tables of common ions and common losses

Tables of Common Ions and Common Losses

•  The empirical rules developed so far, and many more besides, are summarized in the two Tables 5.2 and 5.3.

•  Use Table 5.2 to discover the type of compound that is likely to exhibit a large peak at m/z 91.

•  Like Table 5.3, Table 5.2 is not definite about structural assignments because such tables cannot show every possibility. The stress is on possible inferences.

•  Looking up m/z 91 in the first column of Table 5.2 leads to the likelihood that C7H7

+ from a benzyl compound (C6H5CH2X) or substituted toluene (CH3C6H4X) is responsible for a signal at m/z 91.

•  In fact, the formation of C7H7+ from benzyl compounds is one of the most

characteristic processes in all of mass spectrometry, i.e. almost all benzyl compounds exhibit large (often base) peaks at m/z 91, and a peak at m/z 91 is almost always associated with C6H5CH2X or CH3C6H4X compounds. The prominence of C7H7

+ is due to its great resonance stabilization:


•  An unknown compound afforded peaks at [M - 28]+•, [M - 45]+, [M - 46]+• and [M - 73]+. Using Table 5.3,assign the substance to a possible class of compounds.


The substance would appear to be an ethyl ester. Whilst loss of 28 mass units would suggest a quinone, phenol, aromatic ethyl ether, propyl ketone or ethyl ester according to Table 5.3, the other peaks are more diagnostics. A carboxylic acid or an ethyl ester would normally afford [M - 45]+ ions, and an ethyl ester or an aromatic nitro compound would give [M - 46]+• ions. A peak for [M - 73]+

ions is listed as being characteristic of an ethyl ester alone. The first three peaks individually can be associated with a number of structural types but, taken together, they are alI indicative of RCOOC2H5 It would be easy to be misled by looking up just one of the first three peaks, so always cross-check observations and associated inferences for common structural assignments.

AN APPROACH TO INTERPRETING ELECTRON IONIZATION MASS SPECTRA

•  You now know quite a lot about the behavior of compounds in an EI mass spectrometer, but you need to consider how best to interpret the mass spectrum of an unknown compound.

•  There is no prescribed procedure that will always work, so in what follows only general hints are given on how to tackle the job – not sacrosanct rules.

•  In the examples below, you will not meet any more fragmentations that are new to you but you will be introduced to some key aspects of mass spectrometry, for example, isotope recognition.


•  How do we begin to extract information from a mass spectrum such as that of compound A shown in the Figure?

•  The first step is usually to assign the molecular ion. To be due to M+• ions, a peak must be of highest mass in the spectrum. Looking at the spectrum, a likely candidate for M+• occurs at m/z 60.

•  Molecular ions of organic compounds rarely eject neutral species with masses in the range 5-14, but frequently eject 1 (H•) and 15 (CH3

•) mass units.


•  Our candidate at m/z 60 appears to pass these criteria; it ejects H• and CH3

• to give peaks at m/z 59 and 45 (though the latter is small) and does not afford significant peaks at m/z 46-55.


•  Given that the molecular ion has even mass, comment on the nitrogen content of compound A.


The compound contains either no nitrogen or an even number of nitrogen atoms.


•  With such a small molecule as A (Mw = 60), there cannot be long fragmentation pathways so it is not worth worrying about which peaks are due to primary and which to secondary fragmentation. We should simply note that the major peaks are at m/z 31, 42 ([M - 18]+•; an even-mass, rearrangement ion) and 59 (loss of H•).

•  This behavior, particularly formation of an ion at m/z 31, is highly characteristic of a certain class of compounds. Which?


α-Cleavage of primary alcohols leads to [M - H]+ ions that are usually more abundant than M+• ions, and to CH2OH+ ions at m/z 31.


•  Assuming that we are right, compound A has the partial structure RCH2OH where R must have a mass of 29.

•  Let's now think about the rearrangement ion at m/z 42. For its presence to be consistent with RCH2OH, the alcohol must have a γ- or δ-CH unit for expulsion of H2O to occur.

•  Given that R is 29 mass units, propan-1-ol is the clear favorite.


•  Give two reasons why compound A cannot be propan-2-ol.

α-Cleavage of [CH3CH(OH)CH3]+• would afford a large peak at m/z = 45 (ie. loss of CH3

•) and there is no γ-CH unit so H2O cannot be eliminated

ISOTOPE PATTERNS •  Compound B gives the EI mass spectrum shown in the Figure •  Again, we start by trying to assign the M+• ion. You may regard the

peak at m/z 114 as a candidate or, if you look very closely, the tiny peak at m/z 115. In fact, all four peaks at m/z 112-115 are molecular ions! But don't panic

ISOTOPE PATTERNS •  There's a simple explanation and it concerns isotope

patterns. As a mass spectrometer separates ions on the basis of their m/z values, different isotopes of an element will be detected separately according to their different masses.

ISOTOPE PATTERNS

•  The M+• will not contain satellite peaks if the sample is composed only by elements that show one isotope, like 19F, 127I, 31P and 1H.

•  CH3CH2Cl shows two peaks at m/z 64 and 66, due to CH3CH235Cl and

CH3CH237Cl respectively, with relative intensities 100:32.

•  The two isotopomers CH3CH279Br and CH3CH2

81Br will appear as two peaks of equal intensities at m/z 108 and 110.

ISOTOPE PATTERNS

•  Benzene mass spectrum contains, besides the peak at 78 (100%) due to 12C6H6, a second peak at 79 due to 12C5

13C1H6, the isotopomers in which one of the 6 carbon atoms is constituted by the 13C isotope.

•  As the natural abundance of the latter is 1.11%, the peak at m/z 79 will be 6x1.11%=6,66%.

•  This example shows that also isotopes with low abundance can give peaks M+1 and M+2 of significant intensities if the corresponding atoms are present in a high number.

ISOTOPE PATTERNS

•  The height of the additional peaks can be calculated, in the case of molecules constituted only by elements showing one single isotope, like C, N, O, Si, S, using the equations:

ISOTOPE PATTERNS

( ) ( ) ( ) ( )

( ) ( ) ( )36.346.420.01002

79.006.537.01.11001

×+×+×=×+

×+×+×+×=×+

SiSO

SSiNC

nnnMM

nnnnMM

•  For molecular ions containing elements that show two or more isotopes occurring in nature at comparable quantities (B, Cl, Br, etc), we cannot exclude that a single molecule will contain contemporary two less abundant isotopes.

•  Suppose that you have two chlorines in your molecule. The following table shows the relative abundance of the different isotopomers:

ISOTOPE PATTERNS

Isotopomer Relative abundance

mass

35Cl35Cl 1x1 70(M) 35Cl37Cl 1x0.32 72(M+2) 37Cl35Cl 0.32x1 72(M+2) 37Cl37Cl 0.32x0.32 74(M+4)

•  In general, we can use the following formula:

where a and b are the relative abundance of the two isotopes of m/z 35 and 37, and n is the number of chlorine atoms

•  If the molecule contains both Cl and Br, the formula becomes:

where c and d are the isotopical abundance and m is the number of atoms of the second element.

ISOTOPE PATTERNS

( )3223

22

33 3)n(for

2 2)n(for

babbaa

bababa n

+++=

++=+

( ) ( )mn dcba ++

ISOTOPE PATTERNS

ISOTOPE PATTERNS •  Let’s go back to our spectrum. Can you make the pattern or peaks in

the m/z 112-115 region match any of the ratios given in the table? •  Almost. The peaks at m/z 112 and 114 are in a ratio of 3 : 1 and are

two mass units apart. This corresponds well with the presence of chlorine, whose isotopes occur in a 3 : 1 ratio, two mass units apart.

ISOTOPE PATTERNS •  What about the small peaks at m/z 113 and 115? •  In trying to assign these peaks, remember that we are

dealing with an organic compound; by definition carbon is present. The peaks at m/z 113 and 115 are due to the 13C content.

ISOTOPE PATTERNS

•  There is a 1% chance that a particular carbon atom in a molecule is 13C and not 12C. You will see a small isotope peak due to 13C in most of the spectra.

ISOTOPE PATTERNS •  The identity of compound B is clear once the isotope pattern is assigned to chlorine.

The largest fragment ion at m/z 77 does not contain Cl as the isotope pattern has disappeared. Indeed m/z 112 → 77 corresponds to loss of 35Cl and m/z 114 → 77 to loss of 37Cl.

•  Table 5.2 can be consulted to assign m/z 77 to the phenyl ring, so compound B is chlorobenzene.

•  The peaks at m/z 36 and 38 in a ratio of 3:1 are due to HCI+• (see Table 5.2), and phenyl ions are known to eject ethyne to give C4H3

+ at m/z 51.

ISOTOPE PATTERNS

ISOTOPE PATTERNS

C10

12C10 100%

12C9 13C1

10.8% 12C8 13C2

0.5%

C100

12C100 92.5%

13C1 100%

13C2 53.5%

13C3 18.9% 13C4

5%

13C5 1%

13C6 0.2%

ISOTOPE PATTERNS C500

2.6% 14.3%

38.6%

69.4%

93.2% 100%

89.2%

68.1%

45.4%

26.8%

14.3%

H10 100%

0.2%

ISOTOPE PATTERNS

H192 100%

3.1%

H64 100%

1%

•  Let's put our knowledge of isotope patterns, to the test straight away with compound C, the EI mass spectrum of which is shown in the Figure.

•  The main M+• ion appears to occur at m/z 78 but the size of the peak at m/z 80 is noteworthy.

ISOTOPE PATTERNS

The ratio of relative ion abundances at m/z 78 and 80 is 34%: 1.5%, which can be expressed as 95:4.2.

•  Two of the major fragmentions of compound C occur at m/z 31(CH2OH+ from a primary alcohol?) and m/z 47 (from Table 5.2, CH2SH+ from a primary thiol?).

•  It is also notable that adding the masses of these two fragments gives the mass of the major molecular ion (31 + 47 = 78). Therefore, we can propose that compound C is simply these two units joined together: HOCH2CH2SH.

ISOTOPE PATTERNS

•  At this point in an interpretation, when one or more candidates have been proposed, it is helpful to draw the candidate structure(s) and check for consistency with the spectrum by attempting to predict the behavior of the candidates.

•  Let's do it for HOCH2CH2SH. By noting that there are two sites where charge may be localized (either sulfur or oxygen), you should predict two α-cleavages and, because there is a γ-H, you should expect loss of water.

ISOTOPE PATTERNS

Incidentally, ejection of H2S from compound C might be expected in an analogous process to that for elimination of H2O. Loss of H2S is less favorable than that of H2O and the resulting ion, at m/z 44, gives only a minor peak.

Two organic compounds show the following peaks in the molecular ion region: 151 (100%), 152 (9.2%), 153 (0.7%) 155 (100%), 156 (7.8%), 157 (5.1%), 158 (0.3%) Determine the elementary composition of these two compounds

ISOTOPE PATTERNS

Probability calculation One 13C:

M +1M

% =0.0109*(0.9891)nc−1

0.9891nc*nc*100 = 0.0109

0.9891*nc*100 =1.1*nc

M +1M

% =P(13C1

12Cnc−1)P(12Cnc )

*Nobj *100

P(13C1) = 0.0109; P(12C1) = 0.9891; Nobj = nc

Probability calculation Two 13C:

Nobj=C(n,k) = [n(n-1)(n-2)...(n-k+1)] / k!

With k=2 C(nc,2)=nc*(nc-1)/2!=nc*(nc-1)/2

M + 2M

% = 0.0112 * nc*(nc−1)2

*100 = 0.006*nc*(nc−1)

M + 2M

% =P(13C2

12Cnc−2 )P(12Cnc )

*Nobj *100

M + 2M

% =0.01092 *(0.9891)nc−2

0.9891nc*Nobj *100 =

0.01090.9891⎛

⎝⎜

⎞

⎠⎟

2

*Nobj *100 = 0.0112 *Nobj *100

Accurate Mass Measurement •  Isotopes, apart from 12C, do not have integer relative atomic

masses. The calculated masses of compounds or ions with different elemental compositions may be the same if quoted to the nearest TABLE 2.2 integer.

•  However, if the masses are calculated accurately, they will be slightly different. Since magnetic sector mass spectrometers can measure mass to better than 5 parts in a million (p.p.m.), i.e. a mass of 100.0000 can be differentiated from a mass of 100.0005, it is possible to assign the likely elemental composition of an ion simply from a measurement of its accurate mass.

Accurate Mass Measurement •  Let’s see the spectrum of the following example (Unknown D). •  The obvious candidate for the molecular ion occurs at m/z 149.

There are no conspicuous isotopes present (other than 13C).

Odd number of Ns

C6H5CH2Z CH3C6H4Z

Z weights 58

Accurate Mass Measurement •  At this point in the interpretation of the low-resolution spectrum, life gets

difficult. •  There is a large peak at m/z 43 and an abundant [M - 43]+ ion at m/z 106,

but the species of mass 43 could have several compositions, particularly C3H7, HCNO or C2H3O. This is where high-resolution data really help.

Accurate Mass Measurement

•  The compositions C6H13O4 and C7H9N4 do not contain an odd number of nitrogen atoms. AIso, the former cannot possess a benzylic function as it has only six carbon atoms.

•  One of the possible compositions for the fragment ion at m/z 106 (C4H10O3) is then not tenable because it has more oxygen than the proposed originaI molecule, so the ion can be assigned to C7H8N.

•  Assign an elemental composition to the neutral species ejected in the fragmentation, m/z 149 → 106.

•  C9H11NO - C7H8N = C2H3O. The neutral particle is normally thought of as an acetyl structure, CH3CO•.

•  The ion at m/z 43 is confirmed by accurate mass measurement as being C2H30+, the correct composition for an acetyl unit.


•  It is time to summarize and switch tactics to predictions. The compound consists of the parts shown:

•  These constituents (91 + 14+ 43 + 1= 149 mass units) could be assembled either as structure 2.7 or as 2.8. The first of these seems to fit the data very well, as shown in 2.9, but the disubstituted candidate fits less well.


•  For example, generation of C6H5+ from 2.8 is difficult to envisage,

and ArNHCOCH3 compounds are known to eject 42 mass units, C2H2O, in preference to the acetyl unit (see Tables 5.1 and 5.3).

•  It is good practice, whenever possible, to consult reference spectra. In this case, you would find that the base peak in the mass spectra of all three isomers 2.8 occurs at m/z 107 due to ejection of C2H2O, and that the spectrum of compound 2.7 matches very well with the unknown spectrum.


SUMMARY AND HINTS ON INTERPRETATION OF ELECTRON IONIZATION MASS SPECTRA

1.  Try to locate the molecular ion peak, but remember that there is no guarantee that it will be Iarge or indeed present at all. An odd-mass M+• ion implies an odd number of nitrogens. Note any characteristic isotope patterns. Measure the accurate mass of the M+• ion if possibIe, and hence determine the molecular formula.

2.  Primary losses from M+• ions are most likely to yield fruitful information. 3.  Elimination of a neutral molecule from the molecular ion is necessarily a rearrangement process

and should be noted particularly, for example the ubiquitous six-centre rearrangement of unsaturated species. If a molecule has no nitrogens, rearrangement ions will be readily apparent since they will have even mass.

4.  Once characteristic features of the unknown structure have emerged, it is good practice to draw candidate structures incorporating those features. Then checks for consistency between each candidate and the spectrum should be made. This is best done by predicting the mass spectral behaviour of each possible structure, then comparing predictions with the recorded spectrum. To have confidence in the correctness of a candidate structure, it should provide a satisfactory rationale for all the major peaks in the mass spectrum; it is not necessary for every peak in the spectrum to be rationalized.

5.  In deducing the likely fragmentation behaviour of candidate structures, the charge localization concept may help to identify parts of the molecule where fragmentation is likely to occur. When invoking thermochemical arguments, do not forget the neutral species ejected. Their heats of formation are just as important as those of the ions observed in deciding whether a given fragmentation will occur or not.

6.  Refer to standard mass spectra of all candidate structures, if available, to confirm or eliminate those structures.

tables of common ions and common losses

Documents