You are viewing a free preview of this lesson.
Subscribe to unlock all 14 lessons in this course and every other course on LearningBro.
Where ¹³C NMR gives you a map of the carbon skeleton, ¹H NMR (proton NMR) gives you a map of the hydrogens. Because there are usually more hydrogens than carbons in an organic molecule, and because each hydrogen "feels" its neighbours via spin-spin coupling, ¹H NMR gives more structural information per spectrum than any other technique. Combined with ¹³C, IR and MS, it usually lets you pin down a structure uniquely.
This lesson covers the OCR A-Level Chemistry A (H432) specification points 6.3.2 (b)–(c): interpretation of ¹H NMR spectra including integration, multiplicity (n+1 rule), D₂O exchange, and combined use of NMR with IR and MS.
A ¹H NMR spectrum contains four separate pieces of information for each set of peaks:
Combined, these four give you the full picture of the hydrogen skeleton.
The x-axis of a ¹H spectrum runs from ~0 to ~12 ppm, with TMS at 0 ppm as the reference (same as ¹³C). OCR provides a shift table in the data booklet.
| Proton type | Typical δ (ppm) |
|---|---|
| R–CH₃ (alkyl methyl) | 0.7 – 1.3 |
| R–CH₂–R (alkyl methylene) | 1.2 – 1.4 |
| R₃C–H (alkyl methine) | 1.5 – 2.0 |
| CH₃ adjacent to C=O | 2.0 – 2.5 |
| CH₂ adjacent to C=O | 2.2 – 2.7 |
| CH₃ adjacent to O (ester, ether) | 3.3 – 3.9 |
| CH₂ adjacent to O | 3.3 – 4.5 |
| R–OH (variable — broad, D₂O exchangeable) | 0.5 – 5.0 |
| R–NH₂ (variable — broad, D₂O exchangeable) | 1.0 – 4.5 |
| Alkene =CH | 4.5 – 6.0 |
| Aromatic Ar–H | 6.5 – 8.0 |
| R–CHO (aldehyde H) | 9.0 – 10.0 |
| R–COOH (acid OH) | 10.0 – 13.0 (broad) |
Pattern to remember: Alkyl H around 0–2 ppm, H next to O or N around 3–5 ppm, aromatic around 7 ppm, aldehyde around 9–10 ppm, acid OH around 11–13 ppm.
Use the same equivalence rules as for ¹³C. Equivalent hydrogens (related by symmetry) all appear as a single group with a single chemical shift.
Example: Ethanol, CH₃CH₂OH
Example: 1,4-dimethylbenzene, (CH₃)₂C₆H₄
Modern ¹H NMR spectra are displayed with a step curve or numerical integration values above each peak. The integral represents the area under the peak — not the height — and is proportional to the number of hydrogens giving rise to that peak.
The integrals are usually given as relative numbers (e.g. 3:2:1) — you need to deduce the actual number of Hs from the molecular formula.
Example — ethanol (C₂H₆O): Spectrum shows three peaks with integrals in ratio 3:2:1.
Exam Tip: Integration gives the relative, not absolute, number of hydrogens. Always cross-check against the molecular formula from the mass spectrum or elemental analysis.
This is the crown jewel of ¹H NMR interpretation. Each peak is not a single line but a multiplet split by neighbouring hydrogens.
Key Rule — n+1 Rule: If a hydrogen environment has n equivalent hydrogens on directly bonded neighbouring carbons, its peak is split into (n+1) lines (a multiplet of n+1 peaks).
| Number of neighbours (n) | Number of peaks (n+1) | Name |
|---|---|---|
| 0 | 1 | Singlet (s) |
| 1 | 2 | Doublet (d) |
| 2 | 3 | Triplet (t) |
| 3 | 4 | Quartet (q) |
| 4 | 5 | Quintet |
| 5 | 6 | Sextet |
| 6 | 7 | Septet |
The lines within a multiplet have characteristic relative intensities given by Pascal's triangle: doublet 1:1, triplet 1:2:1, quartet 1:3:3:1.
CH₃CH₂OH has three H environments. Let's predict each multiplicity:
So ethanol gives a triplet + quartet + broad singlet pattern — a diagnostic fingerprint for an ethyl group attached to oxygen.
CH₃COOCH₂CH₃ has:
Subscribe to continue reading
Get full access to this lesson and all 14 lessons in this course.