¹H NMR and Combined Techniques

Spec Mapping — OCR H432 Module 6.3.2 (b)-(c) — Proton (¹H) NMR spectroscopy and combined structure determination, covering interpretation of ¹H NMR spectra in four parallel dimensions — number of peak groups (= number of distinct ¹H environments), chemical shift (delta, 0-12 ppm, identifies environment type via OCR data-booklet table: alkyl 0.7-2.0, CH near O 3.3-4.5, alkene 4.5-6.0, aromatic 6.5-8.0, R-CHO aldehyde 9-10, R-COOH acid OH 10-13), integration (relative number of H per environment, derived as a ratio and cross-checked against molecular formula), and multiplicity from the n+1 splitting rule (n equivalent H on adjacent C give a multiplet with n+1 lines whose relative intensities follow Pascal's triangle); identification of singlets (n=0), doublets (n=1), triplets (n=2, characteristic of -CH₂- next to -CH₃), quartets (n=3, characteristic of -CH₂- next to -CH₃), and the diagnostic triplet+quartet pair of an ethyl group; D₂O exchange experiment for identifying labile -OH and -NH₂ protons (exchangeable peaks vanish after D₂O addition); deuterated NMR solvents (CDCl₃, D₂O, DMSO-d₆); combined structural determination using mass spectrometry (molecular mass, fragmentation), IR (functional groups), ¹³C NMR (carbon framework), and ¹H NMR (hydrogen connectivity) on a single unknown organic compound (refer to the official OCR H432 specification document for exact wording).

Where ¹³C NMR (Lesson 13) gives you a map of the carbon skeleton, ¹H NMR (proton NMR) gives you a much more detailed map of the hydrogens. Because there are usually more hydrogens than carbons in an organic molecule, because hydrogens have ~99.985% natural abundance of the ¹H isotope (compared with ¹³C's 1.1%), and because each ¹H "feels" its neighbours via spin-spin coupling that splits its peak into a characteristic pattern, ¹H NMR gives more structural information per spectrum than any other single analytical technique. A well-recorded ¹H spectrum tells you the number of hydrogen environments (peak groups), the type of each environment (chemical shift), the relative number of H atoms per environment (integration), and the number of H neighbours on adjacent carbons (multiplicity from the n+1 rule). Combined with ¹³C NMR (carbon environments), IR (functional groups), and mass spectrometry (molecular mass and fragmentation), it usually lets you pin down a structure uniquely — the standard A-Level structure-determination workflow. This final lesson develops each of the four ¹H NMR dimensions in turn, introduces the D₂O-exchange experiment that diagnoses labile -OH and -NH₂ protons, and concludes with two full worked examples that take an unknown set of MS, IR and NMR data and produce a confident structural assignment.

Key Definition — ¹H NMR spectroscopy: an NMR technique that observes the ¹H nuclei (protons, spin = 1/2, ~99.985% natural abundance) of an organic molecule in a magnetic field. The output is a plot of intensity against chemical shift (delta, ppm, 0-12 ppm typical range), with each hydrogen environment appearing as a peak group whose chemical shift identifies the environment, whose integration gives the relative number of equivalent H, and whose multiplicity (splitting pattern) reveals the number of H on adjacent carbons via the n+1 rule.

1. What ¹H NMR Tells You — Four Dimensions of Information

A ¹H NMR spectrum contains four separate pieces of information for each set of peaks. Each one is independent of the others, so collectively they tightly constrain the structure of the molecule:

Number of peak groups — tells you the number of different H environments (chemically distinct types of hydrogen).
Chemical shift (δ, ppm) — tells you the kind of environment each group is in (alkyl, near O, aromatic, aldehyde, acid, etc.) via the OCR data-booklet table.
Integration (area under peak) — tells you the relative number of H nuclei in each environment (often given as a step-curve or a numerical label).
Splitting pattern (multiplicity) — tells you the number of neighbouring H on directly bonded adjacent carbons, via the n+1 rule.

Combined, these four give you the full picture of the hydrogen skeleton: how many types of H, what kind each is, how many of each, and what each is next to.

graph TD
  A[1H NMR spectrum] --> B[Count peak groups: number of environments]
  A --> C[Chemical shift ppm: type of environment from data booklet]
  A --> D[Integration: relative number of H per group]
  A --> E[Multiplicity n+1 rule: number of H on adjacent C]
  B --> F[Combine all four to deduce H connectivity]
  C --> F
  D --> F
  E --> F

2. Chemical Shifts in ¹H NMR

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.

2.1 Typical chemical shifts (OCR 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.

2.2 Working out environments

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

3 × CH₃ hydrogens → 1 environment
2 × CH₂ hydrogens → 1 environment
1 × OH hydrogen → 1 environment
Total: 3 environments, 3 groups of peaks.

Example: 1,4-dimethylbenzene, (CH₃)₂C₆H₄

6 × CH₃ hydrogens (both methyls equivalent) → 1 environment
4 × aromatic hydrogens (all equivalent by the C2 symmetry of the ring) → 1 environment
Total: 2 environments, 2 groups of peaks.

3. Integration — Counting the Hydrogens

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.

3.1 How to read integration

The integrals are usually given as relative numbers (e.g. 3:2:1) — you need to deduce the actual number of Hs by cross-referencing against the molecular formula (from MS).

Example — ethanol (C₂H₆O): Spectrum shows three peaks with integrals in ratio 3:2:1.

Total H in molecule = 6.
Ratio 3:2:1 adds to 6.
Peak with integral 3 = 3 H (the CH₃).
Peak with integral 2 = 2 H (the CH₂).
Peak with integral 1 = 1 H (the OH).

Example — para-substituted benzene with an ethyl side chain (e.g. 4-ethylphenol, HO-C₆H₄-CH₂CH₃): Integrals are typically in ratio 1:2:2:2:3 over the five peaks. Total H = 1 + 2 + 2 + 2 + 3 = 10, which matches the molecular formula C₈H₁₀O. The 1H is the phenol -OH; the two 2H groups are the two pairs of equivalent aromatic H (ortho and meta to -OH); the third 2H is the -OCH₂- ... wait, this is 4-ethylphenol so it's -CH₂-, and the 3H is the terminal -CH₃ of the ethyl. The integration-to-formula match is the routine check.

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. If the ratio doesn't add up to the molecular formula in whole numbers, you've miscounted somewhere.

3.2 Diagnostic integration patterns

Some integration patterns are so common they are essentially diagnostic:

9 H singlet at ~0.9-1.0 ppm = tert-butyl (CH₃)₃C-.
6 H singlet at ~3.7 ppm = -OCH₃ pair (e.g. dimethyl ester or two equivalent methoxy groups).
3 H + 2 H triplet/quartet pair = ethyl group (-CH₂CH₃ attached to something).
2 H singlet at ~5 ppm = a -CH₂- between two ester or ether groups (e.g. ArCH₂O-).
6.5-8.0 ppm complex multiplet = aromatic ring.

Recognising these patterns at a glance is part of the practised skill of A-Level NMR interpretation.

4. Multiplicity — The n+1 Rule

This is the crown jewel of ¹H NMR interpretation. Each peak is not a single line but a multiplet split by neighbouring hydrogens.

4.1 The n+1 rule

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.

4.2 Which neighbours count?

Hydrogens on adjacent carbons (one bond away from the carbon bearing the hydrogen you are looking at).
Hydrogens on the same carbon as the observed H do not count (they are equivalent to the observed H, and equivalent H do not split each other).
Hydrogens bonded to O or N are usually not coupled (or appear as a broad singlet) — they exchange rapidly with solvent (see D₂O section below).

4.3 Worked example — ethanol

CH₃CH₂OH has three H environments. Let's predict each multiplicity:

CH₃ peak (~1.2 ppm): The neighbouring carbon (CH₂) has 2 H. So n = 2 → splits into (2+1) = 3 lines → triplet.
CH₂ peak (~3.7 ppm): Neighbour 1 is CH₃ (3 H). Neighbour 2 is O–H (1 H — but this is typically an exchange singlet, see below, so treat as 0 neighbours from OH). Effective n = 3 → (3+1) = 4 lines → quartet.
OH peak (~2.5–5 ppm, variable): Usually a broad singlet because of fast exchange with trace H⁺ — the coupling to CH₂ is averaged out.

So ethanol gives a triplet + quartet + broad singlet pattern — a diagnostic fingerprint for an ethyl group attached to oxygen.

4.4 Worked example — ethyl ethanoate

CH₃COOCH₂CH₃ has three H environments:

Acetyl CH₃ (the CH₃ of the acid part of the ester) — its only neighbours are the carbonyl C, which carries no H. Singlet, 3 H, around 2.0 ppm.
OCH₂ (ester oxygen-bearing CH₂) — its neighbours are the CH₃ of the ethyl group (3 H). n = 3 → Quartet, 2 H, around 4.1 ppm. The downfield shift (~4 ppm) reflects the deshielding effect of the adjacent ester oxygen.
Terminal CH₃ (of the ethyl) — its neighbours are the OCH₂ (2 H). n = 2 → Triplet, 3 H, around 1.3 ppm.

Pattern: singlet (3H at ~2.0) + triplet (3H at ~1.3) + quartet (2H at ~4.1). Integrations 3:3:2.

The triplet + quartet pair at ~1.3 and ~4.1 ppm is the unmistakable signature of an ethyl group attached to an ester oxygen. This pattern is diagnostic, and you should recognise it instantly.

4.5 Worked example — propanal

Propanal CH₃CH₂CHO has three H environments:

CHO aldehyde H — neighbours are the CH₂ (2 H). n = 2 → Triplet, 1 H, around 9.7 ppm (very far downfield because of the aldehyde C=O).
CH₂ — neighbours are CHO (1 H) on one side and CH₃ (3 H) on the other. Combined n = 1 + 3 = 4 → Quintet (5 lines), 2 H, around 2.4 ppm. (In practice this is often seen as a more complex multiplet because the two sets of neighbours have different coupling constants.)
CH₃ — neighbours are CH₂ (2 H). n = 2 → Triplet, 3 H, around 1.1 ppm.

Pattern: triplet (1H at ~9.7) + quintet (2H at ~2.4) + triplet (3H at ~1.1). The aldehyde H at 9.7 ppm is a hallmark feature; together with the C=O peak in IR at ~1720 cm⁻¹, this confirms an aldehyde.

graph LR
  A[Count neighbouring Hs on adjacent C] --> B[n neighbours]
  B --> C[n+1 peaks in multiplet]
  C --> D[0 neighbours = singlet]
  C --> E[1 neighbour = doublet]
  C --> F[2 neighbours = triplet]
  C --> G[3 neighbours = quartet]
  C --> H[4 neighbours = quintet]

5. D₂O Exchange — The OCR-Specific Trick

Hydrogens bonded to O (alcohols, carboxylic acids) or N (amines, amides) are labile — they exchange rapidly with other O–H or N–H hydrogens in solution. This has two consequences:

Their coupling to adjacent CH is usually averaged out (so you see a singlet).
You can identify them chemically using a clever trick called D₂O exchange.

5.1 The D₂O exchange experiment

Record the normal ¹H NMR spectrum of the unknown in CDCl₃.
Add a few drops of D₂O (deuterium oxide) to the NMR tube and shake.
Record the spectrum again.
Compare.

What happens chemically: Any O–H or N–H hydrogen in the sample exchanges with a D from D₂O, giving an O–D or N–D. Deuterium is invisible in ¹H NMR (it has the wrong nuclear spin property).

What you see in the spectrum: The peak(s) corresponding to exchangeable O–H or N–H disappear after D₂O is added. The C–H peaks are unchanged.

5.2 Using D₂O exchange to identify functional groups

Peak disappears after D₂O → it was an O–H or N–H. Probable functional groups: alcohol, phenol, carboxylic acid, amine, amide.
Peak remains → it was a C–H. Not exchangeable.

This is the quickest way to tell an alcohol peak apart from a C–H peak when both happen to fall at similar chemical shifts.

Worked example: A compound shows four peaks at 1.2, 3.6, 4.8 and 7.0 ppm. After D₂O, the 4.8 ppm peak has vanished. Conclusion: the 4.8 ppm signal was an exchangeable O–H or N–H (probably an alcohol, given the chemical shift).

Exam Tip: D₂O exchange is tested specifically by OCR. Know the procedure (add D₂O, compare spectra, exchangeable peak disappears) and the conclusion (identifies O–H or N–H protons).

6. Combined Use of Techniques: IR + MS + NMR

No single technique gives a complete answer. In practice, and in OCR exams, you combine three:

Technique	What it tells you
Mass spectrometry (MS)	Molecular mass (from M⁺), atom count, isotope pattern (halogens), fragmentation pattern
Infrared spectroscopy (IR)	Functional groups (C=O, O–H, N–H, C–H region, fingerprint region)
¹H and ¹³C NMR	Hydrogen/carbon environments, connectivity, neighbours

6.1 A combined problem

Problem: An unknown C₃H₆O₂ gives:

MS: M⁺ = 74. No obvious M+2 peak (no halogens).
IR: Broad absorption 2500–3300 cm⁻¹, strong peak at 1710 cm⁻¹.
¹H NMR: Two peaks. Peak 1 at δ 1.2 ppm, triplet, integral 3. Peak 2 at δ 2.4 ppm, quartet, integral 2. A broad peak at 11.5 ppm (integral 1) that disappears on D₂O addition.
¹³C NMR: Three peaks at δ 9, 28 and 179 ppm.

Reasoning:

MS: Molecular mass 74 consistent with C₃H₆O₂ (12×3 + 6 + 16×2 = 74). No halogens.
IR: Broad 2500–3300 cm⁻¹ = O–H of a carboxylic acid. 1710 cm⁻¹ = C=O of a carboxylic acid.
¹³C NMR: Peak at 179 ppm is a carboxylic acid C=O. Peaks at 9 and 28 ppm are alkyl carbons. Three environments total.
¹H NMR: A 3:2 ratio plus an exchangeable OH. Triplet at 1.2 and quartet at 2.4 is a classic ethyl group: CH₃ next to CH₂ gives a triplet at 1.2, CH₂ next to CH₃ gives a quartet at 2.4, and the CH₂ is shifted downfield because it is next to the C=O.
D₂O: The 11.5 ppm peak disappears → it is the carboxylic acid OH.

Conclusion: The compound is propanoic acid, CH₃CH₂COOH.

The molecular formula, IR functional groups, NMR connectivity and D₂O exchange all reinforce each other. Any one technique alone might leave doubt; all four together give a unique answer.

Lesson 14: ¹H NMR and Combined Techniques