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Wet-chemistry tests for functional groups are the original tools of organic analysis. Long before infrared, NMR, and mass spectrometers existed, chemists identified unknown compounds by adding small drops of standard reagents and observing colour changes, precipitates, or gas evolution. Even today — in the spectrometer age — these tests retain a place because they are fast, cheap, and require no instrumentation. AQA A-Level Chemistry Paper 3 routinely asks candidates to design a sequence of bench tests to distinguish a small set of unknowns or to predict the observations a given compound will produce. This lesson previews the canonical tests: bromine water and acidified manganate(VII) for alkenes; hydrolysis-then-silver-nitrate for halogenoalkanes; acidified potassium dichromate(VI) for primary, secondary, and tertiary alcohols; Fehling’s solution and Tollens’ reagent to distinguish aldehydes from ketones; 2,4-DNPH (Brady’s reagent) for any carbonyl; sodium carbonate or hydrogencarbonate for carboxylic acids; nitrous acid for primary amines; and the tri-iodomethane (iodoform) test for the CH₃CO– / CH₃CH(OH)– motif. We finish with worked test-sequences for unknown mixtures.
Spec mapping (AQA 7405): This lesson maps to §3.3.6 (test-tube reactions used to identify organic functional groups) and cross-references every organic-group lesson in the AQA A-Level series — §3.3.4 (alcohols), §3.3.7 (aldehydes and ketones), §3.3.8 (carboxylic acids and esters), §3.3.5 (haloalkanes), §3.3.9 (amines), §3.3.10 (polymers), §3.3.12 (aromatic chemistry) and §3.3.16 (organic synthesis). Lessons L0–L2 of this analytical course (mass spectrometry, infrared, NMR) develop the spectroscopic alternative; wet tests should be seen as complementary, not redundant. Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: AO1 recall items dominate: candidates must memorise reagent, conditions, and positive observation for each test. AO2 questions present a specific compound (e.g. butan-2-ol, ethanal, 2-methylpropan-2-ol) and ask you to predict the outcome of a stated test. AO3 questions are the discriminator: you are given a set of unknowns and must design a test sequence — typically two or three tests in order — that uniquely identifies each one, and evaluate the selectivity (e.g. “why not use 2,4-DNPH first?”). Paper 3 frequently couples this with practical-skills marks for safe technique, drop counts, and observation timing.
Reagent appearance: orange/yellow aqueous solution.
Method: Add a few drops of bromine water to the unknown in a test tube and shake. No heat or catalyst required.
Positive result: The orange colour disappears — bromine water is decolourised (orange → colourless). The reaction is essentially instantaneous at room temperature.
Equation (ethene as example): CH₂=CH₂ + Br₂ → CH₂BrCH₂Br (1,2-dibromoethane)
Mechanism: Electrophilic addition. The pi electrons of the C=C bond polarise approaching Br₂, generating a Br⁺ electrophile and a bromide ion which then attacks the carbocation intermediate.
Reagent appearance: intense purple.
Method: Add a few drops of dilute acidified KMnO₄ to the sample at room temperature.
Positive result with cold dilute KMnO₄: the purple colour is discharged, going to colourless (Mn²⁺) or a brown deposit of MnO₂ in less acidic conditions. The alkene is oxidised to a diol (1,2-glycol). With hot or concentrated KMnO₄ the C=C is cleaved to two carboxylic acids or ketones depending on substitution.
Exam Tip: Benzene does NOT decolourise bromine water or acidified manganate(VII), despite having C=C bonds in the Kekulé model. The delocalised pi system is too stable to undergo addition. Only alkenes (and phenol, which reacts via electrophilic substitution to give a white precipitate of 2,4,6-tribromophenol) react. This is a key distinction — examiners love it.
Halide ions in solution give characteristic precipitates with silver nitrate. The complication for halogenoalkanes is that the halogen is covalently bonded — you must first hydrolyse the C–X bond to release free halide ion.
Method:
Observations:
| Halide | Precipitate | Colour | Solubility in NH₃(aq) |
|---|---|---|---|
| Cl⁻ | AgCl | White | Soluble in dilute NH₃ |
| Br⁻ | AgBr | Cream | Soluble in concentrated NH₃ |
| I⁻ | AgI | Yellow | Insoluble in NH₃ (even concentrated) |
The NH₃-solubility row gives the confirmatory second test — sometimes a white precipitate is genuinely AgCl, sometimes it is a contaminating carbonate; the ammonia-solubility check resolves the ambiguity.
Exam Tip: The rate of precipitate formation also gives information. C–I has the weakest bond (240 kJ mol⁻¹) and hydrolyses fastest; C–Cl has the strongest bond (340 kJ mol⁻¹) and is the slowest. Rate order: iodoalkane > bromoalkane > chloroalkane. Examiners exploit this by asking which of three halogenoalkanes gives a precipitate first.
Reagent: Acidified potassium dichromate(VI), K₂Cr₂O₇/H₂SO₄.
Reagent appearance: orange.
Method: Add a few drops of the alcohol to the orange reagent in a test tube and warm in a water bath for two to three minutes.
| Alcohol Type | Observation | Oxidation Product |
|---|---|---|
| Primary (R–CH₂OH) | Orange → green on warming | Aldehyde (then carboxylic acid in excess oxidant under reflux) |
| Secondary (R₂CHOH) | Orange → green on warming | Ketone |
| Tertiary (R₃COH) | No colour change (stays orange) | No reaction — the C bearing OH has no H to lose |
The colour change is due to the half-reaction Cr₂O₇²⁻(aq) + 14H⁺ + 6e⁻ → 2Cr³⁺(aq) + 7H₂O. The chromium oxidation state changes from +6 (orange) to +3 (green) as it is reduced. The alcohol is the reducing agent, supplying the electrons.
The K₂Cr₂O₇ colour change alone gives the same result (orange → green) for both primary and secondary alcohols. To distinguish them, you must inspect the oxidation product:
Then run Fehling’s solution or Tollens’ reagent on the carbonyl product: a positive result (brick-red Cu₂O precipitate from Fehling’s, or silver mirror from Tollens’) confirms the carbonyl is an aldehyde → the original alcohol was primary. A negative result confirms a ketone → the alcohol was secondary.
Reagent: Phosphorus pentachloride (PCl₅, solid).
Method: Add a small spatula of solid PCl₅ to a few drops of the unknown.
Positive result: Vigorous reaction; steamy/misty white fumes of HCl gas are produced. Damp blue litmus held over the tube turns red.
Equation: R–OH + PCl₅ → R–Cl + POCl₃ + HCl(g)
This test detects any –OH (alcohol of any class, plus carboxylic acid OH, plus phenolic OH, plus water). It is general, not selective — and PCl₅ reacts violently with water, so the sample must be anhydrous.
Reagent: Tollens’ reagent — ammoniacal silver nitrate solution, [Ag(NH₃)₂]⁺(aq).
Preparation: Add a few drops of dilute NaOH to AgNO₃(aq) to form a brown precipitate of Ag₂O; then add dilute ammonia solution dropwise until the precipitate just dissolves, forming the colourless diamminesilver(I) complex. Tollens’ reagent must be freshly prepared — stored solutions form explosive silver fulminate.
Method: Add a few drops of the unknown to Tollens’ reagent in a scrupulously clean glass test tube and warm gently (~60 °C) in a water bath for one to two minutes.
Positive result (aldehyde): A bright silver mirror deposits on the inner walls of the test tube. (If the tube is not perfectly clean, the silver instead forms a black/grey precipitate.)
Negative result (ketone): No silver mirror; solution remains colourless.
Equation: RCHO + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → RCOO⁻ + 2Ag(s) + 4NH₃ + H₂O
The aldehyde is oxidised to a carboxylate ion; Ag⁺ is reduced to metallic Ag.
Reagent: Fehling’s solution — an alkaline solution of copper(II) complexed with tartrate (or sodium citrate, in Benedict’s variant). The colour is deep blue.
Method: Add a few drops of the unknown and warm in a water bath for two to three minutes.
Positive result (aldehyde): Blue solution produces a brick-red precipitate of Cu₂O.
Negative result (ketone): Solution remains blue — no precipitate.
Equation: RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O(s) + 3H₂O
The aldehyde is oxidised; Cu²⁺ (blue) is reduced to Cu¹ (in Cu₂O, brick-red).
Practical note: Fehling’s and Tollens’ give the same diagnostic information (aldehyde-positive, ketone-negative). Fehling’s is preferred where silver-mirror cleanliness is hard to achieve; Tollens’ is preferred where the brick-red precipitate is hard to see against a dark sample. Aromatic aldehydes (e.g. benzaldehyde) give a positive result with Tollens’ but a negative (or sluggish) result with Fehling’s — the activated aromatic ring stabilises the aldehyde against the milder copper-based oxidant. Examiners may exploit this.
Reagent: 2,4-dinitrophenylhydrazine (a yellow-orange solid) dissolved in a methanol/sulfuric-acid mixture.
Method: Add a few drops of the unknown to Brady’s reagent.
Positive result: An orange/yellow precipitate forms within seconds — a 2,4-dinitrophenylhydrazone.
Mechanism: Nucleophilic addition–elimination. The –NH–NH₂ group adds across the C=O, then water is eliminated to leave a C=N–NH–Ar linkage.
This test is positive for any compound with a C=O that is an aldehyde or ketone — not esters, not amides, not carboxylic acids (those C=O groups are deactivated by the attached –OR / –NR₂ / –OH).
Identification of the specific carbonyl: Filter, wash, and recrystallise the orange precipitate. Measure its melting point. Compare with published tables of 2,4-DNPH-derivative melting points: each aldehyde and ketone gives a derivative with a characteristic, sharp m.p. (typically 100–300 °C). For example, propanone’s 2,4-DNPH melts at 126 °C; ethanal’s at 168 °C; benzaldehyde’s at 237 °C. A sharp match (within ±1–2 °C) identifies the carbonyl.
Exam Tip: 2,4-DNPH confirms a C=O is present (aldehyde or ketone) but does NOT distinguish between them. Always pair it with Tollens’ or Fehling’s to discriminate. A common AO3 question: “a student tests an unknown with 2,4-DNPH and gets a yellow precipitate; what should they do next?” Answer: Tollens’ or Fehling’s. If positive, aldehyde; if negative, ketone.
Reagent: Na₂CO₃(aq) or NaHCO₃(aq).
Method: Add a few drops of the unknown to the carbonate/hydrogencarbonate solution.
Positive result: Immediate effervescence — bubbles of a colourless gas. Bubble the gas through limewater (Ca(OH)₂(aq)) — it turns milky, confirming CO₂.
Equations:
The carboxylic acid is acidic enough (pKa ~ 4–5) to protonate carbonate ion. Universal indicator paper would show pH < 7.
Reagent: Magnesium ribbon (or sodium, less commonly used because too vigorous).
Method: Drop a small piece of clean magnesium into the unknown.
Positive result: Effervescence — bubbles of hydrogen gas, which gives a squeaky pop with a lighted splint.
Equation: Mg + 2RCOOH → (RCOO)₂Mg + H₂↑
The carboxylic acid is acidic enough to displace H₂ from magnesium. Note that alcohols also react with sodium to give H₂ but do not react with Mg — useful selectivity.
Key Point: Phenol is acidic enough to react with Na₂CO₃ (giving sodium phenoxide and CO₂) but only marginally — the reaction is slow and may not be obvious. Phenol does not react appreciably with NaHCO₃. Carboxylic acids react vigorously with both. Hence NaHCO₃ distinguishes a carboxylic acid (effervescence) from phenol (no effervescence). This is a popular AO3 discriminator.
Reagent: Iodine (I₂) in aqueous sodium hydroxide (NaOH), warmed in a water bath at ~60 °C.
Positive result: A pale yellow precipitate of tri-iodomethane (CHI₃, also called iodoform), with a characteristic antiseptic smell.
This test is positive for compounds containing either of the following sub-structures:
Equation (for a methyl ketone): CH₃COR + 3I₂ + 4NaOH → CHI₃↓ + RCOONa + 3NaI + 3H₂O
Mechanism (outline): The three methyl-group H atoms are sequentially replaced by I (acid–base/halogenation cascade in the alkaline medium); the resulting CI₃ group is then cleaved by hydroxide because the three electron-withdrawing iodines stabilise the leaving CI₃⁻ anion (which immediately picks up a proton to give CHI₃).
Exam Tip: Ethanol gives a positive iodoform test because it is first oxidised to ethanal (CH₃CHO) by the alkaline iodine, and ethanal does have the CH₃CO motif (R = H). But methanol (CH₃OH) does NOT give a positive test — oxidation would produce methanal (HCHO), which lacks any CH₃CO motif. Likewise propan-1-ol (CH₃CH₂CH₂OH) gives a negative test — oxidation gives propanal (CH₃CH₂CHO), which has no CH₃CO group (the carbonyl is flanked by CH₃CH₂, not CH₃). This selectivity makes the iodoform test a precise probe for the CH₃CO– / CH₃CH(OH)– motif.
Reagent: Sodium nitrite (NaNO₂) with dilute hydrochloric acid (generates HNO₂ in situ). Reaction is performed in an ice bath, holding the temperature below 10 °C (ideally 0–5 °C).
Method: Mix the unknown with NaNO₂ and dilute HCl over ice.
For aliphatic primary amines: Effervescence — nitrogen gas (N₂) bubbles off as the aliphatic diazonium salt formed (R–N₂⁺) is unstable and decomposes immediately, expelling N₂ and generating an alcohol or carbocation product. Visible positive test.
R–NH₂ + HNO₂ → R–OH + N₂↑ + H₂O
For aromatic primary amines (e.g. phenylamine, C₆H₅NH₂): The aromatic diazonium salt (Ar–N₂⁺) is stable below 10 °C and remains in solution. Coupling this with a phenol or aromatic amine produces a brightly coloured azo dye (orange, red, or yellow) — a positive test that distinguishes aromatic primary amines.
For secondary and tertiary amines, no N₂ evolves — secondary amines give N-nitroso compounds (yellow oil); tertiary amines give complex products. Hence the HNO₂ test distinguishes primary amines from secondary/tertiary.
Reagent: Neutral iron(III) chloride solution, FeCl₃(aq). Positive result: intense purple/violet colour. The purple species is a phenoxide–Fe(III) complex. Aliphatic alcohols give no colour. Phenol also reacts with bromine water (no catalyst) to give a white precipitate of 2,4,6-tribromophenol and decolourise the Br₂ — a useful double-positive test; benzene does not react because the activating –OH lone pair is absent.
Esters have no good single-reagent bench test. Identify by: IR (strong C=O at 1735–1750 cm⁻¹, plus C–O at 1200–1300 cm⁻¹); ¹H NMR (distinctive CH₃–O– singlet at δ ~3.7–3.9 ppm for methyl esters); or hydrolysis + product testing (reflux with aqueous acid/alkali, then run carbonate-effervescence on the acid product and an oxidation test on the alcohol product). At the bench, signpost IR/NMR and note the absence of positives in the other wet tests (no Br₂ decolouration, no NaHCO₃ effervescence, no Tollens’ silver mirror).
AO3 questions ask for the minimum test set that uniquely identifies a group of unknowns. Branch from general to specific:
Two or three carefully chosen tests usually suffice.
You are given a sample of propan-1-ol (CH₃CH₂CH₂OH). Predict the result of each of the following tests, with brief justification.
| Test | Predicted Result | Reason |
|---|---|---|
| Br₂(aq) | No decolouration | No C=C bond. |
| K₂Cr₂O₇/H₂SO₄, warm | Orange → green | Primary alcohol, oxidised to propanal then propanoic acid. |
| 2,4-DNPH on the distilled product | Orange precipitate | Propanal (an aldehyde) has C=O. |
| Tollens’ on the distilled product | Silver mirror | Aldehyde — positive. |
| Iodoform (I₂/NaOH) | NO yellow precipitate | The carbon bearing OH has no CH₃ attached (CH₃CH(OH)– motif absent); oxidation gives propanal which has no CH₃CO motif. |
| NaHCO₃ | No effervescence | Alcohols are not acidic enough to react. |
| PCl₅ | Steamy HCl fumes | The –OH group reacts. |
A common student error: assuming propan-1-ol gives a positive iodoform test “because it has a CH₃ group.” It doesn’t — the CH₃ needs to be on the carbon bearing OH (i.e. CH₃CH(OH)–), which is true for propan-2-ol but not for propan-1-ol.
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