Part 2: Proton NMR: shift, integration and splitting

Treat the number of signals, their positions, integrated areas and splitting patterns as complementary clues, and practise predicting a complete simple spectrum.

A signal is not the same as one line

Equivalent ¹H nuclei produce one chemical signal. In a high-resolution spectrum that signal can be split into several closely spaced lines by coupling to nearby non-equivalent protons. Count chemical environments first, then describe each signal’s splitting; a quartet is one signal with four lines, not four proton environments.

Compare complete surroundings and symmetry. In ethyl ethanoate, CH₃COOCH₂CH₃, the two methyl groups are different: one is adjacent to C=O and one to CH₂. The CH₂ is bonded to O, giving a third environment. In propanone the two methyl groups are equivalent, giving one six-H signal.

For the simple first-order molecules used here, hydrogens within an ordinary CH₃ are equivalent. Some more complex molecules have non-equivalent hydrogens within a CH₂ group; if such detail is supplied, follow the data rather than assuming every CH₂ is always one set.

Locate the hydrogen, not just a nearby functional-group name

A hydrogen bonded directly to an aldehyde carbonyl carbon gives a different shift from hydrogens on the carbon next to C=O. Similarly, O–H is not the same environment as H–C–O. Write the local bond sequence when using the data chart to avoid assigning every proton near oxygen to the same region.

For simple examples, alkyl H is usually at low δ; H–C–O is further downfield; alkene and aromatic H lie further downfield again, and aldehyde H is very downfield. Acid OH commonly occurs at a high, variable δ and may be broad. The exact ranges and overlap are supplied in the Pearson data chart.

Electron-withdrawing atoms reduce local shielding, often raising δ, but chemical shift depends on the whole environment and experimental conditions. Use approximate ranges to propose an assignment, then test it against integration and splitting. A single matching δ value is not unique identification.

The total area of a multiplet counts relative hydrogens

Integrated area is proportional to the number of equivalent H atoms giving the signal. The relevant area includes all lines in the multiplet; the height of its tallest line is not a proton count. Ratios may be scaled: 6:4:6 reduces to 3:2:3.

Use the molecular formula to convert relative ratios into atom counts. If a C₄H₈O₂ molecule has integration ratio 3:2:3, the sum is 8 and these can be actual H counts. A ratio 1:1 alone does not say each signal represents one H; in a six-H molecule it could mean 3 H and 3 H.

Original worked conversion: integrated areas 15, 10 and 15 give a ratio 3:2:3 after dividing by 5. If the formula contains eight observable hydrogens, assign 3 H, 2 H, 3 H. If an OH proton exchanges or is not integrated reliably, the sum can require special care; check the question’s assumptions.

Use n+1 only for an appropriate neighbouring set

For simple first-order coupling to n equivalent adjacent protons that are non-equivalent to the observed protons, a signal is split into n+1 lines. A CH₃ next to CH₂ gives a triplet, while that CH₂ next to CH₃ gives a quartet. Equivalent nuclei do not split one another in this elementary model.

No adjacent coupled H gives a singlet. One gives a doublet; two a triplet; three a quartet. The line-intensity ratios for these simple cases are 1; 1:1; 1:2:1; 1:3:3:1. They describe lines within a signal, not the relative numbers of H in different signals.

When an H set couples to two different non-equivalent neighbouring sets with distinguishable coupling constants, the pattern may be a more complex multiplet. Do not always add all nearby protons and force a single n+1 pattern. Coupling across an O or carbonyl is generally not included in the simple adjacent-carbon rule used for these introductory examples.

A triplet integrating to 3 H and a quartet integrating to 2 H with consistent coupling supports an ethyl fragment. Their shifts establish whether it is an ordinary alkyl ethyl group, an OCH₂CH₃ group or an ethyl group next to C=O. Splitting identifies neighbours; shift locates the fragment.

OH and NH need a separate check

OH and NH chemical shifts can vary with solvent, temperature, concentration and hydrogen bonding. Rapid proton exchange often broadens these signals and removes observable coupling to neighbouring carbon-bound H. In a typical simple ethanol spectrum, the OH may appear as a broad singlet rather than splitting the CH₂ signal; under other conditions coupling can be observed.

Useful extension: adding D₂O can exchange OH/NH protons for deuterium, causing their ¹H signals to disappear or diminish. This can help identify an exchangeable proton but should be used only when the experimental information supports it. A missing OH peak is not, by itself, proof that no OH group exists.

Always separate the structural requirement from the measurement: an alcohol has an OH hydrogen, but a particular spectrum may not display it cleanly. Other signals, IR and formula data can resolve the uncertainty.

Worked prediction: ethyl ethanoate

Draw CH₃–C(=O)–O–CH₂–CH₃ and label the three sets. The acyl methyl has no adjacent carbon-bound H across the carbonyl carbon, so it gives a 3 H singlet near illustrative δ 2.0. OCH₂ has 3 equivalent neighbours in its ethyl CH₃, so it gives a 2 H quartet near δ 4.1. The terminal CH₃ has 2 CH₂ neighbours, so it gives a 3 H triplet near δ 1.2.

Check all eight hydrogens are accounted for and note the reciprocal triplet/quartet pairing. The carbonyl carbon has no proton signal but is essential to interpreting both δ and the molecular formula. Predict three signals, not eight because 3+2+3=8, and not eight lines because 1+4+3=8.

The plotted methyl-propanoate example in the combined-analysis part has the same 3:2:3 ratio and triplet/quartet/singlet types but different shifts and attachments. Similar splitting patterns can therefore belong to different isomers; use every type of NMR evidence.

Explain a deduction in connected steps

A useful answer states signal position, integrated area and multiplicity, then the proposed fragment and why it follows. “δ 4.1, 2 H quartet supports OCH₂ beside CH₃” carries more structural reasoning than a disconnected list of “oxygen, two, four”. Finally combine fragments and confirm the formula.

Pearson 9CH0/02 June 2023 Q7(c)(iii), scheme PDF p.28 and report p.48, distinguishes correctly identified proton environments from carbon labels and recognises symmetry-related groups. When asked about hydrogen environments, label the hydrogens or their clearly defined sets rather than only circling carbon atoms.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources