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Lesson 6 introduced the combined-techniques workflow for relatively simple unknowns: mass spectrometry for Mᵣ and isotope patterns, infrared for functional groups, and ¹H NMR for hydrogen environments and multiplicity. Real exam problems and real laboratory work push further. The molecule may carry two or more functional groups simultaneously (a hydroxy-acid, an amino-alcohol, an ester-containing pharmaceutical); the molecular formula may admit several constitutional isomers that share Mᵣ but differ in connectivity; or the sample may be a polymer whose repeat unit must be inferred from partial-degradation data. This lesson adds two further weapons to the analytical toolkit — ¹³C NMR for counting and classifying carbon environments, and chromatography (TLC / GC / HPLC retention behaviour) for purity checking and comparison against authentic standards — then drills five carefully chosen worked examples that demand all of them. The aim is not memorisation of fragments but a transferable strategy: list candidate molecular formulae, count environments, eliminate isomers systematically, and confirm with one decisive chemical or chromatographic test.
Spec mapping (AQA 7405): This lesson anchors §3.3.15 (nuclear magnetic resonance spectroscopy) and §3.3.16 (chromatography), and extends the combined-techniques approach developed in lesson 6 to AQA's full multi-spectrum exam-style problem. It draws explicitly on the foundations laid in lessons 0–5 of this course (mass spectrometry, IR, ¹H NMR foundations, chemical tests for functional groups). Quantitative analytical methods — calibration curves, Beer–Lambert, percentage purity by titration, gravimetric analysis — are developed in lesson 8. Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: Recall of the combined-technique strategy (which technique tells you what, in what order to use them) is AO1. Application of the strategy to distinguish constitutional isomers of a given molecular formula, or to identify a multi-functional unknown from MS / IR / ¹H NMR / ¹³C NMR data, is AO2 and is examined on every AQA Paper 2 spectroscopy question. AO3 is tested by problems that demand rigorous elimination of candidate structures using multiple independent lines of evidence, not just identification of the most obvious one. The grade-band model answers below illustrate the difference: a C-grade response identifies the correct structure; an A* response also proves why no other isomer fits the data.
Mass spectrometry (MS) gives the molecular ion (M⁺·) at m/z = Mᵣ. M+1/M+2 satellites diagnose Cl (3:1 M:M+2) or Br (1:1). Fragment losses to memorise: 15 (CH₃·), 17 (OH·), 18 (H₂O), 28 (CO), 29 (CHO·), 31 (OCH₃·), 43 (COCH₃· or C₃H₇·), 45 (COOH· or OC₂H₅·), 77 (C₆H₅⁺).
Infrared (IR) diagnoses functional groups: alcohol O–H 3230–3550 cm⁻¹; carboxylic-acid O–H 2500–3300 cm⁻¹ (much broader, dimer); C=O 1715 (ketone), 1725–1740 (aldehyde), 1735–1750 (ester), 1700–1725 (acid), 1630–1690 (amide); N–H pair for primary amines/amides, single for secondary. The fingerprint region (<1500 cm⁻¹) is for reference-spectrum comparison only.
¹H NMR gives three pieces of information per signal: chemical shift δ (environment), integration ratio (relative H count), and multiplicity (n+1 rule — n equivalent neighbours → (n+1)-line pattern).
¹³C NMR exploits the same magnetic-resonance phenomenon as ¹H NMR but observes the ¹³C nucleus, which has natural abundance only 1.1 %. The chemical-shift range is much wider — roughly 0–220 ppm versus 0–12 ppm for ¹H — which means signals rarely overlap, and each chemically distinct carbon usually gives a separate peak. At AQA A-Level the spectra are presented in proton-decoupled form, so every ¹³C signal is a clean singlet; multiplicity is not examined.
What ¹³C gives you that ¹H does not:
| Carbon environment | δ (¹³C) / ppm |
|---|---|
| Saturated alkyl C (CH₃, CH₂, CH) remote from electronegative atoms | 5–50 |
| C–N (amines) | 30–65 |
| C–O (alcohols, ethers, esters O-alkyl) | 50–90 |
| C≡C, C≡N | 65–120 |
| Aromatic / alkene C=C | 110–160 |
| Carboxylic acid, ester, amide C=O | 160–185 |
| Aldehyde, ketone C=O | 190–220 |
A peak at ~170 ppm essentially commits you to a carboxylic-acid / ester / amide carbonyl; a peak at ~200 ppm commits to an aldehyde or ketone; peaks at 110–160 ppm commit to aromatic or alkene carbons. The ¹³C spectrum often resolves an ambiguity that ¹H alone cannot — for example, distinguishing an ester (carbonyl ~170) from a ketone (~205) when the IR C=O bands overlap in the 1715–1740 cm⁻¹ window.
Spectroscopy proposes; a chemical test disposes. The wet-lab confirmations from lessons 1–5 of this course remain decisive when spectra are ambiguous:
One additional chemical test, performed as a confirmation after spectroscopic assignment, is often the question that lifts a Paper-2 answer from band B to band A.
Thin-layer chromatography (TLC) is the simplest separation technique: a thin layer of silica or alumina (the stationary phase) is coated on a glass or plastic plate; a small spot of the sample, dissolved in a volatile solvent, is applied near one edge; the plate is then placed in a sealed tank with a shallow layer of solvent (the mobile phase) and the solvent rises by capillary action, carrying components of the mixture different distances depending on their relative affinities for the two phases.
The diagnostic quantity is the retention factor:
Rf = distance moved by component / distance moved by solvent front
Rf is always between 0 and 1, is reproducible under fixed conditions (same stationary phase, same mobile phase, same temperature), and provides a fingerprint that can be compared against authentic standards run on the same plate. Two species giving the same Rf on multiple solvent systems are very probably identical. A single TLC spot also serves as a purity check: a single Rf spot for the sample alongside a single spot for the pure reference at the same Rf strongly supports identity and purity.
Gas chromatography (GC) and high-performance liquid chromatography (HPLC) extend the same principle to quantitative analysis; the diagnostic quantity becomes retention time, and integrated peak areas give relative amounts. For the structure-determination problems in this lesson, TLC against an authentic standard is the typical confirmatory step.
Before assembling any structure from spectroscopic data, compute the degree of unsaturation (also called the index of hydrogen deficiency, IHD). For a molecular formula C_c H_h N_n O_o X_x (X = halogen):
IHD = (2c + 2 + n − h − x) / 2
Oxygen and sulfur do not appear in the formula. Each ring contributes 1, each C=C or C=O contributes 1, each C≡C or C≡N contributes 2, and a benzene ring contributes 4 (one ring + three formal double bonds).
Worked checks:
| Formula | IHD calculation | Result | Likely features |
|---|---|---|---|
| C₃H₆O₃ | (6 + 2 − 6) / 2 | 1 | one C=O, no ring |
| C₈H₉NO | (16 + 2 + 1 − 9) / 2 | 5 | benzene (4) + C=O (1) |
| C₉H₈O₄ | (18 + 2 − 8) / 2 | 6 | benzene (4) + 2 × C=O (2) |
| C₄H₆O₂ | (8 + 2 − 6) / 2 | 2 | 2 × C=O, or C=C + C=O, or ring + C=O |
| C₆H₅Cl | (12 + 2 − 5 − 1) / 2 | 4 | benzene |
An IHD ≥ 4 should immediately prompt the question "is there a benzene ring?", confirmed by aromatic ¹H signals at δ 6.5–8.0 and aromatic ¹³C signals at δ 110–160.
The problem. Three constitutional isomers share the molecular formula C₂H₄O₂ and Mᵣ = 60. They are ethanoic acid (CH₃COOH), methyl methanoate (HCOOCH₃), and glycolaldehyde (HOCH₂CHO). All three give the same M⁺· at m/z 60 and the same IHD of 1. Distinguish them using IR, ¹H NMR and ¹³C NMR.
Step 1 — IHD. IHD = (4 + 2 − 4)/2 = 1. One C=O (no ring).
Step 2 — IR diagnostics.
The acid is immediately separated by its 2500–3300 cm⁻¹ O–H envelope. To distinguish methyl methanoate from glycolaldehyde, look for the alcohol O–H in glycolaldehyde and the aldehyde C–H near 2720 cm⁻¹.
Step 3 — ¹H NMR diagnostics.
Counting the number of signals alone identifies glycolaldehyde (three) versus the other two (two each); the chemical-shift positions distinguish acid from ester.
Step 4 — ¹³C NMR diagnostics.
The ¹³C carbonyl shift alone discriminates all three: ~178 (acid), ~162 (ester), ~204 (aldehyde).
Answer. Each isomer is uniquely identified by the combination of IR (presence/absence of broad acid O–H, alcohol O–H, aldehyde C–H), ¹H NMR (number of signals; chemical shift of the carbonyl-adjacent group; presence of exchangeable peaks), and ¹³C NMR (carbonyl chemical shift).
Given data.
Step 1 — molecular formula. Mᵣ = 90 is consistent with C₃H₆O₃: 3(12) + 6(1) + 3(16) = 90. IHD = (6 + 2 − 6)/2 = 1. One C=O, no ring.
Step 2 — IR. Two distinct O–H environments are present: the very broad 2500–3300 cm⁻¹ envelope is the dimerised carboxylic-acid O–H; the sharper 3300–3600 cm⁻¹ feature is a separate alcohol O–H. The C=O at 1715 cm⁻¹ supports a carboxylic acid (acids: 1700–1725; aldehydes 1720–1740; esters 1735–1750). So the molecule contains both –COOH and –OH.
Step 3 — ¹H NMR. Four environments. The two exchangeable protons (D₂O test, integrations 1H each) account for the carboxylic OH (~11 ppm) and the alcohol OH (~4.0 ppm). The two triplets at δ 2.6 (2H) and δ 3.8 (2H) are mutually coupled — each CH₂ has two neighbours — and the chemical shifts tell us δ 2.6 is the CH₂ adjacent to the carbonyl (α to C=O) and δ 3.8 is the CH₂ bound to O (HOCH₂–).
Step 4 — ¹³C NMR. Three carbon environments: ~37 ppm (saturated CH₂ α to C=O), ~58 ppm (CH₂ bound to O), ~178 ppm (carboxylic-acid C=O). Three peaks for three distinct carbons matches C₃H₆O₃.
Step 5 — assembly. Connect the fragments: HO–CH₂–CH₂–COOH. This is 3-hydroxypropanoic acid (also called β-hydroxypropanoic acid). It has the correct Mᵣ (90), the correct two distinct OH groups, the correct two coupled CH₂ triplets, the correct three ¹³C environments, and the m/z 73 fragment (loss of OH, 17) and m/z 45 fragment (COOH⁺ or HOCH₂CH₂⁺ — both 45) verify.
Answer. HOCH₂CH₂COOH, 3-hydroxypropanoic acid. The systematic feature here is the simultaneous diagnosis of two functional groups from IR (two different O–H environments) and confirmation from the connectivity revealed by ¹H NMR triplet–triplet coupling.
Context. A condensation polymer is partially hydrolysed by aqueous acid, and the recovered monomer fragments are analysed. The hydrolysate is found to contain two products. Product A: Mᵣ 118; IR shows two very broad O–H envelopes (2500–3300 cm⁻¹) and a strong C=O at 1715 cm⁻¹; ¹H NMR shows δ 7.85 (s, 4H, aromatic) and δ 11 (broad s, 2H, exchangeable). Product B: Mᵣ 62; IR shows broad O–H at 3230–3550 cm⁻¹ but no C=O; ¹H NMR shows a single signal at δ 3.7 (s, 4H) and one exchangeable signal at δ 3.4 (broad s, 2H).
Analysis of A. Mᵣ 118 with IHD = (10 + 2 − 6)/2 = 4 from candidate C₈H₆O₄ suggests an aromatic compound. Two exchangeable protons (2H), aromatic singlet 4H (the symmetric para-disubstituted ring with two equivalent substituents gives a single aromatic signal), and two carboxylic-acid OH groups identify A as benzene-1,4-dicarboxylic acid (terephthalic acid).
Analysis of B. Mᵣ 62 fits C₂H₆O₂: IHD = (4 + 2 − 6)/2 = 0. Two equivalent CH₂ groups (one signal, 4H total) and two equivalent OH groups (2H exchangeable). Structure: HOCH₂CH₂OH — ethane-1,2-diol (ethylene glycol).
Monomer assembly. The polymer is therefore the condensation product of terephthalic acid + ethylene glycol with loss of water at each ester linkage: this is poly(ethylene terephthalate), PET — the polyester of plastic bottles and clothing fibres. The repeat unit is –OOC–C₆H₄–COO–CH₂–CH₂– with formula C₁₀H₈O₄ (Mᵣ of repeat unit 192, equal to 118 + 62 − 2 × 18 for two condensations losing two H₂O).
This worked example signposts directly to the organic-advanced course lessons on condensation polymers and step-growth polymerisation, where the chemistry of esterification and amide-bond formation is treated in detail.
Given data.
Step 1 — formula. Mᵣ 180 and the two distinct C=O peaks (one ester, one carboxylic acid) and the aromatic protons suggest a salicylate-type structure. Try C₉H₈O₄: 9(12) + 8(1) + 4(16) = 180. ✓ IHD = (18 + 2 − 8)/2 = 6 — benzene (4) + two C=O (2). ✓
Step 2 — fragments. The loss of 42 (M⁺· → m/z 138) corresponds to loss of ketene, CH₂=C=O, the diagnostic loss for an acetate ester of an aromatic system (the McLafferty-type rearrangement of an aryl acetate gives the parent phenol cation and ketene). The m/z 138 fragment is therefore the cation of the deacylated parent phenol — and m/z 138 is consistent with C₇H₆O₃ = 2-hydroxybenzoic acid (salicylic acid).
Step 3 — assembly. The molecule is 2-ethanoyloxybenzoic acid (the IUPAC name) — better known as aspirin (acetylsalicylic acid). Structure: an ortho-disubstituted benzene ring bearing –OCOCH₃ at one position and –COOH at the adjacent position.
Step 4 — verification. The ester C=O appears at 1750 cm⁻¹ (correct for an aryl acetate); the carboxylic-acid C=O at 1690 cm⁻¹ (slightly lowered by intramolecular hydrogen bonding to the neighbouring acetoxy oxygen, characteristic of ortho-substituted salicylates); the singlet 3H at δ 2.3 is the –OCOCH₃ acetate methyl; the broad exchangeable singlet at δ 11 is the –COOH; the ortho-disubstituted aromatic pattern accounts for the four aromatic protons. The ¹³C count of 7 environments matches: two C=O, four aromatic C (one of which is the ipso to OCOCH₃, another the ipso to COOH, plus two equivalent pairs), and one CH₃ — actually 2 + 4 + 1 = 7. ✓
Answer. 2-ethanoyloxybenzoic acid (aspirin), C₉H₈O₄, Mᵣ 180.
The scenario. A student attempts the oxidation of butan-1-ol (CH₃CH₂CH₂CH₂OH, Mᵣ 74) with excess acidified K₂Cr₂O₇ under reflux with distillation. They isolate a product that gives the following data.
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