12.6 Introduction to Mass Spectrometry

Home , Bromomethane, Mass spectrum, Tetramethylsilane

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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 wavenumber. Modern IR spectrometers, called Fourier-transform infrared spectrometers (FTIR spectrometers), can provide an IR spectrum in a few seconds. (See Further Exploration 12.2.) The Further Exploration 12.2 FTIR Spectroscopy 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 infrared-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,acompoundisvaporizedinavacuum and bombarded with an electron beam of high energy—typically, 70 eV (electron-volts) (more 1 than 6700 kJ mol_ ). 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 C H bonds.

1 1 L H H

HHC e_ HHC| 2e_ (12.15) 3 H1 3 + 3 H1 8 + The product of this reaction is sometimes abbreviated as follows:

HC| H abbreviated as CH4| 3 H1 8 8

The symbol | means that the molecule is both a radical (a species with an unpaired electron) and a cation—a8 radical cation. The species CH4| is called the methane radical cation. 8 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 559

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.

(12.16) CH4| |CH3 H mass 8 16 + 8 = methyl cation mass 15 = 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: H H

HH"C|H H "C| (12.17) L 8 L + 8 "H "H 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.

(12.18) CH4| CH3 H| mass 8 168 + mass 1 = methyl radical = 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.)

|CH3 |CH2 H (12.19a) mass8 14+ 8 = |CH2 |CH H (12.19b) 8 mass 13+ 8 = |CH C| H (12.19c) mass 8 12+ 8 = 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 reactions. 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: CH4| |CH3, |CH2, |CH, C|, and H|. In the mass spectrometer, the fragment ions are separated 8according8 to their mass-to-charge8 ratio, mÜz (m mass, z the charge of the fragment). Because most ions formed in the electron-impact mass= spectrometer= have unit charge, the mÜz 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 abundance) as a function of the ionic mass (or mÜz). When the ions are produced by electron impact, 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 spectrum of methane shows peaks at mÜz 16, 15, 14, 13, 12, and 1, corresponding to the various = 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 560

560 CHAPTER 12 • INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY

100 m/z = 16 relative m/z abundance 80 1 3.36 12 2.80 60 13 8.09 14 16.10 40 15 85.90 16 100.00 (base peak) 20 relative abundance 17 1.17 0 0 10 20 mass-to-charge ratio m/z

Figure 12.14 The EI mass spectrum of methane. Can you explain why there is an ion at mÜz 17? (See Sec. 12.6B for the answer.) =

ionic species that are produced from methane by electron ejection and fragmentation, as shown in Eqs. 12.16–12.19. The mass spectrum can be determined for any molecule that can be vaporized in a high vacuum, and this includes most organic compounds. (Other techniques for vaporizing less volatile molecules have been developed and are discussed brieﬂy in Sec. 12.6E.) Mass spectrometry is used for three purposes: (a) to determine the molecular mass of an unknown compound, (b) to determine the structure (or a partial structure) of an unknown compound by an analysis of the fragment ions in the spectrum, and (c) to conﬁrm the structures of compounds with known or suspected structures. The ion derived from electron ejection before any fragmentation takes place is known as the molecular ion, abbreviated M. The molecular ion occurs at an mÜzvalueequaltothemolecular mass of the sample molecule.Thus,inthemassspectrumofmethane,themolecularionoccursat mÜz 16. In the mass spectrum of decane (see Fig. 12.15), the molecular ion occurs at mÜz 142.= Except for peaks due to isotopes, discussed in the next section, the molecular ion peak is the= peak of highest mÜz in any ordinary mass spectrum. The base peak is the ion of greatest relative abundance in the mass spectrum—that is, the ion with the largest peak. The base peak is arbitrarily assigned a relative abundance of 100%, and the other peaks in the mass spectrum are scaled relative to it. In the mass spectrum of methane, the base peak is the same as the molecular ion, but in the mass spectrum of decane, the base peak occurs at mÜz 43. In the decane spectrum and in most others, the molecular ion and the base peak are different.=

B. Isotopic Peaks Look again at the mass spectrum of methane in Fig. 12.14. This mass spectrum shows a small but real peak at mÜz 17, a mass that is one unit higher than the molecular mass. This peak is called an M 1 peak,= because it occurs one mass unit higher than the molecular ion (M). This ion occurs+ because chemically pure methane is really a mixture of compounds containing the various isotopes of carbon and hydrogen.

12 13 12 methane CH4, CH4, CDH3, and so on = mÜz 16 17 17 = The isotopes of several elements and their natural abundances are given in Table 12.3 on p. 562. 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 561

12.6 INTRODUCTION TO MASS SPECTROMETRY 561

100 base peak 43 80 57 CH3(CH2)8CH3 60 decane

20 71 relative abundance molecular ion (M) 85 142 0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 mass-to-charge ratio m/z

Figure 12.15 The EI mass spectrum of decane.

13 12 Possible sources of the mÜz 17 peak for methane are CH4 and CDH3. 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- 13 12 tope involved. The abundance of CH4 methane relative to that of CH4 methane is then given by the following equation:

abundance of 13C peak relative abundance ᎏᎏᎏ12 (12.20a) = ΂΃abundance of C peak

natural abundance of 13C (number of carbons) ᎏᎏᎏ12 = X ΂΃natural abundance of C 0.0110 (number of carbons) ᎏᎏ = X ΂0.9890΃ (number of carbons) 0.0111 (12.20b) = X 13 Because methane has only one carbon, the mÜz 17 (M 1) peak due to CH4 is about 1.1% of the mÜz 16, or M, peak. A similar calculation= can be+ made for deuterium. = natural abundance of 2H relative abundance (number of hydrogens) ᎏᎏᎏ1 (12.21) = X ΂΃natural abundance of H 0.00015 (4) ᎏᎏ 0.0006 = X ΂0.99985΃ =

Thus, the CDH3 naturally present in methane contributes 0.06% to the isotopic peak. Because the contribution of deuterium is so small, 13C is the major isotopic contributor to the M 1 peak. (We’ll ignore contributions of 2H 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 13C. 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 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 562

562 CHAPTER 12 • INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY

TABLE 12.3 Exact Masses and Isotopic Abundances of Several Isotopes Important in Mass Spectrometry

Element Isotope Exact mass Abundance, %

hydrogen 1H1.00782599.985 2H* 2.0140 0.015 carbon 12C12.000098.90 13C13.003351.10 nitrogen 14N14.0030799.63 15N15.000110.37 Oxygen 16O15.9949199.759 17O16.999130.037 18O17.999160.204

ﬂuorine 19F18.99840100. silicon 28Si 27.97693 92.21 29Si 28.97649 4.67 30Si 29.97377 3.10

phosphorus 31P30.97376100.

sulfur 32S31.9720795.0 33S32.971460.75 34S33.967874.22

chlorine 35Cl 34.96885 75.77 37Cl 36.96590 24.23

bromine 79Br 78.91834 50.69 81Br 80.91629 49.31

iodine 127I 126.90447 100.

*2H is commonly known as deuterium, abbreviated D.

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; sulfur 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 79Br and 81Br. 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 mÜz 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 13C present naturally in bromomethane. 79 For example, the mÜz 95 peak corresponds to bromomethane containing only Br and one 13 81 13 C. The mÜz 97 peak= arises from methyl bromide that contains only Br and one C. Although isotopes= such as 13C and 18O are normally present in small amounts in organic compounds, it is possible to synthesize compounds that are selectively enriched with these and 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 563

12.6 INTRODUCTION TO MASS SPECTROMETRY 563

100 94 96 isotopic molecular ions 80

60 15 40

20 relative abundance

0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 mass-to-charge ratio m/z

Figure 12.16 The EI mass spectrum of bromomethane. The two molecular ions at mÜz 94 and at mÜz 96 have nearly equal abundance and result from the presence of the two isotopes 79Br and 81Br.= =

other isotopes. Isotopes are especially useful because they provide specific labels at particular atoms without significantly changing their chemical properties. One use of such compounds is to determine the fate of specific atoms in deciding between two mechanisms. Another use is to provide nonradioactive isotopes for biological metabolic studies (studies that deal with the fates of chemical compounds when they react in biological systems). When a compound has been isotopically enriched, isotopic peaks are much larger than normal. Mass spectrometry is used to measure quantitatively the amount of such isotopes present in labeled compounds.

PROBLEMS 12.15 The mass spectrum of tetramethylsilane, (CH3)4Si, has a base peak at mÜz 73. Calculate the relative abundances of the isotopic peaks at mÜz 74 and 75. = 12.16 From the information in Table 12.3, predict the appearance= of the molecular ion peak(s) in the mass spectrum of chloromethane. (Assume that the molecular ion is the base peak.)

C. Fragmentation In EI mass spectrometry, the molecular ion is formed by loss of an electron. If this ion is stable, it decomposes slowly and is detected by the mass spectrometer as a peak of large relative abundance. If this ion is less stable, it decomposes, sometimes completely, into smaller pieces. Two cases of such fragmentation are most commonly observed, and in each case, two products are formed. 1. One fragmentation product can be a radical, in which case the other product is a cation with no unpaired electrons (an even-electron ion). 2. One fragmentation product can be a neutral molecule, in which case the other product must be, like the molecular ion, a radical cation (an odd-electron ion). In either case, the cation is called a fragment ion. Only the ion is detected in the mass spectrum; the radical (case 1) or neutral molecule (case 2) is not detected. As an example of case 1, consider the mÜz 57 ion in the mass spectrum of decane (Fig. 12.15). This is formed in the following way. One= of several possible molecular ions is formed by ejection of an electron from a carbon–carbon bond. 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 564

564 CHAPTER 12 • INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY

– –e . CH3CH2CH2CH2 CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2 CH2CH2CH2CH2CH2CH3 (12.22) decane molecular ion of decane (a radical cation, m/z = 142)

Next, the molecule splits at the site of electron ejection to give a carbocation with mÜz 57 and a radical with mass 85. In this fragmentation, only the cation is detected. = = . . CH3CH2CH2CH2 CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2 + CH2CH2CH2CH2CH2CH3 (12.23) molecular ion of decane a cation a radical (a radical cation, m/z = 142) m/z = 57 (not detected by the (detected by the mass spectrometer) mass spectrometer)

Notice that there is also a peak in Fig. 12.15 at mÜz 85. This does not arise from the radical, but rather from fragmentation of the same bond in= the opposite manner to give the carbocation with mÜz 85 and the radical with mass 57. = = . . CH3CH2CH2CH2 CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2 + CH2CH2CH2CH2CH2CH3 (12.24)

molecular ion of decane a radical a cation (not detected by the m/z = 85 mass spectrometer) (detected by the mass spectrometer) 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.)

...... OH .OH –e– CH3(CH2)4CH2 CH2 CH3(CH2)4CH2 CH2 (12.25) molecular ion of 1-heptanol m/z = 116 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.6

1 1

H OH| H OH| L CH3(CH2)4 "CH "CH2 CH3(CH2)4 CH "CH2

LL LL6

molecular ion of 1-heptanol

.. CH3(CH2)4 CH8 CH| 2 + H OH.. (12.26) LL L odd-electron ion, m/z 98 a neutral molecule (detected by the mass spectrometer)= (not detected by the mass spectrometer) 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_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 565

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 appear 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 relatively 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 12.4 The base peak in the mass spectrum of 2,2,5,5-tetramethylhexane (molecular mass 142) is at = mÜz 57, which corresponds to a composition C4H9. (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:

(CH3)3C CH2CH2 C(CH3)3. L L (a) A fragment with the composition C4H9 could be a tert-butyl cation formed by splitting the compound at the bond to either of the tert-butyl groups:

C4H9 CH3 CH3

H3C "C CH2 CH2 "C CH3 LL L L L "CH3 "CH3 (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 relatively high abundance. (c) To form this cation, one electron is ejected from the C C bond, and the compound fragments so that the unpaired electron remains on the methyleneL carbon (see Eqs. 12.23–12.24):

CH3 CH3 CH3 CH3 electron ejection H3C "C CH2 CH2 "C CH3 H3C "C CH2 CH2|"C CH3 LL L 3 L LL L 8 L "CH3 "CH3 "CH3 "CH3 molecular mass = 142 m/z = 142; molecular ion

fragmentation

CH3 CH3 H C "C CH CH "C CH (12.27) 3 2 8 2 | 3 LL L + L "CH3 "CH3 m/z 57 = Fragmentation might have occurred at the same bond so that the unpaired electron remains associated with the tert-butyl group and a primary carbocation with mÜz 85 is formed. (In other words, a more stable free radical and a less stable carbocation would be= formed.) There is no peak at mÜz 85. That this mode of fragmentation is not observed demonstrates that carbocation stability is= more important than free-radical stability in determining fragmentation patterns. 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 566

566 CHAPTER 12 • INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY

PROBLEMS 12.17 The peak of highest mass in the EI mass spectrum of 2,2,5,5-tetramethylhexane (the molecule discussed in Study Problem 12.4) occurs at mÜz 71 and has about 33% relative abundance. (a) In a structure of the molecule, indicate the bond= at which fragmentation occurs to give this ion. (b) Give a mechanism for this fragmentation. (c) What is the structure of the fragment ion at mÜz 71? (Hint: Apply what you know about carbocations.) = 12.18 Indicate whether the following peaks in the mass spectrum of 1-heptanol are odd-electron or even-electron ions. (Don’t attempt to give their structures.) (a) mÜz 83 (b) mÜz 70 (c) mÜz 56 (d) mÜz 41 = = = = 12.19 The mass spectrum of 2-chlorobutane shows large and almost equally intense peaks at mÜz 57 and mÜz 56. (a)= Classify each= peak as an even-electron or odd-electron ion. (b) What stable neutral molecule can be lost to give the odd-electron ion? (c) Suggest a mechanism for the origin of each fragment ion.

D. The Molecular Ion. Chemical-Ionization Mass Spectra The molecular ion is the most important peak in the mass spectrum for two reasons. First, the mÜz of the molecular ion occurs at the molecular mass, and one of the most important uses of mass spectrometry is the determination of molecular mass. Second, the mass of the molecular ion is the basis for the calculation of losses due to fragmentation. Unfortunately, a peak due to the molecular ion is weak or absent in some mass spectra. Consider, for example, the EI mass spectrum of di-sec-butyl ether shown in Fig. 12.17a. The molecular mass of this ether is 130. A peak at this mass, however, is essentially absent. The three most prominent peaks in the EI mass spectrum of di-sec-butyl ether occur at mÜz 101, mÜz 57, and mÜz 45 (base peak). The mÜz 101 peak correspond to a loss of 29= mass units= (that is, an ethyl= group), and occurs in the= following way. First, the molecular ion is

formed by loss of an electron from the oxygen unshared pair:

.. –e– .

.. CH3CH2CH O.. CHCH2CH3 CH3CH2CH O CHCH2CH3 (12.28)

CH3 CH3 CH3 CH3 di-sec-butyl ether molecular ion of di-sec-butyl ether (m/z = 130)

(Remember from the discussion of Eq. 12.25 that unshared electrons are more easily removed than bonding electrons.) Next, an ethyl radical is lost by a process that mass spectrometrists call a-cleavage and free-radical chemists call b-scission:

a-cleavage (b-scission)

. .

.. CH3CH2 CH O.. CHCH2CH3 CH3CH2 + CH O CHCH2CH3 (12.29)

CH3 CH3 CH3 CH3 molecular ion of di-sec-butyl ether m/z = 101 (m/z = 130)

This ion reacts further by a-elimination of 2-butene (56 mass units) to give the base peak at mÜz 45. = 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 567

12.6 INTRODUCTION TO MASS SPECTROMETRY 567

100 base peak 45 CH3CH2CH O CHCH2CH3 80 57 CH3 CH3 60 di-sec-butyl ether

40 101 almost no 20

relative abundance molecular ion at 130

0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 mass-to-charge ratio m/z (a) EI mass spectrum

100 75 base peak 131 80 (M + 1) 73 60 101

40 115 20 relative abundance

0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 mass-to-charge ratio m/z

(b) CI mass spectrum

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 mÜz 130 is essentially absent in the EI spectrum, whereas the molecular ion (as the protonated ether at mÜz 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.

b-elimination

.. CH3CH O.. CHCH3 CH3CH OH + CHCH3 (12.30)

CHCH3 m/z = 45 CHCH3 H m/z = 101

The mÜz 57 peak is formed by a process called inductive cleavage, which is nothing more than the= radical-cation version of an SN1-like dissociation:

inductive cleavage

. ..

.. . CH3CH2CH O CHCH2CH3 CH3CH2CH + O.. CHCH2CH3 (12.31)

CH3 CH3 CH3 CH3 molecular ion of di-sec-butyl ether (m/z = 130) 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 568

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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):

.. CH3CH2CH O.. CHCH2CH3 + H CH3CH2CH O CHCH2CH3 (12.32) (a gas-phase CH3 CH3 proton source) CH3 CH3 di-sec-butyl ether conjugate acid of di-sec-butyl ether m/z = 131

This conjugate-acid cation is an even-electron ion; it is not a radical cation. In a CI mass spectrum, 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 mÜz 131, which is also the base peak. The relatively small number of fragments+ come from the= loss of various neutral molecules from this ion. For example, the largest fragment peak at mÜz 75 arises from loss of 2-butene in a b-elimination process analogous to the one in Eq. 12.30:=

H H

.. CH3CH2CH O.. CHCH3 CH3CH2CH O H + CHCH3 (12.33)

CH3 CHCH3 CH3 CHCH3 H m/z = 75 conjugate acid of di-sec-butyl ether m/z = 131

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. 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 569

12.6 INTRODUCTION TO MASS SPECTROMETRY 569

PROBLEMS 12.20 (a) The EI mass spectrum of CH3OCH2CH(CH3)2 (methyl isobutyl ether) contains only a trace of a molecular ion and a base peak at mÜz 45, which arises from a-cleavage. Show the a-cleavage process and give the structure= of the ion with mass 45. (b) The mass spectrum of methyl isobutyl ether does not show a peak due to inductive= cleavage, in contrast to the mass spectrum of di-sec-butyl ether (Eq. 12.31). Use what you know about carbocation stability to explain the absence of this peak. (c) What major difference(s) would you expect to ﬁnd when comparing the CI mass spectrum of methyl isobutyl ether with its EI spectrum? 12.21 Show the elimination reactions that account for each of the following fragments in the CI mass spectrum of di-sec-butyl ether (Fig. 12.17b). (Hint: b-Elimination reactions can also form CAO double bonds.) (a) mÜz 101 (b) mÜz 115 (c) mÜz 73 = = =

E. The Mass Spectrometer A mass spectrometer must produce gas-phase ions, sort them by mass, and detect the relative number of ions of each mass. We’ve already learned about two ways of producing ions: electron impact and chemical ionization. Ion sorting in a conventional “magnetic-sector” mass Further Exploration 12.3 spectrometer depends on the fact that the paths of moving ions are bent by a magnetic field. The Mass Spectrometer When subjected to a magnetic field, ions with large mÜz traverse a path of larger radius than ions of smaller mÜz (Fig. 12.18, p. 570). Another type of ion sorting is called “time-of-flight.” A time-of-flight spectrometer differentiates ions by the amount of time it takes them to move through an electric field. Ions of smaller mÜz are accelerated more easily by the electric field, and thus take less time to move through the field, than ions of larger mÜz. Regardless of the ion-sorting method, ions are detected as an ion current. The mass spectra in this text are actu- ally plots of relative ion currents versus mÜz, with the largest ion current (that is, the base peak) assigned a relative value of 100. A modern mass spectrometer is an extremely sensitive instrument and can readily pro- 6 duce a mass spectrum from amounts of material in the range of micrograms (10_ g) to 12 picograms (10_ g). For this reason, the instrument is very useful for the analysis of materi- als available in only trace quantities. It has played a key role in such projects as the analysis of drug levels in blood serum and the elucidation of the structures of insect pheromones (Sec. 14.9) that are available only in minuscule amounts. It is also an important tool in the modern forensics laboratory (“crime lab”). One of the operating characteristics of a mass spectrometer is its resolution—how well it separates ions of different mass. A relatively simple mass spectrometer readily distinguishes, over a total mÜz range of several hundred, ions that differ in mass by one unit. More complex mass spectrometers, called high-resolution mass spectrometers, can resolve ions that are separated in mass by only a few thousandths of a mass unit. Why is such high resolution useful? Suppose an unknown compound has a molecular ion at mÜz 124. Two possible formulas for this ion are C8H12O and C9H16. Both formulas have the same= nominal mass (that is, the same mass to the nearest whole number). However, if the exact mass (the mass to four or more decimal places) is calculated for each formula (using the values of the most abundant isotopes in Table 12.2), then different results are obtained:

C8H12O, exact mass: 124.0888

C9H16, exact mass: 124.1252 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 570

570 CHAPTER 12 • INTRODUCTION TO SPECTROSCOPY. INFRARED SPECTROSCOPY AND MASS SPECTROMETRY

ion source (ions are formed and accelerated here)

to vacuum pump

magnet magnetic field direction B

lower-mass ions

the separated ion beam

ion collector

higher-mass ions analyzer tube

ion exit slit

Figure 12.18 Diagram of a magnetic-sector mass spectrometer.After ionization of the sample by electron bombardment, 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.

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 interest) as well as the elemental analysis of each fragment in the mass spectrum! Because a modern 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 spectrometry. In another technique, nicknamed ESI (electrospray ionization), a solution of the analyte 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 12_BRCLoudon_pgs4-4.qxd 11/26/08 9:01 AM Page 571

ADDITIONAL PROBLEMS 571

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.

KEY IDEAS IN CHAPTER 12

I Spectroscopy deals with the interaction of matter and the CAC stretching absorption are very useful for the electromagnetic radiation. Electromagnetic radiation identiﬁcation of alkenes. The O H stretching ab- is characterized by its energy, wavelength, and fre- sorption is diagnostic for alcohols.L quency, which are interrelated by Eq. 12.3. I In electron-impact (EI) mass spectrometry, a molecule I Infrared spectroscopy deals with the absorption of in- loses an electron to form the molecular ion, a radical frared radiation by molecular vibrations. An infrared cation, which in most cases decomposes to fragment spectrum is a plot of the infrared radiation transmit- ions.The relative abundances of the fragment ions are ted through a sample as a function of the wavenum- recorded as a function of their mass-to-charge ratios ber or wavelength of the radiation. mÜz,which,for most ions,equal their masses.Both molecular masses and partial structures can be de- I The frequency of an absorption in the infrared spec- rived from the masses of these ionic fragments. trum is equal to the frequency of the bond vibration involved in the absorption. I Associated with each peak in a mass spectrum are other peaks at higher mass that arise from the pres- I The wavenumber or frequency of an absorption is ence of isotopes at their natural abundance. Such iso- greater for vibrations involving stronger bonds and topic peaks are particularly useful for diagnosing the smaller atomic masses (Eqs. 12.10 and 12.13). The presence of elements that consist of more than one smaller of two atomic masses involved in a bond vi- isotope with high natural abundance,such as chlorine bration has the greater effect on the frequency of the and bromine. vibration. I Ionic fragments are of two types: even-electron ions, I The intensity of an absorption increases with the which contain no unpaired electrons; and odd-elec- number of absorbing groups in the sample and the tron ions, which contain an unpaired electron. size of the dipole moment change that occurs in the molecule when the vibration occurs. Absorptions I In chemical-ionization (CI) mass spectrometry, mole- that result in no dipole moment change are infrared- cules are ionized by direct protonation in the gas inactive. phase. Because this is a much gentler ionization technique than EI, a CI mass spectrum typically contains a I The infrared spectrum provides information about greater proportion of molecular ion (as its conjugate the functional groups present in a molecule. The acid) than the EI spectrum of the same compound. AC H stretching and bending absorptions and L

ADDITIONAL PROBLEMS

12.22 List the factors that determine the wavenumber of an changes in the infrared spectrum that could be used to infrared absorption. indicate whether the reaction has proceeded as indi- cated? (Your answer can include disappearance as well 12.23 List two factors that determine the intensity of an in- as appearance of IR absorptions.) frared absorption. (a) 1-methylcyclohexene methylcyclohexane LT 12.24 Indicate how you would carry out each of the follow- (b) 1-hexanol 1-methoxyhexane LT ing chemical transformations. What are some of the