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Interpreting the mass
spectrum of 1-bromopropane
CH3CH2CH2Br
[Author
©
Dr
Phil Brown PhD:
Doc Brown's advanced level organic chemistry exam revision notes
suitable for students of UK A level chemistry courses,
IB chemistry & US K12 grade
11, grade 12 and AP honors chemistry courses: Molecular
spectrometry - analysing the mass spectrum of 1-bromopropane
[updated
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mass spectrum of
CH3CH2CH2Br
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LINKS associated
with 1-bromopropane
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The
chemistry of organic halogen compounds
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This is a BIG
website, you need to take time to explore it
Mass spectrometry
- introduction and spectra index
See also the
Isomers of molecular formula
C3H7X (where
X =
F, Cl, Br or I)
Introductory note on the mass spectrum of 1-bromopropane
Students and teachers please note
my explanation of the mass spectrum of 1-bromopropane is designed for
advanced, but pre-university, chemistry courses.
If M represents the
1-bromopropane molecule, the initial ionisation to give the molecular ion is:
M(g) +
high KE e- ==> [M•]+(g) + 2e-
and for fragmentation equations assume [M]+ is the start of the
processes and all species are in a gaseous state.
I've not usually shown an unpaired electron on e.g. an ion or a non-ionised
alkyl radical R e.g.
[M•]+ ==> [X]+ + R•,
but you should be aware this is a more accurate depiction of some
processes.
I've used simplified equations to show how some of
the ions that might be formed in the fragmentation pattern for the
mass spectrum of 1-bromopropane and only the formation of singly charged
positive are considered for the mass spectrum of 1-bromopropane.
I've included a stick diagram and table of m/z ions for the mass spectrum of
1-bromopropane
and doing the mass spectrum analysis under standard conditions,
databases can be compiled based on complex fingerprint patterns, often involving
the relative intensities of many fragment ions, and used to identify compounds including
1-bromopropane.
In selected cases, where two
different fragment ions have the same integer m/z value,
I've pointed out that modern mass spectrometers can measure
relative ion mass to four decimal places. So, using
accurate isotopic masses, I've calculated the accurate ion
masses, BUT strictly speaking, 0.0005 should be deducted
for singly charged ions to account for the loss of the
electron in their formation. I have NOT done this,
but the mass spectrometer software does!
1-bromopropane,
C3H7Br,
CH3CH2CH2Br,
,
The molecular structure and naming of haloalkanes
Interpreting the fragmentation pattern of the mass spectrum of
1-bromopropane
[M]+ is the molecular ion peak (M) with an m/z of
122 and 124 corresponding to [C3H7Br]+, the original 1-bromopropane molecule minus an electron,
[CH3CH2CH2Br]+
There are two molecular ion peaks
because bromine has two isotopes, 50.6% 79Br and 49.4%
81Br.
Their average relative isotopic mass is ~80, so the relative
molecular mass for 1-bromopropane is ~123.
[CH3CH2CH279Br]+
is technically the molecular ion peak M, because it has a
slightly bigger intensity than the
other
[CH3CH2CH281Br]+
peak because of the slightly greater abundance of the lighter 79Br
isotope and is referred to as the M+2 ion peak.
However, this means any fragment carrying a bromine
atom should show up as twin peaks, two mass units apart and
approximately of equal height (intensities), but slightly bigger for
the 79Br molecular or fragment ions.
You might see very tiny M+1 and M+3 peaks at m/z 123 and 125 corresponding to an ionised
1-bromopropane
molecule with one 13C atom in it i.e. an ionised 2-bromopropane molecule of
formula [13C12C2H779/81Br]+
Carbon-13 only accounts for ~1% of all carbon atoms
(12C ~99%), but the more carbon atoms in the molecule,
the greater the probability of observing these 13C M+1
peaks.
1-bromopropane has 3 carbon atoms, so on
average, ~1 in 33 molecules will contain a 13C atom.
The most abundant ion of the molecule under mass
spectrometry investigation (1-bromopropane) is usually given an arbitrary value of
100, called the base ion peak, and all other abundances
('intensities') are measured against it.
The base ion peak for
the mass spectrum of 1-bromopropane is the m/z 43 ion
[C3H7]+
The
parent molecular ions are m/z 122 and 124
[CH3CH2CH279Br]+
and [CH3CH2CH281Br]+
Unless otherwise indicated, assume the carbon atoms in
1-bromopropane are the 12C isotope.
|
m/z value of
[fragment]+ |
124 |
122 |
109 |
107 |
95 |
93 |
82 |
81 |
|
[molecular fragment]+ |
[C3H781Br]+ |
[C3H779Br]+ |
[C2H481Br]+ |
[C2H479Br]+ |
CH281Br |
CH279Br |
[H81Br]+ |
[81Br]+ |
|
m/z value of
[fragment]+ |
80 |
79 |
44 |
43 |
42 |
41 |
39 |
29 |
27 |
15 |
|
[molecular fragment]+ |
[H79Br]+ |
[79Br]+ |
[13C12C2H7]+ |
[C3H7]+ |
[C3H6]+ |
[C3H5]+ |
[C3H3]+ |
[C2H5]+ |
[C2H3]+ |
[CH3]+ |
Analysing and explaining the principal ions in the
fragmentation pattern of the mass spectrum of 1-bromopropane
PLEASE NOTE
I have found it difficult to find 'authentic' equations to explain mass
spectra fragmentation patterns and it is complex chemistry! I've identified
the formulae of the ionised fragments on the mass spectrum diagram, but the
equations are from the internet or my conjecture as to how the ions might be
formed - please take care in using the information, especially for
assignments at university or pre-university level.
Atomic masses: C = 12
(~1/100 is C-13); H = 1; O = 16; Br = 79
or 81
Bond enthalpies = kJ/mol: C-C = 348; C-H = 412;
C-Br 276
Equations to explain the most abundant
m/z ion peaks of
1-bromopropane
Formation of m/z 107 and 109 ions:
[CH3CH2CH2Br]+ ===> [C2H4Br]+
+ CH3
C-C bond scission to lose the end methyl group.
Mass loss = 124 - 15 = 109 and 122 - 15
= 107
Very small peaks since C-C bond scission is much
less likely than C-Br scission - the latter having a significantly
lower bond enthalpy.
Where R is alkyl, the
double RBr peaks of roughly 1 : 1 abundance ratio are characteristic
of organo-bromine compounds (one m/z ion peak is slightly shorter
than the other, technically 50.7 : 49.3).
Formation of m/z 93 and 95 ions:
[CH3CH2CH2Br]+ ===> [CH279Br]+
or [CH281Br]+ +
C2H5
C-C bond scission.
Mass loss = 124 - 29 = 95 and 122 -
29 = 93
Again, very small peaks since C-C bond scission
is much less likely than C-Br scission - the latter having a
significantly lower bond enthalpy.
Again, t he
double RBr peaks of roughly 1 : 1 ratio are characteristic of
organo-bromine compounds (one m/z ion peak is slightly shorter
than the other, technically 50.6 : 49.4).
Formation of m/z 80 and 82 ions:
[CH3CH2CH2Br]+ ===> [H79Br]+
or [H81Br]+ + C3H6
Elimination of hydrogen bromide from the parent
molecular ion.
Although small, t he
double HBr peaks of roughly 1 : 1 ratio are very characteristic
of organo-bromine compounds (one m/z ion peak is slightly
shorter than the other, technically 50.6 : 49.4).
Formation of m/z 79 and 81 ions:
[CH3CH2CH2Br]+ ===> [79Br]+
or [81Br]+ + C3H7
C-Br bond scission, but alkyl group is more likely
to carry the positive charge (see below).
Mass loss = 124 - 43 = 81 and 122 - 43 =
79
Again, although small, t he
double Br peaks of roughly 1 : 1 ratio are very characteristic of
organo-bromine compounds (one m/z ion peak is slightly shorter than
the other, technically 50.6 : 49.4).
Formation of m/z 43 ion:
[CH3CH2CH2Br]+ ===> [C3H7]+
+ Br
C-Br bond fission to give the ionised alkyl ion
(bond weaker than C-C).
Mass loss 124 - 81 = 43 and 122 - 79 =
43
The m/z 43 ion is the
base ion peak, the most
abundant and 'stable' ion fragment.
It undergoes success proton loss e.g. 43 => 42 => 41
=> 40 => 39
Formation of m/z 29 ion:
[CH3CH2CH2Br]+ ===> [C2H5]+
+ CH2Br
C-C bond scission.
Mass loss 124 - 95 = 29 and 122 - 93
= 29
further proton
loss from [C2H3]+
gives m/z ions of 28. and 27
Formation of m/z 15 ion:
[CH3CH2CH2Br]+ ===> [CH3]+
+ C2H4Br
C-C bond scission.
Mass loss 124 - 109 = 15 and 122 -
107 = 15
Summary of the
mass spectrum of 1-bromopropane and extra comments and practice questions
The mass spectrum of 1-bromopropane (CH₃CH₂CH₂Br) with
precision, structure, and exam-focused clarity.
Overview of Mass
Spectral Features of the mass spectrum of 1-bromopropane
1-bromopropane is a haloalkane containing bromine, which
introduces distinct isotopic peaks due to 79Br
and 81Br. Fragmentation occurs via C–C and C–Br bond
cleavage, producing alkyl and acylium-type cations.
Prominent m/z Ions
and Their Origins in the mass spectrum of 1-bromopropane
|
m/z |
Ion Formula |
Fragment Origin |
Notes |
| 122 |
C3H779Br⁺ |
Molecular ion with 79Br |
One of two molecular ion peaks |
| 124 |
C3H781Br⁺ |
Molecular ion with 81Br |
~Equal intensity to m/z 122 due to Br
isotopes |
| 43 |
C3H7⁺ |
Propyl cation |
Base peak
— most stable fragment |
| 41 |
C3H5⁺ |
Loss of H2 from
C3H7⁺ |
|
| 29 |
C2H5⁺ |
Ethyl cation |
Typical alkane fragment |
| 15 |
CH3⁺ |
Methyl cation |
Seen in most hydrocarbon spectra |
| 79/81 |
Br⁺ |
Bromine isotopes |
Diagnostic for Br-containing compounds |
The m/z 122 and 124 peaks are nearly equal in intensity
due to the natural abundance of 79Br (~50.5%) and 81Br
(~49.5%).
Common
Misconceptions about the mass spectrum of 1-bromopropane
(see also below)
- Assuming only one molecular ion peak: Bromine gives
two peaks (m/z 122 and 124), not just one.
- Mistaking m/z 43 for CHO⁺: In haloalkanes, it is
usually C3H7⁺, not an aldehyde fragment.
- Ignoring isotopic patterns: The M and M+2 peaks are
crucial for identifying halogens — especially Br and Cl.
- Expecting base peak to be molecular ion: In
1-bromopropane, the base peak is m/z 43, not 122 or 124.
Exam Tips for
questions involving the mass spectrum of 1-bromopropane
(see also above)
- Look for M and M+2 peaks: A 1:1 ratio strongly suggests
bromine.
- Use base peak to identify stable fragments: m/z 43 is a
classic alkyl cation.
- Compare with similar haloalkanes: e.g. 1-chloropropane
shows M and M+2 in a 3:1 ratio.
- Don't confuse fragment ions with functional groups:
e.g. CHO+ versus C3H7+ for m/z
43
Practice isotope logic: Br gives 2 peaks, Cl gives 2
peaks with different ratios, I gives one (for halogens).
Practice Question:
Mass Spectrum of 1-Bromopropane
Compound: 1-bromopropane (C3H7Br)
Context: A student obtains the mass spectrum of
1-bromopropane. The spectrum shows two prominent molecular ion peaks at m/z
122 and m/z 124, along with several smaller fragment peaks.
Question:
- Explain why the molecular ion peak of 1-bromopropane appears as a pair
at m/z 122 and m/z 124.
- Identify the fragment ion responsible for the peak at m/z 43
and explain its formation.
- Predict the relative intensities of the m/z 122 and m/z
124 peaks and justify your answer using isotopic abundances.
- Suggest one reason why the base peak (most intense peak) might occur at
m/z 43 rather than at the molecular ion peak.
Model Answer
a) Isotopic Pair at m/z 122 and 124
- Bromine has two major isotopes: ⁷⁹Br and ⁸¹Br.
- 1-bromopropane contains one bromine atom, so the molecular ion exists in
two forms:
- C3H7⁷⁹Br → m/z 122
-
C3H7⁸¹Br → m/z 124
- These peaks are two mass units apart due to the isotopic difference.
b) Fragment Ion at m/z 43
- The m/z 43 peak corresponds to the propyl cation (C3H7⁺).
- It forms when the Br atom is lost as a neutral radical:
-
C3H7Br →
C3H7⁺ (m/z 43) + Br•
- This is a common fragmentation pathway due to the relatively weak C–Br
bond.
c) Relative Intensities of m/z 122
and 124
- Natural abundance of bromine isotopes:
- ⁷⁹Br ≈ 50.6%
- ⁸¹Br ≈ 49.4%
- Therefore, the m/z 122 and m/z 124 peaks will be
approximately equal in intensity, forming a near 1:1 ratio.
d) Base Peak at m/z 43
- The base peak is the most stable and abundant fragment ion.
- The propyl cation (C3H7⁺)
is relatively stable due to charge delocalization.
- Fragmentation to form this ion is energetically favorable, leading to
high abundance.
Key words & phrases: C3H7Br CH3CH2CH2Br image diagram on how to interpret and explain the mass spectrum of
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1-bromopropane equations showing the
formation of the ionised fragments in the mass spectrum of
1-bromopropane
what does the mass spectrum tell you about the structure and
properties of the 1-bromopropane molecule?
Links associated
with
1-bromopropane
The infrared spectrum of
1-bromopropane
The H-1 NMR spectrum of 1-bromopropane
The C-13 NMR spectrum of 1-bromopropane
The chemistry of HALOGENOALKANES (haloalkanes)
revision notes INDEX
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