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In an examination or in a real analytical laboratory, no single spectroscopic technique is sufficient on its own to identify an unknown organic compound with confidence. Mass spectrometry (MS) supplies the relative molecular mass and, often, key fragmentation clues, but it does not directly tell you about bond types or hydrogen environments. Infrared (IR) spectroscopy reveals which functional groups are present from characteristic absorption frequencies but offers no information about the carbon skeleton. Proton (¹H) NMR spectroscopy reports on the number of chemically distinct hydrogen environments, their integration, and their splitting patterns, yet it cannot establish the molecular mass directly. Each technique is one piece of a jigsaw. The skill expected at A-Level — and the skill that earns the highest marks on AQA Paper 3 — is the ability to combine those three streams of evidence into a single, self-consistent structural proposal. This lesson develops that skill on a deliberately simple class of compounds: monofunctional molecules of modest molecular mass, where one careful pass through MS → IR → ¹H NMR reveals the structure without elaborate fragmentation chemistry. Complex polyfunctional compounds and ambiguous cases are deferred to lesson 7.
Spec mapping (AQA 7405): This lesson maps to §3.3.15 (NMR spectroscopy) and §3.3.16 (chromatography — referenced for context only), with cross-references to §3.3.13 (mass spectrometry of molecules) and §3.3.14 (infrared spectroscopy). It builds directly on L0 (mass spectrometry foundations), L1 (IR spectroscopy), L2 (¹H NMR), and L3 (¹³C NMR), assembling those individual techniques into a single deductive workflow. The harder, polyfunctional and aromatic-system extensions appear in L7 (combined techniques II); quantitative high-resolution MS and isotope-ratio analysis appear in L8. Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: Recalling the systematic order in which evidence is extracted (MS → IR → ¹H NMR → propose → check) is AO1. Extracting the relative molecular mass from a molecular ion peak, identifying functional groups from IR wavenumbers, and reading chemical shift / multiplicity / integration data are AO2 skills. Combining all three streams of evidence to deduce a consistent structure — and rejecting alternative candidates with explicit reasoning — is the AO3 hallmark of a strong response.
A reliable order of attack avoids two of the most common A-Level errors: jumping to a structure too quickly, and failing to check that every piece of data is accounted for. The recommended sequence is:
Locate the molecular ion peak (M⁺˳) at the highest m/z value (ignoring isotope peaks such as M+1 or M+2 unless they carry diagnostic information about chlorine or bromine). The integer m/z of the molecular ion is the relative molecular mass Mᵣ. From Mᵣ, propose plausible molecular formulae. For purely C/H/O compounds the saturation rule CₙH₂ₙ₊₂Oₘ (no rings, no π-bonds) and the index of hydrogen deficiency (IHD) help narrow down the options. Watch for M+2 peaks of roughly equal height (Br) or one-third the height of M (Cl) to detect halogens.
Concentrate on the functional-group region (1500–4000 cm⁻¹) rather than the fingerprint region (below 1500 cm⁻¹). The fingerprint region is diagnostic only when compared to a reference spectrum and is rarely useful for unknown identification at A-Level. Within the functional-group region, the key absorptions are:
| Wavenumber (cm⁻¹) | Bond | Functional group |
|---|---|---|
| 3200–3600 (broad) | O–H | alcohol (hydrogen-bonded) |
| 2500–3300 (very broad) | O–H | carboxylic acid |
| 3300–3500 (one or two peaks, medium) | N–H | amine (one peak) or primary amide (two peaks) |
| 1680–1750 (sharp, strong) | C=O | aldehyde / ketone / acid / ester / amide |
| 1000–1300 (strong) | C–O | alcohols, ethers, esters |
| 1450–1650 | C=C | alkene or aromatic |
Within the C=O range, distinguish: aldehyde and ketone ~1715–1730; ester ~1735–1750; carboxylic acid ~1700–1720 (often accompanied by a very broad O–H); amide ~1630–1690.
For each NMR peak record three pieces of data:
OH and NH protons appear as broad singlets and disappear on a D₂O shake — a powerful diagnostic when given.
Combine the evidence to propose a structure. Then run a final consistency check: every peak in every spectrum must be accounted for. If you cannot account for one of them, your structure is wrong (or incomplete).
Practical-skills box — strategy: Do not try to identify the compound in one go. Process each spectrum individually first, building up partial information — "Mᵣ = 60", "alcohol (broad O–H)", "three environments in 3:2:2 with the CH₂ at 3.6 ppm" — and then assemble. Examiners report that candidates who try to "see" the structure straight away from the NMR alone frequently propose a compound inconsistent with the IR or the molecular mass and then fail to notice. The systematic approach is slower for one or two problems but pays for itself on a long Paper 3 question.
Data:
Step 1 — MS. Mᵣ = 60. For CₙH₂ₙ₊₂O the saturated alcohol formula is C₃H₈O (Mᵣ = 36 + 8 + 16 = 60). ✔ IHD = (2×3 + 2 − 8)/2 = 0 — no rings, no double bonds.
Step 2 — IR. Broad O–H at 3200–3600 cm⁻¹ → alcohol. C–O stretch at 1050 cm⁻¹ confirms a single-bonded C–O. No C=O absorption (1680–1750), so not an aldehyde, ketone, acid, ester, or amide.
Step 3 — ¹H NMR. Four environments with integration ratio 3:2:2:1, total 8H — matches C₃H₈O.
Step 4 — Assemble. A propyl chain CH₃–CH₂–CH₂– with an OH on the end: propan-1-ol, CH₃CH₂CH₂OH (Mᵣ = 60 ✔).
Rejection of alternatives: propan-2-ol [(CH₃)₂CHOH] would show only three environments (two equivalent CH₃ doublets, a septet for CH, and an exchangeable OH). The four-environment pattern with the sextet at δ ≈ 1.5 rules it out. Answer: propan-1-ol. ✔
Data:
Step 1 — MS. Mᵣ = 74. Plausible formulae include C₃H₆O₂ (36 + 6 + 32 = 74). IHD = (6 + 2 − 6)/2 = 1 — one ring or π-bond.
Step 2 — IR. C=O at 1735 cm⁻¹ is squarely in the ester range (1735–1750). C–O at 1180 cm⁻¹ confirms the C–O–C single bond of an ester. No broad O–H → not a carboxylic acid.
Step 3 — ¹H NMR. Total integration 3+3 = 6 H → matches C₃H₆O₂. Two singlets means each CH₃ has no hydrogens on the adjacent atom.
Step 4 — Assemble. CH₃CO–O–CH₃: methyl ethanoate (methyl acetate), CH₃COOCH₃ (Mᵣ = 74 ✔).
Reasoning aside — why not ethyl methanoate, HCOOCH₂CH₃? Ethyl methanoate has the same molecular formula C₃H₆O₂ and the same Mᵣ = 74, and is a perfectly valid candidate from MS alone. The ¹H NMR is decisive: ethyl methanoate would show three environments — a singlet ~8.0 ppm (1H, the formate H–CO–), a quartet ~4.2 ppm (2H, OCH₂), and a triplet ~1.3 ppm (3H, CH₃). The two-singlet pattern given is incompatible with ethyl methanoate. Answer: methyl ethanoate. ✔
This example illustrates the discipline of the systematic method: MS narrows to two candidates of identical Mᵣ; IR confirms the family (ester); NMR distinguishes between the two isomers conclusively.
Data:
Step 1 — MS. Mᵣ = 58. For CₙH₂ₙO the formula C₃H₆O fits (36 + 6 + 16 = 58). IHD = (6 + 2 − 6)/2 = 1.
Step 2 — IR. C=O at 1715 cm⁻¹ → ketone or aldehyde (the two overlap at this end of the range). No broad O–H → not an acid or alcohol.
Step 3 — ¹H NMR. A single peak at δ = 2.1 ppm integrating for 6H, no splitting. All six hydrogens are in identical chemical environments — a strikingly symmetric molecule. The chemical shift 2.1 ppm is typical of a CH₃ next to C=O.
Step 4 — Assemble. C₃H₆O with one C=O and six equivalent hydrogens demands two equivalent CH₃ groups attached to the carbonyl: (CH₃)₂C=O. Answer: propan-2-one (acetone, propanone), CH₃COCH₃ (Mᵣ = 58 ✔).
MS verification. Loss of Mᵣ − 43 = 15 (CH₃˙) leaves CH₃CO⁺ at m/z = 43 — the acylium cation, the signature fragment of a methyl ketone. The peak at m/z = 15 is the methyl cation CH₃⁺. Every peak accounted for. ✔
Rejection of propanal. Propanal CH₃CH₂CHO has the same Mᵣ = 58 and the same IR (C=O at ≈ 1730). It is rejected by the NMR — propanal would show three environments (triplet ~1.0 ppm, multiplet ~2.4 ppm, triplet ~9.8 ppm for CHO), not a single singlet. The aldehyde CHO at δ ≈ 9.8 ppm would be unmissable. Confirmed: propan-2-one. ✔
Data:
Step 1 — MS. Mᵣ = 60. Possible formulae: C₃H₈O (propanol; tried in Example 1) or C₂H₄O₂ (24 + 4 + 32 = 60). IHD for C₂H₄O₂ = (4 + 2 − 4)/2 = 1.
Step 2 — IR. The very broad absorption from 2500–3300 cm⁻¹ is unmistakable: a carboxylic acid O–H (hydrogen-bonded dimer, exceptionally broad). C=O at 1715 cm⁻¹ is on the low side of carbonyl, consistent with an acid. So the compound is a carboxylic acid — the C₃H₈O alcohol from Example 1 is eliminated.
Step 3 — ¹H NMR. Integration ratio 3:1 → 4H total, matching C₂H₄O₂.
Step 4 — Assemble. CH₃–COOH: ethanoic acid (acetic acid), CH₃COOH (Mᵣ = 60 ✔).
MS verification. Loss of 15 (CH₃) from Mᵣ gives COOH⁺ at m/z = 45 — the carboxyl acylium cation. Loss of OH (17) from Mᵣ would give 43 (CH₃CO⁺), often also seen in acid spectra. The 15 peak is CH₃⁺. ✔ Confirmed: ethanoic acid.
The discrimination between propan-1-ol (Example 1) and ethanoic acid (Example 4) is a textbook example of why IR is essential: both have Mᵣ = 60, but the broad 2500–3300 cm⁻¹ envelope versus a 3200–3600 cm⁻¹ O–H instantly distinguishes the two functional groups.
Data:
Step 1 — MS. Mᵣ = 122. Trial formula C₇H₆O₂ = 84 + 6 + 32 = 122. ✔ IHD = (14 + 2 − 6)/2 = 5. Five degrees of unsaturation = a benzene ring (which contributes 4: three C=C plus the ring itself) + one further π-bond, almost certainly a C=O.
Step 2 — IR. Very broad 2500–3300 cm⁻¹ → carboxylic acid O–H. C=O at 1690 cm⁻¹ is slightly lower than a simple aliphatic acid because the carbonyl is conjugated with the aromatic ring, which lowers the C=O stretching frequency.
Step 3 — ¹H NMR. Integration ratio 5:1 → 6H total, matching C₇H₆O₂.
Step 4 — Assemble. C₆H₅– attached to –COOH: benzoic acid, C₆H₅COOH (Mᵣ = 122 ✔).
MS verification.
Every peak in every spectrum is accounted for. Confirmed: benzoic acid.
This example also illustrates an important point of arithmetic discipline. A passing thought "phenylethanone (acetophenone) C₆H₅COCH₃ has Mᵣ = 120, close to 122" is a trap: 120 ≠ 122. The molecular ion locks the formula. Acetophenone is also rejected by the NMR — it would show a 3H singlet near δ ≈ 2.6 ppm (the CH₃) and no D₂O-exchangeable proton. The data only fit benzoic acid.
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