Spin-Spin Splitting and Analytical Techniques Revision Notes | A Level AQA Chemistry

1. Why NMR Signals Split — The Core Idea

You already know that protons in different chemical environments resonate at different chemical shifts in an NMR spectrum. But look closely at a real spectrum and you will notice something striking: many of those signals are not single peaks — they are clusters of two, three, or four peaks sitting very close together.

This is spin-spin splitting, and it carries a huge amount of structural information. The splitting happens because protons on adjacent carbon atoms influence each other's magnetic environment. Each neighbouring proton can align either with or against the applied magnetic field, and those tiny differences shift the resonance frequency of the proton you are observing — producing multiple peaks instead of one.

The key constraint is that only non-equivalent protons on adjacent carbons cause splitting. Protons on the same carbon as the one you are observing do not split its signal, and neither do protons that are equivalent to it by symmetry.

2. The n+1 Rule — Predicting the Number of Peaks

Here is the rule you need to commit to memory: if a proton has n non-equivalent neighbouring protons on adjacent carbons, its NMR signal will be split into n+1 peaks. That is the n+1 rule, and it is the engine behind reading splitting patterns at A-Level.

At AQA A-Level, you need to be confident with three specific outcomes of this rule: the doublet (n = 1, so 2 peaks), the triplet (n = 2, so 3 peaks), and the quartet (n = 3, so 4 peaks). These three patterns appear repeatedly in aliphatic compounds, and recognising them quickly is a genuine exam skill.

Notice that a singlet (one peak, no splitting) arises when there are no non-equivalent neighbouring protons — for example, the protons of a methoxy group (–OCH₃) attached directly to a carbonyl carbon. Recognising singlets is just as important as recognising doublets, triplets, and quartets.

3. Applying the n+1 Rule — Ethanol as a Model Case

Ethanol (CH₃CH₂OH) is the classic teaching example for spin-spin splitting, and for good reason — it contains three distinct proton environments that each produce a different splitting pattern.

Work through each environment in turn. The –OH proton appears as a singlet under standard A-Level conditions because rapid exchange prevents coupling. The –CH₂– protons are adjacent to the CH₃ group, which has 3 protons, so n = 3 and the signal splits into a quartet. The –CH₃ protons are adjacent to the CH₂ group, which has 2 protons, so n = 2 and the signal splits into a triplet.

Notice the mutual relationship: the CH₃ group splits the CH₂ signal into a quartet, and the CH₂ group splits the CH₃ signal into a triplet. This reciprocal splitting is a hallmark of the ethyl group (–CH₂CH₃) and you will encounter it in many exam questions.

4. A Step-by-Step Worked Example — Assigning Splitting Patterns

Exam questions often give you a structural formula and ask you to predict the splitting pattern for each proton environment, or they give you a spectrum and ask you to assign peaks to environments. The method is the same in both directions: identify each distinct proton environment, count the non-equivalent protons on adjacent carbons, and apply n+1.

Worked example: Predicting NMR Splitting Patterns for 1-Bromopropane (CH₃CH₂CH₂Br)

1-Bromopropane has the structure CH₃–CH₂–CH₂Br. Predict the splitting pattern for each of the three proton environments and justify your answer using the n+1 rule.

1. Environment A: –CH₃ (3H, at C1) Environment B: –CH₂– (2H, at C2, between CH₃ and CH₂Br) Environment C: –CH₂Br (2H, at C3, adjacent to Br) — Three different environments exist because each carbon has a different set of neighbours. Protons on the same carbon are equivalent to each other but non-equivalent to protons on other carbons.
2. –CH₃ at C1 is adjacent only to –CH₂– at C2. C2 carries 2 protons. n = 2 — Only the protons on the directly adjacent carbon count. The CH₂Br group at C3 is two carbons away from CH₃ and does not cause splitting at A-Level.
3. n + 1 = 2 + 1 = 3 peaks Pattern: triplet — The –CH₃ signal appears as a triplet. This is the same pattern seen for the CH₃ group in ethanol.
4. C2 is adjacent to C1 (–CH₃, 3 protons) and C3 (–CH₂Br, 2 protons). Total non-equivalent neighbours = 3 + 2 = 5 n = 5 — Environment B is flanked by two different groups, so you add the proton counts from both adjacent carbons. This gives a more complex multiplet — a sextet (6 peaks) — which is beyond the standard AQA doublet/triplet/quartet set but is the correct prediction for this environment.
5. C3 is adjacent only to C2 (–CH₂–, 2 protons). n = 2 — Bromine does not carry protons, so it contributes nothing to the splitting. Only the 2 protons on C2 matter.
6. n + 1 = 2 + 1 = 3 peaks Pattern: triplet — The –CH₂Br signal also appears as a triplet. Both terminal groups in this molecule happen to give triplets, but they appear at different chemical shifts — the CH₃ triplet near 0.9 ppm and the CH₂Br triplet near 3.4 ppm.

Answer: Environment A (–CH₃): triplet (n = 2). Environment B (–CH₂–): sextet (n = 5, beyond standard AQA set but correct). Environment C (–CH₂Br): triplet (n = 2). The two triplets are distinguished by their different chemical shifts.

The worked example above highlights an important point: when a proton environment is flanked by two different adjacent groups, you add the proton counts from both sides. At AQA A-Level, questions are typically designed so that the environments you are asked about produce doublets, triplets, or quartets — but you should be aware that more complex multiplets exist.

5. Reading Splitting Patterns in Reverse — Structure Determination

In the exam, you will often work in the opposite direction: a spectrum is given and you must deduce structural features from the splitting patterns. The logic is the same n+1 rule, applied backwards. If you see a quartet, the proton producing it has 3 neighbours. If you see a triplet, it has 2 neighbours.

The combination of a quartet and a triplet in the same spectrum is a strong indicator of an ethyl group (–CH₂CH₃). The quartet arises from the –CH₂– protons (3 neighbours on CH₃) and the triplet from the –CH₃ protons (2 neighbours on CH₂). This pattern is so characteristic that experienced chemists recognise it immediately.

6. Combining Analytical Techniques — The Full Picture

NMR splitting patterns are powerful, but no single technique gives a complete structural answer on its own. Scientists have developed a range of analytical techniques that, used together, allow the structure of a new compound to be confirmed with confidence. This is a key appreciation point in the AQA specification.

Think of each technique as providing a different layer of information. Mass spectrometry gives the molecular mass (from the molecular ion peak, M⁺) and fragmentation patterns that suggest which bonds are present. Infrared spectroscopy identifies functional groups through characteristic bond stretching frequencies — for example, a broad O–H stretch around 2500–3300 cm⁻¹ for a carboxylic acid, or a sharp C=O stretch near 1700–1750 cm⁻¹. ¹H NMR spectroscopy then reveals the number of distinct proton environments (from the number of signals), the relative number of protons in each environment (from integration), and the connectivity of the carbon skeleton (from splitting patterns).

The reason scientists use multiple techniques is that each one can confirm or rule out structural possibilities that the others leave ambiguous. For example, mass spectrometry might give a molecular mass consistent with two different molecular formulae, but IR spectroscopy can distinguish between them by revealing which functional groups are present. NMR then pins down the exact arrangement of those groups.

7. Exam Technique — Maximising Marks on NMR Questions

NMR questions at A-Level follow predictable patterns, and knowing the examiner's expectations lets you write answers that collect every available mark. The most common errors are not conceptual — they are presentational. Here is what the mark scheme is looking for.

When asked to predict a splitting pattern, always state both the name of the pattern (doublet, triplet, or quartet) and the reasoning using n+1 explicitly. Writing 'triplet because n = 2, so n+1 = 3' is a complete answer. Writing 'triplet' alone may not earn the reasoning mark if one is available.

When asked to use the spectrum to deduce structure, work systematically: chemical shift first (which environment?), then integration (how many protons?), then splitting pattern (how many neighbours?). Examiners reward structured reasoning, not guesswork.

Apply the n+1 rule carefully: count the equivalent neighbouring protons, then add one. A CH2 group next to two equivalent CH3 groups has six neighbouring protons, so its signal is split into seven peaks.