Advanced Organic Chemistry: Mass spectrum of 2-bromopropane CH3CHBrCH3

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Interpreting the mass spectrum of 2-bromopropane CH3CHBrCH3

[Author ©  Dr Phil Brown 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 2-bromopropane [updated Mar 12th 2026 *]

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Introductory note on the mass spectrum of 2-bromopropane

Students and teachers please note my explanation of the mass spectrum of 2-bromopropane is designed for advanced, but pre-university, chemistry courses.

If M represents the 2-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 2-bromopropane and only the formation of singly charged positive are considered for the mass spectrum of 2-bromopropane.

I've included a stick diagram and table of m/z ions for the mass spectrum of 2-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 2-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!

C3H7Br CH3CHBrCH3 mass spectrum of 2-bromopropane fragmentation pattern of m/z m/e ions for analysis and identification of isopropyl bromide image diagram doc brown's advanced organic chemistry revision notes 

2-bromopropane, C3H7Br, CH3CHBrCH3(c) doc b , (c) doc b

The molecular structure and naming of haloalkanes

Interpreting the fragmentation pattern of the mass spectrum of 2-bromopropane (a secondary haloalkane)

[M]+ is the molecular ion peaks (M) with an m/z of 122 and 124 corresponding to [C3H7Br]+, the original 2-bromopropane molecule minus an electron, [CH3CHBrCH3]+

There are two molecular ion peaks because bromine as two isotopes, 50.6% 79Br and 49.4% 81Br.

m/z ion 122 [CH3CH79BrCH3]+ is the slightly higher intensity M peak because of the slightly higher % of the bromine-79 isotope, and considered the M ion

m/z ion 124 [CH3CH81BrCH3]+ is the slightly lesser intensity M+2 peak because of the slightly lesser % of the bromine-81 isotope, and considered the M+2 ion

Their average relative isotopic mass is ~80, so the relative molecular mass for 2-bromopropane is ~123.

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), .

You might see very tiny M+1 and M+3 peaks at m/z 123 and 125, corresponding to an ionised 2-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+1/3 peaks.

2-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 (2-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 2-bromopropane is the m/z 43 ion [C3H7]+

Identifying the species giving the most prominent peaks (apart from M) in the fragmentation pattern of 2-bromopropane.

The parent molecular ion peaks are m/z 122 and 124 ions: [CH3CH79BrCH3]+ and [CH3CH81BrCH3]+

Unless otherwise indicated, assume the carbon atoms in 2-bromopropane are the 12C isotope.

m/z value of [fragment]+ 124 122 109 107 82 81 80 79
[molecular fragment]+ [C3H781Br]+ [C3H779Br]+ [C2H481Br]+ [C2H479Br]+ [H81Br]+ [81Br]+ [H79Br]+ [79Br]+
m/z value of [fragment]+ 44 43 42 41 39 27 15
[molecular fragment]+ [13C12C2H7]+ [C3H7]+ [C3H6]+ [C3H5]+ [C3H3]+ [C2H3]+ [CH3]+

Analysing and explaining the principal ions in the fragmentation pattern of the mass spectrum of 2-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% 13);  H = 1; O = 16;  Br = 79 or 81 (ratio ~1 : 1)

Bond enthalpies kJ/mol: C-C = 348;  C-H = 412;  C-Br 276

Equations to explain the most abundant ion peaks of 2-bromopropane

Formation of m/z 107 and 109 ions:

[CH3CHBrCH3]+  ===>  [C2H4Br]+  +  CH3

C-C bond scission to lose an 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.

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 50.6 : 49.4).

Formation of m/z 80 and 82 ions:

[CH3CHBrCH3]+  ===>  [H79Br]+  or  [H81Br]+  +  C3H6

Elimination of hydrogen bromide from the parent molecular ion.

Although small, the 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:

[CH3CHBrCH3]+  ===>  [79Br]+  or  [81Br]+  +  C3H7

C-Br bond scission, but the alkyl fragment is more likely to carry the positive charge (see below).

Again, although small, the 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:

[CH3CHBrCH3]+  ===>  [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 peak ion, the most abundant and 'stable' ion fragment.

It undergoes success proton loss e.g. 43 => 42 => 41 => 40 => 39

Formation of m/z 41 ion:

[C3H7]+  ===>  [C3H5]+  +  H2

Loss of hydrogen, mass change 43 - 2 = 41

Formation of m/z 27 ion:

[C2H4Br]+  ===>  [C2H3]+  +  HBr

Elimination of hydrogen bromide.

Mass loss 109 - 82 = 27  and  107 - 80 = 27

Formation of m/z 15 ion:

[CH3CHBrCH3]+  ===>  [CH3]+  +  C2H4Br

C-C bond scission.

Mass loss 124 - 109 = 15  and  122 - 107 = 15


Summary of the mass spectrum of 2-bromopropane and extra comments

The mass spectrum of 2-bromopropane (CH3–CHBr–CH3) with exam precision and clarity.


Key Features of 2-Bromopropane of relevance to its mass spectrum

  • Molecular formula: C3H7Br
  • Relative molecular mass: ~123
  • Contains bromine, which has two major isotopes: ⁷⁹Br and ⁸¹Br (nearly 1:1 ratio)
  • Fragmentation: Typical alkyl cleavage and halogen loss

Prominent m/z Ions and Their Origins in the mass spectrum of 2-bromopropane

m/z Ion Origin / Fragmentation Notes
122 [M]⁺ (with ⁷⁹Br) Molecular ion with ⁷⁹Br isotope One of the twin molecular ion peaks
124 [M+2]⁺ (with ⁸¹Br) Molecular ion with ⁸¹Br isotope Equal intensity to m/z 122
43 C3H7 Propyl cation from cleavage of C–Br Often the base ion peak
41 C3H5 Allyl-type fragment (loss of H2) Resonance-stabilized
27 C2H3 Vinyl cation or rearranged fragment Seen in many halogenoalkanes

Common Misconceptions about the mass spectrum of 2-bromopropane (see below too)

  • Confusing M and M+2 peaks: Students may expect a single molecular ion peak. Bromine causes two peaks (m/z 122 and 124) of equal intensity due to isotopes.
  • Expecting a peak at average mass (~123): Mass spectrometry shows actual isotopic masses, not averages.
  • Misidentifying base peak: The most intense peak is often not the molecular ion but a stable fragment like C3H7⁺ (m/z 43).
  • Overinterpreting small peaks: Peaks like m/z 27 may be misattributed unless fragmentation pathways are understood.

Exam Tips for questions involving the mass spectrum of 2-bromopropane (see above too)

  • Always mention bromine isotopes: “Twin molecular ion peaks at m/z 122 and 124 due to ⁷⁹Br and ⁸¹Br.”
  • Identify base peak clearly: “The base peak at m/z ~43 corresponds to the propyl cation.”
  • Use correct ion notation: [M]⁺ for molecular ion, not just “parent peak.”
  • Avoid average mass references: Quote actual m/z values, not calculated averages.
  • Link fragmentation to structure: “Loss of Br gives a stable alkyl cation at m/z ~43.”

Practice Question: Mass Spectrum of 2-Bromopropane

Compound: 2-bromopropane (C3H7Br)

A student analyses the mass spectrum of 2-bromopropane. The spectrum shows two prominent molecular ion peaks at m/z 122 and m/z 124, and a base peak at m/z 43.

Question:

  1. Explain why the molecular ion peak appears as a pair at m/z 122 and m/z 124.
  2. Identify the fragment ion responsible for the base peak at m/z 43 and explain its formation.
  3. Suggest why the molecular ion peak is less intense than the base peak.
  4. Predict the relative intensities of the m/z 122 and m/z 124 peaks and justify your answer using isotopic abundances.
  5. Explain how the fragmentation pattern helps distinguish 2-bromopropane from its isomer 1-bromopropane.

Model Answer

a) Isotopic Pair at m/z 122 and 124

  • Bromine has two major isotopes: ⁷⁹Br and ⁸¹Br.
  • 2-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 isopropyl cation (CH(CH3)2⁺).
  • It forms when the Br atom is lost as a neutral radical:
    • C3H7Br → CH(CH3)2⁺ (m/z 43) + Br•
  • This is the most stable carbocation formed from 2-bromopropane due to tertiary-like stabilization.

c) Molecular Ion Peak Intensity

  • The molecular ion peak is less intense because:
    • The C–Br bond is relatively weak and easily cleaved.
    • The isopropyl cation is highly stable, so fragmentation is favoured.
    • Many molecules fragment before reaching the detector intact.

d) 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.

e) Distinguishing from 1-Bromopropane

  • Both compounds show molecular ion peaks at m/z 122 and 124 due to Br isotopes.
  • However, their fragmentation differs:
    • 1-bromopropane forms a straight-chain propyl cation (m/z 43) with less stability.
    • 2-bromopropane forms a branched isopropyl cation (m/z 43) which is more stable and dominant.
  • The relative intensity and stability of the m/z 43 peak is higher in 2-bromopropane, making it the base peak.

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Links associated with 2-bromopropane

The infrared spectrum of 2-bromopropane

The H-1 NMR spectrum of 2-bromopropane

The C-13 NMR spectrum of 2-bromopropane

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