13.5 Structure from Combined Spectra: IR, NMR and Mass
★★★ 2021–2025: 26 parts, 253 marks (Civil Services 8 parts / 90 m · IFoS 9 / 90 m · Geo-Scientist 9 / 73 m). Geo-Scientist Paper III sets one such problem almost every year, usually for 10 marks.
Syllabus lines. Civil Services: spectroscopy, “principle and applications in structure elucidation”: vibrational (“specific frequencies of functional groups in polyatomic molecules”), ¹H NMR (“chemical shift and spin-spin interaction and coupling constants”), mass spectrometry (“parent peak, base peak, metastable peak, McLafferty rearrangement”). IFoS: the same, plus “application of ¹H NMR to simple organic molecules” and “fragmentation of simple organic molecules; α-cleavage”. Geo-Scientist: “Spectroscopic methods of analysis”: IR “characteristic frequencies of organic molecules and interpretation of spectra”, proton and carbon-13 NMR, mass spectrometry “isotope abundances; the molecular ion, metastable ions. McLafferty rearrangement”. UV and the Woodward–Fieser rules have their own notes (13.1).
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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes
13.5.1 The six-step route ★★★
- Get the formula (given, or from M⁺ and the isotope peaks).
- Work out the DBE. DBE ≥ 4 with ¹H signals at 6.5–8 ppm: a benzene ring.SN
- Use IR to say which groups are present and absent (no 3300 band: no O–H/N–H).
- Turn every ¹H NMR signal into a fragment (shift → neighbour, integral → H count, multiplicity → H next door).SN
- Count ¹³C signals for symmetry; use the C=O shift and UV for conjugation.
- Assemble, then check the formula, DBE and the main mass fragments; name a near-miss isomer and say which signal rules it out. That last line is what separates 6/10 from 9/10.SN
13.5.2 Formula, DBE and the molecular ion ★★★
- Count halogens (X) like H; ignore O and S. Benzene ring = 4; C=O, C=C = 1 each; C≡N, C≡C = 2. C₄H₅NO₂: (8 + 2 + 1 − 5)/2 = 3 = C≡N + C=O CSE 2025 Q8(c)(ii) 10m.SN
- Nitrogen rule: an odd M⁺ means an odd number of N atoms. M⁺ 179 → one N (phenacetin) Geo 2025 Q9(c) 10m; 185/187 → one N and one Br Geo 2024 Q9(a) 10m.SN
- Isotope peaks: M : M+2 ≈ 1 : 1 → one Br (⁷⁹Br, ⁸¹Br); ≈ 3 : 1 → one Cl; two Br give 1 : 2 : 1. 166/168 of equal height: C₄H₇BrO₂ CSE 2025 Q8(b)(i) 10m Geo 2023 Q10(d) 10m.SN
- No formula given? Work from M and the tests: M = 108 with C, H, O → C₇H₈O (DBE 4). O–H at 3401 cm⁻¹, aromatic =C–H at 3077, CH₂ at 2899; FeCl₃ negative (not a phenol), oxidised by KMnO₄ (a CH₂OH on a ring): benzyl alcohol Geo 2025 Q10(b)(ii) 5m.
13.5.3 IR: which groups are there ★★★
| C=O compound | ν(C=O) / cm⁻¹ | Also look for |
|---|---|---|
| Anhydride | two bands, ≈ 1820 and 1760 | cyclic: 1850, 1780 |
| Acid chloride | ≈ 1800 | – |
| Ester | 1735–1750 (vinyl or aryl ester 1760–1770; conjugated 1715–1730) | C–O 1150–1300 strong |
| Aldehyde | 1725 (conjugated ≈ 1700) | C–H 2720, 2820 |
| Ketone | 1715 (conjugated 1680–1690) | – |
| Carboxylic acid | 1710 | O–H 2500–3300, very broad |
| Amide | 1650–1690 | N–H 3200–3400 |
- Conjugation lowers ν(C=O) by about 25 cm⁻¹; an O-vinyl or O-aryl ester raises it. Ethyl acrylate (CH₂=CH–CO₂Et, C=O ≈ 1725) vs vinyl propionate (EtCO₂–CH=CH₂, ≈ 1760) IFoS 2023 Q7(e) 5m.SN
- Aldehyde vs ketone (propanal vs propanone): the 2720/2820 pair settles it, not the small C=O shift IFoS 2023 Q6(c) 5m.
- C≡N 2250 (2220 when conjugated with a ring); C₃H₃N with 2250, 1650 and 3100 cm⁻¹ and no 3300 band: acrylonitrile CSE 2025 Q6(b) 15m.SN
- A C=C with no change in dipole (tetrasubstituted, symmetric) is IR-inactive: 2,3-dimethylbut-2-ene shows no C=C band and one ¹H singlet; the 1-ene shows =CH₂ at 3080, 1645, 890 cm⁻¹ CSE 2024 Q8(c)(i) 10m.
- Ring C–H bends: monosubstituted 750 + 700; ortho 750; meta 780 + 690; para 800–850; 1,2,4-trisubstituted 820 + 880.
13.5.4 ¹H NMR: from signals to fragments ★★★
| Pattern | Fragment |
|---|---|
| q (2H) + t (3H) | ethyl; O–CH₂CH₃ when the q is at δ 4.0–4.4 |
| septet (1H) + d (6H) | isopropyl, (CH₃)₂CH– |
| s (9H) ≈ 1.0–1.3 | tert-butyl |
| s (3H) 3.6–3.9 | OCH₃ (ester or aryl ether) |
| s (3H) 2.0–2.6 | CH₃–C=O (ketone, acetate, acetamide); Ar–CH₃ ≈ 2.3 |
| two t (2H each) | –CH₂–CH₂– between two groups with no H |
| two d (2H each), J ≈ 8 Hz | para-disubstituted benzene (AA′BB′) |
| s (5H) ≈ 7.3 | C₆H₅ with an alkyl group |
| two d (1H each), J ≈ 16 Hz | trans CH=CH (cis ≈ 10 Hz) |
| signal lost on D₂O shake | O–H or N–H |
- A signal is split by the H on the next carbon (n + 1 rule); H with no neighbouring H gives a singlet. C₄H₅NO₂ with only two singlets 3 : 2: OCH₃ and CH₂ isolated by CN and C=O, methyl cyanoacetate CSE 2025 Q8(c)(ii).SN
- Same formula, different order of groups: C₁₀H₁₂O₂ with CH₃ singlet 2.0 + two triplets (2.93, 4.3) + C₆H₅ = 2-phenylethyl acetate; with q 4.13 + t 1.23 + singlet 3.72 (2H) = ethyl phenylacetate. The O–CH₂ is always the downfield one CSE 2024 Q7(c)(i) 10m.SN
- C₅H₁₀O isomers giving DNP: t + q → pentan-3-one; two singlets (9H, 1H at δ 9.5) → pivalaldehyde, which has no α-H and gives two products with conc. KOH (Cannizzaro: neopentyl alcohol + pivalate); d + s + septet → 3-methylbutan-2-one CSE 2022 Q7(c)(ii) 10m.
- M = 102, only d (6H) 1.1 and septet (1H) 3.55: diisopropyl ether; 1 mol HI → isopropanol (O–H 3330 cm⁻¹, exchangeable δ 4.4) + 2-iodopropane (septet at 4.25); excess HI gives only 2-iodopropane CSE 2022 Q7(c)(i) 10m.
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