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In the previous lesson, you learned that ¹H NMR reveals the number of hydrogen environments and the relative number of hydrogens in each. Now we explore two further layers of information: chemical shift values (which tell you what type of environment each hydrogen is in) and spin-spin splitting (which tells you about neighbouring hydrogens).
The chemical shift (δ) at which a hydrogen signal appears is determined by the electron density around that hydrogen. Electronegative groups withdraw electron density, deshielding the hydrogen and moving it to higher δ (downfield). Electron-donating groups increase shielding and move the signal to lower δ (upfield).
You must learn these ranges for the exam:
| Type of proton | δ / ppm | Example compound | Typical δ |
|---|---|---|---|
| R–CH₃ (alkyl) | 0.7–1.2 | Ethane CH₃CH₃ | 0.9 |
| R–CH₂–R (alkyl) | 1.2–1.4 | Propane CH₃CH₂CH₃ | 1.3 |
| R₃CH (alkyl) | 1.4–1.7 | 2-Methylpropane | 1.6 |
| C=C–CH₃ (allylic) | 1.6–2.0 | Propene CH₃CH=CH₂ | 1.8 |
| C=O–CH₃ or C=O–CH₂ | 2.0–2.5 | Propanone CH₃COCH₃ | 2.1 |
| N–CH₃ or N–CH₂ | 2.2–2.8 | Trimethylamine | 2.3 |
| O–CH₃ or O–CH₂ | 3.3–4.3 | Methanol CH₃OH | 3.4 |
| Cl–CH or Br–CH | 3.0–4.2 | Chloromethane CH₃Cl | 3.1 |
| O–CO–CH₂ (ester O–CH) | 3.7–4.3 | Ethyl ethanoate –OCH₂– | 4.1 |
| C=C–H (alkene) | 4.5–6.0 | Ethene CH₂=CH₂ | 5.3 |
| Ar–H (aromatic) | 6.5–8.0 | Benzene C₆H₆ | 7.3 |
| R–CHO (aldehyde) | 9.0–10.0 | Ethanal CH₃CHO | 9.8 |
| R–COOH (acid) | 10.0–12.0 | Ethanoic acid | 11.4 |
| R–OH (alcohol) | 0.5–5.0 (variable) | — | Variable |
| R–NH₂ (amine) | 1.0–4.5 (variable) | — | Variable |
The trend in chemical shifts can be rationalised by electronegativity and electron density:
One of the most powerful features of ¹H NMR is that peaks are often split into characteristic patterns called multiplets. This splitting arises because neighbouring non-equivalent hydrogens interact through the bonding electrons — a phenomenon called spin-spin coupling.
The fundamental rule for predicting splitting patterns is:
A signal is split into (n+1) peaks, where n is the number of equivalent neighbouring hydrogens (on adjacent carbon atoms).
| n (neighbours) | Number of peaks | Pattern name | Intensity ratio (Pascal's triangle) |
|---|---|---|---|
| 0 | 1 | Singlet (s) | 1 |
| 1 | 2 | Doublet (d) | 1:1 |
| 2 | 3 | Triplet (t) | 1:2:1 |
| 3 | 4 | Quartet (q) | 1:3:3:1 |
| 4 | 5 | Quintet (quin) | 1:4:6:4:1 |
| 5 | 6 | Sextet (sext) | 1:5:10:10:5:1 |
| 6 | 7 | Septet (sept) | 1:6:15:20:15:6:1 |
| Pattern | What it means | Common structural fragment |
|---|---|---|
| Triplet (3H) + Quartet (2H) | Ethyl group CH₃CH₂– | Ethyl ester, ethyl ether, ethanol |
| Doublet (6H) + Septet (1H) | Isopropyl group (CH₃)₂CH– | Isopropyl alcohol, isopropyl ester |
| Doublet (3H) + Quartet (1H) | CH₃CH< fragment | Lactic acid, alanine |
| Two singlets | No adjacent H on either group | CH₃COOCH₃ (methyl ethanoate) |
| Singlet (9H) | tert-Butyl group (CH₃)₃C– | tert-Butyl alcohol, neopentane |
Ethanol (CH₃CH₂OH):
1,1-Dichloroethane (CHCl₂CH₃):
Propan-1-ol (CH₃CH₂CH₂OH):
The coupling constant (J), measured in Hz, is the distance between adjacent peaks in a multiplet. Coupled nuclei share the same J value — if the CH₃ triplet has J = 7 Hz, the adjacent CH₂ quartet will also have J = 7 Hz.
Typical J values:
| Coupling type | Relationship | J value (Hz) |
|---|---|---|
| Vicinal (H–C–C–H) | Through 3 bonds | 6–8 |
| Geminal (H–C–H) | Through 2 bonds | 12–15 |
| Aromatic (ortho) | Adjacent ring positions | 6–9 |
| Aromatic (meta) | One position apart | 1–3 |
| Aromatic (para) | Opposite positions | 0–1 |
| Trans alkene | H–C=C–H trans | 12–18 |
| Cis alkene | H–C=C–H cis | 6–12 |
Hydroxyl (O–H) and amine (N–H) protons typically appear as broad singlets rather than showing coupling to adjacent CH protons. This is because they undergo rapid chemical exchange — the hydrogen rapidly jumps between different molecules. This exchange is much faster than the NMR timescale, which averages out any coupling.
In very pure, dry samples at low temperature, O–H coupling to adjacent CH protons can sometimes be observed, but under normal conditions it is not.
A compound with molecular formula C₃H₆O₂ shows:
The triplet (3H) and quartet (2H) pattern with J values around 7 Hz is classic for an ethyl group (CH₃CH₂–). The signal at δ = 8.1 (1H singlet) is in the aldehyde region — but this is actually a formate ester: the compound is ethyl methanoate (HCOOC₂H₅). The H at δ = 8.1 is the formate hydrogen, and the ethyl group appears as the characteristic triplet-quartet pattern.
Two compounds both have the molecular formula C₃H₆O and both show C=O in the IR. How does ¹H NMR distinguish them?
Propanal (CH₃CH₂CHO):
Propanone (CH₃COCH₃):
The difference is unmistakable: propanal gives three signals with splitting, while propanone gives a single singlet. The aldehyde hydrogen at δ ≈ 9.8 is also diagnostic.
Mistake 1: Applying the n+1 rule to non-adjacent hydrogens. Splitting only occurs between hydrogens on adjacent carbons (or geminal H on the same carbon). Hydrogens separated by two or more carbons do not normally split each other.
Mistake 2: Forgetting that equivalent hydrogens do not split each other. The 6 equivalent hydrogens in propanone do not cause splitting — they appear as a singlet, not a septet.
Mistake 3: Assuming all multiplets follow simple n+1 patterns. When a proton has two sets of non-equivalent neighbours with different coupling constants, the pattern can be more complex (e.g., a doublet of triplets). At A-Level, such complex patterns are typically described as "multiplet."
Mistake 4: Ignoring the coupling constant. If a triplet has J = 7 Hz, the coupled quartet must also have J = 7 Hz. If the J values do not match, the signals are not coupled to each other.
Mistake 5: Treating the variable OH peak as a reliable indicator of chemical shift. The OH position changes with concentration and temperature. Use the D₂O shake test to confirm its identity rather than relying on its chemical shift value.
Chemical shifts tell you the type of environment (alkyl, adjacent to C=O, adjacent to O, aromatic, aldehyde, or acid). Splitting patterns tell you how many neighbouring hydrogens are present (n+1 rule). Coupling constants confirm which signals are coupled to each other. Together, chemical shifts, integration, and splitting provide enough information to determine molecular connectivity in most organic compounds.
Edexcel 9CH0 specification, Topic 19 — Modern Analytical Techniques II, sub-strand 19.1 requires candidates to interpret ¹H NMR spectra using: (i) chemical-shift values from the data booklet (typical ranges: 0.5–2.0 alkyl C–H; 2.0–3.0 alpha-to-C=O; 3.0–4.5 OCH/OCH₂; 4.5–6.0 alkene; 6.0–8.5 aromatic; 9–10 aldehyde C–H; 9–13 carboxylic acid OH); (ii) the n + 1 rule for spin–spin coupling, where a proton with n equivalent neighbours appears as a multiplet of n + 1 lines (singlet, doublet, triplet, quartet, quintet, sextet); and (iii) the principle that equivalent protons do not split each other (refer to the official specification document for exact wording). Examined on Paper 2 and Paper 3.
Question (8 marks):
Predict the ¹H NMR spectrum of ethyl ethanoate CH₃COOCH₂CH₃ in CDCl₃, including chemical shift, integration, and splitting for each peak.
(a) Identify the number of ¹H environments. (1)
(b) For each environment, give the expected chemical shift, integration, splitting pattern, and explanation. (6)
(c) State why the three protons of CH₃CO– appear as a singlet rather than coupling to neighbouring protons. (1)
Solution with mark scheme:
(a) Three environments: CH₃CO–, –OCH₂–, –CH₂CH₃. A1.
(b)
(c) The CH₃CO– carbon is bonded to C=O and to its three methyl protons; the next-nearest H is across the ester oxygen and four bonds away. Coupling typically only occurs through 2 or 3 bonds (^3J coupling), so the methyl Hs do not couple measurably to any neighbour and appear as a singlet. A1.
Total: 8 marks (M3 A5).
Question (6 marks): An unknown organic compound C has the following ¹H NMR data:
The compound has Mr = 88 and the molecular formula contains C, H, and O.
(a) Calculate the molecular formula. (1)
(b) Identify the functional groups present from the ¹H NMR data. (2)
(c) Suggest a structure for C and justify each peak assignment. (3)
Mark scheme decomposition by AO:
(a)
(b)
(c)
Total: 6 marks split AO1 = 0, AO2 = 3, AO3 = 3.
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