Detect a CH₃CO– group with the iodoform test
Use alkaline aqueous iodine to identify the methyl carbonyl group in an aldehyde or ketone.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Distinguish aldehydes from ketones by structure and evidence; state reagent, condition and observation. Written practice is self-checked
Before you begin
- Use carbonyl structures, alcohol oxidation and reduction from Topics 13 and 16, nucleophiles and curly arrows, and the organic reaction patterns from Topic 15.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which group is detected in an aldehyde or ketone by the carbonyl iodoform test?
What you will learn
- Recognise CH₃CO–R
- State alkaline I₂(aq) conditions
- Interpret yellow CHI₃ and carboxylate formation
The test detects a methyl carbonyl structure
An aldehyde or ketone containing CH₃CO–R reacts with alkaline aqueous iodine to give a yellow precipitate of triiodomethane, CHI₃, and a carboxylate ion, RCO₂⁻.
Propanone, CH₃COCH₃, gives CHI₃ and ethanoate. Ethanal also contains the group with R = H and can give CHI₃ and methanoate.
Read the group on the carbonyl side
The methyl group must be directly attached to the carbonyl carbon. A carbonyl elsewhere in the chain without CH₃CO– does not satisfy the test pattern.
Alkaline conditions mean the coproduct is a carboxylate, not the neutral carboxylic acid. The yellow solid is CHI₃, not iodine or a silver compound.
Pause and explain
What yellow solid forms in a positive carbonyl iodoform test?
Worked example
Predict the iodoform result for butan-2-one and identify the carboxylate.
Show the worked solution
- Butan-2-one contains CH₃CO–CH₂CH₃.
- It has the required methyl carbonyl group.
- R is ethyl, so RCO₂⁻ is propanoate.
Answer Positive: yellow CHI₃ precipitate and propanoate, CH₃CH₂CO₂⁻.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Does ethanal give a positive iodoform test? Identify R and the carboxylate.
Give me a hint
Write ethanal as CH₃CO–R.
Compare my reasoning
- Ethanal is CH₃CHO = CH₃CO–H.
- It has the methyl carbonyl group with R = H.
- The carboxylate product is methanoate.
Yes: yellow CHI₃ forms and RCO₂⁻ is HCO₂⁻ (methanoate).
Look for these in your work
- Group identified
- R assigned
- Products named
Does pentan-3-one contain CH₃CO–R?
Give me a hint
Inspect the groups attached directly to its carbonyl carbon.
Compare my reasoning
- Pentan-3-one is CH₃CH₂COCH₂CH₃.
- Both groups attached to C=O are ethyl.
- No methyl group is directly bonded to the carbonyl carbon.
No; pentan-3-one is not a methyl ketone and lacks CH₃CO–R.
Look for these in your work
- Structure written
- Substituents checked
- Conclusion justified
A yellow solid forms with alkaline iodine from an unknown ketone. What can you infer and what is the solid?
Give me a hint
Use the diagnostic group and product name.
Compare my reasoning
- The solid is CHI₃.
- A CH₃CO–R group is supported.
- The exact full structure still requires more evidence.
Yellow CHI₃ supports a CH₃CO–R group; it does not identify one unique molecule.
Look for these in your work
- Solid identified
- Group inferred
- Inference limited
Common mistakes
- Looking for CH₃CH(OH)– instead of CH₃CO–
- Saying every ketone gives a positive result
- Naming the coproduct as a neutral acid in alkaline solution
What can you explain now?
Which group is detected in an aldehyde or ketone by the carbonyl iodoform test?
Compare with the explanation
CH₃CO–R
The methyl group is directly attached to the carbonyl carbon.
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Make it stick
How do the structure CH₃CO–R, alkaline iodine and yellow CHI₃ connect?
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