Answers to Mass Spectra Unknowns
1.
The parent peak is at m/z = 16 amu (m/z = mass / charge, amu=atomic mass
unit). The tiny peak at 17 amu is not considered the parent peak because it
comes from the presence of the rare (1.1% natural abundance) carbon isotope,
13C.
What could have a mass of 16 amu? CH4+ ion created in the mass
spectrometer has a mass of 12 amu, therefore the unknown compound is most
likely CH4. The peak at 15 amu is CH3+ and the peak at 14 amu is CH2+, etc.
Thus one sees not only the parent peak, but also fragments of CH4+ that form
as it looses H. Although it is not obvious from the spectrum, the height of the
M+1 peak (at 17 amu) is about 1.1% the height of the M parent peak (at 16
amu).
H
H C H
H
What you should have learned from #1:
The parent peak gives the molecular weight.
The peak due to a rare isotope is not the parent peak.
Molecules fragment in the mass spectrometer and because these fragments
turn up in the mass spectrum the spectra are complicated.
2.
Interpretation of this spectrum is complicated by the fact that the tallest peak
is not the one furthest on the right. The tallest peak is at m/z = 28 amu, but
the furthest significant peak on the right (again ignoring rare isotopes) is at
m/z = 30 amu. Therefore the parent peak is at 30. What could have a mass of
30 amu? C2H6 (2x12 + 6x1) = ethane. The fragment at 28 is what forms when
ethane looses 2 Hs namely ethene (C2H4), but you don’t need to analyze the
fragmentation pattern to get the right answer. If you could measure the ratio
of the heights of the M+1 peak (at 31) to the height of the M peak (at 30) it
would be about 2.2%. This tells you that 2 Cs are present, but you knew that
already from the molecular weight.
H H
H C C H
H H
What you should have learned from #2:
Finding the parent peak is not always simple. It is not necessarily the
largest peak.
The fragmentation pattern is very complex and not easy to interpret. Focus
on the parent peak, then make an educated guess (hypothesis).
3.
Again the parent peak is not the largest peak (at 29), but rather the furthest
peak of significance on the right, i.e. at m/z = 44 amu. What could have a mass
of 44 amu? C3H8 is consistent with a mass of 44 amu (3x12 + 8x1) = propane.
Mass Spectra Unknowns
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The big peak at 29 amu is an ethyl fragment that is left after a methyl group
has broken off of propane. Again, you needn’t interpret the fragmentation
pattern to get to the right answer. The height of the M+1 (at 45 amu), if you
could find it, would be 3.3% of the M peak (at 44 amu).
What you should have learned from #3:
Finding the parent peak is not always simple. It is not necessarily the
largest peak.
The fragmentation pattern is very complex and not easy to interpret. Focus
on the parent peak, then make an educated guess (hypothesis).
4.
Again the parent peak is not the largest peak (43 amu), but rather the furthest
peak of significance on the right, i.e. at m/z = 72 amu. What could have a mass
of 72 amu? C5H12 is consistent with a mass of 72 amu (5x12 + 12x1) = pentane.
The big peak at 43 amu is an propyl fragment that is left after an ethyl group
has broken off of propane, but again, you needn’t interpret the fragmentation
pattern to get to the right answer. In principle the height of the M+1 (at 73
amu), could tell you about the carbons but it is vanishingly small on this
spectrum. If you could find it, its abundance would be only 5.5% of the
abundance of the M peak (at 72 amu).
What you should have learned from #4:
Finding the parent peak is not always simple. It is not necessarily the
largest peak.
The fragmentation pattern is very complex and not easy to interpret. Focus
on the parent peak, then make an educated guess (hypothesis).
5.
The parent peak is at m/z = 72. Your first guess is that maybe this compound
is pentane (as in #4), but visual inspection shows that spectra 4 and 5 are not
the same, therefore it can not be pentane (or #4 is not pentane – they can’t both
be pentane). If I tell you that the height of the M+1 (73 amu) is 4.4% of the M
peak (72 amu), then you know there are only 4 carbons in this molecule. What
4 carbon molecule has a MW of 72? C4H10O. Note that this molecule is
deficient in hydrogen; a saturated molecule has a molecular formula CnH2n+2.
The missing hydrogens means there is a double bond or ring present. One
possibility is:
O
Mass Spectra Unknowns
2
The molecule is called 2-butanone, but you haven’t learned this nomenclature
yet, so don’t worry about how to name it, focus on the structure. Note that it
has one π bond, which is consistent with one degree of unsaturation ( = 2 Hs
missing).
What you should have learned from #5:
There is usually more than one MF (molecular formula) for a given MW
(molecular weight).
Each molecule has its own unique mass spectrum (its like a finger print).
Different molecules with the same MW will have unique fragmentation
patterns, but generally interpretation of that fragmentations is too
complicated.
The M+1 abundance, if present, can tell the # of Cs, and could therefore
distinguish between isomers.
6.
Hopefully at this point you are getting good at finding the parent peak. It
should be the furthest peak of significance on the right, in this case at m/z = 86
amu. What could have a mass of 86 amu? C6H14 is consistent with a mass of
86 amu (6x12 + 14x1) = hexane. You are not responsible for interpreting the
fragmentation pattern, but in this case it is straightforward (pentyl fragment
at 71 amu, butyl at 57 amu, propyl at 43 amu, and ethyl at 29 amu).
What you should have learned from #6:
Finding the parent peak is not always simple. It is not necessarily the
largest peak.
The fragmentation pattern is very complex and not easy to interpret. Focus
on the parent peak, then make an educated guess (hypothesis).
7.
This mass spectrum has a parent peak identical with #6, namely at m/z = 86
amu. Since its mass spectrum looks different it probably is. If I tell you the
abundance of the M+1 is 5.5% of the parent peak abundance, you know there
are 5 Cs and unknown #7 can not be hexane. What has a mass of 86, but only
5 Cs? After some tinkering you could come up with C5H10O. This molecule has
one degree of unsaturation, which means 2 Hs are missing. That means a
double bond or ring is present. There is more than one isomer consistent with
this information, among others are:
O
O
Mass Spectra Unknowns
3
These two are hard to distinguish based on just the parent peak, and so we will
simply leave it at that. (The correct answer turns out to be 3-pentanone, but
you can’t know that unless you know how to interpret the fragmentation
pattern.)
What you should have learned from #7:
There is usually more than one MF for a given MW.
Each molecule has its own unique mass spectrum (its like a finger print).
Different molecules with the same MW will have unique fragmentation
patterns, but generally interpretation of that fragmentations is too
complicated.
The M+1 abundance, if present, can tell the # of Cs, and could therefore
distinguish between isomers.
8.
This mass spectrum has a parent peak identical with #6 and #7, namely at m/z
= 86 amu. Since its mass spectrum looks different it probably is. If I tell you
the abundance of the M+1 is 5.5% of the parent peak abundance, you know
there are 5 Cs and unknown #8 can not be hexane, but is more likely to be
C5H10O. As in #7 more than one structure is consistent with this information
and it is hard to distinguish between the isomers suggested in #7 (2-pentanone
and 3-pentanone). In either case it is not the same as #7. Which ever #7 is, #8
is probably the other isomer. (The correct answer turns out to be 2-pentanone,
but you can’t know that unless you know how to interpret the fragmentation
pattern.)
O
What you should have learned from #7:
Sometimes isomers are so similar they are not easily distinguished by MS.
Analysis of the fragmentation pattern can distinguish compounds, but it is
complex.
9.
A parent peak at m/z = 17 amu is clearly present. This is a very light molecule.
Virtually no peak at M+1 is to be seen indicating that C may not even be
present. What could have a MW of 17? NH3 (1x14 + 3x1). The fragments at
16, 15, and 14 could be NH2+, NH+, and N+, respectively. One curious thing
about this mass spectrum is that the MW is odd. Most common atoms in
organic compounds have even masses (12C, 14N, and 16O). Although hydrogen
has an odd mass it is usually found in pairs because carbon is tetravalent and
oxygen divalent. The one exception is N. Although its mass is even, N is
trivalent, so an odd number of Hs gives rise to an odd MW. This is known as
the nitrogen rule: there is a high probability that odd MW compounds have a
nitrogen (or more correctly an odd number of Ns).
Mass Spectra Unknowns
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H N H
H
What you should have learned from #9:
If the parent peak has an odd mass-to-charge ratio (m/z), N is present.
10. The parent peak is at m/z = 31 amu has an odd MW and so nitrogen is likely to
be present. What could have a MW of 31? CH5N (1x12 + 5x1 + 1x14). Note too
that the N–H readily breaks in the “harsh” conditions of the mass spectro–
meter, so a strong M–1 peak is often seen.
H
H C N H
H H
What you should have learned from #10:
If the parent peak has an odd mass-to-charge ratio (m/z), N is present.
11. The parent peak is at m/z = 45 amu has an odd MW and so nitrogen is likely to
be present. What could have a MW of 45? C2H7N (2x12 + 7x1 + 1x14). (Note
the presence of a strong M–1 fragment, which would be consistent with the
presence of nitrogen.) Unknown #11 is ethyl amine:
NH2
What you should have learned from #11:
If the parent peak has an odd mass-to-charge ratio (m/z), N is present.
12. Like #11 the parent peak of #12 is at m/z = 45 amu has an odd MW and so
nitrogen is likely to be present. The mass spectrum however looks different
from #11. This could be an isomer #11, namely dimethyl amine:
H
N
What you should have learned from #12:
If the parent peak has an odd mass-to-charge ratio (m/z), N is present.
Sometimes isomers are so similar they are not easily distinguished by MS.
13. The parent peak at m/z = 59 amu has an odd MW and so nitrogen is likely to be
present. What could have a MW of 59? C3H9N (3x12 + 9x1 + 1x14). One
possibility is propyl amine:
Mass Spectra Unknowns
5
NH2
What you should have learned from #13:
The parent peak may be quite weak.
If the parent peak has an odd mass-to-charge ratio (m/z), N is present.
14. Like #13 the parent peak is at m/z = 59 amu, so #14 is likely to be an isomer of
#13. There are several possibilities, all of which would difficult to distinguish
by mass spectrometer except for by their fragmentation “fingerprint”. #14
turns out to be trimethyl amine:
N
What you should have learned from #14:
If the parent peak has an odd mass-to-charge ratio (m/z), N is present.
Isomers can only be distinguished by their fragmentation “fingerprint”.
15. This unknown presents a new pattern: two strong peaks separated by two
mass units, in this case at m/z = 50 and 52 amu. (Take a moment to review
spectra #1–14 and convince yourself that this is so.) What could this be? It is
tempting to assign m/z = 52 as the parent peak and m/z = 50 as some fragment,
but this turns out not to be the case. The peak at 52 is not due to some rare
isotope like 13C, 15N, or 2H because this would give a weak peak at M+1, not
M+2. There is however an atom with an abundant heavy isotope containing
two extra neutrons, namely chlorine. The natural abundances of 35Cl and 37Cl
are 75.8% and 24.2%, which means the M peak should be about three times as
high as the M+2 as indeed it is for #15. This then becomes of telltale signature
for chlorine. If chlorine is present and the MW is 50, what could #15 be? (MW)
– (35 amu for 35Cl) = 15. What has a mass of 15? A methyl group. Therefore
#15 could be methyl chloride:
H
H C Cl
H
What you should have learned from #15:
Chlorine shows an unusual isotopic signature with strong peaks at M and
M+2 (in a ratio of about 3:1).
16. This unknown presents another new pattern. As in #15 there are two strong
peaks separated by two mass units, in this case at m/z = 94 and 96 amu. What
could this be? Again it is tempting to assign m/z = 96 as the parent peak and
m/z = 94 as some fragment, but this turns out not to be the case. The peak at
Mass Spectra Unknowns
6
96 is not due to some rare isotope like 13C, 15N, or 2H because this would give a
weak peak at M+1, not M+2. But the ratio is not correct for 35Cl/37Cl; it is not
~2:1. It turns out that another of the halogens, namely bromine, that has an
abundant heavy isotope containing two extra neutrons.
The natural
79
81
abundances of Br and Br are 50.9% and 49.3%, which means the M peak
should be about the same height as the M+2, as indeed it is for #16. This then
becomes of telltale signature for bromine. If bromine is present and the MW is
94, what could #16 be? (MW) – (79 amu for 35Cl) = 15. What has a mass of 15?
A methyl group. Therefore #16 could be methyl bromide:
H
H C Br
H
What you should have learned from #16:
Bromine shows an unusual isotopic signature with strong peaks at M and
M+2 (in a ratio of about 1:1).
17. This unknown shows the unusual pattern of a strong M and M+2 (m/z = 64 and
66 amu). Therefore you hypothesize that either chlorine or bromine is present.
The ratio of heights (abundances) is closer to 3:1, and not 1:1, so chlorine must
be present. What chlorine containing compound has a mass of 64 amu? (MW
of 64) – (35 for 35Cl) = 29. What has a mass of 29? An ethyl group. Therefore
#17 could be ethyl chloride:
Cl
What you should have learned from #17:
Chlorine shows an unusual isotopic signature with strong peaks at M and
M+2 (in a ratio of about 3:1).
18. Look carefully at the scale on this mass spectrum. The two peaks furthest on
the right look close together, but are in fact separated by two mass units (at
m/z = 108 and 110 amu). The peak heights (abundances) are about 1:1, so
bromine is probably present. What bromine containing compound has a mass
of 108 amu? (MW of 108 – 79 for 79Br) = 29. What has a mass of 29? An ethyl
group. Therefore #18 could be ethyl bromide:
Br
What you should have learned from #18:
Be sure to check the scale on the mass spectrum carefully.
Bromine shows an unusual isotopic signature with strong peaks at M and
M+2 (in a ratio of about 1:1).
Mass Spectra Unknowns
7
19. Careful inspection of this mass spectrum indicates two peaks of nearly equal
abundances at m/z = 122 and 124, so bromine is probably present. What
bromine containing compound has a mass of 122 amu? (MW of 122 – 79 for
79Br) = 43. What has a mass of 43? A propyl group. Therefore #19 could be
propyl bromide:
Br
What you should have learned from #18:
Be sure to check the scale on the mass spectrum carefully.
Bromine shows an unusual isotopic signature with strong peaks at M and
M+2 (in a ratio of about 1:1).
20. This spectrum is virtually indistinguishable from #19 (except for the absence of
a very weak peak at 39 amu), therefore this is like to be an isomer of #19.
Therefore #20 could be 2-propyl bromide:
Br
What you should have learned from #18:
Be sure to check the scale on the mass spectrum carefully.
Bromine shows an unusual isotopic signature with strong peaks at M and
M+2 (in a ratio of about 1:1).
Isomers are often very difficult to distinguish in the mass spectrum.
Mass Spectra Unknowns
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