 (M + 1)
base peak
140
mass-to-charge ratio m/z
(a) EI mass spectrum
(b) CI mass spectrum
CH 3 CH 2 CH O CHCH 2 CH 3
CH 3 CH 3
di- sec -butyl ether
Figure 12.17 Mass spectra of di- sec -butyl ether. (a) Electron-impact (EI) mass spectrum. (b) Chemical-ionization
(CI) mass spectrum.The molecular ion at m Ü z = 130 is essentially absent in the EI spectrum, whereas the molecu-
lar ion (as the protonated ether at m Ü z = 131) is the base peak in the CI spectrum. Notice that the CI spectrum has
a smaller number of fragment ions that the EI spectrum.
568 CHAPTER 12 • INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY
The decomposition mechanisms shown here— a -cleavage, b -elimination, and inductive
cleavage—are very common decomposition mechanisms in the mass spectra of molecules
containing atoms with unshared electron pairs. These processes lead to relatively stable
cations, and this is why the molecular ion does not survive.
This example illustrates the point that we cannot be sure in many cases whether the ion of
highest mass in a compound of unknown structure is the molecular ion or a fragment ion. The
question is, then, how can one determine with certainty the molecular mass of an unknown
compound?
Recall that molecular ions in EI mass spectra are formed by a highly energetic electron-
bombardment process. When a molecular ion has a very high energy, it is likely to dissipate
that energy by fragmentation. However, if we could form a molecular ion by a “softer” (less
energetic) method, the tendency of the ion to undergo fragmentation would be decreased. An
ionization method commonly used for this purpose is called chemical ionization , and mass
spectra derived from chemical ionization are called chemical-ionization mass spectra, or CI
mass spectra for short.
In chemical ionization, the molecule of interest in the gas phase is not bombarded with
high-energy electrons. Rather, it is treated with a source of gas-phase protons. If the molecule
contains a basic site, it is protonated to give its conjugate acid. In the case of di- sec -butyl ether,
the basic site is the oxygen lone pairs, and the ionization process is protonation of this oxygen
(Sec. 8.7):
This conjugate-acid cation is an even-electron ion; it is not a radical cation. In a CI mass spec-
trum, the peak for this ion necessarily occurs one mass unit higher than the molecular mass of
the molecule itself because of the added proton. Because this ion is formed in a relatively low-
energy process, it does not fragment so readily as the molecular ion in the EI mass spectrum.
The CI mass spectrum of di- sec -butyl ether is shown in Fig. 12.17b. This shows a prominent
M + 1 ion at m Ü z = 131, which is also the base peak. The relatively small number of frag-
ments come from the loss of various neutral molecules from this ion. For example, the largest
fragment peak at m Ü z = 75 arises from loss of 2-butene in a b -elimination process analogous
to the one in Eq. 12.30:
Typically, the mass spectroscopy of a compound with unknown structure is investigated by
running both its EI and CI mass spectra. The CI mass spectrum typically gives a strong
M + 1 peak that reveals the molecular mass M. The richer fragmentation pattern of the EI
spectrum can then be used to deduce other aspects of the structure.
O .. (12.33)
CH 3
CH 3 CH 2 CH
H
conjugate acid of di- sec -butyl ether m/z = 131
CHCH 3
CHCH 3
CHCH 3
CHCH 3
H
O ..
CH 3
CH 3 CH 2 CH
H
H +
m/z = 75
CHCH 2 CH 3 (12.32)
CH 3
O ..
CH 3
CH 3 CH 2 CH CHCH 2 CH 3
CH 3
O ..
CH 3
+ H CH 3 CH 2 CH
H
(a gas-phase
proton source)
di- sec -butyl ether conjugate acid of di- sec -butyl ether m/z = 131
570 CHAPTER 12 • INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY
The difference of 0.0364 mass units is easily resolved by a high-resolution mass spectrometer.
Computers used with such instruments can be programmed to work backward from the exact
mass and provide an elemental analysis of the molecular ion (and therefore the compound of in-
terest) as well as the elemental analysis of each fragment in the mass spectrum! Because a mod-
ern high-resolution mass spectrometer with its associated computer and other accessories can
cost several hundred thousand dollars, it is generally shared by a large number of researchers.
Before a compound can be analyzed by mass spectrometry, it must be vaporized. This pre-
sents a difficult problem for large molecules that have negligible vapor pressures. Research in
mass spectrometry has focused on novel ways to produce ions in the gas phase from large non-
volatile molecules, many of which are of biological interest. In one technique, nicknamed
MALDI (matrix-assisted laser desorption ionization), the material to be analyzed (analyte) is
co-crystallized with a material, termed a matrix , that can absorb radiation from a laser. In a
process that is not fully understood, bombarding the matrix–analyte mixture with light from
the laser ultimately produces gas-phase ions of the analyte, which are analyzed by mass spec-
trometry. In another technique, nicknamed ESI (electrospray ionization), a solution of the an-
alyte is atomized in highly charged droplets, much as we might atomize perfume in a sprayer.
This process results in the formation of highly charged molecules in the gas phase, and these
ion source (ions are formed and accelerated here)
to vacuum pump
magnetic field direction
lower-mass ions
the separated ion beam
ion collector
ion exit slit
analyzer tube
higher-mass ions
magnet
B
Figure 12.18 Diagram of a magnetic-sector mass spectrometer. After ionization of the sample by electron bom- bardment, the ions are accelerated by a high voltage and are passed into a magnetic field B along a path perpen- dicular to the field. The field bends the paths of the ions; the paths of lower-mass ions ( red ) are bent more than those of higher-mass ions ( blue ). (See Further Exploration 12.3.) As the field is progressively increased, ions of in- creasingly higher mass attain exactly the correct path to enter the detection slit.
12.22 List the factors that determine the wavenumber of an
infrared absorption.
12.23 List two factors that determine the intensity of an in-
frared absorption.
12.24 Indicate how you would carry out each of the follow-
ing chemical transformations. What are some of the
changes in the infrared spectrum that could be used to
indicate whether the reaction has proceeded as indi-
cated? (Your answer can include disappearance as well
as appearance of IR absorptions.)
(a) 1-methylcyclohexene LT methylcyclohexane
(b) 1-hexanol LT 1-methoxyhexane
ADDITIONAL PROBLEMS 571
! Spectroscopy deals with the interaction of matter and
electromagnetic radiation. Electromagnetic radiation
is characterized by its energy, wavelength, and fre-
quency, which are interrelated by Eq. 12.3.
! Infrared spectroscopy deals with the absorption of in-
frared radiation by molecular vibrations. An infrared
spectrum is a plot of the infrared radiation transmit-
ted through a sample as a function of the wavenum-
ber or wavelength of the radiation.
! The frequency of an absorption in the infrared spec-
trum is equal to the frequency of the bond vibration
involved in the absorption.
! The wavenumber or frequency of an absorption is
greater for vibrations involving stronger bonds and
smaller atomic masses (Eqs. 12.10 and 12.13). The
smaller of two atomic masses involved in a bond vi-
bration has the greater effect on the frequency of the
vibration.
! The intensity of an absorption increases with the
number of absorbing groups in the sample and the
size of the dipole moment change that occurs in
the molecule when the vibration occurs. Absorptions
that result in no dipole moment change are infrared-
inactive.
! The infrared spectrum provides information about
the functional groups present in a molecule. The
ACLH stretching and bending absorptions and
the CAC stretching absorption are very useful for the
identification of alkenes. The OLH stretching ab-
sorption is diagnostic for alcohols.
! In electron-impact (EI) mass spectrometry, a molecule
loses an electron to form the molecular ion, a radical
cation, which in most cases decomposes to fragment
ions.The relative abundances of the fragment ions are
recorded as a function of their mass-to-charge ratios
m Ü z , which, for most ions, equal their masses. Both
molecular masses and partial structures can be de-
rived from the masses of these ionic fragments.
! Associated with each peak in a mass spectrum are
other peaks at higher mass that arise from the pres-
ence of isotopes at their natural abundance. Such iso-
topic peaks are particularly useful for diagnosing the
presence of elements that consist of more than one
isotope with high natural abundance, such as chlorine
and bromine.
! Ionic fragments are of two types: even-electron ions,
which contain no unpaired electrons; and odd-elec-
tron ions, which contain an unpaired electron.
! In chemical-ionization (CI) mass spectrometry, mole-
cules are ionized by direct protonation in the gas
phase. Because this is a much gentler ionization tech-
nique than EI, a CI mass spectrum typically contains a
greater proportion of molecular ion (as its conjugate
acid) than the EI spectrum of the same compound.
KEY IDEAS IN CHAPTER 12
are analyzed by mass spectrometry. These techniques have made possible the analysis of ma-
terials with molecular masses in excess of 100,000, such as proteins, nucleic acids, and syn-
thetic polymers. For their discovery and development of these techniques, John P. Fenn, of
Virginia Commonwealth University, and Koichi Tanaka, of the Shimadzu Corporation in
Tokyo, shared part of the 2002 Nobel Prize in Chemistry.
ADDITIONAL PROBLEMS
