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12.6 INTRODUCTION TO MASS SPECTROMETRY, Study notes of Optics

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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 Fou r i er - t ra n s fo r m i nf r a re d s p ec t ro m e te r s (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 (CHCl3),
its isotopic analog (CDCl3), and methylene chloride (CH2Cl2) 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 CH48
| is called the methane radical cation.
3
1
1
C|
H H abbreviated as
H
H
8|
8
CH4
33 1
1
CHH
e_
H
H
+
3
1
1
C|
HH
2e_
H
H
+
8
Further Exploration 12.2
FTIR Spectroscopy
12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 558
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe

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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

..

  • e

.

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