Identify the CH₃CH(OH)– group with the iodoform test
Use alkaline aqueous iodine to produce yellow triiodomethane from a methyl carbinol group.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. State reagents and conditions, follow the stated reaction pathway, and distinguish each alcohol class and oxidation product. Written practice is self-checked
Before you begin
- Use functional-group formulae and reaction types from Topic 13, alkene reactions from Topic 14, and halogenoalkane substitution from Topic 15.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which group gives the alcohol iodoform test?
What you will learn
- Recognise CH₃CH(OH)–R
- State alkaline I₂(aq) conditions
- Interpret yellow CHI₃ and carboxylate products
The test detects a specific alcohol structure
An alcohol containing CH₃CH(OH)–R reacts with alkaline aqueous iodine to form a yellow precipitate of triiodomethane, CHI₃, and a carboxylate ion, RCO₂⁻.
For propan-2-ol, CH₃CH(OH)CH₃, the group is present and the carboxylate product is ethanoate. The test depends on the structural group, not simply on being an alcohol.
Observe the yellow solid and identify the ion
A positive iodoform test gives yellow CHI₃(s). In alkaline solution, the other organic product is a carboxylate, not a neutral carboxylic acid.
Ethanol also contains the required arrangement with R = H and can give CHI₃ plus methanoate. Do not confuse this with AgNO₃, which detects halide ions.
Pause and explain
What yellow precipitate forms in a positive iodoform test?
Worked example
Predict whether butan-2-ol gives a positive iodoform test and identify the yellow precipitate.
Show the worked solution
- Write the local structure at the OH-bearing carbon: CH₃CH(OH)CH₂CH₃.
- It contains CH₃CH(OH)–R with R = ethyl.
- The yellow solid is CHI₃.
Answer Positive test: a yellow precipitate of CHI₃ forms; the carboxylate product is propanoate.
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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.
Predict the iodoform-test result for ethanol and name the carboxylate product.
Give me a hint
Treat ethanol as CH₃CH(OH)–H.
Compare my reasoning
- Ethanol contains the required methyl carbinol arrangement.
- Alkaline iodine gives yellow CHI₃.
- For R = H, RCO₂⁻ is methanoate.
Positive: yellow CHI₃ precipitate and methanoate, HCO₂⁻.
Look for these in your work
- Structural group recognized
- Precipitate correct
- Carboxylate correct
Does 2-methylpropan-1-ol give a positive alcohol iodoform test? Explain from its structure.
Give me a hint
Inspect the carbon bearing OH and its attached groups.
Compare my reasoning
- The OH-bearing carbon is CH₂OH.
- It is not CH₃CH(OH)–R.
- The diagnostic group is absent.
No; it has CH₂OH rather than the required CH₃CH(OH)–R group.
Look for these in your work
- OH-bearing carbon identified
- Pattern compared
- Conclusion justified
A positive test gives a yellow solid and a carboxylate. Distinguish the test from the halogenoalkane silver nitrate test.
Give me a hint
Compare the reagent and precipitate identity.
Compare my reasoning
- The iodoform test uses alkaline I₂(aq) and gives CHI₃.
- The halogenoalkane test uses AgNO₃ in ethanol and gives AgX.
- Yellow iodoform does not identify a halogen.
Alkaline iodine detects CH₃CH(OH)–R by yellow CHI₃; ethanolic silver nitrate detects released halide as AgX.
Look for these in your work
- Reagents compared
- Solids distinguished
- Inference bounded
Common mistakes
- Claiming every alcohol gives a positive test
- Naming iodine as the yellow precipitate
- Reporting the carboxylate as a carboxylic acid in alkaline solution
What can you explain now?
Which group gives the alcohol iodoform test?
Compare with the explanation
CH₃CH(OH)–R
The methyl group next to the OH-bearing carbon is the key pattern.
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Make it stick
What structural group, reagent and yellow product define a positive alcohol iodoform test?
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