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Mass spectrometry (MS) is the analytical chemist's molecular weighing scale — and far more besides. By ionising a sample, separating the resulting cations by mass-to-charge ratio (m/z), and recording their relative abundances, the technique returns a fingerprint from which both molecular identity and structural detail can be reconstructed. For an organic chemist, three pieces of information dominate: the molecular ion peak M⁺˳ (which gives the relative molecular mass Mᵣ directly), the fragmentation pattern (which betrays connectivity through the way the molecular ion breaks apart), and the isotope pattern (which fingerprints chlorine and bromine instantly and counts carbons in higher-resolution work). This lesson develops each of those threads with worked spectra, then signposts the contrast between hard ionisation methods such as electron impact and the soft ionisation techniques (ESI, MALDI) that have transformed proteomics and pharmacology.
Spec mapping (AQA 7405): This lesson maps to §3.3.15 (mass spectrometry of organic compounds) and connects back to §3.1.1.2 (atomic mass spectrometry, treated in atomic-structure lesson 1). Cross-references run forward to lesson 1 of this course (infrared spectroscopy), lesson 2 (¹H and ¹³C NMR), and the combined-techniques problem-solving in lessons 6 and 7. The instrumental cycle (ionisation → acceleration → deflection → detection) is identical to that taught for atomic mass spectrometry in atomic-structure L1; the new layer is the chemistry of fragmentation. Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: AO1 recall covers the four instrumental stages, the definition of the molecular ion as a radical cation, and the natural-abundance isotope ratios for Cl, Br and ¹³C. AO2 application is dominant — reading m/z values from a printed spectrum, identifying the molecular ion, computing losses to neighbouring peaks, and predicting fragment masses for a given structure. AO3 evaluation appears in deduction questions: proposing a structure that fits a given mass spectrum, choosing between two isomers on fragmentation evidence, and recognising when isotope ratios force the presence of a particular halogen. Combined-technique questions (MS + IR + NMR) are weighted heavily on Paper 2 and Paper 3.
The full instrumental treatment of mass spectrometry — vaporisation, ionisation, acceleration, deflection in a magnetic field, detection — was developed in atomic-structure lesson 1 and is not re-derived here. The key results carry over: ions of mass-to-charge ratio m/z are separated and recorded with their relative abundances on the y-axis, and the most intense peak is conventionally rescaled to 100% to give the base peak. For organic work, the dominant ionisation method is electron impact (EI) at ~70 eV, which knocks one electron out of the molecule to give a radical cation:
M(g) + e⁻ → M⁺˳(g) + 2e⁻
The product M⁺˳ is both positively charged (a cation) and bears one unpaired electron (a radical) — hence the symbol with both a "+" and a "·" superscript. Examiners insist on this notation, and an answer that writes the molecular ion as a simple cation M⁺ forfeits the AO1 mark for the definition. EI is a hard ionisation method: the 70 eV electrons deposit far more energy than a typical bond enthalpy (~3–5 eV), so most molecular ions fragment further before they reach the detector. The resulting cascade of daughter ions is what makes EI mass spectra so structurally informative; soft ionisation methods (introduced at the end of this lesson) give cleaner molecular ions at the cost of fragmentation detail.
The molecular ion peak appears at the highest m/z value in the spectrum (ignoring the small M+1 and M+2 isotope satellites discussed below). Because the charge on M⁺˳ is +1, m/z is numerically equal to the relative molecular mass Mᵣ. This is the single most important reading you take from any organic mass spectrum, and the conventional first step in any deduction question is to circle the molecular ion peak and write its m/z value.
Some compounds fragment so readily that the molecular ion peak is weak or even absent. Highly branched alkanes (such as 2,2,4-trimethylpentane) lose a methyl radical from a tertiary or quaternary carbon almost as soon as they ionise, leaving a tiny M⁺˳ peak buried under a forest of fragments. Alcohols frequently lose water (loss of 18) so efficiently that the M⁺˳ peak is dwarfed by [M−18]⁺˳. When in doubt, the M+1 isotope satellite (described below) can often be used to locate a weak molecular ion: it sits exactly one mass unit higher and carries a predictable intensity (~1.1n% of M⁺˳ for n carbons).
The nitrogen rule provides an elegant cross-check. Most stable organic molecules have an even Mᵣ. The exception: compounds containing an odd number of nitrogen atoms have an odd Mᵣ. This follows because the most common isotope of every common element except nitrogen has an even mass divided by an even valency (or odd mass divided by an odd valency), while ¹⁴N is uniquely even-mass / odd-valency (3). One nitrogen therefore shifts the molecular mass parity by one. A mass spectrum showing M⁺˳ at an odd m/z (e.g. 59, 73, 87) almost always contains one (or three, etc.) nitrogen atoms; methylamine CH₃NH₂ (Mᵣ 31), ethylamine C₂H₅NH₂ (Mᵣ 45) and propan-1-amine C₃H₇NH₂ (Mᵣ 59) are textbook examples.
Three elements give isotope patterns visible in routine organic mass spectra: ¹³C, ³⁵Cl/³⁷Cl, and ⁷⁹Br/⁸¹Br.
Carbon (¹²C/¹³C): Natural abundance of ¹³C is 1.1%. For every carbon in the molecule, there is a 1.1% chance that it is the heavier isotope, so the M+1 isotope peak has an intensity approximately 1.1n% of M⁺˳, where n is the number of carbons. A compound with M⁺˳ at m/z 78 and M+1 at 5.0% would have n ≈ 5/1.1 ≈ 4.5, suggesting 4 or 5 carbons; benzene (C₆H₆, n = 6, predicted M+1 ≈ 6.6%) and pyridine (C₅H₅N, n = 5, predicted M+1 ≈ 5.5%) are both plausible. This carbon-counting trick is heavily used in undergraduate organic chemistry and occasionally appears as an A* exam discriminator.
Chlorine (³⁵Cl/³⁷Cl): The natural abundance is ³⁵Cl 75.8%, ³⁷Cl 24.2% — almost exactly a 3:1 ratio. A compound with one chlorine therefore shows two molecular ion peaks: M⁺˳ (with ³⁵Cl) and M+2 (with ³⁷Cl) in a 3:1 height ratio. With two chlorines the pattern becomes triplet 9:6:1 (³⁵Cl₂, ³⁵Cl³⁷Cl, ³⁷Cl₂); with three chlorines it is a quartet 27:27:9:1. The doublet for monochloro compounds is diagnostic: any organic spectrum with two peaks separated by 2 m/z in a 3:1 ratio almost certainly contains one chlorine atom.
Bromine (⁷⁹Br/⁸¹Br): Almost exactly equal abundances — ⁷⁹Br 50.7%, ⁸¹Br 49.3% — give a 1:1 doublet for monobromo compounds. With two bromines the pattern is a triplet 1:2:1; with three bromines a quartet 1:3:3:1. A 1:1 doublet separated by 2 m/z is a near-perfect bromine fingerprint, and the visual contrast against chlorine's 3:1 doublet is one of the easiest discriminations in the whole of analytical chemistry.
Exam Tip: Memorise the two isotope-ratio fingerprints: chlorine = 3:1, bromine = 1:1, both separated by 2 m/z. The 3:1 vs 1:1 distinction is worth 1–2 marks on most papers and is a frequent A* discriminator when combined with fragment-mass analysis.
The molecular ion produced by EI typically carries 5–20 eV of excess internal energy — more than enough to break covalent bonds. Bond cleavage produces a smaller cation (which is detected) and a neutral radical or molecule (which is not). Schematically:
M⁺˳ → A⁺ + B˳ (A⁺ detected; B˳ invisible)
or
M⁺˳ → A˳ + B⁺ (B⁺ detected; A˳ invisible)
The cation that survives is, by Stevenson's rule, usually the one with lower ionisation energy — equivalently, the more thermodynamically stable cation. Tertiary > secondary > primary carbocations; allylic and benzylic cations are especially favoured; acylium ions R−C≡O⁺ are stabilised by π-delocalisation; the tropylium ion C₇H₇⁺ (m/z 91) is aromatically stabilised. The relative intensities of the daughter peaks therefore encode information about which fragments are thermodynamically favoured, and by extension about the structure of the parent.
| Mass Lost | Fragment Lost | Likely Functional Group / Origin |
|---|---|---|
| 15 | CH₃˳ | Methyl group |
| 17 | OH˳ | Hydroxyl group (alcohols, carboxylic acids) |
| 18 | H₂O | Alcohols (dehydration) |
| 28 | CO or N₂ | Carbonyl loss; phenols, anilines |
| 29 | CHO˳ or C₂H₅˳ | Aldehyde; ethyl group |
| 31 | CH₃O˳ (OCH₃) | Methyl ester / methoxy group |
| 43 | CH₃CO˳ or C₃H₇˳ | Methyl ketone; propyl group |
| 45 | COOH˳ or OC₂H₅˳ | Carboxylic acid; ethyl ester / ethoxy |
| 77 | C₆H₅˳ | Phenyl (loss of benzene minus H) |
| m/z | Ion | Structure / Origin |
|---|---|---|
| 15 | CH₃⁺ | Methyl cation |
| 29 | CHO⁺ or C₂H₅⁺ | Formyl (aldehydes); ethyl cation |
| 43 | CH₃CO⁺ or C₃H₇⁺ | Acylium (methyl ketones); propyl cation |
| 57 | C₄H₉⁺ or C₃H₅CO⁺ | Butyl cation; propanoyl cation |
| 77 | C₆H₅⁺ | Phenyl cation (substituted benzenes) |
| 91 | C₇H₇⁺ (tropylium) | Aromatic 7-ring cation from methylbenzene derivatives |
| 105 | C₆H₅CO⁺ | Benzoyl cation (aryl ketones, aromatic acids) |
Key Definition: The base peak is the most intense peak in the mass spectrum (rescaled to 100% relative abundance). The base peak is not necessarily the molecular ion peak — in many spectra a stabilised daughter ion (e.g. CH₃CO⁺ at m/z 43 for methyl ketones, or tropylium C₇H₇⁺ at m/z 91 for methylbenzenes) is more intense than M⁺˳ itself.
The benzylic and tropylium ions deserve special mention. When toluene (methylbenzene) ionises, the molecular ion fragments by loss of one hydrogen radical to give a cation initially of structure C₆H₅CH₂⁺ (benzyl, m/z 91). This rearranges almost instantaneously to the seven-membered aromatic tropylium ion C₇H₇⁺, in which the positive charge is delocalised over all seven carbons. The aromatic stabilisation is large enough that the tropylium peak is the base peak in toluene's spectrum, and a strong peak at m/z 91 is one of the most reliable indicators of a methylbenzene moiety anywhere in a structure.
Ethanol (CH₃CH₂OH) and methoxymethane (CH₃OCH₃) are structural isomers, both with molecular formula C₂H₆O and Mᵣ 46. Their mass spectra are immediately distinguishable.
Ethanol: M⁺˳ at m/z 46 (moderate intensity); strong peak at m/z 31 (CH₂OH⁺ — α-cleavage of the C–C bond next to the OH, very characteristic of primary alcohols); peak at m/z 45 (loss of one H from M); peak at m/z 29 (CHO⁺ after further loss); peak at m/z 27 (C₂H₃⁺ after loss of water).
Methoxymethane: M⁺˳ at m/z 46 (weak); base peak at m/z 45 (loss of CH₃˳ to give CH₃O−CH₂⁺ equivalent, or methoxymethyl cation); peak at m/z 15 (CH₃⁺) is prominent; no significant peak at m/z 31 because the symmetrical ether does not produce a stabilised CH₂OH-type fragment in the same way.
The discriminator is the m/z 31 peak: it is strong for ethanol, weak or absent for methoxymethane. This is a textbook MS deduction worth two marks in a typical structure-determination question.
A spectrum shows M⁺˳ at m/z 72, a base peak at m/z 43, and additional peaks at m/z 57 and m/z 29.
The compound is butanone, CH₃COCH₂CH₃. The dominant α-cleavage on each side of the C=O accounts for both base peak (m/z 43) and m/z 57 simultaneously.
1-chloropropane and 2-chloropropane share the molecular formula C₃H₇Cl (Mᵣ 78/80, 3:1 chlorine doublet). Their fragmentation patterns separate them:
Both isomers show the 3:1 chlorine doublet on M⁺˳, confirming a single Cl. The m/z 63/65 doublet (or its absence) discriminates them.
Chlorobenzene C₆H₅Cl: M⁺˳ at m/z 112/114 in 3:1 ratio (Cl fingerprint); strong peak at m/z 77 (C₆H₅⁺, loss of Cl˳); peak at m/z 51 (C₄H₃⁺ after further loss of C₂H₂).
Bromobenzene C₆H₅Br: M⁺˳ at m/z 156/158 in 1:1 ratio (Br fingerprint); strong peak at m/z 77 (C₆H₅⁺, loss of Br˳); peak at m/z 51 as above.
Both spectra share the m/z 77 phenyl-cation peak — the giveaway that an aromatic ring is present and a single substituent has been lost. The isotope ratio on M⁺˳ instantly distinguishes chlorine from bromine.
| Compound class | M⁺˳ strength | Diagnostic fragments | Notes |
|---|---|---|---|
| Straight-chain alkanes | Visible | Loss of 14 (CH₂) series; base peak often m/z 43 or 57 | "Picket fence" of CₙH₂ₙ₊₁⁺ peaks |
| Branched alkanes | Weak | Strong stabilised cations at branch points | M⁺˳ may be invisible |
| 1° alcohols | Often weak (dehydration) | m/z 31 (CH₂OH⁺); [M−18] from H₂O loss | Loss of 18 is diagnostic |
| Aldehydes | Visible | m/z 29 (CHO⁺); [M−29] from CHO loss | Both signals from C–CHO cleavage |
| Methyl ketones | Visible | m/z 43 (CH₃CO⁺) often base peak | α-cleavage either side of C=O |
| Carboxylic acids | Often weak | m/z 45 (COOH⁺); [M−17] from OH loss | Loss of 17 is diagnostic |
| Methyl esters | Visible | m/z 59 (CO₂CH₃⁺); [M−31] from OCH₃ loss | Loss of 31 is diagnostic |
| Methylbenzenes | Strong | m/z 91 (tropylium) typically base peak | Aromatic stabilisation |
| Aryl halides | Strong + isotope doublet | m/z 77 (C₆H₅⁺) | Cl 3:1; Br 1:1 |
| Amines (1°) | Odd-mass M⁺˳ | m/z 30 (CH₂=NH₂⁺) for primary amines | Nitrogen rule confirms N |
Electron-impact ionisation is destructive to large or fragile molecules — peptides, proteins, oligonucleotides and many pharmaceuticals fragment so completely under EI that no molecular ion survives. Two soft ionisation methods avoid this problem:
Electrospray ionisation (ESI) sprays a dilute solution of the analyte through a charged metal capillary at atmospheric pressure. The solvent evaporates from the resulting charged droplets, leaving multiply-charged molecular ions in the gas phase. For small organics ESI typically gives a [M+H]⁺ pseudo-molecular ion at m/z = Mᵣ + 1; for proteins it gives a characteristic envelope of [M+nH]⁻⁺ peaks at multiple charge states, from which Mᵣ can be deconvoluted to ~1 Da precision on molecules of 50 kDa or more. ESI is the workhorse of LC-MS in pharmaceutical and clinical analysis.
Matrix-assisted laser desorption/ionisation (MALDI) co-crystallises the analyte with a UV-absorbing matrix and then fires a pulsed laser at the spot. The matrix absorbs the laser energy and gently transfers a single proton to the analyte, again giving [M+H]⁺ with minimal fragmentation. MALDI is dominant in proteomics and polymer analysis.
For AQA A-Level, the practical implication is simple: for an EI spectrum, M⁺˳ = Mᵣ; for an ESI or MALDI spectrum labelled [M+H]⁺, Mᵣ = m/z − 1. Examiners occasionally present a soft-ionisation spectrum of a large molecule and check that you recognise the +1 offset.
Practical-skills box — Interpreting an unknown spectrum: (1) Locate the highest-m/z significant peak; this is M⁺˳ (or [M+H]⁺ if the ionisation method is soft). (2) Read Mᵣ. (3) Inspect the M+2 peak: 3:1 ratio = one Cl; 1:1 ratio = one Br; 9:6:1 triplet = two Cl, etc. (4) Use M+1 intensity to estimate carbon count (~1.1n% of M⁺˳). (5) For each major daughter peak, compute the neutral loss M⁺˳ − fragment and match against the standard table (15, 17, 18, 28, 29, 31, 43, 45). (6) Identify the most intense cation peak and match against the standard fragment table (43, 77, 91, 105). (7) Combine with IR (functional groups) and NMR (H and C environments) to converge on a unique structure.
Question 1. [13 marks total]
A student records the mass spectrum of an unknown liquid X. Significant peaks (with relative abundances in parentheses) are observed at m/z = 78 (32%), m/z = 80 (10%), m/z = 77 (5%), m/z = 51 (8%), m/z = 49 (24%), m/z = 51 (re-labelled CH₂Cl region), m/z = 49 (8% Cl), m/z = 43 (100%), m/z = 27 (45%), m/z = 15 (12%). [For clarity assume that the peaks at m/z 78 and 80 are the molecular ion doublet, and that m/z 49 and 51 are a separate doublet.]
(a) Identify the molecular ion peak(s) and determine Mᵣ. Suggest a molecular formula for X consistent with the data and explain how the data support it. [4 marks]
(b) Identify the fragment ions at m/z = 43, m/z = 27 and m/z = 15. Hence propose a structure for X and justify it using both fragmentation and isotope data. [5 marks]
(c) Explain why the molecular ion peaks at m/z = 78 and 80 appear in a 3:1 ratio. [2 marks]
(d) Pure ethanol and pure methoxymethane have the same relative molecular mass (Mᵣ = 46) but distinguishable mass spectra. State the most diagnostic fragment peak that allows a chemist to identify ethanol, and give the formula of the cation responsible. [2 marks]
(a) Molecular ion and formula [4 marks, AO2]
(b) Fragments and structure [5 marks, AO2 + AO3]
(c) 3:1 isotope ratio [2 marks, AO1]
(d) Ethanol vs methoxymethane discriminator [2 marks, AO2]
The three responses below cover the meaningful A-Level range: Grade C (the borderline-pass floor), Grade B (solid mark-scheme coverage), and Grade A* (top-band synthesis). No Grade D or E responses are shown — no A-Level student is aiming for those bands, and modelling failure adds nothing pedagogically. The commentary after each response (editorial, not a real examiner report) names the marks earned and the specific moves that differentiate from adjacent bands.
(a) The molecular ion peak is at m/z = 78 because it is the highest mass peak. So Mᵣ = 78. The peaks at 78 and 80 are in a 3:1 ratio, which means the compound contains one chlorine atom. Taking 35 (for Cl) away from 78 gives 43, which is C₃H₇. So the formula is C₃H₇Cl.
(b) The peak at m/z 43 is C₃H₇⁺ (propyl cation), formed by loss of Cl from the molecular ion. The peak at m/z 27 is C₂H₃⁺. The peak at m/z 15 is CH₃⁺. The structure is 2-chloropropane (CH₃CHClCH₃) because losing the chlorine gives a propyl cation, and a secondary cation is more stable than a primary one.
(c) Chlorine has two isotopes, ³⁵Cl (about 75%) and ³⁷Cl (about 25%). The molecular ion containing ³⁵Cl is at m/z 78 and the one containing ³⁷Cl is at m/z 80. The ratio of abundances (75:25 = 3:1) is reflected in the heights of the two peaks.
(d) Ethanol has a strong peak at m/z = 31 which is missing from methoxymethane. This peak is the cation CH₂OH⁺.
Editorial commentary (Grade C): Hits every mark-scheme bullet at a competent but unembellished level. Correctly identifies M⁺˳, applies the chlorine fingerprint, justifies 2-chloropropane on cation-stability grounds, and answers part (d) crisply. To progress to B, the response could acknowledge that 1-chloropropane is also consistent with the data (the m/z 43 peak does not uniquely identify either isomer without supporting evidence), and could compute the M+1 isotope intensity to confirm C₃H₇.
(a) The highest-m/z significant peak is at m/z = 78. This is the molecular ion M⁺˳, so Mᵣ = 78. The peak at m/z 80 (10% relative to 32% at m/z 78, a 3:1 ratio) is the M+2 isotope peak indicative of one chlorine. Subtracting ³⁵Cl (35) from Mᵣ 78 gives a residue of 43 = C₃H₇. The molecular formula is C₃H₇Cl — propan-1-yl chloride or propan-2-yl chloride.
(b) The base peak at m/z 43 corresponds to C₃H₇⁺ (propyl cation) — loss of Cl˳ (mass 35) from M⁺˳ gives the residue. The peak at m/z 27 is C₂H₃⁺ (vinyl cation, arising from further fragmentation, e.g. loss of CH₄ from C₃H₇⁺). The peak at m/z 15 is CH₃⁺. The mass spectrum is consistent with either 1-chloropropane or 2-chloropropane; 2-chloropropane is the better fit because loss of Cl directly yields a secondary carbocation (more stable than the primary cation from the 1-isomer) and this stabilisation explains why m/z 43 is the base peak.
(c) Naturally-occurring chlorine consists of ³⁵Cl (75.8%) and ³⁷Cl (24.2%). The molecular-ion population therefore divides between molecules containing ³⁵Cl (m/z 78) and ³⁷Cl (m/z 80) in a ratio that mirrors the natural-abundance ratio of the isotopes: 75.8 : 24.2 ≈ 3:1. This is the characteristic isotope fingerprint of one chlorine atom.
(d) Ethanol shows a strong peak at m/z = 31, which is weak or absent in methoxymethane. The cation responsible is CH₂OH⁺ (or equivalently [CH₂=OH]⁺), produced by α-cleavage of the C–C bond next to the hydroxyl group. Methoxymethane has no such C–C bond adjacent to oxygen and therefore cannot generate CH₂OH⁺ directly.
Editorial commentary (Grade B): Now A-Level-rigorous: cites natural-abundance values to 0.1%, distinguishes M⁺˳ from M+2 by relative intensity, names the α-cleavage mechanism in part (d), and explains why methoxymethane cannot reproduce the m/z 31 peak. To progress to A*, the response could compute the M+1 isotope peak intensity from carbon count (~3.3% for n = 3 carbons), comment on the rearrangement of primary to secondary cations that complicates the 1-chloropropane / 2-chloropropane discrimination, and acknowledge that the C₃H₇⁺ peak does not uniquely identify the carbon skeleton.
(a) The highest-m/z significant peak is at m/z 78 — the molecular ion M⁺˳ — so Mᵣ = 78. The 3:1 intensity ratio of m/z 78 to m/z 80 (32% : 10%) precisely matches the natural abundance of ³⁵Cl (75.8%) to ³⁷Cl (24.2%), confirming exactly one chlorine atom. Subtracting ³⁵Cl (35) from 78 gives 43, exactly C₃H₇. The molecular formula is C₃H₇Cl. As an internal cross-check, the M+1 isotope peak would be expected at ~3.3% of M⁺˳ (1.1% × 3 carbons) — consistent with C₃ within the precision of the data.
(b) m/z 43 = C₃H₇⁺, formed by α-cleavage of the C–Cl bond and loss of Cl˳. m/z 27 = C₂H₃⁺ (vinyl cation), formed by further loss of CH₄ from the propyl cation or by an E1-type elimination. m/z 15 = CH₃⁺. The data are formally consistent with both 1-chloropropane and 2-chloropropane: in both cases the base peak at m/z 43 corresponds to the propyl cation. However, 2-chloropropane is the more probable assignment because (i) cleavage of C–Cl in 2-chloropropane gives directly the secondary isopropyl cation (CH₃)₂CH⁺, whereas 1-chloropropane gives the primary n-propyl cation that must rearrange to secondary before being detected; (ii) the secondary cation's greater stability explains the dominance of m/z 43 as the base peak; (iii) Stevenson's rule predicts the lower-IE fragment (secondary > primary) survives. Confirmation would require complementary ¹H NMR — 2-chloropropane shows a septet at δ 4.0 (CH) and a doublet at δ 1.5 (2×CH₃); 1-chloropropane shows three sets of signals (CH₂Cl triplet, central CH₂ multiplet, terminal CH₃ triplet). The MS alone leaves a small ambiguity that NMR resolves.
(c) Chlorine has two stable isotopes, ³⁵Cl and ³⁷Cl, with natural abundances of 75.8% and 24.2% respectively (1:0.319 = ~3:1). For a single-chlorine molecule the molecular ion population partitions exactly along the isotope-abundance ratio, producing M⁺˳ (m/z 78, ³⁵Cl-containing) and M+2 (m/z 80, ³⁷Cl-containing) in approximately 3:1 height ratio. The pattern is independent of all other isotopes (¹³C contributes only to M+1 not M+2) and is therefore a clean, almost unique fingerprint for one Cl atom — with two chlorines the pattern becomes 9:6:1 (triplet at M, M+2, M+4), and the visual contrast is one of the most rapid discriminations in routine mass spectrometry.
(d) Ethanol shows a strong peak at m/z = 31 which is weak/absent in methoxymethane. The cation is CH₂OH⁺ ([CH₂=OH]⁺, a stabilised oxocarbenium-type cation), formed by α-cleavage of the C–C bond in CH₃CH₂OH to give CH₃˳ + CH₂OH⁺. Methoxymethane (CH₃OCH₃) has no C–C bond α to the oxygen, so cannot generate CH₂OH⁺ by direct α-cleavage — instead its base peak appears at m/z = 45 (CH₃OCH₂⁺, loss of H˳ from the molecular ion, methoxymethyl cation). The presence/absence of m/z 31 is therefore the single most diagnostic fragment for distinguishing the two isomers, and would be reinforced by ¹H NMR (ethanol shows three environments at δ 1.2, 3.7, 5.4; methoxymethane shows one singlet at δ 3.3) and by IR (ethanol's broad O–H 3200–3550 cm⁻¹ is absent in the ether).
Editorial commentary (Grade A):* A genuine A* response. The internal M+1 cross-check (3.3% for 3 carbons) is the kind of synoptic reasoning Edexcel and AQA mark schemes love. The Stevenson's-rule justification for 2-chloropropane, the acknowledgement of cation rearrangement that complicates the discrimination, the comparison to ¹H NMR for unambiguous structure assignment, and the alternative base peak (m/z 45) for methoxymethane all demonstrate understanding well beyond the textbook treatment. The response would score full marks and additional examiner credit for synthesis across techniques.
Three undergraduate-adjacent extensions for the curious A-Level student:
This lesson is aligned to the AQA A-Level Chemistry 7405 specification, §3.3.15, and is designed to prepare you for Paper 2 and Paper 3 structure-determination questions where mass spectrometry is combined with IR and NMR. Refer always to the official AQA specification document for the canonical statement of what is examinable.