mass spectrum of 1-bromobutane C4H9Br CH3CH2CH2CH2Br fragmentation pattern of m/z m/e ions for analysis and identification of n-butyl iodide image diagram doc brown's advanced organic chemistry revision notes

Advanced pre-university Organic Chemistry: Mass spectrum of 1-bromobutane CH₃CH₂CH₂CH₂Br

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Interpreting and explaining the mass spectrum of 1-bromobutane

[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 analysis of 1-bromobutane (mass spectra) [spectra page updated RE-EDIT]

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and Isomers of molecular formula C₄H₉X (where X = F, Cl, Br or I and basic data on NMR chemical shifts)

Practise exam questions based on the mass spectrum of 1-bromobutane

Introductory note on the mass spectrum of 1-bromobutane

Students and teachers please note my explanation of the mass spectrum of 1-bromobutane is designed for advanced, but pre-university, chemistry courses.

If M represents the 1-bromobutane molecule, the initial ionisation to give the molecular ion is:

M_(g) + high KE e^- ==> [M•]⁺_(g) + 2e^- and 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 are formed in the fragmentation pattern for 1-bromobutane.

I've included stick diagram and table of m/z ions for the mass spectrum of 1-bromobutane and conducting the mass spectrum analysis under standard conditions, a database can be built up based complex fingerprint patterns, often involving relative intensities of many fragment ions, that can be used to identify compounds including 1-bromobutane.

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 and compared the accurate ion masses if appropriate for 1-bromobutane. 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 for 1-bromobutane, but the mass spectrometer software does!

mass spectrum of 1-bromobutane C4H9Br CH3CH2CH2CH2Br fragmentation pattern of m/z m/e ions for analysis and identification of n-butyl iodide image diagram doc brown's advanced organic chemistry revision notes

1-bromobutane, C₄H₉Br, CH₃CH₂CH₂CH₂Br, CH₃-CH₂-CH₂-CH₂-Br

The molecular structure and naming of haloalkanes

Interpreting the fragmentation pattern of the mass spectrum of 1-bromobutane

[M]⁺ is the parent molecular ion peaks (M) with m/z values of 136 and 138 corresponding to [C₄H₉Br]⁺, the original 1-bromobutane molecule minus an electron, [CH₃CH₂CH₂CH₂Br]⁺.

There are two possibilities because bromine has two isotopes, ⁷⁹Br and ⁸¹Br in the ratio ~1 : 1.

Therefore the molecular ion can be [CH₃CH₂CH₂CH₂⁷⁹Br]⁺ and [CH₃CH₂CH₂CH₂⁸¹Br]⁺, which should, and do show up, as a double peak of ~equal heights (~equal abundance).

These are referred to as the M and M+2 peaks respectively, emphasising the two mass unit difference due to the bromine isotopes in the two molecular ions of 1-bromobutane.

The two bromine isotopes also account for the 'twin peaks' of m/z ions 93 & 95 and 107 & 109 (details in the analysis below).

% isotopic composition of isotopes in the naturally occurring element:

⁷⁹Br = 78.9183 (50.7%) and ⁸¹Br = 80.9163 (49.3%)

The peak for molecular/fragment ions with the heavier, slightly less abundant ⁸¹Br isotope, are slight shorter (slightly smaller intensity).

Bromine consists of two isotopes, ⁷⁹Br and ⁸¹Br in roughly equal proportions, therefore any molecular ion or fragment containing a bromine atom will show up as a double peak of similar height (abundance) two mass units apart e.g. m/z ions 93 and 95, and 107 and 109, plus the molecular ion peaks of m/z values 136 and 138 (but still, all pairs of ~equal height!) in the mass spectrum of 1-bromobutane.

The very small M+1 and M+3 peaks at m/z 137 and 139, corresponds to an ionised 1-bromobutane molecule with one ¹³C atom in it i.e. an ionised 1-bromobutane molecule of formula [¹³C¹²C₃H₉Br]⁺

Unless otherwise stated, C means a ¹²C atom, if not, the isotopic carbon atom ¹³C will be indicated.

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-bromobutane has 4 carbon atoms, so on average, ~1 in 25 molecules will contain a ¹³C atom.

The most abundant ion of the molecule under mass spectrometry investigation (1-bromobutane) 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-bromobutane is m/z ion 57 [C₄H₉]⁺

Identifying the species giving the most prominent peaks (apart from M) in the fragmentation pattern of 1-bromobutane.

Unless otherwise indicated, assume the carbon atoms in 1-bromobutane are the ¹²C isotope.

Some of the possible positive ions, [molecular fragment]⁺, formed in the mass spectrometry of 1-bromobutane.

The parent molecular ion M of 1-bromobutane is m/z 136: [CH₃CH₂CH₂CH₂⁷⁹Br]⁺

but the M+2 m/z 138 molecular ion [CH₃CH₂CH₂CH₂⁸¹Br]⁺ will also fragment in the same way.

m/z value of [fragment]⁺	138	136	109	107	95	93	58, with ¹³C atom	57, all ¹²C atoms
[molecular fragment]⁺	[C₄H₉⁸¹Br]⁺	[C₄H₉⁷⁹Br]⁺	[C₂H₄⁸¹Br]⁺	[C₂H₄⁷⁹Br]⁺	[CH₂⁸¹Br]	[CH₂⁷⁹Br]	[C₄H₉]⁺	[C₄H₉]⁺

m/z value of [fragment]⁺	56	55	79	80	81	82	m/z ions of 79 to 82 have a very low abundance, just tiny peaks in the mass spectrum of 1-bromobutane.
[molecular fragment]⁺	[C₄H₈]⁺	[C₄H₇]⁺	[⁷⁹Br]⁺	[H⁷⁹Br]⁺	[⁸¹Br]⁺	[H⁸¹Br]⁺

m/z value of [fragment]⁺	43	42	41	40	39	29	28	27	26	15
[molecular fragment]⁺	[C₃H₇]⁺	[C₃H₆]⁺	[C₃H₅]⁺	[C₃H₂]⁺	[C₃H₃]⁺	[C₂H₅]⁺	[C₂H₄]⁺	[C₂H₃]⁺	[C₂H₂]⁺	[CH₃]⁺

Analysing and explaining the principal ions in the fragmentation pattern of the mass spectrum of 1-bromobutane

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); Br = 79 or 81 (~1:1 isotope abundance ratio)

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

Possible equations to explain some of the most abundant ion peaks of 1-bromobutane (tabulated above)

Formation of m/z 107 and 109 ions:

[CH₃CH₂CH₂CH₂Br]⁺ ===> [C₂H₄Br]⁺ + C₂H₅

C-C bond scission in the parent molecular ion,

mass change 136/138 - 29 = 107/109 (M-29 'twin' ion peaks)

The C-Br bond is the weakest bond in the molecule, hence the most likely bond scission is C-Br with 1-bromobutane (see below m/z 57 ion).

Note the twin ~1:1 peaks due to the two bromine isotopes.

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:

[CH₃CH₂CH₂CH₂Br]⁺ ===> [CH₂Br]⁺ + C₃H₇

C-C bond scission in the parent molecular ion,

mass change 136/138 - 43 = 93/95 (M-43 'twin' ion peaks)

The C-Br bond is the weakest bond in the molecule, hence the most likely bond scission is C-Br with 1-bromobutane (see below m/z 57 ion).

Note the twin ~1:1 peaks due to the two bromine isotopes.

Again, the 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 57 ion:

[CH₃CH₂CH₂CH₂Br]⁺ ===> [C₄H₉]⁺ + Br

The [C₄H₉]⁺ ion is likely to be the more stable secondary carbocation [(CH₃)₃C]⁺ rather than the linear butyl carbocation [CH₃CH₂CH₂CH₂]⁺.

This alternative ionisation compared to above is much more likely, C-Br bond (weakest) scission in the parent molecular ion,

mass change 136/138 - 79/81 = 57

The m/z 57 ion is the base peak ion, the most abundant and 'stable' ion fragment.

The m/z 58 ion is likely to be [¹³C¹²C₃H₉]⁺ i.e. as above but with a ¹³C atom in the hydrocarbon fragment.

The m/z 57 ion can lose a hydrogen atom/molecule to give m/z ions 56 and 55.

There is a low probability that the bromine atom can also be ionised to give m/z 79 and 81 ions - you can just about make out the tiny twin peaks.

Formation of m/z 56 ion:

[CH₃CH₂CH₂CH₂Br]⁺ ===> [C₄H₈]⁺ + HBr

Elimination of hydrogen bromide from the parent molecular ion can also give the m/z 56 ion.

Mass change 136/138 - 80/82 = 56.

There is a low probability that the hydrogen bromide molecule can also be ionised to give m/z 80 and 82 ions - you can just about make out the tiny twin peaks.

Formation of m/z 41 and 39 ions:

Possible reactions include:

m/z 41: [C₄H₈]⁺ ===> [C₃H₅]⁺ + CH₃

m/z 39: [C₃H₅]⁺ ===> [C₃H₃]⁺ + H₂

Formation of m/z 29, 28, 27 and 26 ions:

Possible reactions include:

[CH₃CH₂CH₂CH₂Br]⁺ ===> [C₂H₅]⁺ + CH₂CH₂Br

From bond scission in the parent molecular ion.

m/z 27: [C₄H₈]⁺ ===> [C₂H₃]⁺ + C₂H₅

m/z 28: [C₄H₈]⁺ ===> [C₂H₄]⁺ + C₂H₄

m/z 29: [C₄H₈]⁺ ===> [C₂H₅]⁺ + C₂H₃

You can also get m/z ions 26 to 28 from proton loss from the m/z 29 ion.

QUESTIONS

Advanced A-level chemistry - practise exam questions on the mass spectrum of 1-bromobutane

This is a joint AI-doc b experiment!

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. In the mass spectrum of 1-bromobutane, which statement about the molecular ion region is most accurate?

A. There is a single molecular ion peak at $m / z = 136$

B. There are two molecular ion peaks at $m / z = 136$ and $m / z = 138$ of similar intensity

C. There are two molecular ion peaks at $m / z = 136$ and $m / z = 138$ with a 3:1 intensity ratio

D. There is a single molecular ion peak at $m / z = 80$

Q2. Which feature in the mass spectrum of 1-bromobutane most clearly indicates the presence of bromine?

A. A single intense peak at $m / z = 44$

B. Two peaks 2 m/z units apart with a 3:1 intensity ratio

C. Two peaks 2 m/z units apart with approximately equal intensity

D. A broad unresolved cluster around $m / z = 100$

Q3. In the mass spectrum of 1-bromobutane, peaks at $m / z = 79$ and $m / z = 81$ are observed. These are best assigned to:

A. $C 3 H 7 +$ B. $B r +$ C. $C 4 H 9 +$ D. $C 2 H 5 B r +$

Q4. In many teaching spectra of 1-bromobutane, the base peak (most intense peak) is at $m / z = 57$ . Which fragment does this correspond to?

A. $C 2 H 5 +$ B. $C 3 H 7 +$ C. $C 4 H 9 +$ D. $B r +$

Q5. Using $C = 12$ , $H = 1$ , $Br = 79$ or 81, which pair of nominal m/z values corresponds to the molecular ion peaks of 1-bromobutane?

A. 79 and 81

B. 137 and 139

C. 43 and 45

D. 136 and 138

Q6. You are given two spectra: one for 1-bromobutane and one for 1-chlorobutane. Which observation identifies the bromobutane spectrum?

A. Molecular ion doublet 2 m/z apart with a 3:1 intensity ratio

B. Molecular ion doublet 2 m/z apart with a 1:1 intensity ratio

C. Single molecular ion peak with no isotopic splitting

D. Cluster of peaks 1 m/z apart with equal intensity

Q7. Which description best explains the formation of the butyl cation ( $C 4 H 9 +$ , $m / z = 57$ ) from 1-bromobutane?

A. Loss of $a$ $C H 3$ radical from the molecular ion

B. Cleavage of the C–Br bond and rearrangement to a secondary carbocation on the third carbon

C. Loss of $a B r +$ from the molecular ion

D. Loss of $a H +$ from the molecular ion

Q8. Molecular ion versus base peak

A student says: “The highest m/z peak in any mass spectrum is always the base peak.” For 1-bromobutane, why is this statement wrong?

A. The highest m/z peak is always a fragment, never the molecular ion

B. The base peak is the most intense peak, not necessarily the one with the highest m/z

C. The base peak always occurs at $m / z = 44$ for organic molecules

D. The base peak must be the Br⁺ peak at $m / z = 79$

Q9. How does high-resolution mass spectrometry (HRMS) help confirm that the compound giving the 1-bromobutane spectrum is indeed $C 4 H 9 B r$ ?

A. HRMS increases the intensity of the molecular ion peak

B. HRMS measures exact m/z values to distinguish between different formulas with the same nominal mass

C. HRMS removes isotopic peaks, leaving only the most abundant isotope

D. HRMS only detects fragment ions, not molecular ions

Q10. Interpreting a given spectrum

A spectrum shows:

Molecular ion doublet at $m / z = 136$ and $m / z = 138$ (1:1 ratio)
Strong peaks at $m / z = 43$ , 79, and 81

Which conclusion is most reasonable?

A. The compound is an unhalogenated C₄ alkane

B. The compound contains chlorine and a propyl fragment

C. The compound contains bromine and a propyl fragment, consistent with 1-bromobutane

D. The compound must be an alcohol due to the strong peak at 43

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

Comparing the infrared, mass, 1H NMR and 13C NMR spectra of the 4 halogenoalkane isomers of C₄H₉Br NOTE: The images are linked to their original detailed spectral analysis pages AND can be doubled in size with touch screens to increase the definition to the original 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane image sizes. These four molecules are structural isomers of molecular formula C₄H₉Br and exemplify the infrared, mass, 1H NMR and 13C NMR spectra of lower aliphatic halogenoalkanes (haloalkanes, alkyl halides, bromoalkanes, alkyl bromides).


INFRARED SPECTRA (above): Apart from the significant differences in the fingerprint region at wavenumbers 1500 to 400 cm^-1, there are no other great striking differences, but each could be identified from its infrared spectrum.


MASS SPECTRA (above): All four give the parent molecular ions of m/z 136 and 138, but it is only a relatively tiny peak for 2-bromobutane and 2-bromo-2-methylpropane. All four give the base ion peak of m/z 57. All four give prominent peaks for m/z ions 27, 29, 39 and 41 and all give a tiny peak from an ionised iodine atom at m/z 127. They look quite similar to me and lack a clear fingerprint fragmentation pattern. There are small differences in the relative abundances (peak heights) for pairs of ions involving ⁷⁹Br/⁸¹Br isotopes e.g. m/z 93/95, 107/109 and 121/123. 1-bromo-2-methylpropane is the only one of the four to have a prominent peak for the m/z 43 ion.


1H NMR SPECTRA (above): The 1H NMR spectra of all four molecules give different integrated proton ratios i.e.1-bromobutane four peaks of ratio 3:2:2:2; 2-bromobutane four peaks of ratio 3:3:2:1, 1-bromo-2-methylpropane three peaks of ratio 6:2:1 and 2-bromo-2-methylpropane gives just one peak '1' (effectively no ratio involved), so all four molecular structures can be distinguished from each other by their ¹H NMR spectra proton ratios, numbers of peaks and (n+1) rule splitting patterns.


13C NMR SPECTRA (above): The 13C NMR spectra of the four molecules show various numbers of carbon-13 chemical environments i.e 1-bromobutane and 2-bromobutane show four 13C NMR resonances, 1-bromo-2-methylpropane three 13C NMR resonances and 2-bromo-2-methylpropane only two 13C resonances. Therefore 1-bromo-2-methylpropane and 2-bromo-2-methylpropane can be distinguished from the other three by their number of resonances in their 13C NMR spectra, but 1-bromobutane and 2-bromobutane cannot be distinguished from each other from their number of 13C NMR resonance lines - other data would be required.

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The infrared spectrum of 1-bromobutane (n-butyl bromide)

The H-1 NMR spectrum of 1-bromobutane (n-butyl bromide)

The C-13 NMR spectrum of 1-bromobutane (n-butyl bromide)

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ANSWERS

Advanced A-level chemistry - practise exam questions on the mass spectrum of 1-bromobutane

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. In the mass spectrum of 1-bromobutane, which statement about the molecular ion region is most accurate?

A. There is a single molecular ion peak at $m / z = 136$

B. There are two molecular ion peaks at $m / z = 136$ and $m / z = 138$ of similar intensity

C. There are two molecular ion peaks at $m / z = 136$ and $m / z = 138$ with a 3:1 intensity ratio

D. There is a single molecular ion peak at $m / z = 80$

Correct answer: B

Explanation:

1-bromobutane has formula $C 4 H 9 B r$ .

Bromine has two major isotopes, ⁷⁹Br and ⁸¹Br, in roughly 1:1 abundance.

This gives a doublet molecular ion: two peaks 2 m/z apart (136 and 138) with similar heights.

Common misconception:

C: confusing bromine (1:1) with chlorine (3:1).
D: mixing up the Br⁺ fragment (79/81) with the molecular ion.

Q2. Which feature in the mass spectrum of 1-bromobutane most clearly indicates the presence of bromine?

A. A single intense peak at $m / z = 44$

B. Two peaks 2 m/z units apart with a 3:1 intensity ratio

C. Two peaks 2 m/z units apart with approximately equal intensity

D. A broad unresolved cluster around $m / z = 100$

Correct answer: C

Explanation:

Bromine: ⁷⁹Br and ⁸¹Br ≈ 1:1 → two peaks 2 m/z apart, equal height.
This pattern appears in the molecular ion and in Br-containing fragments (e.g. Br⁺ at 79/81).

Common misconception:

B: again, chlorine’s 3:1 pattern.
A: assuming any strong peak is a “signature” of a particular element.

Q3. In the mass spectrum of 1-bromobutane, peaks at $m / z = 79$ and $m / z = 81$ are observed. These are best assigned to:

A. $C 3 H 7 +$ B. $B r +$ C. $C 4 H 9 +$ D. $C 2 H 5 B r +$

Correct answer: B

Explanation:

The pair at 79 and 81 with ~1:1 intensity is characteristic of Br⁺ (⁷⁹Br⁺ and ⁸¹Br⁺).
Hydrocarbon fragments (C₂, C₃, C₄) appear at lower m/z (e.g. 29, 43, 57).

Common misconception:

D: overcomplicating and assuming a brominated fragment rather than the simple halogen ion.

Q4. In many teaching spectra of 1-bromobutane, the base peak (most intense peak) is at $m / z = 57$ . Which fragment does this correspond to?

A. $C 2 H 5 +$ B. $C 3 H 7 +$ C. $C 4 H 9 +$ D. $B r +$

Correct answer: C

Explanation:

This is a relatively stable carbocation fragment formed from C-I bond scission and forms the base peak.

Common misconception:

D: assuming the halogen fragment must be the strongest peak.

Q5. Using $C = 12$ , $H = 1$ , $Br = 79$ or 81, which pair of nominal m/z values corresponds to the molecular ion peaks of 1-bromobutane?

A. 79 and 81

B. 137 and 139

C. 43 and 45

D. 136 and 138

Correct answer: D

Explanation:

$C 4 H 979 B r$ : $4 \times 12 + 9 \times 1 + 79 = 136$
$C 4 H 981 B r$ : $4 \times 12 + 9 \times 1 + 81 = 138$
In nominal, teaching-style spectra, these appear as 136 and 138 (electron mass + rounding), separated by 2 m/z.

Common misconception:

A: confusing the Br⁺ fragment with the molecular ion.
B: using raw arithmetic but not recognising how exam spectra label nominal m/z.

Q6. You are given two spectra: one for 1-bromobutane and one for 1-chlorobutane. Which observation identifies the bromobutane spectrum?

A. Molecular ion doublet 2 m/z apart with a 3:1 intensity ratio

B. Molecular ion doublet 2 m/z apart with a 1:1 intensity ratio

C. Single molecular ion peak with no isotopic splitting

D. Cluster of peaks 1 m/z apart with equal intensity

Correct answer: B

Explanation:

Chlorine: ³⁵Cl/³⁷Cl → 3:1 ratio, 2 m/z apart.
Bromine: ⁷⁹Br/⁸¹Br → 1:1 ratio, 2 m/z apart.
So the 1:1 doublet belongs to 1-bromobutane.

Common misconception:

A: classic chlorine pattern misapplied to bromine.

Q7. Which description best explains the formation of the butyl cation ( $C 4 H 9 +$ , $m / z = 57$ ) from 1-bromobutane?

A. Loss of $a$ $C H 3$ radical from the molecular ion

B. Cleavage of the C–Br bond and rearrangement to a secondary carbocation on the third carbon

C. Loss of $a B r +$ from the molecular ion

D. Loss of $a H +$ from the molecular ion

Correct answer: B

Explanation:

Ionisation causes C–Br bond cleavage, forming a butyl cation that can rearrange to a more stable secondary butyl cation.
This gives the intense $m / z = 57$ peak.

Common misconception:

C: assuming bromine leaves as Br⁺; in reality, Br usually leaves as a neutral radical, and the positive charge stays on the carbon fragment.

Q8. Molecular ion versus base peak

A student says: “The highest m/z peak in any mass spectrum is always the base peak.” For 1-bromobutane, why is this statement wrong?

A. The highest m/z peak is always a fragment, never the molecular ion

B. The base peak is the most intense peak, not necessarily the one with the highest m/z

C. The base peak always occurs at $m / z = 44$ for organic molecules

D. The base peak must be the Br⁺ peak at $m / z = 79$

Correct answer: B

Explanation:

Base peak = tallest peak (highest intensity).
Highest m/z = peak at largest mass-to-charge value.
For 1-bromobutane, the molecular ion region (136/138) is at high m/z, but the base peak is typically at $m / z = 43$ .

Common misconception:

A: overcorrecting and denying molecular ions.
D: assuming the halogen fragment must dominate, it would also be a ~1:1 doublet.

Q9. How does high-resolution mass spectrometry (HRMS) help confirm that the compound giving the 1-bromobutane spectrum is indeed $C 4 H 9 B r$ ?

A. HRMS increases the intensity of the molecular ion peak

B. HRMS measures exact m/z values to distinguish between different formulas with the same nominal mass

C. HRMS removes isotopic peaks, leaving only the most abundant isotope

D. HRMS only detects fragment ions, not molecular ions

Correct answer: B

Explanation:

HRMS gives exact mass (to several decimal places).
This allows you to confirm that the molecular ion matches the exact mass expected for $C 4 H 9 B r$ , ruling out other formulas with similar nominal mass.

Common misconception:

C: thinking resolution “simplifies” by deleting isotopes; in fact, it resolves them more clearly.

Q10. Interpreting a given spectrum

A spectrum shows:

Molecular ion doublet at $m / z = 136$ and $m / z = 138$ (1:1 ratio)
Strong peaks at $m / z = 43$ , 79, and 81

Which conclusion is most reasonable?

A. The compound is an unhalogenated C₄ alkane

B. The compound contains chlorine and a propyl fragment

C. The compound contains bromine and a propyl fragment, consistent with 1-bromobutane

D. The compound must be an alcohol due to the strong peak at 43

Correct answer: C

Explanation:

137/139 (1:1) → bromine in the molecular ion.
79/81 → Br⁺ fragment.
43 → propyl cation.
Together, these strongly support a brominated C₄ chain, i.e. 1-bromobutane.

Common misconception:

B: misreading the isotopic ratio as chlorine-like.
A: ignoring the clear bromine pattern.

Q1. In the mass spectrum of 1-bromobutane, which statement about the molecular ion region is most accurate?

Q2. Which feature in the mass spectrum of 1-bromobutane most clearly indicates the presence of bromine?

Q3. In the mass spectrum of 1-bromobutane, peaks at m/z=79 and m/z=81 are observed. These are best assigned to:

Q4. In many teaching spectra of 1-bromobutane, the base peak (most intense peak) is at m/z=57. Which fragment does this correspond to?

Q5. Using C=12, H=1, Br=79 or 81, which pair of nominal m/z values corresponds to the molecular ion peaks of 1-bromobutane?

Q6. You are given two spectra: one for 1-bromobutane and one for 1-chlorobutane. Which observation identifies the bromobutane spectrum?

Q7. Which description best explains the formation of the butyl cation (C4H9+, m/z=57) from 1-bromobutane?

Q8. Molecular ion versus base peak

Q9. How does high-resolution mass spectrometry (HRMS) help confirm that the compound giving the 1-bromobutane spectrum is indeed C4H9Br?

Q10. Interpreting a given spectrum

Q1. In the mass spectrum of 1-bromobutane, which statement about the molecular ion region is most accurate?

Q2. Which feature in the mass spectrum of 1-bromobutane most clearly indicates the presence of bromine?

Q3. In the mass spectrum of 1-bromobutane, peaks at m/z=79 and m/z=81 are observed. These are best assigned to:

Q4. In many teaching spectra of 1-bromobutane, the base peak (most intense peak) is at m/z=57. Which fragment does this correspond to?

Q5. Using C=12, H=1, Br=79 or 81, which pair of nominal m/z values corresponds to the molecular ion peaks of 1-bromobutane?

Q6. You are given two spectra: one for 1-bromobutane and one for 1-chlorobutane. Which observation identifies the bromobutane spectrum?

Q7. Which description best explains the formation of the butyl cation (C4H9+, m/z=57) from 1-bromobutane?

Q8. Molecular ion versus base peak

Q9. How does high-resolution mass spectrometry (HRMS) help confirm that the compound giving the 1-bromobutane spectrum is indeed C4H9Br?

Q10. Interpreting a given spectrum

Q3. In the mass spectrum of 1-bromobutane, peaks at $m / z = 79$ and $m / z = 81$ are observed. These are best assigned to:

Q4. In many teaching spectra of 1-bromobutane, the base peak (most intense peak) is at $m / z = 57$ . Which fragment does this correspond to?

Q5. Using $C = 12$ , $H = 1$ , $Br = 79$ or 81, which pair of nominal m/z values corresponds to the molecular ion peaks of 1-bromobutane?

Q7. Which description best explains the formation of the butyl cation ( $C 4 H 9 +$ , $m / z = 57$ ) from 1-bromobutane?

Q9. How does high-resolution mass spectrometry (HRMS) help confirm that the compound giving the 1-bromobutane spectrum is indeed $C 4 H 9 B r$ ?

Q3. In the mass spectrum of 1-bromobutane, peaks at $m / z = 79$ and $m / z = 81$ are observed. These are best assigned to:

Q4. In many teaching spectra of 1-bromobutane, the base peak (most intense peak) is at $m / z = 57$ . Which fragment does this correspond to?

Q5. Using $C = 12$ , $H = 1$ , $Br = 79$ or 81, which pair of nominal m/z values corresponds to the molecular ion peaks of 1-bromobutane?

Q7. Which description best explains the formation of the butyl cation ( $C 4 H 9 +$ , $m / z = 57$ ) from 1-bromobutane?

Q9. How does high-resolution mass spectrometry (HRMS) help confirm that the compound giving the 1-bromobutane spectrum is indeed $C 4 H 9 B r$ ?