2-MS interp
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Transcript of 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
0
50
100Lets review the basiccomponents of asimple mass spectrum.
Friday, August 6, 2010
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.
Friday, August 6, 2010
A simple mass spectrum
Mass / Charge
0
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.
Friday, August 6, 2010
A simple mass spectrum
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.
Friday, August 6, 2010
A simple mass spectrum
Mass / Charge
0
50
100
Types of information Mass of fragments Presence of isotope lines Gaps between lines.
We’ll look as some of the basics
Friday, August 6, 2010
A simple mass spectrum
Mass / Charge
0
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
Friday, August 6, 2010
A simple mass spectrum
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
Friday, August 6, 2010
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
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
Atomic masses and isotope
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
Friday, August 6, 2010
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%)
Friday, August 6, 2010
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.
Friday, August 6, 2010
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
Friday, August 6, 2010
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.
Friday, August 6, 2010
Obtaining a ‘good’ spectrum
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.
Friday, August 6, 2010
Obtaining a ‘good’ spectrum
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.
Friday, August 6, 2010
Obtaining a ‘good’ spectrum
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.
Friday, August 6, 2010
Obtaining a ‘good’ spectrum
Proper spectrum
Thresholdtoo low forsome or allof the scans
Saturationof the twolargest lines
Friday, August 6, 2010
Obtaining a ‘good’ spectrum
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.
Friday, August 6, 2010
Obtaining a ‘good’ spectrum
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.
Friday, August 6, 2010
Obtaining a ‘good’ spectrum
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
Working with the molecular ion
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.
Friday, August 6, 2010
Working with the molecular ion
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?
Friday, August 6, 2010
Working with 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).
Friday, August 6, 2010
Atomic masses and isotope
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
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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
Friday, August 6, 2010
Normalization table
Next, calculate the normalized abundances for your isotope lines - A+1 in this example, + 10% relative error
m/e rel.abund
Normabund NF Cn-1 Cn Cn+1
A 44 31440 100 .00318
A+1 45 1026 3.26+.33
A+2
Friday, August 6, 2010
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
Normabund NF Cn-1 Cn Cn+1
A 44 31440 100 .00318 24 36 48
A+1 45 1026 3.26+.33 2.2 3.3 4.4
A+2
Friday, August 6, 2010
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
Normabund NF Cn-1 Cn Cn+1
A 44 31440 100 .00318 24 36 48
A+1 45 1026 3.26+.33 2.2 3.3 4.4
A+2
Friday, August 6, 2010
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)
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
Carbon - an A+1 element
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.
Friday, August 6, 2010
Carbon - an A+1 element
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
Chlorine
Cl Cl2 Cl3 Cl4
By the time you have 4 chlorine, the A+2 lineis significantly larger than the A line.
Friday, August 6, 2010
Bromine
Br Br2 Br3 Br4
Bromine, which is almost a 1:1 mixture of79Br and 81Br also has a distinctive pattern.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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
Friday, August 6, 2010
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
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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
Friday, August 6, 2010
Saturated hydrocarbons
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.
Friday, August 6, 2010
Saturated hydrocarbons
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
Unsaturated hydrocarbons
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
Friday, August 6, 2010
Unsaturated hydrocarbons
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+
Friday, August 6, 2010
Unsaturated hydrocarbons
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+
Friday, August 6, 2010
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.
Friday, August 6, 2010
n-hexane
0 20 40 60 80
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.
Friday, August 6, 2010
n-hexane vs. 1-hexene
0 20 40 60 80
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.
Friday, August 6, 2010
All three hexenes
0 20 40 60 80
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.
Friday, August 6, 2010
1,3-hexadiene
0 20 40 60 80
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.
Friday, August 6, 2010
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
Friday, August 6, 2010
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
Friday, August 6, 2010
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
Friday, August 6, 2010
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?
Friday, August 6, 2010
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.
+
Friday, August 6, 2010
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.
Friday, August 6, 2010
Rings and double bonds
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.
Friday, August 6, 2010
Rings and double bonds
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.
Friday, August 6, 2010
Rings and double bonds
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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.
Friday, August 6, 2010
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!
Friday, August 6, 2010
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.
Friday, August 6, 2010
Logical losses example
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
Friday, August 6, 2010
Logical losses example
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
Friday, August 6, 2010
Logical losses example
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.
Friday, August 6, 2010
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)
Friday, August 6, 2010
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.
Friday, August 6, 2010
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
Friday, August 6, 2010
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.
Friday, August 6, 2010