2-MS interp

Mass spectral interpretationfor chromatographers

This material is designed to help GC/MS beginners who want to understand a few mass spectral interpretation basics.

It is not intended to be a complete treatment of mass spectral interpretation.

We’ll not be introducing any complex fragmentation or rearrangements.

Friday, August 6, 2010

What is a mass spectrum

Mass / Charge

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50

100Lets review the basiccomponents of asimple mass spectrum.


A simple mass spectrum

Mass / Charge

0

50

100Mass spectra are typicallyrepresented as a ‘bar’ typegraph. Each line is used torepresent a specific M/Z.



Mass / Charge

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50

100

X axis - Mass / Charge - (M/Z)

This format is used becausea M/Z of 28 can representa mass of 28 with a singlecharge or might be 56 witha charge of 2, .....

A single charge is the mostcommon so the X axis best equates to mass.



Mass / Charge

0

50

100

Y axis - abundance.In a standard spectrum, the ismade relative to the largest line(the base peak) so it will run from0-100%.

Absolute abundance are typicallyused in GC/MS spectra. They willrun from 0 to the highest recordedline intensity.



Mass / Charge

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50

100

Types of information Mass of fragments Presence of isotope lines Gaps between lines.

We’ll look as some of the basics



Mass / Charge

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50

100

The masses in this example should look familiar.

This is a spectrum for air - a mixture.

Nitrogen - N2

15N14N - an isotope peak. Its small because 15N is not very common in nature



Mass / Charge

0

50

100

Well we know what most of the lines are due to butwhat could 16 and 17 be from?

O2CO2

Water?

N2


Fragmentation pattern

Mass / Charge

0

50

100

18 - H2O17 - OH16 - O

Water shows fragmentationwhere it can loose 1 or 2 H.

You’ll see that fragmentationpatterns are very importantin MS interpretation


Atomic masses and isotopes

The atomic weights that we find on the periodic table represent an average for all naturally occurring isotopes.

Example Chlorine - atomic weight = 35.453 A mixture of 35Cl - 75.77%, 37Cl - 24.23%

We must not only need to account for these isotopes, we can exploit them for determining the number and types of elements that are present.


Atomic masses and isotope

To help with some of our calculations, we commonly work with relative abundances of the isotopes.

The most abundant isotope is set to 100% and the the other isotopes are normalized to it.

For most elements in organic molecules, the most abundant element is also the lightest.

Lets look at some of the more common elements.



Mass RB Mass RB Mass RB Class

H 1.0078 100 A

C 12.0000 100 13.0034 1.1 A+1

N 14.0031 100 15.0001 0.37 A+1

O 15.9949 100 17.9992 0.2 A+2

F 18.9984 100 A

S 31.9720 100 32.9715 0.8 33.9679 4.4 A+2

Cl 34.9989 100 36.9659 32.5 A+2

Br 78.9183 100 80.9163 98 A+2

RB = relative abundance


Isotope classes

The A, A+1, A+2 class system is used to indicate the types of isotopes present.

A - Only a single isotope. This may mean that the element only exists as a single isotope (like fluorine)

The abundance of all but one isotope is too small to use. (Example. deuterium - 0.0015%)


Isotope classes

A+1 Two isotopes. Examples. C and N

A+2 At least two isotopes with the highest

mass isotope being +2 from the lowest mass isotope. Examples. O, S, Cl, Br.

In organic MS, the most abundant isotopes are also the lowest mass.


General steps inmass spectral interpretation

Obtain a ‘good’ spectrum

Identify structural features

Evaluate the general appearance

Find the molecular ion if possible

Determine the elemental composition

Propose possible structures


Obtaining a ‘good’ spectrum

With GC/MS, your mass spectra will vary across a peak.

In addition, there can be artifacts resulting from the carrier gas and the column.

A few simple steps must be taken to insure that the spectrum you evaluate is a ‘good’ one.



You must have scanned over the entire mass range of the component.

At least 2 mass units higher than the ‘molecular ion’ mass.

If chlorine or bromine are present, you should scan 2 x the number of each higher.

For high MW samples, you don’t typically need to scan below 50.



Most systems have some sort of user based ‘threshold.’ Lines below this value are not stored.

You need your threshold to be low enough to have recorded less intense lines. This is typically about 0.1% relative abundance.

Remember, you molecular ion cluster is often one of the least intense line clusters you have.



Averaging and skewing of spectra.

We commonly take an average spectrum by averaging several scans across a peak.

This can cause some problems.

At the peak top, large lines may have saturated the detector.

At the start/end of a peak, smaller lines may not have exceeded the threshold.



Proper spectrum

Thresholdtoo low forsome or allof the scans

Saturationof the twolargest lines



So what can you do.

First, make sure you have at least 7 spectra being collected / peak. You need this to obtain any sort of reasonable integration anyway.

Produce a series of average spectra The first few scans The last few scans The center scans.



All scans should give similar results. If the threshold is too high, you can simply

use the ‘center’ average.

If saturation occurs, avoid the center of the peak - use one side or the other.

This test will also indicate if the peak is ‘pure.’ The presence of new major lines indicates co-elution.



Finally, GC/MS is subject to background problems that you can account for.

Air, water, other contaminates in the carrier gas.

Column bleed.

The effects of background can be minimized by calculating and subtracting an average background spectrum - near the peak.


Mass defect

Our current system of assigning atomic masses is based on 12C = 12.00000000.

As a result of this and the mass lost to binding energy, the mass of every other element is NOT a whole number.

With high resolution MS, we can use this mass defect to specifically identify the elemental composition of each line.

With unit resolution MS, this can present a problem.


Mass defect

In organic MS, we are dealing primarily with H, C, N, and O.

The most common isotopes of H and N both weigh more than their assigned ‘whole number’ atomic mass.

H - 1.0078N - 14.0031

Hydrogen presents the greatest problem because our compounds contain many.


Mass defect

Our unit resolution MS may miss assign the mass if a peak if it is > 0.5.

We can predict how many hydrogen must be present for this to cause a problem.

# Hydrogen = 0.5 / (1.0078 - 1) = 64 hydrogen

128 hydrogen would give us a +1 shift.


Mass defect

An alkane would represent the worst case so lets determine how large it must be to hit the 0.5 limit.

alkane = CNH2N+2

2N + 2 = 64 N = 31 MW = 31 * 12 + 64 = 434

So its only a real problem at high mass and most systems can correct for it.


Working with the molecular ion

One of the most powerful features of MS is the ability to determine the molecular formula from the molecular ion.

The molecular ion must: Be the highest mass ion - exclusive of

isotope related ions which also must be measurable.

Be an ‘odd electron ion’

Agree with the rest of the spectrum.



With GC/MS data, even if the molecular ion is present, it may be hard to work with.

For now, lets assume that the data is good and we already know what the molecular or parent ion is.

We’ll deal with some of the additional rules.

For now, we’ll simply show how to work with the molecular ion to determine the molecular formula.



Lets assume that you have obtained the following mass spectral data:

m/e abund m/e abund 26 3472 41 14044 27 20593 42 7575 28 48287 43 27536 29 79823 44 31440 39 14675 45 1026

Which line would be the molecular ion?



A good guess would be 44. Why? m/e 44 is too big compared to 43 m/e 45 is too small compared to 44

Also, remember that most of our elements consist of a mixture of isotopes.

Masses greater that the molecular ion are to be expected (hoped for).



Mass RB Mass RB Mass RB Class

H 1.0078 100 AC 12.0000 100 13.0034 1.1 A+1

N 14.0031 100 15.0001 0.37 A+1O 15.9949 100 17.9992 0.2 A+2F 18.9984 100 A

S 31.9720 100 32.9715 0.8 33.9679 4.4 A+2Cl 34.9989 100 36.9659 32.5 A+2

Br 78.9183 100 80.9163 98 A+2

RB = relative abundance


Normalization

The next step is to normalize your data.

Raw GC/MS spectra are typically unnormalized.

Standard spectra are normalized to the largest (base) peak which may not be the molecular ion.

There are a few simple steps that you can follow that will help.


Normalization

Select a potential molecular ion along with any lines of higher mass.

Normalize these lines so that the potential molecular ion is 100.

The error of the isotope lines is estimated as being +0.2 absolute or +10% relative.

Using these values, you can begin constructing a normalization table.


Normalization

Select a potential molecular ion m/e abund 44 31440 (potential molecular ion) 45 1026

The values can now be plugged into your normalization table.

Normalization table - a simple way to track your calculations.


Normalization table

m/e rel.abund

Normabund NF Cn-1 Cn Cn+1

A 44 31440 100 .00318

A+1 45 1026

A+2

Start by putting the masses and abundancesin the first two columns.

Then calculate your normalization factor (NF)where NF = 100 / rel. abund of A


Normalization table

Next, calculate the normalized abundances for your isotope lines - A+1 in this example, + 10% relative error

m/e rel.abund


A 44 31440 100 .00318

A+1 45 1026 3.26+.33

A+2


Normalization table

Based on the A+1 peak, there appears to be three carbons. We’ll look a 3+1 carbons.

In the next columns, calculate the mass due to C and the A+1 abundances.

m/e rel.abund


A 44 31440 100 .00318 24 36 48

A+1 45 1026 3.26+.33 2.2 3.3 4.4

A+2


Normalization table

Three carbons looks like a good choice.

Now we need to account for the remaining mass and come up with a formula.

m/e rel.abund


A 44 31440 100 .00318 24 36 48

A+1 45 1026 3.26+.33 2.2 3.3 4.4

A+2


Finishing up

Mass of parent ion 44Mass of 3 carbon 36

Remaining mass 8

A mass of 8 is pretty small. Most likely it is due to hydrogen.

Potential formula - C3H8 (propane)


Using isotopic abundances

How did I know there were three carbons?

No, I didn’t just make it up.

The key to using the A+1 and A+2 lines is the known abundances of naturally occurring isotopes.

In our simple example, all that was present was H and C so lets consider them first.


Carbon - an A+1 element

When we normalize to the molecular ion, the A+1 element can help us determine the number of carbons in our compound.

Roughly speaking, for every 100 carbons we measure, one of them will be a 13C.

When normalized to the molecular ion, this equates to an increase in the A+1 peak of 1.1 for every carbon.



As the number of carbons increase, the probability of there being a 13C also goes up. This is an additive process.

For our propane exampleH3C - CH2 - CH3 1.1

H3C - CH2 - CH3 1.1

H3C - CH2 - CH3 1.1

total 3.3

For hydrocarbons, theintensity of the A+1 line

tells us the # carbon.2H has too small of anintensity to be seen.

Other A+1 elements willalso contribute to theintensity of this line.



There is also the chance that a molecule might contain more than one 13C.

The probability of this happening is pretty rare - #C *(1.1)2 - 0.000364% for 3 C

You instrument will not be able to see this. Its less intense than the contribution from deuterium.


Nitrogen - the other A+1 element

Nitrogen will also contribute to the A+1 line intensity (15N).

A nitrogen contributes 0.37 to the A+1 line.

For organic GC/MS work, we typically need to rely on another method to determine nitrogen (Nitrogen rule).

0.37/N is pretty hard to see. Compounds containing many N are not

very volatile so are not often seen.


diethylamine

10 30 50 70

30

44

58

73

In this example, you can see that the molecular ion is odd (73) and the other major ions are even.

This is because the loss of odd mass fragments is most common.mass / charge


The A+2 elements

Several A+2 elements are seen in organic GC/MS work.

16O 100 18O 0.2 32S 100 33S 0.8 34S 4.4 35Cl 100 37Cl 32.5 79Br 100 81Br 98

18O is of too small of an abundance to be very useful in GC/MS work.

Sulfur has an A+1 and A+2 isotope so you will need to correct the A+1 line when sulfur is present.


The A+2 elements

Chlorine and bromine are both common in organic compounds.

They also produce very characteristic patterns due to the relatively high abundance of there isotopes.

Lets look at some of their patterns. You need to be able to identify them if you ever plan of finding the molecular ion of a brominated or chlorinated compound.


Chlorine

Cl Cl2 Cl3 Cl4

By the time you have 4 chlorine, the A+2 lineis significantly larger than the A line.


Bromine

Br Br2 Br3 Br4

Bromine, which is almost a 1:1 mixture of79Br and 81Br also has a distinctive pattern.


Chlorine and Bromine

The cluster patterns of polychlorinated and polybrominated species are useful in helping to identify this type of material.

Some elements, like tin can give patterns similar to polyhalogenated species. So another factor to consider is the weight of the proposed molecular ion.


Chlorine and Bromine

minimum weight of # molecular ion Cl Br 1 35 79 2 70 158 3 105 267 4 140 316 5 175 395 6 210 474

These do not includethe weights for any

other elements.

If your proposed molecular ion doesn't

have this minimumweight, then you

should re-think yourproposed assignment.


Another example

0 20 40 60 80

CH2Cl2

Dichloromethane - a common GC/MS solvent

Note the Cl2

pattern at 84and the Clpattern at 49

84

49


General appearance

Knowing the molecular formula is nice but only part of the job.

We would like to know exactly what the molecule is. This requires that we look at the rest of the data.

There are a few basic rules and methods that can be used to help interpret simpler mass spectra.

Let’s review them


General appearance

The overall appearance can often give you an idea as to what type of material you are dealing with.

Some factors to consider. Degree of fragmentation Presence of clusters General shape Odd or even major lines in clusters.


General appearance

Lets look at a few examples so you get an idea what we’re talking about.

First, lets look at some hydrocarbons Straight chain Branched Unsaturated Rings

Each factor will have an effect on the overall appearance.


Saturated hydrocarbons

This is a classic hydrocarbon pattern

Note the regular spacing of theclusters at intervals of 14 mass units.

Also, the largest line in each cluster isan odd value.

n-hexadecane

43

29

57

71

85

99



5-methylpentadecane43

57

71

85

167

57 169

85 141

When a branch is present,we see lines associate withfragmentation on either sideof the branch producinglines of higher intensity.



normal hydrocarbondistribution

Presence of a singlebranch.

Deviations from the normal distribution may indicatethe presence of a branch. As branching increases,the number of lines goes up.

As a result, this approach is only of limited use.


Unsaturated hydrocarbons

Alkenes Spectra are similar to alkanes.

Favored ionization process is ionization the the pi bond.

Results in reduced fragmentation and increased parent peak intensity.

Fragments that contained the double bond are offset by -2.



Note the increased intensityof the parent ion.

The major ions are at 41, 55, 69, 83, 97 ......

For an alkane they would be at 43, 57, 71, 85, 99, .....

11197

83

69

55

41

29

1-hexadecene



Cycloalkanes Compared to alkanes:

Reduced fragmentation is also observed.

The is an increase in the parent ion.

Reason: Fragmentation requires that at least two bonds be broken.

EIC - C - C - C - C - C+



Aromatics Aromatics are easy to ionize and produce

stable molecular ions.

This results in reduced fragmentation and abundant parent peaks.

Unfortunately, it is difficult or impossible to determine ring substitution patterns

EI+


Hydrocarbon examples

Lets look at a series of simple examples.

All will be C6 hydrocarbons

The spectra will only show the major lines to make it easier to view the changes.

First we’ll look at some linear C6 hydrocarbons. Then we’ll evaluate some cyclic ones.


n-hexane

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rela

tive

ab

un

da

nc

e

M/Z

41

27

43

57

71

86

Note the molecular ion at 86.

We have a typical hydrocarbon pattern with 57 being the most intense line.


n-hexane vs. 1-hexene

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n-hexane

0 20 40 60 80

1-hexene

Note how the molecularion for 1-hexene (at 84)is more intense.

Also, in each cluster,the CnH2n-1 and CnH2n

ions are of increasedintensity.


All three hexenes

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0 20 40 60 80

0 20 40 60 80

1-hexene

2-hexene

3-hexene

Here you can see that theis not a large differenceregardless of where the double bond is.

The tendency for the bondto migrate duringionization tends to make the spectra look quitesimilar.


1,3-hexadiene

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0 20 40 60 80

n-hexane

0 20 40 60 80

1-hexene

1,3-hexadiene

With the addition of asecond double bond, youcan see that the mass ofthe molecular ion isreduced by two to 82.

In addition, there is lessfragmentation.


cyclohexane

0 20 40 60 80

0 20 40 60 80

For cycloalkanes, we see a largeincrease in the intensity of themolecular ion.

This is because at least two bondsmust be broken to form a fragment

n-hexane

cyclohexane

+CH2-CH2-CH2-CH2-CH2-CH2

EI


Unsaturated rings

As the degree ofunsaturation increases,the relative intensity ofthe molecular ion willincrease.

Fragmentation is alsoreduced.

cyclohexene

1,3-cyclohexadiene

benzene

0 20 40 60 80

0 20 40 60 80

0 20 40 60 80


Aromatic systems

Aromatic species are easy to ionize and tend to produce stable molecular ions.

In the case of benzene, you have three double bonds and one ring for 4 degrees of unsaturation.

With the exception of some ortho isomers, most substitution patterns can not be distinguished.

+EI


Some other helpful tools

Some simple tools can be used to help interpret your spectra.

The odd electron rules. Used to confirm if a spectral line could be due to a molecular ion.

The nitrogen rule. Helpful if an odd number of N are present.

Logical losses. Do the other lines make sense based on your molecular ion?


Odd-electron ions

Ionization of our sample occurs by the loss of an electron.

The species we form is a ‘radical’

M M

Our original molecule had an even number of electrons. A molecular ion will have an odd number.

This is a useful in identifying molecular ions.

+


Rings and double bonds

Assume that you have identified a spectral line as a possible molecular ion.

We can use a simplistic relationship to determine it could be molecular ion.

Its based on the valences of the elements involved. Only ‘odd’ electron ions could be a molecular ion.

After determining the formula for an ion, this is a good test to run.



Rings + Double bonds = X - 1/2 Y + 1/2 Z + 1

Where X = # of C and Si Y = # of H, F, Cl, Br and I Z = # of N and P

It the calculated value is an integer, it can only come from an odd electron ion.

All molecular ions are odd electron ions but not all odd electron ions are molecular ions.



Examples Assume that you have determined the

following formula for specific lines based on isotopic abundances.

C3H9NO

C5H5N

C7H5N2

C8H16Cl

Determine if any could be a molecular ion.



C3H9NO R+DB = 3 - 9/2 + 1/2 + 1 = 0

Odd electron ion - could be molecular ion.

C5H5N R+DB = 5 - 5/2 + 1/2 + 1 = 4

Odd electron ion - could be molecular ion.

C7H5N2 R+DB = 7 - 5/2 + 2/2 + 1 = 6.5

Even electron ion - not molecular ion.

C8H16Cl R+DB = 8 - 17/2 + 1 = 0

Even electron ion - not molecular ion.


The nitrogen rule

In organic compounds, the is a relationship between the valance and the mass of the most common isotope for most elements.

Even elements have an even valance. Odd elements have an odd valance.

This leads to the ‘nitrogen rule.’ It assumes that we are limiting our elements to C, H, halogens, O and N.


The nitrogen rule

A compound containing only C, H, O or X will have an even molecular weight

A compound with an odd number of nitrogens will have an odd molecular weight.

A compound with an even number of nitrogens will have an even molecular weight.

So if the parent ion is odd, look for nitrogen.


Logical losses

Only a limited number of neutral fragments of low mass are commonly lost.

1 H 27 C2H3

15 CH3 29 C2H5 or CHO

16 NH2 31 OCH3 or CH2OH

17 OH 35 Cl 19 F 43 OC2H5 or COOH

26 CN 46 NO2

These are for single bond cleavage - radical losses.


Logical losses

We can also have neutral losses. It requires that at least two bonds are broken so they are less common.

2 H2 27 HCN 36 HCl

17 NH3 28 CO/C2H2 44 CO2

18 H2O 30 CH2O 74 C3H6O2

20 HF 34 H2S 80 HBr

We can use these ‘losses’ to help piece thing together.


Logical loss tables

The last two tables only represented partial lists of logical losses.

More complete lists of these losses and common fragments can be obtained from the main menu for this course material.

It is advised that you print out copies of these tables before attempting to work the example mass spectra.


Illogical losses

If the losses you encounter make no sense, then you most likely have misidentified the parent ion.

Some illogical neutral losses 3 H3 11 B

4 He 12 BH or C 6 ? 14 N or CH2

9 Be 24 C2

These simply don’t occur!


Logical losses example

Lets see how logical losses can help.

First, here is the information needed for the parent ion.

M/e Rel. Abund 57 100.00 84 0.10 85 0.40 86 15.51 87 1.00

86 appears to be theparent ion.

It’s even numberedso we either have noN or an even number.



Now to renormalize and determine the molecular formula.

m/e rel.abund

Normabund NF C5 C6 C7

A 86 15.5 100 6.5 72

A+1 87 1.0 6.5 6.6

A+2

86 - 72 = 14 C6H14



0

10

20

30

40

50

60

70

80

90

100

0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88

15

C2H5

86

71

57

43

29

C3H7

C4H9

C5H11

CH3

Re

lati

ve A

bu

nd

an

ce

M / Z



CH3 - CH2 - CH2 - CH2 - CH2 - CH3

71

57

43

29

15

So n-hexane appears to be the likely material.Mass, elements, lines and losses all make sense.


Another logical loss example - CH2Cl2

You can also seehow we have a Cl2

pattern at 84 and aCl pattern at 49

This also confirmsthe loss of a Cl.

0 20 40 60 80

84

49

- Cl(35)

- CH2Cl(49)


Common ions

A number of organic functional groups result in some common ions being produced.

Observing these lines can help - Identify possible functional groups - When developing SIM based GC/MS methods.


Some common ions

Compound type Common lines

Alkanes 29, 43, 57, 71, 85, 99Alkenes/cycloalkanes 27, 41, 55, 69, 83, 97Aliphatic alcohols 31, 45, 59, 73, 87, 101Aromatics 38, 39, 50-2, 63-5, 75-8Acids/esters 45, 59, 73, 87, 101Alkyl amines 30, 44, 58, 72, 86, 100Chloroalkyl 49, 63, 77, 91, 105Alkyl silanes 31, 45, 59, 73, 87, 101

Alkyl silanes are a common component of column bleed in capillary column GC/MS data


Conclusion

Remember, we’ve only covered a few of the rules for working with mass spectra.

We’ve totally left out any of the mechanisms of ion fragmentation.

On the other hand, the assumption was that you were working with a GC/MS system.

The chromatographic information will also help solve many problems.


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