Everything so far told you what compounds do. Organic analysis runs the film backwards: given an unknown, how do you prove what it is? You have three tools — test-tube reactions that reveal a functional group, mass spectrometry that pins down the molecular formula, and infrared spectroscopy that fingerprints the bonds. Top marks go to a systematic deduction, not one lucky guess.
Identifying functional groups by test-tube reactions & Required Practical 6
Four functional groups, four reliable wet tests — this is Required practical 6. For each, learn the reagent and the exact observation, because the marks are in the observation, not the name of the test.
| Functional group | Reagent | Positive observation |
|---|---|---|
| Alkene (C=C) | bromine water | orange solution → colourless solution (decolourised) |
| Carboxylic acid (–COOH) | Na2CO3 / NaHCO3 (aq) | effervescence (CO2; turns limewater cloudy) |
| Aldehyde (–CHO) | Tollens’ reagent | silver mirror |
| Aldehyde (–CHO) | Fehling’s / Benedict’s solution | blue solution → brick-red precipitate |
| 1° / 2° alcohol | acidified K2Cr2O7 | orange solution → green solution |
Two of these do double duty. The carbonate test is the sure way to spot a carboxylic acid (only it is acidic enough to release CO2). And Tollens’/Fehling’s separate an aldehyde from a ketone: the aldehyde is oxidised and so reduces the reagent (silver mirror / brick-red precipitate), while a ketone does nothing.
bromine water · shake
The alkene adds Br2 across its C=C and uses it up — say decolourised, never “clear”
sodium carbonate solution
Only a carboxylic acid is acidic enough to free CO2 from a carbonate; the gas turns limewater cloudy
tollens’ reagent · warm gently
Fehling’s does the same job: blue solution → brick-red precipitate with the aldehyde, stays a blue solution with the ketone
acidified K2Cr2O7 · warm
1°/2° alcohols (and aldehydes) reduce orange Cr2O72− to green Cr3+; a 3° alcohol resists oxidation
Interactive — the test-tube identifier
You carry these four out on unknown liquids and record the observations. Two exam habits win marks: give the solution colour or precipitate (not “it reacts”), and give the negative result too (“stays an orange solution”, “no visible change”) so the pair is distinguished. An incorrect reagent scores nothing for the observations that follow. Practical skills are examined across all three papers — see the required practicals page.
These are the same tests from alkenes (bromine water) and alcohols (dichromate, Tollens’/Fehling’s) — gathered here as an identification toolkit.
🧪 Exam-style questions
Give a reagent that could be added separately to each of CH3CH2CH2CHO and CH3CH2CH(OH)CH3 to distinguish between them. State what is observed in each case.
Show answer
Reagent: Tollens’ reagent (ammoniacal silver nitrate). 1 mark
With CH3CH2CH2CHO (aldehyde): silver mirror. 1 mark
With CH3CH2CH(OH)CH3 (secondary alcohol): no reaction / no visible change. 1 mark
Fehling’s (brick-red precipitate vs no change) is an equally good alternative. Do not accept acidified dichromate — it oxidises both.
Give a reagent that could be added separately to cyclohexane and cyclohexene to distinguish between them. State what is observed in each case.
Show answer
Reagent: bromine water. 1 mark
With cyclohexane: stays an orange solution / no visible change. 1 mark
With cyclohexene: decolourised (orange solution → colourless solution). 1 mark
Acidified KMnO4 (purple solution → colourless solution with the alkene) also scores. Ignore “clear” for colourless.
Three bottles contain either propan-1-ol, propanal or propanone. Each is warmed with Fehling’s solution. Identify the liquid that reacts with Fehling’s and give the observation, then suggest a further test to distinguish the remaining two liquids, with the observation for the one that reacts.
Show answer
Reacts with Fehling’s: propanal (aldehyde) — blue solution gives a brick-red precipitate. 1 mark
Further test: warm with acidified potassium dichromate(VI). 1 mark
Propan-1-ol (alcohol) turns the orange solution green; propanone gives no change. 1 mark
Source: AQA A-Level Chemistry past papers.
Mass spectrometry: finding the molecular formula
From atomic structure you know a mass spectrometer weighs ions. Run a whole molecule through it and the peak at the highest mass/charge value — the molecular ion peak, M — gives the relative molecular mass of the compound. (Lower peaks come from the molecule fragmenting.)
High-resolution mass spectrometry
Ordinary (nominal) masses are whole numbers, so different compounds can share one. High-resolution mass spectrometry uses precise atomic masses to several decimal places, and that pins down the molecular formula — because compounds with the same nominal Mr have different precise Mr.
Using precise masses 1H = 1.0078, 12C = 12.0000, 16O = 15.9949:
C4H8O: 4(12.0000) + 8(1.0078) + 15.9949 = 72.0573
C3H4O2: 3(12.0000) + 4(1.0078) + 2(15.9949) = 72.0210
Both are “72” on a nominal scale, but a high-resolution instrument reads 72.0573 versus 72.0210 — enough to say which formula it is.
There is one thing mass spectrometry cannot do: tell isomers apart. Isomers share a molecular formula, so they share a precise Mr — identical molecular-ion peaks. To separate isomers you need infrared (the fingerprint region) or, at A2, NMR.
Interactive — precise-mass distinguisher
All five compounds below have a nominal Mr of 72. Pick two, then compare their precise masses (using 1H = 1.0078, 12C = 12.0000, 16O = 15.9949) to see whether a high-resolution mass spectrometer could tell them apart.
🧪 Exam-style questions
High-resolution mass spectrometry gives precise relative molecular masses. Which compound has a precise Mr different from that of butanone (C4H8O)? Tick (✓) one box.
The precise Mr of C5H12 is 72.1416 and of C5H10 is 70.1260. Use these data to find the precise Ar of hydrogen and of carbon (to 4 decimal places), then the precise Mr of C6H6.
Show answer
Difference of two H atoms = 72.1416 − 70.1260 = 2.0156, so Ar(H) = 1.0078. 1 mark
5 C = 72.1416 − 12(1.0078) = 60.0480, so Ar(C) = 12.0096. 1 mark
Mr(C6H6) = 6(12.0096) + 6(1.0078) = 78.1044. 1 mark
Answers to 4 dp (penalised once if not). ECF carries through.
0.500 g of a hydrocarbon is analysed and found to contain 0.450 g of carbon. Calculate the empirical formula of this hydrocarbon.
Show answer
Mass of H = 0.500 − 0.450 = 0.050 g. 1 mark
Moles: C = 0.450 ÷ 12.0 = 0.0375; H = 0.050 ÷ 1.0 = 0.050. 1 mark
Ratio C : H = 0.0375 : 0.050 = 3 : 4, so the empirical formula is C3H4. 1 mark
Source: AQA A-Level Chemistry past papers.
Infrared spectroscopy
Bonds behave like springs: they absorb infrared radiation and vibrate more. Because each type of bond absorbs at its own characteristic wavenumber, an IR spectrum tells you which bonds — and therefore which functional groups — are present. You don’t memorise the numbers; they are given in Table A of the Data Booklet (reproduced below — the exact data you get in the exam). Your job is to use it.
| Bond | Wavenumber / cm−1 | Tells you |
|---|---|---|
| N–H (amines) | 3300–3500 | amine |
| O–H (alcohols) | 3230–3550 (broad) | alcohol |
| C–H | 2850–3300 | almost every organic molecule |
| O–H (acids) | 2500–3000 (very broad) | carboxylic acid |
| C≡N | 2220–2260 | nitrile |
| C=O | 1680–1750 | aldehyde, ketone, acid or ester |
| C=C | 1620–1680 | alkene |
| C–O | 1000–1300 | alcohol, ester |
| C–C | 750–1100 | fingerprint region (not diagnostic) |
Reading a spectrum is a triage: is there a strong C=O around 1700? Is there a broad O–H? A sharp peak near 1700 with a very broad 2500–3000 hump means a carboxylic acid; a C=O with no broad O–H points to an aldehyde or ketone; a broad O–H at 3230–3550 with no C=O is an alcohol.
The fingerprint region
Below about 1500 cm−1 is the fingerprint region — a dense pattern of peaks that is unique to each molecule. You don’t interpret individual peaks here; you match the whole pattern to a database of known spectra. That is how two compounds with the same functional groups (even isomers) are told apart, and how impurities are spotted.
Interactive — IR peak explorer
Infrared and global warming
The same physics explains the greenhouse effect. Bonds in carbon dioxide, methane and water vapour absorb infrared radiation given off by the Earth; the polar bonds vibrate and re-emit that energy, trapping heat in the atmosphere. Say the bonds are polar and absorb IR — not that “CO2 is a polar molecule” (it isn’t).
Press play to follow the energy: short-wavelength radiation in from the Sun, long-wavelength infrared back out, and some of it trapped. Drag the slider to add greenhouse gas.
🧪 Exam-style questions
Which method can distinguish between (CH3)2CHCH2CHO and (CH3)3CCHO? Tick (✓) one box.
A compound with molecular formula C4H8O gives an infrared spectrum with a strong absorption at about 1715 cm−1 and no broad absorption in the range 3230–3550 cm−1. Which functional group does it contain? Tick (✓) one box.
The infrared spectrum of methane shows absorptions around 3000 cm−1. Use this to explain why methane acts as a greenhouse gas.
Show answer
Methane absorbs infrared radiation (its C–H bonds, 2850–3300 cm−1, absorb IR and vibrate), so it traps heat in the atmosphere. 1 mark
Must convey absorbing / taking in IR. Penalise reference to bonds other than C–H for methane.
The infrared spectrum below is of an organic compound. Which compound could produce this spectrum? Tick (✓) one box.
Compounds A and B both have the molecular formula C4H8O2. Use their infrared spectra below to deduce a possible structural formula for compound A and for compound B.
Show answer
A shows a strong C=O absorption (1680–1750 cm−1) and a broad O–H (alcohols) absorption (3230–3550 cm−1), so it is a hydroxy-carbonyl — e.g. CH3COCH2CH2OH (or CH3CH(OH)CH2CHO). 1 mark
B shows the broad O–H (alcohols) absorption (3230–3550 cm−1) and a C=C absorption (1620–1680 cm−1) but no C=O at 1680–1750 cm−1 — e.g. HOCH2CH=CHCH2OH (but-2-ene-1,4-diol). 1 mark
Any C4H8O2 structure that fits the peaks is accepted; B must contain the C=C. Do not accept a carboxylic acid for A (its O–H would be the very broad 2500–3000 cm−1 band). Quote every peak in full — bond, group and Data Booklet range, e.g. “O–H (alcohols), 3230–3550 cm−1” — a bare number identifies nothing.
Source: AQA A-Level Chemistry past papers.
Deducing a structure — systematically
The hardest marks on this topic come from an unknown and a stack of data. Weak answers spot one clue and guess; strong answers work a sequence and rule structures in or out. Use this order:
- Molecular formula — from the mass spectrum (molecular ion / precise mass) or from combustion/empirical data.
- Functional groups — from the infrared spectrum (C=O? broad O–H? C≡N?).
- Confirm — with a test-tube reaction (silver mirror, orange solution→green solution, effervescence…).
- Consistency check — does one structure fit all the evidence? Rule the others out.
- Examiner reports flag that top marks here need a systematic elimination, not one spotted peak — connect every piece of data to the structure.
- Give observations, never bare test names: “Tollens’ → silver mirror”, not just “Tollens’”.
- Remember what each tool can’t do: mass spec can’t separate isomers; a single IR peak names a group but not the whole molecule.
Interactive — build the mark-scheme answer
“A compound X has a precise Mr of 72.0573 and could be C4H8O (72.0573) or C3H4O2 (72.0210). Its infrared spectrum shows a strong absorption at 1715 cm−1 and no broad absorption in the range 3230–3550 cm−1. When X is warmed with Tollens’ reagent, no silver mirror forms. Identify X, justifying every step.” [5 marks] Select every statement that earns a mark — and nothing that doesn’t. Order doesn’t matter: AQA credits each point on its own.
🧪 Capstone question
A student has samples of butanoic acid, 2-methylpropanal, 2-methylpropanoic acid and 2-methylpropan-1-ol but does not know which is which. Two can be identified by simple chemical tests; the other two (which share a functional group) can then be distinguished by a spectroscopic technique. Describe how to identify all four. This question is marked using levels of response.
Show answer
Stage 1 — the aldehyde: add Tollens’ (or Fehling’s). Only 2-methylpropanal gives a silver mirror (or brick-red precipitate). Stage 1
Stage 2 — the alcohol: warm with acidified potassium dichromate(VI). Only 2-methylpropan-1-ol turns the orange solution green; done in a sensible order so the aldehyde is already identified. Stage 2
Stage 3 — the two acids: butanoic acid and 2-methylpropanoic acid share the –COOH group, so tests match. Use infrared spectroscopy: compare the fingerprint regions against known spectra to match each one. Stage 3
Levels of response: full marks need all three stages, communicated in a logical sequence. (A carbonate test also separates the two acids from the aldehyde/alcohol as an alternative route.)
Source: AQA A-Level Chemistry past papers.
At A2 you add two more instruments to this toolkit: NMR spectroscopy (3.3.15) for the carbon–hydrogen skeleton, and chromatography (3.3.16) for separating mixtures first.
- Test-tube tests: alkene → bromine water orange solution→colourless solution; carboxylic acid → carbonate effervescence; aldehyde → Tollens’ silver mirror / Fehling’s brick-red precipitate; 1°/2° alcohol → dichromate orange solution→green solution (ketone & 3° alcohol: no reaction).
- Mass spectrometry gives the molecular formula: the molecular ion peak (M) = Mr; high-resolution precise masses separate compounds with the same nominal Mr but different formulae.
- Isomers have the same molecular formula, so the same precise Mr — mass spectrometry cannot tell them apart.
- Infrared: bonds absorb IR at characteristic wavenumbers (Data Booklet). Key: O–H alcohol 3230–3550 (broad); O–H acid 2500–3000 (broad); C=O 1680–1750; C≡N 2220–2260.
- The fingerprint region (< 1500 cm−1) is unique to each molecule — match it to a known/database spectrum to confirm identity (or spot an impurity).
- IR & global warming: bonds in CO2, CH4 and H2O absorb IR (the bonds are polar and vibrate), trapping heat — the greenhouse effect.
- Deduce systematically: molecular formula (mass spec) → functional groups (IR) → confirm with tests → consistency check.