Carbon-13 and proton NMR: VCE Chemistry U4A2-04

Loav · VCE Chemistry · Unit 4 / Area of Study 2

U4A2-04 · Expanded original 2026 teaching guide · Study guide

An environment is not just an element

Nuclear magnetic resonance (NMR) provides information about atoms in different chemical environments. Two carbon atoms can produce one carbon-13 signal if they are chemically equivalent by the molecular symmetry. Conversely, two CH₃ groups can have different environments if their neighbours differ. The number of signals is therefore usually the number of distinct environments, not the total number of atoms. Real overlap can hide signals, so use the simplified assumptions supplied with a problem.

Chemical shift, δ, is measured in parts per million (ppm) relative to a reference, commonly TMS at 0 ppm. Different electronic surroundings change shielding and shift. A higher ppm value is described as downfield or more deshielded. Electronegative atoms, carbonyl groups and aromatic systems affect shifts, but a chemical-shift table gives ranges rather than a universal exact value. Solvent and other conditions matter. NMR axes conventionally decrease in ppm from left to right.

Carbon-13: count, then classify

The prescribed low-resolution carbon-13 analysis asks you to use the number and positions of carbon signals. Start with a candidate structure and label equivalent carbons. Then check whether the predicted environment count and types match the spectrum. Do not use ordinary carbon-13 peak height or area as a reliable count of carbons: signal intensities depend on factors beyond the atom count. One signal may represent several symmetry-equivalent carbons.

The 2026 data book places many saturated alkyl carbons below about 60 ppm, alcohol-linked CH₂ at 50–90 ppm, alkene and aromatic carbons at 110–150 ppm, carboxylic-acid carbonyl carbons at 160–185 ppm, ester carbonyls at 165–175 ppm, aldehyde carbonyls at 190–200 ppm and ketone carbonyls at 205–220 ppm. Overlapping ranges mean that position alone is not unique identification. Use the specific table entry appropriate to the proposed connectivity.

Worked example 1: symmetry changes the count

Compare propan-1-ol, CH₃CH₂CH₂OH, and propan-2-ol, CH₃CH(OH)CH₃. In propan-1-ol, the three carbon positions have different surroundings, giving three carbon environments. In propan-2-ol, the two terminal CH₃ groups are equivalent by symmetry, and the central carbon is distinct, giving two environments. Original schematic carbon-13 data at 10, 25 and 65 ppm therefore fit propan-1-ol better than propan-2-ol, assuming an adequately resolved pure-sample spectrum. The 65 ppm signal supports a carbon attached to OH; the environment count provides the key isomer distinction.

Proton NMR: three separate information channels

Position: chemical shift suggests the surroundings of a proton set. In the 2026 book an ordinary R–CH₃ group is typically 0.9–1.0 ppm, a CH₃ next to an ester carbonyl is around 2.0 ppm, an alcohol-linked CH₂ is 3.3–4.5 ppm, and an ester OCH₂ is 3.7–4.8 ppm. OH and NH signals vary substantially: alcohol OH is given as 1–6 ppm. An aldehyde proton is much farther downfield at 9.4–10.0 ppm. Use these values as evidence in combination, not as exact identity labels.

Integration: the step height of an integration curve is proportional to the area under a resonance signal or multiplet. Ratios of these increases give relative numbers of equivalent protons. The line height of a peak is not its integral. An integral ratio 3:2:1 becomes an atom count only when combined with a formula or another constraint; the same raw ratio could in principle be multiplied by a common factor. Sum all lines in a multiplet when evaluating that environment's integral.

Splitting: in simple first-order cases, a set of equivalent protons with n equivalent neighbouring protons gives n + 1 lines. A neighbouring CH₃ can split a CH₂ signal into a quartet; neighbouring CH₂ can split CH₃ into a triplet. The signal tells you about the protons producing it; its splitting tells you about the neighbours. Do not count the protons in the observed set itself as n.

The simple rule is restricted. Equivalent protons do not split one another in the elementary model. Neighbours that are not equivalent can produce more complicated patterns, so do not automatically add all nearby hydrogen atoms into a single n. Rapidly exchanging OH and NH protons often appear broad and do not show predictable neighbouring coupling. Their behaviour depends on conditions; do not promise that every OH signal is always a sharp singlet. Coupling constants are not required.

Low and high resolution

A low-resolution proton spectrum distinguishes the principal resonance environments and their integrated areas without resolving fine splitting. A high-resolution spectrum may resolve the multiplet structure. Three environments can yield many visible lines in a high-resolution spectrum. Counting each line as an environment overestimates the number. Draw a bracket over each multiplet before counting signals.

Worked example 2: account for every hydrogen

Original schematic data for C₂H₆O show a triplet near 1.0 ppm integrating to 3, a quartet at 3.6 ppm integrating to 2, and a broad exchangeable signal at 2.0 ppm integrating to 1. The 3:2 pattern and reciprocal triplet/quartet splitting support CH₃CH₂. The downfield CH₂ is compatible with attachment to oxygen; the exchangeable signal supports OH. Ethanol, CH₃CH₂OH, explains all six H atoms. The OH shift is an illustrative value within its broad allowed range, not a fixed fingerprint.

Check a rival: methoxymethane, CH₃OCH₃, has equivalent methyl groups and no OH. In the simple model it would give one proton environment representing six H, so it does not fit these data. This explicit comparison is stronger than recognising a familiar-looking pair of peaks.

Shift: surroundingsIntegral: H ratioSplitting: neighbours
Original conceptual diagram: Treat chemical shift, integrated area and splitting as different constraints on one candidate structure.

Practice with worked answers

1. Predict carbon environments for butane and 2-methylpropane.

Butane has two environments: equivalent terminal CH₃ carbons and equivalent internal CH₂ carbons. 2-Methylpropane also has two: three equivalent methyl carbons and one central CH carbon. Equal counts alone do not distinguish them; shifts or proton data are needed.

2. Integration steps are 12, 8 and 4 units. A molecular formula has six H. Assign proton counts.

Divide by four to get 3:2:1. Their sum is six, agreeing with the formula, so the sets represent three, two and one H. Do not treat the raw step height 12 as twelve hydrogens.

3. A CH₃ signal neighbours a CH₂ group in a simple first-order system. What pattern and integral are expected?

The three equivalent H give an integral proportional to 3. The two neighbouring equivalent H give n + 1 = 3 lines: a triplet. Three lines do not mean three proton environments.

4. A molecule has five carbons but only three carbon-13 signals. Must the sample be wrong?

No. Symmetry can make multiple carbons equivalent, producing fewer environments than carbon atoms. Overlap can also reduce observed signal count. Draw candidate equivalences and use shifts and other spectra before deciding.

Finish with an audit

For each candidate, check molecular formula, carbon environment count, proton environment count, shift ranges, integral sum and compatible splitting. Reject a candidate if one decisive feature cannot be explained rather than ignoring it. All numerical data here are original schematic values, not experimental spectra. Spectroscopy problems can be solved from supplied data; any real sample handling belongs in a supervised, risk-assessed school activity.

Go deeper: build a connectivity argument

For each proton signal, keep three separate labels: the protons represented, their integrated number, and the neighbouring protons suggested by splitting. These are easily confused. A quartet integrating to two H describes a two-H environment coupled to three equivalent neighbours. It does not mean four H in that environment or two neighbouring groups. A reciprocal triplet with integral three makes a CH₂CH₃ fragment persuasive; a quartet by itself does not locate that fragment in the molecule.

Use chemical shift to attach the fragment to the rest of the structure. An oxygen-linked CH₂ and a carbonyl-adjacent CH₂ may both be quartets with two-H integrals, but their electronic surroundings differ. Carbon-13 then offers a separate check on carbon types. This is why shift reasoning remains valid even when one particular assessment dataset is better resolved by splitting or integration. There is no universal rule that one kind of NMR evidence always outranks another.

Worked example 3: a singlet need not represent one carbon

An original schematic unknown has formula C₄H₁₀O, a broad OH feature in IR, two carbon-13 environments, and proton signals with integral ratio 9:1. The nine-H signal is a singlet; the smaller signal is broad and exchangeable. Rather than assigning nine different H sites, ask what symmetry could make nine H equivalent. Three equivalent CH₃ groups provide nine H, leaving one H on OH.

The structure (CH₃)₃COH, 2-methylpropan-2-ol, fits: three methyl carbons form one carbon environment and the central OH-bearing carbon forms the other. The central carbon has no attached H, so the methyl groups have no neighbouring carbon-bound protons to split them in the elementary model. The OH contribution need not split them under rapid exchange. No exact shift is supplied because the environment-count and integration constraints are sufficient for this candidate analysis.

Check a competing structure: butan-1-ol has four different carbon environments and several carbon-bound proton environments. It cannot account for the supplied two-carbon-environment and 9:1 pattern as simply. Notice the division of labour: the carbon spectrum establishes the symmetry constraint, the proton integral accounts for the H atoms, and splitting tests neighbouring H. A central carbon can appear in carbon-13 even when it has no attached proton and therefore no corresponding proton signal.

Know when the classroom shortcut ends

A CH₂ group next to CH₃ on one side and CH on the other has two different neighbouring sets. Simply adding three plus one and declaring a quintet is not generally justified; separate couplings can create a more complex multiplet. The course does not require coupling-constant analysis, so say that the simple single-set n + 1 rule is insufficient unless the problem supplies an appropriate simplifying assumption. The two H on a CH₂ are also not guaranteed equivalent in every molecular setting. Use the simple equivalences given or justified by the problem rather than extending them universally.

A spectrum can contain reference and solvent signals in addition to sample signals. TMS at 0 ppm is a reference, not a carbon or proton environment of the unknown. Use supplied labels before counting. Overlap can merge resonances, and exchangeable-proton integrals may be less dependable in real data. In schematic problems, use the stated ideal assumptions, but mention these limitations when evaluating a real sample.

Transfer practice

5. A molecular formula contains twelve H; non-overlapping integrated signals have relative areas 3:1. How many H does each represent?

The ratio sums to four units, so each unit corresponds to 12/4 = 3 H. The signals represent nine H and three H. They need not represent one methyl and one single proton. Check whether the proposed structure can actually supply two such equivalent sets.

6. A carbon-13 spectrum shows three sample signals plus TMS. A student draws a four-carbon chain because there are four lines. Give two independent corrections.

Exclude the labelled TMS reference, leaving three sample environments. Also, three environments do not necessarily mean three carbon atoms: symmetry can make multiple carbons equivalent. Use the molecular formula and candidate symmetry to establish the carbon total.

7. In a simple ethyl fragment, the CH₂ quartet has integral two and the CH₃ triplet integral three. If the CH₂ quartet shifts downfield, must its integral or splitting change?

No. A changed electronic environment can move its chemical shift without changing the number of protons in that set or its three equivalent neighbours. The integral can remain two and the pattern a quartet. This is why shift, area and splitting carry different information.