This mini-lesson covers AQA 3.3.6 Organic analysis, 3.3.15 NMR spectroscopy and 3.3.16 Chromatography: functional group tests, infrared spectra, the molecular ion peak in mass spectrometry, ¹³C and ¹H NMR (chemical shift, integration and the n+1 rule), why TMS is used, the D₂O shake, and TLC, column and gas chromatography.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Everything here is A-level standard — the maths is done properly, not skipped. Press Start when you're ready.
Analysis · tests and IR
Functional group tests, infrared and mass spectrometry
Test-tube tests you must know:
Alkene — decolourises bromine water (orange → colourless).
Carboxylic acid — adds to a carbonate to give effervescence (CO₂). Alcohols and phenols do not.
Aldehyde — gives a silver mirror with Tollens’ reagent, and a brick-red precipitate with Fehling’s. A ketone gives neither.
Primary/secondary alcohol — turns acidified dichromate from orange to green. A tertiary alcohol does not.
Halogenoalkane — warm with NaOH(aq), acidify with HNO₃, then add AgNO₃: white (Cl), cream (Br), yellow (I).
Infrared spectroscopy — bonds absorb IR radiation and vibrate. Each bond absorbs at a characteristic wavenumber (cm⁻¹). Use the data sheet, but recognise:
The complex region below 1500 cm⁻¹ is the fingerprint region, unique to each compound — it is matched against a database to make an exact identification.
Mass spectrometry — the peak at the highest m/z is the molecular ion (M⁺) peak, and it gives the Mr directly. The molecular ion also fragments: a peak at m/z 15 is CH₃⁺, 29 is C₂H₅⁺ or CHO⁺, and 43 is C₃H₇⁺ or CH₃CO⁺.
Calculate
Your turn
1The mass spectrum of a compound has its highest-m/z peak at m/z = 46. State the Mr of the compound.
Hint: The molecular ion peak (highest m/z) gives the Mr directly. (This compound is ethanol.)
Quick check
Quick check
?An unknown compound gives effervescence with sodium carbonate and a very broad IR absorption at 2500–3000 cm⁻¹. What is it?
NMR · the basics
NMR: ¹³C and ¹H
Nuclei with an odd mass number (¹H, ¹³C) behave like tiny magnets. In a strong magnetic field they can align with or against it; a radio-frequency photon of exactly the right energy flips them. The frequency required depends on the electron environment of the nucleus — so chemically different atoms absorb at different frequencies.
Chemical shift, δ, is measured in parts per million (ppm) relative to a standard. The more electronegative the neighbouring atoms, the more the nucleus is deshielded and the larger the shift.
¹³C NMR — the easy one. The number of peaks = the number of different carbon environments. That is essentially all you need (there is no splitting to worry about).
¹H NMR — three pieces of information from one spectrum:
Number of peaks → the number of different proton environments.
Integration (the relative area under each peak) → the relative number of hydrogens in each environment.
Splitting pattern → how many hydrogens are on the adjacent carbon(s).
The n + 1 rulea proton with n hydrogens on the ADJACENT carbon(s) is split into n + 1 peaks: 0 → singlet, 1 → doublet, 2 → triplet, 3 → quartet
Why TMS ((CH₃)₄Si) is the standard, defined as δ = 0: all 12 of its hydrogens are equivalent, so it gives a single sharp peak; it is chemically inert and non-toxic; it is volatile, so it is easily removed from the sample; and it absorbs upfield of virtually every other proton, so its peak does not overlap with the sample.
Calculate
Your turn
2How many peaks does the ¹³C NMR spectrum of ethanol, CH₃CH₂OH, have?
peaks
Hint: Count the different CARBON environments — the CH₃ carbon and the CH₂ carbon.
Calculate
Your turn
3How many peaks does the ¹H NMR spectrum of ethanol, CH₃CH₂OH, have?
peaks
Hint: Count the different PROTON environments: CH₃, CH₂ and OH.
Calculate
Your turn
4In ethanol, the CH₃ protons sit next to a CH₂ group. Using the n + 1 rule, into how many lines is the CH₃ peak split?
lines
Hint: n = 2 hydrogens on the adjacent carbon, so the peak has n + 1 lines (a triplet).
Sort it
Sort each proton by its chemical shift
Tap a proton environment, then tap the chemical shift range it appears in.
🟩 δ 0.7–1.6 (alkyl)
🟪 δ 3.1–4.2 (H on C next to O or halogen)
🟦 δ 9.0–12.0 (CHO or COOH)
Quick check
Quick check
?A ¹H NMR spectrum shows a quartet and a triplet with integration 2 : 3. What fragment is present?
Quick check
Quick check
?A sample is shaken with D₂O and one peak in the ¹H NMR spectrum disappears. What does this tell you?
Chromatography
TLC, column and gas chromatography
All chromatography separates a mixture by the balance between two phases:
the stationary phase (the silica or alumina on a TLC plate; the packing in a column; the liquid coating in a GC column) — a component that adsorbs strongly to it moves slowly;
the mobile phase (the solvent, or the carrier gas) — a component that is more soluble in it moves quickly.
Thin-layer chromatography (TLC) — measure the retention factor:
Rf = distance moved by the spot ÷ distance moved by the solvent frontR_f is always between 0 and 1, and it depends on the solvent — so you must run a known reference on the SAME plate
Colourless spots are visualised under UV light (using a fluorescent plate) or with a locating agent such as ninhydrin for amino acids.
Gas chromatography (GC) — the sample is vaporised and carried by an inert gas through a long column. Each component has a characteristic retention time. The area under each peak is proportional to the amount of that component, so GC is quantitative.
The limitation, and the fix: two different compounds can have the same Rf or the same retention time, so chromatography alone can never prove an identity. That is why it is coupled to a second technique — GC–MS, where each separated component goes straight into a mass spectrometer for a definitive identification.
Match it
Match the technique to what it tells you
Tap a technique on the left, then what it tells you on the right.
Technique
What it tells you
Calculate
Your turn
5On a TLC plate, a spot moves 4.5 cm while the solvent front moves 9.0 cm. Calculate the Rf value.
Hint: R_f = distance moved by the spot ÷ distance moved by the solvent front = 4.5 ÷ 9.0.
Quick check
Quick check
?Why can a chromatogram alone never prove the identity of an unknown compound?
Analysis · exam traps
Reading a spectrum without losing marks
The M⁺ peak is the one at the HIGHEST m/z (ignore the tiny M+1 peak from ¹³C). It gives the Mr directly.
The n + 1 rule counts the hydrogens on the ADJACENT carbon(s) — not the hydrogens in the peak itself. A CH₃ next to a CH₂ is a triplet (2 + 1), even though it has three hydrogens.
Integration gives RATIOS, not absolute numbers. A 2 : 3 ratio could be 2 : 3 or 4 : 6 — use the Mr to decide.
Equivalent protons do not split each other. The three hydrogens of a CH₃ are in the same environment, so they give one peak.
The four reasons TMS is the standard — a single sharp peak (all 12 H equivalent), non-toxic and inert, volatile (easily removed from the sample), and it absorbs upfield of virtually everything else, so its peak does not overlap. Quote at least two.
Quick check
Quick check
?How many peaks are there in the ¹H NMR spectrum of methyl ethanoate, CH₃COOCH₃?
Calculate
Your turn
6How many peaks are there in the ¹³C NMR spectrum of propanone, CH₃COCH₃?
peaks
Hint: The two CH₃ carbons are EQUIVALENT (identical environments), and the C=O carbon is different.
Recap
The big ideas to know
Mass spec: the M⁺ peak (the highest m/z) gives the Mr; the fragments tell you the pieces
Infrared: O–H (acid) 2500–3000 broad · O–H (alcohol) 3230–3550 · C=O 1680–1750 · N–H 3300–3500. Use the data sheet
¹³C NMR: the number of peaks = the number of different carbon environments
¹H NMR: number of peaks = number of proton environments · integration = how many H in each · splitting = n + 1
n+1 rule: a peak split into n + 1 lines has n hydrogens on the adjacent carbon(s)
TMS: the reference at δ = 0: it gives a single sharp peak, is non-toxic and inert, is volatile (easily removed) and absorbs upfield of almost everything else
Chromatography: TLC and paper (Rf) · column · gas (retention time). Separation depends on the balance between the stationary and mobile phases
That is AQA 3.3.6, 3.3.15 and 3.3.16 — and the last of the 21 topics. Press Finish.
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