Loav · VCE Chemistry · Unit 4 / Area of Study 2
U4A2-01 · Expanded original 2026 teaching guide · Study guide
What does a chemical test actually tell you?
A qualitative test asks what is present, rather than how much is present. Its useful output is an observation: a colour disappears, a gas forms, or a solid melts over a measured interval. An inference comes afterwards. For example, “bubbles were observed” is an observation; “a carboxyl group may be present” is an inference. Connecting them requires a chemical explanation and consideration of competing explanations.
No single colour test identifies every possible organic molecule uniquely. A good investigation uses comparison samples, an appropriate blank, controlled conditions and more than one independent line of evidence. In a mixture, different molecules can be responsible for different observations. Never assume that a positive alkene test and a positive acid test must come from one molecule unless the sample is known to be pure.
Three functional-group questions
Carbon–carbon double bonds: a suitable bromine reagent loses its orange/brown colour when bromine reacts with an alkene. The reaction is addition across C=C: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br. One double bond consumes one molecule of bromine in this simplified addition model. A colour change under specified test conditions supports a reactive unsaturated group. Other reactive compounds can also consume bromine, so the result is not an absolute proof. Benzene does not behave as three ordinary independent alkene bonds in this test.
Alcohol hydroxyl groups: oxidation provides evidence for oxidisable primary or secondary alcohols. Acidified dichromate(VI), where used in a teacher-managed demonstration, changes from orange to green as chromium(VI) is reduced to chromium(III). A primary alcohol can be oxidised to an aldehyde and then a carboxylic acid; a secondary alcohol gives a ketone. A tertiary alcohol generally does not give the same oxidation under these conditions. Therefore a negative oxidation test does not prove that OH is absent. Aldehydes and other reducing substances can also give positive results.
Reaction with a reactive metal is sometimes used as evidence for an O–H-containing compound: 2CH₃CH₂OH + 2Na → 2CH₃CH₂ONa + H₂. This is not selective for alcohols because water and acids also react. It is hazardous and is discussed here only as chemical evidence, not a practical procedure. The strongest answer says exactly which class of compounds a test can distinguish within the stated candidate list.
Carboxyl groups: a carboxylic acid reacts with hydrogencarbonate to release carbon dioxide. The ionic equation is RCOOH + HCO₃⁻ → RCOO⁻ + CO₂ + H₂O. Effervescence supports an acid reactive enough under the test conditions. Ordinary alcohols do not give this response. Carbon dioxide identification provides stronger evidence than bubbles alone. Other acids can react too; hydrogencarbonate is not a universal molecule-identification reagent.
Purity: use both the range and the temperature
A pure crystalline substance typically melts over a narrow temperature interval close to its reference melting point, measured under comparable conditions. A soluble impurity commonly lowers and broadens the melting range. The width is evidence about purity; the position relative to a reference contributes evidence about identity. “It melts sharply” and “it melts near the expected temperature” are different claims. A sharp melting point alone does not identify a substance because unrelated compounds may have similar melting points.
Worked example 1: interpret supplied melting data
An original classroom dataset gives reference compound X a melting range of 121–122 °C. Sample A melts at 120–122 °C; sample B at 110–118 °C. A is more consistent with relatively pure X because its interval is narrow and near the reference. B's depressed, wider interval suggests impurity. Do not calculate “percentage purity” from these temperatures: there is no supplied calibration relating melting behaviour to composition. Nor can A be declared 100% pure.
Distillation separates using volatility
When a liquid mixture partially vaporises, its vapour is usually richer in the more volatile component. Condensing that vapour gives a collected distillate of changed composition. Simple distillation is useful for separating a volatile liquid from non-volatile dissolved material, or for a suitable pair of liquids with sufficiently different boiling behaviour. It is not a chemical reaction: the intended molecules remain chemically unchanged.
Fractional distillation provides repeated vaporisation–condensation stages in a fractionating column. The successive equilibrations improve separation of suitable miscible liquids with closer boiling temperatures. It does not guarantee perfect separation: some mixtures, including azeotropic mixtures, impose limitations. Discuss this limitation qualitatively; detailed azeotrope calculations are not needed here. A boiling-temperature plateau helps monitor a fraction but is not conclusive proof of purity.
Worked example 2: choose a separation
A solution contains water and a non-volatile dissolved salt. If the target is collected water, simple distillation is appropriate: water vaporises and condenses while most salt remains. For two miscible volatile liquids with relatively close boiling points, choose fractional distillation and explain the repeated enrichment stages. “Use a column because it looks more advanced” gives no chemical justification. If the desired product decomposes before boiling, reconsider the separation instead of assuming stronger heating solves the problem.
Try, explain, then check
1. A candidate list contains hexane, hex-1-ene and hexanoic acid. Which supplied observations distinguish them?
Under suitable controlled conditions, bromine decolourisation supports hex-1-ene; hydrogencarbonate effervescence supports hexanoic acid. Hexane gives neither expected response. The conclusion depends on this limited candidate list and suitable controls.
2. A student says a negative dichromate test proves no hydroxyl groups. Repair the claim.
It indicates no readily oxidisable group was detected under those conditions. A tertiary alcohol can contain OH but fail the oxidation test. Additional evidence such as an appropriate IR band is needed.
3. A crystal melts sharply at 89–90 °C; the desired product reference is 135–136 °C. Is it a pure sample of the desired product?
A narrow interval suggests relative purity, but its large disagreement with the reference argues against the desired identity. It may be a comparatively pure different compound. Repeat/verify the measurement and use another identification method.
4. Why does obtaining one distillation fraction not prove molecular identity?
Different compounds can have similar volatility, and components can co-distil. A fraction may still be a mixture. Compare independent evidence such as IR, NMR or chromatography with suitable reference data.
Safety and final check
Use supplied observations or school-supervised work only. Bromine, chromium(VI), reactive metals, hot glass and flammable solvents require a teacher-approved risk assessment, current safety data, appropriate ventilation, personal protection and segregated waste. Never heat a sealed distillation system. A conceptual diagram is not an apparatus assembly or operating protocol. Before finishing an answer, underline the observation, the chemical inference and its limitation separately.
Go deeper: design a decision, not a list of tests
Before choosing a test, write the possible identities and predict their responses. A useful test divides the candidates into different groups. If every candidate would decolourise the reagent, a positive result does not discriminate between them. Prefer the least hazardous approved test that answers the remaining question, using supplied observations where practical work is unnecessary. Use fresh separate sample portions for independent tests: a reagent introduced by one test may interfere with the next.
Ask what controls would make a negative result meaningful. A known positive comparison shows that the reagent and conditions can produce the expected response; a reagent blank helps identify a background change. If the positive comparison fails, the unknown's negative result cannot safely exclude the target group. Inadequate mixing, a poorly soluble sample or insufficient reaction time can also create an apparent negative. This is why “no visible change under these conditions” is more defensible than “the molecule cannot react”.
Solubility is useful supporting evidence but is not an on/off detector for polar functional groups. An OH group enables hydrogen bonding with water, while a long hydrocarbon portion can make the whole molecule only sparingly soluble. Two liquids forming layers does not prove that both are non-polar. Likewise, a single clear phase does not show that a chemical reaction occurred. Separate the physical observation from the molecular explanation.
Worked example 3: distinguish evidence failure from a surprising identity
Original supplied observations concern a pure unknown selected from propan-1-ol, propan-2-ol and propanoic acid. It gives no hydrogencarbonate effervescence, and its acidified oxidant changes colour. Valid positive controls are supplied. First exclude the carboxylic-acid candidate using the negative acid test, with the control establishing a functioning test. Both alcohol candidates can be oxidised, so oxidation does not distinguish primary from secondary here. Reporting propan-1-ol would exceed the evidence.
What next? Obtain carbon-13 NMR data: propan-1-ol has three environments, whereas symmetric propan-2-ol has two. Three well-resolved sample signals therefore favour propan-1-ol within this list. If instead the original sample were a mixture, this reasoning would need revision because different components could produce the test and spectral signals. The result illustrates a general strategy: identify the surviving alternatives and choose new evidence that separates those particular alternatives.
Transfer practice: explain the decision
5. A sample and a known alkene both fail to decolourise a bromine reagent. Can you exclude C=C in the sample?
No. The failed positive comparison undermines the test. Check reagent suitability and the stated conditions using the approved school method; do not reinterpret a failed control as structural evidence. An independent spectrum may help while the test is investigated.
6. Two narrow-boiling fractions have similar boiling temperatures. Give one reason this does not prove equal identity, and one useful next measurement.
Different compounds can have similar boiling points; some components also co-distil. Compare a diagnostic IR/NMR pattern or chromatography against authenticated references. Choose a measurement that distinguishes the plausible compounds rather than simply repeating the same temperature measurement.
7. A desired crystal has reference melting range 148–149 °C. A sample initially melts at 137–145 °C; after purification it melts at 147–149 °C. Evaluate both improvement and remaining uncertainty.
The interval narrows from 8 °C to 2 °C and moves toward the reference, supporting improved purity and consistency with the desired product. It does not quantify percentage purity or uniquely identify the compound. Independent structural evidence remains useful. Measuring both samples with comparable heating and calibrated equipment makes the comparison more valid.
Explain why separation can improve purity without proving it: distillation enriches a volatile component because liquid and vapour have different compositions, but enrichment is a relative change, not a guarantee of a single compound. Assess the collected material independently. A process label such as “purified” records what was attempted; a defensible purity claim records what was measured and what the measurement could miss.
