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Interpreting and explaining the mass
spectrum of 1-chloropropane CH3CH2CH2Cl
[Author
©
Dr Phil Brown GRIC, PhD:
Doc Brown's advanced level organic chemistry exam revision notes
suitable for students of UK A level chemistry courses & US K12 grade
11, grade 12 and AP honors chemistry courses: Molecular
spectrometry - analysing the mass spectrum of 1-chloropropane
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mass spectrum
CH3CH2CH2Cl
LINKS associated
with 1-chloropropane
The
chemistry of organic halogen compounds
This is a BIG website, please take time to explore it
Mass spectrometry
- introduction and spectra index
Isomers of molecular formula
C3H7X (where
X =
F, Cl, Br or I)
A
practise exemplar question on the mass spectrum of 1-chloropropane
plus
questions for you to do
with answers
and explanations
Introductory note on the mass spectrum of 1-chloropropane
Students and teachers please note
my explanation of the mass spectrum of 1-chloropropane is designed for
advanced, but pre-university, chemistry courses.
If M represents the
1-chloropropane 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-chloropropane and only the formation of singly charged
positive are considered for the mass spectrum of
1-chloropropane.
I've included a stick diagram and table of m/z ions for the mass spectrum of
1-chloropropane
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-chloropropane.
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-chloropropane C3H7Cl
The molecular structure and naming of haloalkanes
Interpreting the fragmentation pattern of the mass spectrum of
1-chloropropane
[M]+ are the molecular ion peaks (M) with a m/z
values of 78 and 80 corresponding to
[C3H7Cl]+, the original 1-chloropropane molecule minus an electron
(shown below)
m/z 78 ion is
[CH3CH2CH235Cl]+
is the higher intensity peak because of the greater % abundance of the
chlorine-35 isotope and is considered the
M ion peak.
m/z ion 80 is
[CH3CH2CH237Cl]+
is the lower intensity peak because of the
lesser % abundance of the chlorine-37 isotope and is considered the
M+2 ion peak.
Since chlorine has two common isotopes of 35Cl
and 37Cl in the ratio 3 : 1, you should observe double peaks
in the intensity ratio 3 : 1, two mass units apart for molecular
fragments containing a chlorine atom from the fragmentation pattern of
1-chloropropane.
This also applies to the molecular ion, so two
molecular ions are observed at m/z 78 and 80.
Unless otherwise stated, C means a
12C atom, if not, the isotopic carbon atom 13C
will be indicated.
You might see tiny M+1 and M+3 peaks at m/z 79
and m/z 81, corresponding to an ionised
1-chloropropane
molecule with one 13C atom in it i.e. an ionised 1-chloropropane molecule of
formula [13C12C2H735/37Cl]+
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 this 13C M+1
peak.
1-chloropropane 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-chloropropane) is usually given an arbitrary abundance 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-chloropropane is the m/z 42 ion
[C3H6]+
Identifying the species giving the most prominent peaks
(apart from M) in the fragmentation pattern of 1-chloropropane.
P arent
molecular ion peaks are m/z 78 ion
[CH3CH2CH235Cl]+
and
m/z ion 80
[CH3CH2CH237Cl]+
i
Unless otherwise indicated, assume the carbon atoms in
1-chloropropane are the 12C isotope.
|
m/z value of
[fragment]+ |
65 |
63 |
51 |
49 |
43 |
42 |
|
[molecular fragment]+ |
[CH2CH237Cl]+ |
[CH2CH235Cl]+ |
[CH237Cl]+ |
[CH235Cl]+ |
[CH3CH2CH2]+ |
[C3H6]+ |
|
m/z value of
[fragment]+ |
41 |
40 |
39 |
29 |
28 |
27 |
26 |
|
[molecular fragment]+ |
[C3H5]+ |
[C3H4]+ |
[C3H3]+ |
[C2H5]+ |
[C2H4]+ |
[C2H3]+ |
[C2H2]+ |
Analysing and explaining the principal ions in the
fragmentation pattern of the mass spectrum of 1-chloropropane
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: H = 1; C = 12
(~1% 13); Cl = 35 or 37 (ratio ~3:1)
Bond enthalpies kJ/mol: C-C = 348; C-Cl = 338;
C-H = 412
Equations to explain the most abundant ion peaks of
1-chloropropane
Formation of m/z 63 and 65 ions:
[CH3CH2CH2Cl]+ ===> [CH2CH235Cl]+
or [CH2CH237Cl]+
+ CH3
C-C bond fission in
the parent molecular ion, end methyl group has
broken off.
Low probability because of the strong C-C bond
enthalpy.
Note the expected 3:1 ratio of chlorine containing
fragments, this is very diagnostic for s chlorine atom in the
molecule.
The double RCl m/z ion
peaks of roughly 3 : 1 abundance ratio are characteristic of
organo-chlorine compounds i.e. caused by the 3 : 1 isotope ratio of
35Cl : 37Cl.
Formation of m/z 49 and 51 ions:
[CH3CH2CH2Cl]+ ===> [CH235Cl]+ or
[CH237Cl]+ + CH3CH2
C-C bond fission in
the parent molecular ion, end ethyl group has
broken off.
Low probability because of the strong C-C bond
(high bond enthalpy).
Again, the double RCl
m/z ion peaks of roughly 3 : 1 abundance ratio are characteristic of
organo-chlorine compounds i.e. caused by the 3 : 1 isotope ratio of
35Cl : 37Cl.
Formation of m/z 43
ion:
[CH3CH2CH2Cl]+ ===> [CH3CH2CH2]+
+ Cl
Bond scission of the C-Cl bond, the weakest bond in
the molecule.
Formation of m/z 42 ion:
[CH3CH2CH2Cl]+ ===> [C3H6]+
+ HCl
Elimination of hydrogen chloride to give a positive
ion of propene, which is relatively stable (resonance stabilised)
compared to other ions.
Note that this
fragmentation pathway also involves the breaking of the weakest bond
in the 1-chlorpropane molecule.
The m/z 42 ion is the base peak ion, the most
abundant and 'stable' ion fragment.
The m/z 42 and 43 ions can lose protons to give m/z
ions of 42 ==> 39
Formation of m/z 41 ion:
[?]+ ===> [C3H5]+
+ ?
Formation of m/z 39 ion:
[?]+ ===> [C3H3]+
+ ?
Formation of m/z 35
& 37 and 36 & 38 ions:
(not shown in
table)
Ionisation of chlorine after C-Cl bond
scission gives [35Cl]+ or [37Cl]+
Elimination of hydrogen chloride gives [H35Cl]+
or [H37Cl]+
Low probability of formation, and of little
importance, the alkyl fragments tend to carry the positive
charge - the more electronegative chlorine tends to retain its
electrons.
Formation of m/z 29 ion:
[CH3CH2CH2Cl]+ ===> [CH3CH2]+
+ CH2Cl
C-C bond scission
Formation of m/z 28 ion:
[CH2CH2Cl]+
===> [C2H4]+
+ Cl
C-Cl bond scission
Formation of m/z 27 ion:
[CH3CHCl]+ ===> [C2H3]+
+ HCl
Elimination of hydrogen chloride from the m/z 63
and 65 ions.
Summary of the
mass spectrum of
1-chloropropane and extra comments
The mass spectrum of 1-chloropropane
(CH3CH2CH2Cl) with a structured breakdown, key fragment ions,
common pitfalls, and exam-focused insights. This haloalkane offers a great
case study in isotopic patterns and fragmentation logic.
Overview of Fragmentation Behavior
for the mass spectrum of 1-chloropropane
1-chloropropane undergoes electron ionization (EI),
producing:
- A molecular ion (M⁺) and its M⁺ + 2
isotopic counterpart due to chlorine isotopes.
- Fragment ions from cleavage of C–C and C–Cl bonds.
- A base peak from a particularly stable carbocation.
Prominent m/z Ions and Origins
for
the
mass spectrum of 1-chloropropane
|
m/z |
Ion Formula |
Origin / Fragmentation |
Notes |
| 78 |
C3H735Cl⁺ |
Molecular ion with ³⁵Cl |
Main M⁺ peak |
| 80 |
C3H737Cl⁺ |
Molecular ion with ³⁷Cl |
M⁺ + 2 peak (³⁷Cl isotope) |
| 43 |
C3H7⁺ |
Propyl cation from C–Cl cleavage |
Often base peak |
| 42 |
C3H6⁺ |
Allyl cation from elimination (loss of
HCl) |
Resonance-stabilized, base ion peak |
| 39 |
C3H3⁺ |
Loss of Cl radical |
Alkene fragment |
| 29 |
C2H5⁺ |
Ethyl cation from C–C cleavage |
Common alkyl fragment |
Common Misconceptions
about
the
mass spectrum of 1-chloropropane
(see also below)
- Ignoring isotopic peaks: Chlorine gives a 3:1
ratio of M⁺ (m/z 78) to M⁺ + 2 (m/z 80). Students often overlook
this diagnostic clue.
- Assuming base peak is always M⁺: In haloalkanes, the
most intense peak is often a fragment ion, not the
molecular ion.
- Misidentifying fragment ions: m/z 43 is not
always from acylium ions (RCO⁺); in this case, it’s a propyl cation.
- Confusing allyl and propyl ions: m/z 41 (C3H5⁺) is
resonance-stabilized and often misattributed.
Exam Tips for
questions involving
the
mass spectrum of 1-chloropropane
(see also above)
- Always comment on isotopic pattern: Chlorine’s ³⁵Cl and
³⁷Cl give a clear M⁺/M⁺+2 signature — a key ID feature.
- Use fragmentation logic: Show how bond cleavage leads
to stable ions (e.g. propyl or allyl cations).
- Compare with known spectra: Especially useful in
multi-technique questions (IR + MS + NMR).
- Don’t overinterpret low-intensity peaks: Focus on base
peak, molecular ion, and isotopic pattern.
- Mention stability: Allyl cation (m/z 41) is stabilized
by resonance — a great point for top marks.
Compound: 1-chloropropane (C3H7Cl)
A student analyses the mass spectrum of a compound suspected to be a
straight-chain chloroalkane.
The spectrum shows two molecular ion peaks at m/z 78 and m/z
80, and a base peak at m/z 43.
Question:
- Explain why the molecular ion peak appears as a pair at m/z 78
and m/z 80.
- Predict the relative intensities of the m/z 78 and m/z
80 peaks and justify your answer using isotopic abundances.
- Identify the fragment ion responsible for the base peak at m/z
43 and explain its formation.
- Deduce the structure of the compound from the fragmentation pattern and
justify your answer.
- Explain how the mass spectrum could help distinguish 1-chloropropane
from its isomer 2-chloropropane.
Model Answers
a) Isotopic Pair at m/z 78 and 80
- Chlorine has two major isotopes: ³⁵Cl and ³⁷Cl.
- The molecular ion exists in two forms:
-
C3H7³⁵Cl → m/z 78
-
C3H7³⁷Cl → m/z 80
- These peaks are two mass units apart due to the isotopic difference.
b) Relative Intensities of m/z 78 and
80
- Natural abundance of chlorine isotopes:
- ³⁵Cl ≈ 75.8%
- ³⁷Cl ≈ 24.2%
- Therefore, the m/z 78 peak will be about three times more
intense than the m/z 80 peak, forming a 3:1 ratio.
c) Fragment Ion at m/z 43
- The m/z 43 peak corresponds to the propyl cation (C₃H₇⁺).
- It forms when the Cl atom is lost as a neutral radical:
-
C3H7Cl →
C3H7⁺ (m/z 43) + Cl•
- This is a common fragmentation pathway due to the relatively weak C–Cl
bond.
d) Structural Deduction
- The presence of a strong m/z 43 peak suggests a straight-chain
propyl cation.
- The molecular ion peaks at m/z 78 and 80 confirm the presence
of one chlorine atom.
- The fragmentation pattern supports a primary chloroalkane with the Cl at
the end of a straight chain.
- Therefore, the structure is CH3CH2CH2Cl
— 1-chloropropane.
e) Distinguishing from 2-Chloropropane
- Both compounds show molecular ion peaks at m/z 78 and 80 due to
Cl isotopes.
- However, their fragmentation differs:
- 1-chloropropane forms a straight-chain propyl cation (m/z
43), which is the base peak
and has much more abundant m/z 29 C2H5+
ion.
- 2-chloropropane forms a branched isopropyl cation (m/z 43),
but the fragmentation pattern may differ in intensity and stability. .
- The relative abundance and stability of the m/z 43 peak, along
with other minor peaks, helps distinguish the isomers.
Below
are more questions for you to do for yourself
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QUESTIONS
Advanced A-level chemistry - practise exam questions on
the mass spectrum
of 1-chloropropane
Jot
down your responses and check out the answers:
ANSWERS
If you think there are
any errors, please email me asap at
chem55555@hotmail.com
I don't mind if students/teachers do a selected printout
of these questions and answers.
Q1
What features in the mass spectrum of 1-chloropropane
would clearly indicate that molecule contained a single
chlorine atom.
ANSWER
Chlorine has two principal isotopes, 35Cl
and 37Cl in the ratio 3:1. Therefore look for
pairs of peaks in the ratio 3:1 e.g. m/z ion values for
R35Cl and R37Cl where R is part of
the original organic molecule.
Q2
Identify the structure of the following ions of m/z
values 15, 29, 42, 43, 49, 51, 63 and 65 and which would
you expect to observe in a 3:1 ratio and why?
ANSWERS
15 [CH3]+,
29 [C2H5]+,
42 [C3H6]+,
43 [C3H7]+,
49 [CH235Cl]+,
51 [CH237Cl]+,
63 [CH2CH235Cl]+,
65 [CH2CH237Cl]+.
You would expect to 3:1 ratios for m/z values 48/51 and
63/65 because of the answer to Q1.
Q3 The ions of
m/z values 42 and 43 are amongst the most prominent
peaks in the mass spectrum of 1-chloropropane. Given the
bond enthalpies kJ/mol: C-C = 348; C-Cl = 338;
C-H = 412, suggest reasons for their relative prominence and suggest
some fragmentation pathways to explain their occurrence.
ANSWERS
The C-Cl bond is the
weakest in the 1-chloropropane molecule and likely to be
the first broken in the ionisation process in the mass
spectrometer.
for m/z
43: [CH3CH2CH2Cl]+ ===> [CH3CH2CH2]+
+ Cl
Bond scission of the C-Cl bond, the weakest bond in
the molecule.
for m/z 42: [CH3CH2CH2Cl]+ ===> [C3H6]+
+ HCl
Elimination of hydrogen chloride to give a positive
ion of propene and this fragmentation pathway also involves the
breaking of the weakest bond in the 1-chlorpropane molecule.
Jot
down your responses and check out the answers:
ANSWERS
If you think there are
any errors, please email me asap at
chem55555@hotmail.com
|
Key words & phrases: C3H7Cl CH3CH2CH2Cl image diagram on how to interpret and explain the mass spectrum of
1-chloropropane m/z m/e base peaks, image and diagram of the mass spectrum of
1-chloropropane, details of the mass spectroscopy of 1-chloropropane, low and high resolution mass
spectrum of 1-chloropropane, prominent m/z peaks in the mass spectrum of
1-chloropropane, comparative
mass spectra of 1-chloropropane, the molecular ion peak in the mass spectrum of
1-chloropropane,
analysing and understanding the fragmentation pattern of the mass spectrum
of 1-chloropropane, characteristic pattern of peaks in the mass spectrum of
1-chloropropane, relative
abundance of mass ion peaks in the mass spectrum of 1-chloropropane, revising the mass
spectrum of 1-chloropropane, revision of mass spectroscopy of 1-chloropropane, most abundant ions in the
mass spectrum of 1-chloropropane, how to construct the mass spectrum diagram for abundance
of fragmentation ions in the mass spectrum of 1-chloropropane, how to analyse the mass
spectrum of 1-chloropropane, how to describe explain the formation of fragmented ions in the
mass spectra of 1-chloropropane equations for explaining the formation of the positive ions
in the fragmentation of the ionised molecule of 1-chloropropane recognising the
base ion peak of 1-chloropropane interpreting
interpretation the mass spectrum of 1-chloropropane
n-propyl chloride How do you interpret the mass spectrum of
1-chloropropane How to interpret
the mass spectrum of 1-chloropropane Explanatory diagram of the mass spectrum of the
1-chloropropane molecule in
terms of its molecular structure.
Listing data of the prominent main peaks in the mass spectrum of
1-chloropropane. How to explain the mass spectrum of 1-chloropropane. The m/z value of the
molecular ion peak in the mass spectrum of 1-chloropropane. Identifying
1-chloropropane from
its mass spectrum pattern. The m/z m/e peak analysis of the mass
spectrum of the 1-chloropropane molecule. The uses of the mass spectrum of the
1-chloropropane molecule. The distinctive features of the mass spectrum of
the 1-chloropropane molecule explained. explaining the fragmentation pattern of the mass spectrum of
1-chloropropane equations showing the
formation of the ionised fragments in the mass spectrum of
1-chloropropane
what does the mass spectrum tell you about the structure and
properties of the 1-chloropropane molecule? Data table of ionised
fragments in the mass spectrum of 1-chloropropane and equations for their
formation in the fragmentation of 1-chloropropane molecules
Links associated
with
1-chloropropane
The
infrared spectrum of 1-chloropropane
The
H-1 NMR spectrum of 1-chloropropane
The
C-13 NMR spectrum of 1-chloropropane
The chemistry of HALOGENOALKANES (haloalkanes)
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the spectroscopy of 1-chloropropane - its mass spectrum,
detailed analysis, diagnostic features, data analysed, useful spectra comments are
suitable for use of pre-university students studying AQA advanced level
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ANSWERS
Advanced A-level chemistry - practise exam questions on
the mass spectrum
of 1-chloropropane
If you think there are
any errors, please email me asap at
chem55555@hotmail.com
Q1
What features in the mass spectrum of 1-chloropropane
would clearly indicate that molecule contained a single
chlorine atom.
ANSWER
Chlorine has two principal isotopes, 35Cl
and 37Cl in the ratio 3:1. Therefore look for
pairs of peaks in the ratio 3:1 e.g. m/z ion values for
R35Cl and R37Cl where R is part of
the original organic molecule.
Q2
Identify the structure of the following ions of m/z
values 15, 29, 42, 43, 49, 51, 63 and 65 and which would
you expect to observe in a 3:1 ratio and why?
ANSWERS
15 [CH3]+,
29 [C2H5]+,
42 [C3H6]+,
43 [C3H7]+,
49 [CH235Cl]+,
51 [CH237Cl]+,
63 [CH2CH235Cl]+,
65 [CH2CH237Cl]+.
You would expect to 3:1 ratios for m/z values 48/51 and
63/65 because of the answer to Q1.
Q3 The ions of
m/z values 42 and 43 are amongst the most prominent
peaks in the mass spectrum of 1-chloropropane. Given the
bond enthalpies kJ/mol: C-C = 348; C-Cl = 338;
C-H = 412, suggest reasons for their relative prominence and suggest
some fragmentation pathways to explain their occurrence.
ANSWERS
The C-Cl bond is the
weakest in the 1-chloropropane molecule and likely to be
the first broken in the ionisation process in the mass
spectrometer.
for m/z
43: [CH3CH2CH2Cl]+ ===> [CH3CH2CH2]+
+ Cl
Bond scission of the C-Cl bond, the weakest bond in
the molecule.
for m/z 42: [CH3CH2CH2Cl]+ ===> [C3H6]+
+ HCl
Elimination of hydrogen chloride to give a positive
ion of propene and this fragmentation pathway also involves the
breaking of the weakest bond in the 1-chlorpropane molecule.
If you think there are
any errors, please email me asap at
chem55555@hotmail.com
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