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Iodoform tests: structures and limits

Key idea: Alkaline iodine gives pale-yellow iodoform with ethanal, methyl ketones, ethanol and alcohols of the form CH₃CH(OH)R. The result identifies this structural pattern rather than one unique compound.

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  • GCE A-Level H2 Chemistry 2027
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A pale-yellow precipitate of triiodomethane (iodoform), CHI₃(s), on warming with alkaline aqueous iodine supports a particular structural pattern. It does not uniquely identify ethanol, a ketone or any one molecule.

Start with the structure that reacts, then ask whether the compound already has it or can form it by oxidation under the test conditions.

Check your understanding 1 Compare the methyl positions

Propanone and pentan-3-one are both ketones and contain methyl groups. Must they both give the usual iodoform precipitate?

Predicted result
Show answer

Propanone has a methyl group directly bonded to the carbonyl carbon. Pentan-3-one has ethyl groups on both sides, so it lacks the methyl-ketone pattern.

The carbonyl pattern and its alcohol precursors

The reacting carbonyl pattern is CH₃C(= O)R, where R is hydrogen, an alkyl group or an aryl group. The methyl group must be directly attached to the carbonyl carbon.

  • If R is hydrogen, the compound is ethanal, CH₃CHO. It is the aldehyde that gives this test.
  • If R is an alkyl or aryl group, the compound is a methyl ketone, such as propanone. Not every ketone has this pattern.

An alcohol with the pattern CH₃CH(OH)R can be oxidised during the test to the matching carbonyl compound. Ethanol is the case where R is hydrogen; appropriate secondary alcohols have an alkyl or aryl R group. Methanol and primary alcohols other than ethanol do not give the usual positive result.

Two entry routes to an iodoform-forming carbonyl patternAn alcohol with formula CH3CH(OH)R can be oxidised during the test to CH3C(=O)R. A compound already containing that carbonyl pattern enters at the second stage. Warm alkaline iodine then gives pale-yellow CHI3. R is hydrogen, an alkyl group or an aryl group.Alcohol precursorCH3CH(OH)ROxidation during the testReacting carbonyl patternCH3C(=O)ROr start with this patternWarm alkaline iodinePale-yellow precipitateCHI3 (iodoform)
An alcohol precursor can form the reacting carbonyl pattern during the test. R means hydrogen, an alkyl group or an aryl group: hydrogen gives the ethanol → ethanal case. These are formula routes, not a reaction mechanism or molecular geometry.

The map describes which structures can fit, not a detailed reaction mechanism. A methyl group somewhere else in a molecule is insufficient. A carboxylic acid or ester containing CH₃CO is also outside this aldehyde/ketone pattern: the group attached to the carbonyl carbon is different.

Check the neighbouring carbon before naming the class

Worked example 1

Two ketones can give different results

Problem

Predict the usual iodoform result for propanone, CH₃COCH₃, and pentan-3-one, CH₃CH₂COCH₂CH₃. Explain the structural difference.

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  1. Find the carbonyl neighbours

    Method

    Check the groups directly bonded to the carbonyl carbon.

    Reason

    The test requires a methyl group directly beside that carbon, not merely somewhere in the molecule.

    Working

    Answer

    Propanone has methyl groups adjacent to the carbonyl; pentan-3-one has ethyl groups.

  2. Predict the result

    Method

    Predict a pale-yellow iodoform precipitate for propanone, but not for pentan-3-one.

    Reason

    Only propanone has the required methyl-ketone pattern.

    Working

    Answer

    “Ketone” alone is too broad to predict the test.

Combine two tests without losing a possibility

Guided practice 2

Retain ethanal as well as a methyl ketone

About 6 min

Problem

The possible compounds are ethanal, propanone and ethanol. Supplied observations show an orange precipitate with 2,4-DNPH and a pale-yellow precipitate with warm alkaline iodine. Which possibilities remain? What additional supplied test could distinguish them?

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Hints

Hint 1

use the broad class test

Which possibilities can give the usual 2,4-DNPH precipitate?

Hint 2

inspect the neighbouring group

Among those possibilities, check for a methyl group directly next to the carbonyl. Must such a compound be a ketone?

Show solution step by step

  1. Combine the positive results

    Method

    Retain ethanal and propanone; exclude ethanol.

    Reason

    Both carbonyl possibilities react with 2,4-DNPH and give iodoform. Ethanol gives iodoform but not the usual 2,4-DNPH precipitate.

    Working

    Answer

    Two positives still leave two possibilities.

  2. Choose a selective comparison

    Method

    Use a supplied Tollens’ result to separate the aldehyde and ketone.

    Reason

    Ethanal gives a silver mirror; propanone does not under the specified test conditions.

    Working

    Answer

    Use carbonyl-test reasoning, rather than repeating a shared positive test.

Check your understanding 2 Check both positive organic tests

The only possible compounds are ethanal, propanone and ethanol. Supplied tests give an orange DNPH precipitate and pale-yellow iodoform. Check which compounds remain and what extra test separates them.

Possibilities after both tests
Additional distinguishing evidence
Show answer

Ethanal or propanone; Tollens’ result. A silver mirror with Tollens’ reagent supports ethanal; a reliable negative result supports propanone within these possibilities.

Check your understanding 3 Interpret an alcohol-compatible pattern

An unknown does not give the usual 2,4-DNPH precipitate, but gives pale-yellow iodoform with warm alkaline iodine. Assuming reliable supplied results and ordinary alcohol/aldehyde/ketone possibilities, suggest one possible alcohol and explain why its structure fits.

Show hint

Which alcohol can be oxidised to a methyl ketone, and which primary alcohol can be oxidised to ethanal?

Show answer

Propan-2-ol is one possibility: its CH₃CH(OH)CH₃ pattern can be oxidised to propanone during the test. Ethanol is another: it is oxidised to ethanal. The negative 2,4-DNPH result excludes the ordinary aldehyde/ketone possibilities under the stated conditions, but the iodoform result does not choose one alcohol uniquely.

Check your understanding 4 Move the alcohol group and reconsider the result

Predict the usual iodoform results for butan-1-ol, CH₃CH₂CH₂CH₂OH, and butan-2-ol, CH₃CH(OH)CH₂CH₃. Both contain a methyl group. Why is that not enough to give the same prediction?

Show answer

Butan-2-ol has a methyl group directly attached to the carbon carrying the hydroxyl group, so oxidation gives the methyl ketone butan-2-one and the iodoform result is positive. Butan-1-ol is oxidised initially to butanal, which lacks the required methyl-carbonyl pattern, so it does not give the usual positive result. The position of the groups matters.

Use the supervised method and its handling instructions; the lesson’s supplied observations do not authorise preparing extra reagents or heating an improvised mixture.

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