Jones M., Fleming S.A. Organic Chemistry

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15.3 Mass Spectrometry (MS) 699

M + e

––

2 e + M

The in-coming

electron and the

ejected electron

The residual

radical cation

A high-energy

electron

FIGURE 15.3 When bombarded

with high-energy electrons,

molecules can lose an electron to

give a radical cation.

Anode

Slit

70-eV

cathode

Ion

beam

Radical

cation M

Magnet

Ion detector

High molecular weight

ions are deflected too

little and hit the wall

Low molecular weight

ions are deflected too

much and hit the wall

Sample M

Negatively

charged

plate

Slit

FIGURE 15.4 A schematic view of a

mass spectrometer.

stationary phase coated on the walls. They often have immense separating powers.

Yet, for the practicing chemist, the old-style simple packed GC columns are still

very important because they allow for the collection of the components of the mix-

ture as they pass out of the column. Not only can the mixtures be separated, but

the pure components can be isolated for identification by the simple means of

attaching a cooled receiver to the end of the chromatograph (Fig. 15.2).

A particularly useful modification of a gas chromatograph is to attach it to a mass

spectrometer (GC/MS) or infrared spectrometer (GC/IR) so that the various com-

ponents can be analyzed directly as they emerge from the column. In the following

sections, we will see what kinds of information MS and IR spectroscopy can give

about molecules.

One thing these very high-energy electrons can do is eject an electron from a

molecule, M, to give the corresponding radical cation,M



(Fig. 15.3). Now what

can happen to the newly formed radical cation? In a typical mass spectrometer, it

is first accelerated toward a negatively charged plate (Fig. 15.4). There is a slit in

the plate, and some of the accelerated ions pass through the slit to form a beam.

15.3 Mass Spectrometry (MS)

First of all, never make the mistake of calling it “mass spectroscopy.” Spectroscopy

involves the absorption of electromagnetic radiation and as we will see later, mass

spectrometry (MS) is different.The mass spectrometrists sometimes get upset if you

confuse the issue.

15.3a The MS Spectrometer In MS, a compound is bombarded with very

high-energy electrons. Typically, electrons of 70 eV are used.

WORKED PROBLEM 15.2 How many kilocalories per mole does 70 eV represent?

See page 4.

ANSWER One electron volt (eV) is 23.06 kcal/mol, so 70 eV is 1614 kcal/mol.

The point of this problem is to show that the incoming electron is carrying a lot

of energy!

700 CHAPTER 15 Analytical Chemistry: Spectroscopy

The beam of ions then follows a curved path between the poles of a magnet. In a

magnetic field, a moving charged particle, the radical cation in this case, is deflected

by an amount proportional to the strength of the magnetic field. Too little deflec-

tion (too weak a magnetic field) results in the ion hitting the far wall of the tunnel

and in loss of the ion. If there is too much deflection (too strong a magnetic field),

the ion impinges on the near wall and is also lost. If, however, the strength of the

magnetic field is just right, the ion will follow a curved path to the detector and be

“seen.” The magnetic field can be tuned so that ions of different mass are curved

just the right amount to be detected. For ions of low mass, a relatively low magnetic

field will be necessary; for ions of higher mass, the field will have to be stronger to

do the required bending of the path.

Mass spectra are reported as graphs of the mass-to-charge ratio,m/z, versus inten-

sity. As long as we are dealing with singly charged ions, the positions of the signals

will correspond to the molecular weights of the ions detected because the charge,

z, will be 1.

So far, we have a device that can detect M

•

ions formed by bombardment of

a molecule M. This species, M

•

, is called the molecular ion. In our detector, we

should see an ion with the mass m corresponding to the molecular weight of M.

For cyclohexane, we should see a molecular ion of 84 mass units, because the

molecular weight of cyclohexane (C

) is 84 (Fig. 15.5). In the mass spectrum of

cyclohexane, there is a small peak above the molecular ion, at m/z  85. It is only

about 7% of the molecular ion, and therefore one must look closely in order to see it.

Cyclohexane

) M = 84

3,4-Dihydro-

2H-pyran

O) M = 84

70 eV

O O

= 84

FIGURE 15.5 Many different ions

can have a molecular weight of 84.

Two possibilities are the molecular

ions of cyclohexane and 3,4-dihydro-

2H-pyran.

This M ⴙ 1 peak is not the result of an impurity, however; it is always there no mat-

ter how carefully the cyclohexane is purified. This M  1 peak exists because the

carbon and hydrogen have naturally occurring isotopes, and the mass spectrometer

can detect molecules containing these heavy isotopes. Table 15.1 gives the relative

natural abundance of some common isotopes.

TABLE 15.1 Some Common Isotopes and Their Abundances

in Nature

Isotope Abundance (%) Isotope Abundance (%)

H 99.985

O 0.038

H (D) 0.015

O 0.200

C 98.90

Cl 75.77

C1.10

Cl 24.23

N 99.63

Br 50.69

N0.37

Br 49.31

O 99.762

127

I 100

For cyclohexane, there is a 1.1% probability that each carbon will be a

C,and there-

fore about 7% probability that the mass of cyclohexane will be one unit high (M  1).

15.3 Mass Spectrometry (MS) 701

Deuterium (

H or D) has a natural abundance of only 0.015%,and therefore contributes

relatively little to the M  1 peak. Sometimes the presence of isotopes is useful in the

diagnosis of structure.An analysis of isotope peaks is particularly useful for organic chlo-

rides and organic bromides, as we will see at the end of this section.

But how do we know that a mass spectrum with m/z  84 is really cyclohexane,

? If the starting material is of unknown structure, we probably do not know

the formula of the molecule.The detection of a molecular ion of m/z  84 only tells

us that the molecular weight of the unknown is 84, nothing more. Any other

molecule of m  84 could give the same molecular ion. In this case, we might be

looking at 3,4-dihydro-2H-pyran, for example (C

O  84, Fig. 15.5).

M of C

C = 6  12.000 = 72.000

H = 12  1.008 = 12.096

M of C

C = 5  12.000 = 60.000

H = 8  1.008 = 8.064

O = 15.995 = 15.995

84.096

84.059

FIGURE 15.6 If molecular weights are

measured carefully enough, mass

spectrometry can differentiate

between different molecules of the

nominal molecular weight of 84.

TABLE 15.2 Some Atomic Weights

Element Atomic Weight Element Atomic Weight

H 1.008

F 19.998

H 2.010

Si 27.977

H (av) 1.008

Si 28.975

B 10.013

Si 29.974

B 11.009 Si (av) 28.086

B (av) 10.810

P 30.974

C 12.000

S 31.972

C 13.003

S 32.972

C (av) 12.011

S 33.968

N 14.003

S 35.967

N 15.000 S (av) 32.066

N (av) 14.007

Cl 34.969

O 15.995

Cl 36.966

O 16.999 Cl (av) 35.453

O 17.999

Br 78.918

O (av) 15.999

Br 80.916

Br (av) 79.904

127

I 126.905

These days, mass spectrometers are so accurate that we can actually tell C

from C

O. The two species do not have the same molecular weight if we

measure molecular weights accurately enough (Fig. 15.6). Table 15.2 gives the accurate

atomic weights for a variety of isotopes as well as the average values.

PROBLEM 15.3 Find other possible structures that might fit the observed m/z of 84.

15.3b MS Measurements When a molecular formula is calculated from

combustion data, it is the average values in Table 15.2 that must be used, because

unless there has been a deliberate introduction of an isotope during the synthesis of

702 CHAPTER 15 Analytical Chemistry: Spectroscopy

the material, it will be the natural

isotopic mixture that appears in the

compound. The mass spectrome-

ter, on the other hand, is able to

separate ions containing different

isotopes. Each peak represents a

particular set of isotopes, for exam-

ple,

O. So, a

good mass spectrometer can give

us the molecular formula of an

unknown, as long as we can detect

the molecular ion. The full mass

spectra of our two m  84 mole-

cules are shown in Figure 15.7.

One can see that the mass spectra

are not simple composites of

molecular ion and M  1 peaks.

There are many additional lower

molecular weight ions detected

that give a characteristic array of

ions known as the fragmentation

pattern for each molecule.

The original radical cation,



, can undergo many fragmen-

tation reactions before it is accelerated into the magnetic field and passed on to the

detector. The fragmenting ion is called the parent ion (p). It can react to give many

daughter ions, and these ions appear as peaks in the mass spectrum—the fragmen-

tation pattern. Often the most intense or largest peak is not the molecular ion. In all

cases, the intensities of mass spectral peaks are given as percentages of the largest peak

in the spectrum, which may be the molecular ion, but usually is not.The largest peak

is called the base peak of the spectrum. You can see in Figure 15.7 that the base peak

for cyclohexane is 56 and the base peak for 3,4-dihydro-2H-pyran is 55. Remember

that the molecular ion, M



, is the furthest to the right on the spectrum, except for

the small isotope peaks.

Is there any way to tell how a given molecular ion will fragment? Yes and no.There

is no way to predict the detailed pattern of daughter ions. Yet we can make general

predictions by applying what we know about reaction mechanisms and the stabili-

ties of ions.Fragmentations usually take place to give the most stable possible cations,

and we are now able to make some reasonable guesses as to what these might be.

First of all, note that there are two ways for a typical parent radical cation to pro-

duce two fragments (Fig. 15.8). When the radical cation

•

produces A and

B by fragmentation, the charge may go with either A or B. The detector can see

only positively charged ions, and so any neutral species formed are invisible. In the

fragmentation process, the charge will generally wind up where it is more stable,

and we are often able to predict where it will go. For example, tert-butyl chloride

gives large amounts of the tert-butyl cation,m/z  57,in its mass spectrum,not Cl



m/z  35 (Fig. 15.8). The charge is more easily borne on the tertiary cation than

on the relatively electronegative chlorine. In fact, the cleavage of tert-butyl chloride

is so facile, the molecular ion is not even detected. Nevertheless, it must be admitted

that predicting is not always a straightforward process. High energies are involved,

and there is plenty of energy available to do a lot of bond breaking. Rearrangements

are also common, especially if a particularly stable ion can be reached.

100

Cyclohexane

Relative abundance

m/z

100 110 1202010 30 40 50 60 70 80

m/z

100

Relative abundance

100 110 1202010 30 40 50 60 70 90

3,4-Dihydro-

2H-pyran

WEB 3D

FIGURE 15.7 The mass spectra of cyclohexane and 3,4-dihydro-2H-pyran.

15.3 Mass Spectrometry (MS) 703

Some other examples will probably be useful. Figure 15.9 shows the mass spectrum

of toluene.The molecular ion is visible as a strong peak at m/z  92, but it is not the base

peak.The largest peak in the spectrum is the M1 peak at m/z 91.What can this be?

THE GENERAL CASE

A SPECIFIC EXAMPLE

ClC

CCl

ClC

][

m/z

100

Relative abundance

100 110 1202010 30 40 50 60 70 90

m/z = 57

m/z = 35m/z = 92

(not observed)

ClC

tert-Butyl chloride

FIGURE 15.8 Two possible

fragmentations for a general

molecule and the mass

spectrum of tert-butyl chloride.

FIGURE 15.9 The mass spectrum

of toluene.

m/z

100

Relative abundance

100 110 1202010 30 40 50 60 70 90

Toluene

70 eV

7 ,

m/z = 91 m/z = 92

–

FIGURE 15.10 The molecular ion of

toluene loses a hydrogen atom to give

the benzyl cation.

A little thought brings us quickly to a reasonable answer. If a hydrogen atom (not a

proton,which would be positively charged and leave a neutral fragment behind) is lost

from the parent ion,what is left behind is a resonance-stabilized benzyl ion (Fig. 15.10).

704 CHAPTER 15 Analytical Chemistry: Spectroscopy

Accordingly, loss of the hydrogen atom from the first-formed radical cation is

relatively easy, and the largest peak in the spectrum belongs to the ion C



m/z  91, the benzyl cation.

PROBLEM 15.4 There is some evidence that the m/z  91 peak may not be the

benzyl cation but another species. Can you (a) think of a C



species that

might be even more stable than the benzyl cation, and (b) write an arrow formal-

ism for the conversion of the benzyl cation into this new species? Hint: See p. 587.

The mass spectra of compounds containing carbon–oxygen double bonds (car-

bonyl compounds) often have large peaks for the resonance-stabilized acylium ions

(Fig. 15.11). For example, the molecular ion of 2,2,4,4-tetramethylpentan-3-one

(di-tert-butyl ketone) ionizes to give large amounts of the acylium ion of m/z  85.

Acylium ions can break down further to give carbon monoxide and a carbocation.

70 eV

O O

70 eV

m/z = 142 m/z = 85 m/z = 57

(CH

)

C C(CH

)

(CH

)

(CH

)

C(CH

)

(CH

)

C(CH

)

m/z

100

Relative abundance

100 110 120 130 140 1502010 30 40 50 60 70 90

(CH

)

C(CH

)

WEB 3D

Acylium ion

FIGURE 15.11 Carbonyl compounds usually give large amounts of a resonance-stabilized acylium ion.

15.3 Mass Spectrometry (MS) 705

70 eV

2,2-Dimethylpropane

(neopentane)

m/z = 15

m/z = 57

100

Relative abundance

100 110 1202010 30 40 50 60 70 90

C(CH

)

m/z

m/z = 72

(not observed)

FIGURE 15.12 As illustrated

by the mass spectrum of

2,2-dimethylpropane, alkanes

fragment to give the most stable

carbocation possible.

Alkanes cleave so as to give the most stable possible carbocation (Fig. 15.12).

For example, 2,2-dimethylpropane (neopentane) fragments in a mass spectrometer

to give an ion of m/z  57, the tert-butyl cation, much more easily than it gives the

far less stable methyl cation, m/z  15.

70 eV

m/z = 56

OH OH

m/z

100

Relative abundance

100 110 1202010 30 40 50 60 70 90

(CH

)

m/z = 74

(not observed)

FIGURE 15.13 In the mass

spectrometer, alcohols such as

1-butanol can eliminate water to

give alkene radical cations.

There are other fragmentation reactions that resemble solution chemistry. For

example, an important peak in the mass spectrum of 1-butanol is at m/z  56.

High-resolution analysis of this peak shows that its formula is C

. Apparently,

the parent ion eliminates water to give a butene radical cation (Fig. 15.13). This

behavior is typical for alcohols and has clear roots in the solution chemistry

described in our sections on elimination reactions (p. 298).

706 CHAPTER 15 Analytical Chemistry: Spectroscopy

PROBLEM 15.5 Draw a schematic molecular orbital diagram for the π orbitals of

the butene radical cation, C

•

WORKED PROBLEM 15.6 What do you suggest as a structure for the ion that cor-

responds to the base peak (m/z  31) in the mass spectrum of 1-butanol shown

in Figure 15.13?

ANSWER The base peak is m/z  31. The mass of 1-butanol (C

O) is 74, so

some quite serious fragmentation must have taken place. A mass of 31 can only

have two heavy (nonhydrogen) atoms, so it must be either a C

(m  24) or CO

(m  28) compound. If the ion contained two carbons, it would have to be C

an impossible formula, so it can only be CH

O. The problem now is to find a

reasonably stable ion of this formula that can be made from a fragmentation of

1-butanol. A simple carbon–carbon bond breaking leads to a resonance-stabilized

cation that certainly is a good candidate.

70 eV

m/z

= 31m/z = 74

Molecular ion

As mentioned earlier, molecules containing chlorine or bromine produce easily

characterized spectra. This observation is one of the most useful aspects of mass

spectrometry for organic chemists.Table 15.1 lists the natural isotopic ratio for

Cl (75.77:24.23), which is essentially a 3:1 ratio. The ratio for

Br to

(50.69:49.31) is close to a 1:1 ratio. Chlorine and bromine are the only elements

frequently found in organic molecules that have significant M  2 isotopes.There

are other elements (osmium and mercury, for example) that have M  2 isotopes,

but we are less likely to deal with mass spectra of such compounds. Therefore, if a

mass spectrum shows an M  2 peak that is one-third the intensity of the molec-

ular ion (M

•

) as shown in Figure 15.14a, then we know there is a chlorine in the

molecule. If a spectrum has an M  2 peak that is the same intensity as the molec-

ular ion, we can conclude that the molecule has a bromine (see Fig. 15.14b).

(a) (b)

Data from National Institute of Standards and Technology, http://webbook.nist.gov

m/z

100

Relative abundance

2010 30 40 50 60 70

Chloroethane

m/z

100 110 1202010 30 40 50 60 70 90

Bromoethane

Cl Br

108

110

FIGURE 15.14 (a) The mass spectrum of chloroethane. If there is a chlorine in the molecule, then an M  2 peak with

one-third the height of the M peak will be observed. (b) The mass spectrum of bromoethane. When there is bromine

in a molecule there is a 1:1 ratio of the M and the M  2 peaks.

15.4 Infrared Spectroscopy (IR) 707

There are still plenty of mysteries in

mass spectrometry, however. For example,

the Jones research group first made the

molecule bicyclo[4.2.2]deca-2,4,7,9-

tetraene (m  130; Fig. 15.15) over 40

years ago, and were startled then to find

that the base peak in the mass spectrum

was at m/z  115. It seemed likely that

this ion (m  15) represented the loss of

a methyl group from the parent ion. We

have been thinking about this process (not constantly!) over the ensuing three decades

and still have no reasonable explanation for the loss of a methyl group. Any ideas?

Despite occasional mysteries,mass spectrometry is still highly useful,even in the

bare-bones form described here. It gives us a way to determine molecular composi-

tion through detection of a molecular ion, and can give hints as to structure from

the fragment ions produced. Even though it is difficult to predict the fragmenta-

tion patterns in more than a general way, it is possible to compare a fragmentation

pattern with that of a known compound in order to prove identity.

Summary

Mass spectrometry usually provides the molecular weight of an unknown com-

pound. If run at sufficiently high resolution,MS can even allow you to determine

the molecular formula of a compound. Some information can be gathered about

structure from the ions formed on fragmentation.

15.4 Infrared Spectroscopy (IR)

Infrared (IR) spectroscopy is true spectroscopy—it informs us about the electromag-

netic radiation absorbed by a molecule.In Chapter 12,we discussed UV/vis spectroscopy,

and IR spectroscopy is qualitatively similar. A schematic picture of the electromag-

netic spectrum (Fig. 15.16) identifies the regions used for UV/vis, IR, and the spec-

troscopy we will study in the second half of this chapter, nuclear magnetic resonance.

70 eV

Bicyclo[4.2.2]deca-

2,4,7,9-tetraene

m/z = 115 ???

m/z = 130

WEB 3D

Visible

10

8

6

4

2

Energy (kcal/mol)

Frequency,ν (Hz)

Wavelength, λ (cm)

Cosmic

rays

Rays X Rays Ultraviolet Infrared Microwaves

Radio waves

(NMR)

> 300 300–30 30 –1 ~10

4

~10

6

FIGURE 15.16 The electromagnetic

spectrum.

FIGURE 15.15 Bicyclo[4.2.2]deca-

2,4,7,9-tetraene apparently loses

a methyl group in the mass

spectrometer. How?

We begin our discussion of IR spectroscopy by recalling certain properties of light

and energy (p. 526). The wavelength of light, λ, is the length of a wave as measured

from peak to peak or from trough to trough (Fig. 15.16). In IR spectroscopy we will

express wavelength in centimeters (cm).Wavelength is related to frequency as shown

in Eq. (15.1). The frequency (ν) is the number of cycles (or waves) passing a given

point per unit time, usually seconds. So, the units of frequency (ν) are cycles per

second (usually written as reciprocal seconds, s

1

), or more simply, hertz (Hz).

the speed of light in

centimeters per second (cm/s)

(15.1)

The quantity is called the wavenumber and its units are reciprocal cen-

timeters (cm

1

). Physically, is the number of wavelengths contained in one cen-

timeter (cm

1

) and is related to the more familiar frequency, ν,by Eq. (15.2). Signals

from IR spectra are reported in wavenumbers (cm

1

(15.2)

Absorption of IR light involves energies of roughly 1–10 kcal/mol, and is

associated with molecular vibrations. A chemical bond can be approximated by a

picture of two objects connected by a vibrating spring. Hooke’s law, which is shown

in Eq. (15.3), describes this situation.

(15.3)

The frequency of vibration, ν, is related by a constant k to the masses of the

objects (m

and m

) and another constant,f, which is the force constant of the spring.

Chemical bonds do behave rather like classical springs. So, like a real spring, the

frequency of vibration in a bond depends on the masses of the two atoms involved

and the strength of the bond between them. Bond strength turns out to be the

dominant feature, although the masses of the atoms do make a difference. We

would expect strong bonds, with higher force constants, to absorb at higher

wavenumbers. Carbon–carbon single bonds ( 90 kcal/mol, 1450 cm

1

), double

bonds ( 168 kcal/mol, 1650 cm

1

),and triple bonds ( 230 kcal/mol, 2180 cm

1

)

illustrate this point well.

Not all molecular vibrations can be detected.For a vibration to be infrared active,

it must produce a net change in the dipole moment of the molecule, which means

that symmetrical vibrations are either weak or invisible in the IR region of the

electromagnetic spectrum.For example, ethylene shows no IR signal for the carbon-

carbon double-bond stretch. However, Raman spectroscopy, which is related to IR

spectroscopy, does detect symmetrical vibrations, and shows a signal at 1623 cm

1

for ethylene. Signals in IR and Raman are usually broad and actually extend over

several wavenumbers. As a result the signal is often referred to as a band.

15.4a The IR Spectrometer In practice, an IR spectrum is obtained by pass-

ing IR radiation (light) through a sample. If no light is absorbed at a given wave-

length, transmission is 100%. If light is absorbed by the sample at a certain

wavelength, transmission is less than 100%. A plot of absorption versus wavenum-

ber (cm

1

) is thus generated and constitutes the IR spectrum of the sample. Spectra

of gases are measured in cells with nonabsorbing windows. For liquids, spectra can

be obtained either as pure samples or in solution.Pure samples are usually described

as “neat.”To measure the spectrum of a solid, the sample is usually compressed into

''''

ν = k2f (m

+ m

)/m

ν =

ν = cν

1/λ = ν

c =ν =

708 CHAPTER 15 Analytical Chemistry: Spectroscopy