Spectroscopy questions have a reputation. A student opens the paper, sees a jagged IR spectrum next to a forest of NMR peaks, and freezes. But here’s what I tell every A‑level student: spectroscopy is detective work, and the clues follow rules. Learn the rules and these become some of the most satisfying — and reliable — marks on the paper.
Let’s break down the three techniques you need: mass spectrometry, infrared (IR), and NMR.
First, why we use spectroscopy
All three techniques answer one question: “What is this molecule?” Each gives you a different type of clue:
- Mass spectrometry → the molecule’s mass (and fragments).
- Infrared (IR) → which functional groups are present.
- NMR → the carbon–hydrogen skeleton — how the atoms are arranged.
Combine the clues and you can identify an unknown compound. That’s exactly what the exam asks you to do.
Mass spectrometry: find the mass
The key feature is the molecular ion peak (M⁺) — the peak at the highest mass, which tells you the relative molecular mass of the whole molecule.
- Look for the peak furthest to the right (highest m/z).
- Fragmentation peaks to its left come from the molecule breaking apart. A loss of 15 suggests a CH₃ group; a loss of 29 suggests CHO or C₂H₅.
- A small M+1 peak (from carbon‑13) can even tell you the number of carbon atoms.
Start here: the molecular mass narrows down what the compound could be.
Infrared (IR): identify the functional groups
IR tells you which bonds (and therefore functional groups) are present, because different bonds absorb infrared radiation at characteristic frequencies (wavenumbers, cm⁻¹).
You’ll be given a data sheet in the exam — you don’t memorise the numbers, you learn to use the table. The key absorptions to recognise:
| Bond / group | Approx. wavenumber (cm⁻¹) | Clue | |—|—|—| | O–H (alcohol) | 3200–3550 (broad) | broad dip = alcohol | | O–H (carboxylic acid) | 2500–3300 (very broad) | very broad = acid | | C=O (carbonyl) | 1680–1750 (sharp, strong) | strong dip = aldehyde/ketone/acid/ester | | N–H (amine) | 3300–3500 | | | C–H | ~2850–3100 | present in almost everything |
How to read it: scan for the tell‑tale peaks. A strong, sharp dip near 1700 shouts “C=O”. A broad dip near 3000–3500 says “O–H”. You’re looking for the presence or absence of these signature groups, not every wiggle.
Ignore the busy fingerprint region (below ~1500 cm⁻¹) for identification — it’s unique to each molecule but not something you interpret peak‑by‑peak at A‑level.
NMR: work out the skeleton
NMR is the richest clue and where students gain or lose the most marks. You’ll meet two types:
¹³C NMR (the simpler one — start here)
- **Number of peaks = number of different carbon environments.**
- The chemical shift (position, in ppm) tells you what kind of carbon each is (using your data sheet).
- Quick and powerful: count the peaks, count the carbon environments.
¹H NMR (proton NMR)
Read it in three steps:
- Number of peaks = number of different hydrogen environments.
- Integration (the ratio/area under each peak) = the number of hydrogens in each environment. A 3:2 ratio might mean a CH₃ next to a CH₂.
- Splitting (the n+1 rule) = how many hydrogens are on the neighbouring carbon. A peak split into a triplet (3 lines) means 2 neighbouring hydrogens (n+1 = 3, so n = 2); a quartet means 3 neighbours. This is how you piece adjacent groups together — the classic CH₃CH₂ (ethyl) group shows a triplet and a quartet.
Putting it all together: the detective method
In a structure‑determination question, work through the clues in order:
- Mass spec → molecular mass (what’s the whole thing weigh?).
- IR → which functional groups are present (is there a C=O? an O–H?).
- NMR → how the carbons and hydrogens are arranged (the skeleton).
- Combine → assemble a structure consistent with all the clues, then check it fits the molecular formula.
Treat it like assembling evidence, and the answer emerges logically rather than by guesswork.
Examiner’s tip
Two things win marks here. First, always use the data sheet — don’t try to recall exact wavenumbers or shifts; the exam gives them to you. Second, on ¹H NMR questions, explicitly state your reasoning: “a triplet means two neighbouring hydrogens (n+1 rule).” Examiners reward the interpretation, not just the final structure. Show the detective work.
The bottom line
Spectroscopy is a step‑by‑step process, not a memory test:
- Mass spec gives the mass.
- IR identifies functional groups (use the data sheet).
- NMR reveals the skeleton — count environments, read integration, apply the n+1 rule.
- Combine the clues into one consistent structure.
Practise a handful of full structure‑determination questions and the process becomes second nature.
If combined‑spectra questions still feel overwhelming, working through them live — decoding each clue together on the whiteboard — is one of the fastest ways to build confidence. It’s a core part of how I teach A2 chemistry.
👉 Book a free intro call and let’s turn spectroscopy into easy marks.
