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Organic Qualitative Tests: What to Know

Learn and apply Organic Qualitative Tests: What to Know in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Organic Qualitative Tests: Orientation

Interpretation is broader than carry-out

The supplied notes list several organic reactions, so their observations may be interpreted. In the practical examination, however, candidates are not required to carry out tests involving 2,4-DNPH, phosphorus(V) chloride or phenol.

Organic tests in Paper 4 are scored on specific observations. If you write only “positive test”, you usually lose the mark.

Use this with Paper 4 Skills: Planning, MMO, PDO, ACE and the Practical and QA (A Level) hub so method, data, and evaluation marks stay aligned.

Definitions (Must Know)

A. Bromine water test (alkene, C=C)

Add bromine water. A positive result is orange → colourless (addition across C=C).

B. 2,4-DNPH test (carbonyl, C=O)

Add 2,4-DNPH (Brady’s reagent). A positive result is an orange/yellow precipitate (carbonyl present: aldehyde or ketone).

C. Tollens’ test (aldehyde)

Warm with Tollens’ reagent (ammoniacal AgNO₃(aq)). A positive result is a silver mirror / grey Ag(s) (aldehyde present).

D. Fehling’s test (aldehyde)

Warm with Fehling’s solution. A positive result is blue → brick-red precipitate of Cu₂O(s) (aldehyde present).

E. Iodoform test

Warm with iodine in alkali (e.g. I₂/NaOH). A positive result is a yellow precipitate of CHI₃(s). This is consistent with:

  • a methyl carbonyl group, CH₃COR (methyl ketone)
  • a secondary alcohol group, CH₃CH(OH)R (oxidises to a methyl carbonyl)
  • ethanol (oxidises to ethanal)

Detailed Explanations

A. Workflow: identify the minimum set of tests

  1. If “unsaturation” is mentioned, do bromine water first (quickest).
  2. If carbonyl is suspected, do 2,4-DNPH.
  3. If 2,4-DNPH is positive, do Tollens’ or Fehling’s to decide aldehyde vs ketone.
  4. If iodoform is mentioned, run it only when the question asks (and interpret carefully).

Mini example:

  • 2,4-DNPH positive + Tollens’ positive → aldehyde present.

B. Why the observations mean what they mean (because…therefore…)

Because bromine adds across a C=C bond, therefore bromine water is decolourised. Because an aldehyde is oxidised (and reduces Ag⁺ to Ag(s)), therefore Tollens’ gives a silver mirror for aldehydes but not ketones.

C. Worked method you can reuse in unknown analysis

  1. Start with the observation the question gives first. Do not invent extra tests.
  2. Match observation to one functional-group claim only.
  3. Add one discrimination test if needed (for example carbonyl → Tollens’/Fehling’s).
  4. End with exam-safe wording:
    • “This result is consistent with…”
    • “This test alone does not distinguish…”
Safety

Bromine water and iodine are irritants. Tollens’ reagent should be freshly prepared and not stored (silver compounds can be hazardous when dry). Follow lab instructions and avoid inhaling fumes.

D. Complete supplied-note map

Organic classSupplied-note observation
alkenedecolourises orange bromine water
chloro-, bromo-, iodoalkaneheat with NaOH(aq), acidify with dilute HNO₃, then add Ag⁺: white, pale-cream or yellow precipitate respectively
primary or secondary alcoholdecolourises acidified KMnO₄ on heating
CH₃CH(OH)R alcohol or methyl carbonylpale-yellow iodoform precipitate on warming with alkaline iodine
phenol or phenylaminedecolourises bromine water and forms a white precipitate
aldehyde or ketoneorange precipitate with 2,4-DNPH
aldehydesilver mirror with Tollens’ reagent; aliphatic aldehyde also gives red-brown precipitate with Fehling’s solution
carboxylic acidliberates CO₂ with Na₂CO₃(aq)
primary amideliberates NH₃ on heating with NaOH(aq)

Do not claim a unique identity when two classes share an observation; choose a discriminating second result.

Worked Examples

Modelled example 1

Infer Two Functional Groups from Supplied Results

Core

Problem

Supplied results show that an unknown decolourises bromine water and gives an orange precipitate with 2,4-DNPH. What functional groups are supported?
Study the worked solution

  1. Interpret bromine water

    Method

    Use orange-to-colourless decolourisation to infer a carbon–carbon double bond.

    Reason

    Bromine adds across an alkene C = C bond under the test conditions.

    Working

    Result supports an alkene.

  2. Interpret 2,4-DNPH

    Method

    Use the orange precipitate to infer a carbonyl group.

    Reason

    A positive 2,4-DNPH result is consistent with an aldehyde or ketone.

    Working

    Result supports C = O; it does not yet distinguish aldehyde from ketone.

  3. Combine without overclaiming

    Method

    Conclude that the unknown contains alkene and carbonyl functionality.

    Reason

    Each supplied observation supports one claim, while carbonyl type still needs selective evidence.

    Working

    Supported groups: alkene + aldehyde/ketone carbonyl.

Guided practice 2

Distinguish a Ketone from an Aldehyde

About 6 min

Problem

Supplied observations show an orange precipitate with 2,4-DNPH, but negative Tollens’ and Fehling’s results. Identify the carbonyl type and justify the evidence chain.

Try this before viewing the solution

2,4-DNPH supports
Negative aldehyde tests support

Hints

Hint 1: broad test
Use 2,4-DNPH to establish the carbonyl class only.
Hint 2: selective tests
Ask which carbonyl type would give a silver mirror or red-brown precipitate.
View solution step by step

  1. Establish the carbonyl

    Method

    State that the orange 2,4-DNPH precipitate supports an aldehyde or ketone.

    Reason

    The reagent is a broad carbonyl test and does not distinguish the two classes.

    Working

    Candidate set: aldehyde or ketone.

  2. Exclude the aldehyde

    Method

    Use both negative selective results to support a ketone.

    Reason

    An aldehyde would reduce Tollens’ reagent and, for an aliphatic aldehyde, Fehling’s solution under the stated tests.

    Working

    Inference: ketone.

Common misconception 3

Correct “2,4-DNPH Identifies an Aldehyde”

Find and correct the mistake

Learner inference

A learner sees an orange 2,4-DNPH precipitate and concludes that the unknown is an aldehyde. Correct the inference and name the minimum additional evidence needed.

Try this before viewing the solution

What the result establishes
Discriminating evidence

View solution step by step

  1. Limit the first inference

    Method

    State that the orange precipitate supports a carbonyl group in an aldehyde or ketone.

    Reason

    Both classes react positively with 2,4-DNPH.

    Working

    One broad test leaves two candidates.

  2. Add selective evidence

    Method

    Use a supplied Tollens’ silver-mirror result or the appropriate Fehling’s red-brown precipitate result.

    Reason

    A positive selective aldehyde result distinguishes an aldehyde; a negative result with confirmed carbonyl supports a ketone.

    Working

    Broad class test → selective discrimination.

Examiner practice 4

Identify a Carboxylic Acid

4 marks

Examination question

State a reagent, observation, confirmatory gas test and inference for identifying a carboxylic acid functional group in an unknown organic compound. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Add the reagent

    1 mark

    Method

    Add aqueous sodium carbonate to a fresh portion of the unknown.

    Reason

    A carboxylic acid reacts with carbonate to release carbon dioxide.

    Working

    Reagent: Na₂CO₃(aq).

  2. Record and test the gas

    2 marks

    Method

    Record effervescence and pass the gas through limewater, which turns milky.

    Reason

    Effervescence shows gas evolution; milky limewater identifies carbon dioxide.

    Working

    Observation: bubbles; limewater → milky.

  3. Make the inference

    1 mark

    Method

    Conclude that the result supports a carboxylic acid group.

    Reason

    Acid–carbonate reaction accounts for the confirmed carbon dioxide.

    Working

    Inference: -CO₂H present.

Challenge 5

Infer a Bromoalkane from a Multi-Step Test

Minimal support

Organic transfer

An unknown is heated with aqueous sodium hydroxide. The mixture is acidified with dilute nitric acid, then aqueous silver nitrate is added. A pale-cream precipitate forms. Identify the halogenoalkane class and explain the purpose of hydrolysis.

Try this before viewing the solution

Halide from precipitate
Original class

Hints

Hint 1: precipitate colour
White, pale-cream and yellow silver halides support chloride, bromide and iodide respectively.
Hint 2: release the ion
The covalently bonded halogen must first be converted into an aqueous halide ion.
View solution step by step

  1. Explain hydrolysis

    Method

    Heat with aqueous hydroxide to substitute the halogen and release halide ions into solution.

    Reason

    Silver nitrate tests aqueous halide ions, not a halogen atom still covalently bonded to carbon.

    Working

    Halogenoalkane → alcohol + X-(aq).

  2. Interpret the precipitate

    Method

    Use the pale-cream silver bromide precipitate to infer bromide and hence a bromoalkane.

    Reason

    The supplied silver-halide colour pattern distinguishes bromide from chloride and iodide.

    Working

    Ag + (aq) + Br-(aq) → AgBr(s); inference: bromoalkane.

Mind Stretchers

Mind stretcher 1Extension

An unknown does not react with 2,4-DNPH, but gives a yellow precipitate of CHI₃(s) on warming with I₂/NaOH. Suggest one possible functional group and one possible compound.

Show Hint

A positive broad test identifies a class; a second selective test must distinguish the candidates.

Show Answer

Mark scheme (examples):

  • Functional group consistent with iodoform: CH₃CH(OH)R (secondary alcohol that oxidises to a methyl carbonyl), or ethanol.
  • One possible compound: propan-2-ol (fits CH₃CH(OH)R where R = CH₃).

Mind stretcher 2: Designing the minimum discrimination sequenceExtension

Question. Two supplied results show that an unknown decolourises bromine water and contains a carbonyl group. Propose the minimum additional observation needed to decide whether the carbonyl is an aldehyde or ketone, and state the practical-scope caveat.

Show Hint

A positive broad test identifies a class; a second selective test must distinguish the candidates.

Show Answer

Use a selective aldehyde result such as a silver mirror with Tollens’ reagent or a red-brown precipitate with Fehling’s solution on warming. A positive result supports an aldehyde; a negative result, alongside confirmed carbonyl evidence, supports a ketone. Candidates may interpret such results, but 9476 does not require them to carry out the 2,4-DNPH test in Paper 4.