Difficulty: Intermediate | Prerequisites: General chemistry, basic organic chemistry (functional groups, bonding), familiarity with electromagnetic radiation and molecular structure. Chapter 5 notes on degrees of unsaturation recommended.
Proton NMR (¹H NMR) spectroscopy is the single most powerful tool organic chemists use to work out the structure of an unknown compound. It tells you how many distinct types of hydrogen are in a molecule, how many of each type there are, and what their chemical environment looks like. This chapter sits at the heart of Organic Chemistry II: everything from reaction mechanisms to synthesis planning depends on being able to read a spectrum and connect peaks to structure. If you can confidently interpret chemical shift, integration, and splitting patterns, you can reverse-engineer a molecular structure from a squiggly line on a chart.
¹H NMR spectroscopy shows you the hydrogen environment of an organic molecule. Each peak (or group of peaks) corresponds to a set of equivalent protons. You read three things from each signal: where it appears (chemical shift, in ppm), how tall it is relative to others (integration, which gives the ratio of protons), and how many lines it splits into (multiplicity, which tells you about neighbouring protons). Master those three and you can assign every peak in a spectrum to a specific part of the molecule.
Chemical shift (δ)
The position of an NMR signal on the x-axis, measured in parts per million (ppm) relative to a reference compound (TMS). In simple terms, this tells you what kind of chemical environment a proton sits in: near an electronegative atom, on an aromatic ring, next to a carbonyl, and so on.
Tetramethylsilane (TMS)
The standard reference compound in NMR, set to δ = 0 ppm. Think of it as the zero mark on your ruler. All other chemical shifts are measured relative to TMS.
Integration
The area under a peak, which is proportional to the number of protons producing that signal. In simple terms, if one peak integrates for 2 and another for 3, you have a 2:3 ratio of hydrogens in those two environments.
Splitting (multiplicity)
The pattern of lines within a single NMR signal, caused by neighbouring non-equivalent protons. A proton next to n equivalent neighbours splits into (n + 1) lines. Think of it as the proton's way of telling you who its neighbours are.
Singlet (s)
A single line, meaning the proton has no adjacent non-equivalent protons (or its neighbours are equivalent to itself in certain symmetrical cases).
Doublet (d)
Two lines of equal intensity, meaning the proton has one neighbouring non-equivalent proton.
Triplet (t)
Three lines in a 1:2:1 ratio, meaning the proton has two equivalent neighbouring protons.
Quartet (q)
Four lines in a 1:3:3:1 ratio, meaning the proton has three equivalent neighbouring protons.
Coupling constant (J)
The distance (in Hz) between adjacent lines in a multiplet. Protons that couple to each other share the same J value. In simple terms, this is the fingerprint that links two sets of protons: if they have the same J, they are talking to each other.
Shielding and deshielding
Shielded protons are surrounded by electron density, which shifts them upfield (lower ppm, toward TMS). Deshielded protons sit near electronegative atoms or in anisotropic fields (such as aromatic rings), which shifts them downfield (higher ppm). Think of it as: more electron density around the proton means more protection from the external magnetic field.
Downfield / upfield
Downfield means higher ppm (left on the spectrum), upfield means lower ppm (right on the spectrum). Downfield signals come from more deshielded protons.
Equivalent protons
Protons in identical chemical environments that produce the same NMR signal. In simple terms, if you can swap two hydrogens by a symmetry operation and the molecule looks the same, those hydrogens are equivalent.
Each distinct signal (peak or group of peaks) represents a unique proton environment in the molecule.
A molecule with high symmetry will have fewer signals than you might expect, because equivalent protons give one signal.
If you see five signals, you have five chemically distinct types of hydrogen.
δ 0 to 1 ppm: Alkyl protons (CH₃, CH₂ on saturated carbons far from electronegative groups).
δ 1 to 3 ppm: Protons on carbons near mildly electronegative groups, or allylic/benzylic protons.
δ 2 to 2.5 ppm: Protons alpha to a carbonyl (C=O).
δ 3.5 to 4.5 ppm: Protons on carbons bonded to oxygen or nitrogen (OCH₂, OCH₃, NCH₂).
δ 4.5 to 6.5 ppm: Vinyl (alkene) protons.
δ 6.5 to 8.5 ppm: Aromatic protons.
δ 9 to 10 ppm: Aldehyde protons (H-C=O).
δ 10 to 13 ppm: Carboxylic acid protons (COOH).
These ranges overlap, so always use them as starting points, not absolute rules.
The integral (the step-curve drawn over each peak) is proportional to the number of protons in that signal.
Read the ratio of the step heights. If they come out as, say, 1 : 1 : 2 : 3, the simplest interpretation is 1H : 1H : 2H : 3H, but it could also be 2H : 2H : 4H : 6H if the molecular formula allows.
Always cross-check with the molecular formula.
Apply the n + 1 rule: a proton with n equivalent neighbouring protons splits into (n + 1) lines.
0 neighbours → singlet
1 neighbour → doublet
2 neighbours → triplet
3 neighbours → quartet
Equivalent protons do not split each other.
The coupling constant J (in Hz) matches between coupled protons, which helps you pair signals.
Combine chemical shift (what kind of proton), integration (how many), and splitting (how many neighbours) to assign each signal to a specific group in the structure.
Draw out partial structures and connect them.
Verify: does your proposed structure account for every signal, every proton, and the molecular formula?
This molecule has a phenyl ring, a benzylic CH₂ bonded to oxygen, an ether linkage to another CH₂, another CH₂ bonded to the hydroxyl, and an OH group. Here is how the spectrum breaks down:
Aromatic H (δ ~7 ppm): Five aromatic protons from the monosubstituted benzene ring. They appear as a cluster of peaks in the aromatic region. Integration corresponds to 5H.
Benzylic OCH₂ (δ ~4.5 ppm): The two protons on the carbon directly attached to both the ring and the oxygen. Appears as a singlet because there are no adjacent protons on the ring side, and the oxygen separates them from the next CH₂. Integration: 2H.
OCH₂CH₂ (δ ~3.5 to 3.7 ppm): The two sets of CH₂ protons between the two oxygens. These are close enough in shift that they can overlap, appearing as a cluster around δ 3.5 to 3.7. The lecture notes annotate this region as "looks like a quartet," which reflects the coupling between the two adjacent CH₂ groups. Integration: 4H total for both CH₂ groups.
OH (δ ~2.5 to 3 ppm): The hydroxyl proton. Its position and shape vary. In very dry solvent, the OH can show coupling to the adjacent CH₂, producing a visible triplet. In typical conditions, exchange broadens it into a rounded singlet. Integration: 1H.
Key lecture note: The OH may show coupling if the NMR solvent is very dry. Rapid proton exchange with trace water normally washes out OH splitting, but in anhydrous conditions you can see coupling to the neighbouring CH₂.
Methacryaldehyde (also called methacrolein) is a small α,β-unsaturated aldehyde. The molecule has three distinct proton environments:
Methyl group, label 1 (δ ~1.8 ppm): Three equivalent protons on the CH₃ attached to the alkene carbon. This is the tallest peak in the spectrum. It appears at relatively low chemical shift because the methyl is bonded to an sp² carbon but not directly attached to the electronegative oxygen. Integration: 3H.
Vinyl protons, label 2 (δ ~6 ppm): Two protons on the terminal =CH₂. These are not equivalent to each other (one is cis to the aldehyde, the other is trans), so they appear as two closely spaced signals in the vinyl region. The lecture marks them with a squiggle indicating complexity. Integration: 2H total.
Aldehyde proton, label 3 (δ ~9.4 ppm): One proton on the H-C=O group. Aldehyde protons are strongly deshielded by the carbonyl and appear far downfield. Integration: 1H.
This is an aromatic ester with a methyl substituent ortho to the ester group. The structure is a benzene ring bearing a -COOCH₂CH₃ group and a -CH₃ group at the ortho position.
Aromatic H (δ ~7 to 8 ppm): Four protons on the disubstituted ring. They appear as a complex multiplet because the substitution pattern (ortho) makes several of the ring protons non-equivalent, each coupling to its neighbours. Integration: 4H.
OCH₂ (ester CH₂), label 2 (δ ~4.3 ppm): Two protons on the -OCH₂- of the ethyl ester. These couple to the adjacent CH₃ (3 neighbours), producing a quartet. Integration: 2H.
ArCH₃, label 1 (δ ~2.2 ppm): Three protons on the methyl group attached directly to the aromatic ring. Appears as a singlet because there are no adjacent protons on the ring to couple with (the ring protons are too far through bonds for typical coupling). Integration: 3H.
Ester CH₃, label 3 (δ ~1.3 ppm): Three protons on the terminal CH₃ of the ethyl ester. Coupled to the adjacent OCH₂ (2 neighbours), producing a triplet. Integration: 3H.
"Integration gives the exact number of protons." It gives a ratio, not an absolute count. A 1:3 ratio could be 1H and 3H or 2H and 6H. You need the molecular formula to pin down the actual numbers.
"Equivalent protons split each other." They do not. Equivalent protons have the same chemical shift and cannot produce observable splitting between themselves. Only non-equivalent neighbouring protons cause splitting.
"The OH proton always shows coupling." In most routine ¹H NMR spectra, OH protons exchange rapidly with trace water in the solvent, which washes out coupling. You typically see a broad singlet. Coupling to adjacent CH₂ is visible only in very dry solvents.
"More peaks always means more proton environments." A quartet is four lines but represents one proton environment. Count signals (groups of lines), not individual lines.
"Chemical shift tells you exactly what functional group a proton is on." Chemical shift ranges overlap. A proton at δ 3.8 could be on a carbon bonded to oxygen or on an allylic carbon. Always use integration and splitting to narrow it down.
⚠️ You will almost certainly be asked to assign peaks in an NMR spectrum to a given structure. Practice matching chemical shift, integration, and splitting to specific groups in the molecule.
⚠️ Aromatic protons in substituted benzene rings produce complex multiplets. Know that monosubstituted benzene gives a 5H aromatic signal, while di-substituted rings give 4H with varied patterns depending on substitution (ortho, meta, para).
⚠️ The ethyl ester pattern (quartet + triplet, always in a 2:3 integration ratio) is a classic NMR fingerprint. Recognise it on sight.
⚠️ Aldehyde protons appear far downfield (δ 9 to 10 ppm) and integrate for 1H. This is a reliable diagnostic for the -CHO group.
⚠️ Be prepared for questions about how solvent conditions (wet vs. dry) affect the appearance of exchangeable protons such as OH and NH.
True or false: A proton with two equivalent neighbours will appear as a triplet. (True)
Fill in the blank: Aldehyde protons typically appear between δ ______ and δ ______ ppm. (9 and 10)
True or false: Integration gives you the exact number of protons producing a signal. (False, it gives a ratio)
Fill in the blank: The n + 1 rule states that a proton with n equivalent neighbours splits into ______ lines. (n + 1)
True or false: OH protons always appear as sharp, well-defined multiplets. (False, they are typically broad due to exchange)
Q: In the ¹H NMR spectrum of ethyl o-toluate, why does the OCH₂ signal appear as a quartet and the terminal CH₃ as a triplet?
A: The OCH₂ has three equivalent neighbouring protons (the CH₃), so it splits into 3 + 1 = 4 lines (quartet). The CH₃ has two equivalent neighbouring protons (the OCH₂), so it splits into 2 + 1 = 3 lines (triplet).
Q: A compound has the molecular formula C₅H₁₀O. Its ¹H NMR shows a singlet at δ 9.8 (1H), a septet at δ 2.4 (1H), and a doublet at δ 1.1 (6H). Propose a structure.
A: The signal at δ 9.8 is an aldehyde (CHO). The 6H doublet suggests two equivalent CH₃ groups split by one neighbour. The septet at δ 2.4 is a CH with six equivalent neighbours (the two CH₃ groups). The structure is 2-methylbutanal is not quite right for the septet: this pattern fits (CH₃)₂CHCHO, which is 2-methylpropanal (isobutyraldehyde, C₄H₈O). Check the formula: if the exam gives C₅H₁₀O, consider 3-methylbutanal instead, but the NMR pattern described points to the isopropyl-aldehyde motif.
Q: Why might the OH peak of 2-(benzyloxy)ethanol appear as a broad singlet in one spectrum and a triplet in another?
A: In most solvents, the OH proton undergoes rapid exchange with trace water, which averages out coupling and produces a broad singlet. In very dry solvent, exchange slows down enough for coupling to the adjacent CH₂ (two neighbours) to become visible, giving a triplet.
Q: An NMR spectrum shows a 5H multiplet near δ 7.3 and a 2H singlet near δ 4.6. What structural fragment do these signals suggest?
A: The 5H aromatic cluster indicates a monosubstituted phenyl ring (C₆H₅-). The 2H singlet near δ 4.6, in the range typical of protons on carbon bonded to oxygen, suggests a benzylic CH₂ next to an oxygen (PhCH₂O-). Together, these signals point to a benzyloxy group.
Q: In the spectrum of methacryaldehyde, the two vinyl protons appear as separate signals rather than one peak. Why?
A: The two protons on the terminal =CH₂ are geometrically non-equivalent: one is cis to the CHO group and the other is trans. Because they have different spatial relationships to the rest of the molecule, they experience different degrees of deshielding and appear at slightly different chemical shifts.
NMR ties directly to molecular symmetry and degrees of unsaturation (Chapter 5): knowing the degree of unsaturation tells you whether to expect aromatic rings, double bonds, or rings before you even look at the spectrum. It also connects to IR spectroscopy, which identifies functional groups by bond vibrations, and mass spectrometry, which gives the molecular formula. In practice, organic chemists use all three techniques together. Understanding NMR is also essential for following reaction mechanisms in later chapters, because confirming that a reaction produced the expected product relies on reading the product's NMR spectrum.
Proton NMR, hydrogen NMR, ¹H NMR, H-1 NMR, nuclear magnetic resonance spectroscopy, chemical shift, delta, ppm, parts per million, TMS, tetramethylsilane, integration, splitting pattern, multiplicity, coupling constant, J value, singlet, doublet, triplet, quartet, multiplet, shielding, deshielding, downfield, upfield, aromatic protons, vinyl protons, aldehyde proton, exchangeable proton, OH NMR, n+1 rule, equivalent protons, non-equivalent protons, spectrum interpretation, spectrum annotation, organic chemistry II, NMR problem solving, ethyl ester NMR, benzyloxy NMR, methacryaldehyde NMR, ethyl o-toluate NMR