Haloform Reaction and Iodoform Test: Which Compounds Give a Yellow Precipitate?
Chemistry · Aldehydes, Ketones And Carboxylic Acid · NEET
The iodoform test is positive only for compounds that have a CH3CO- group (methyl ketone) or a CH3CH(OH)- group (a secondary alcohol with a methyl next to the -OH). With I2 and NaOH they give a pale yellow solid, iodoform (CHI3), that smells like a hospital. Memory hook: "Methyl next to C=O or next to CH-OH = yellow bell rings."
The haloform reaction: a CH3 attached to C=O (or to a CH-OH after oxidation) is replaced, leaving as yellow iodoform (CHI3) and giving an acid salt with one carbon fewer.
Your doubts, answered
Which exact groups give a positive iodoform test?
Only two groups pass. (1) A methyl ketone: CH3-CO-R, meaning a CH3 directly attached to the C=O carbon. (2) A CH3-CH(OH)- group, meaning a secondary alcohol where the carbon carrying -OH also carries a CH3. The second one works because I2/NaOH first oxidises the CH3-CH(OH)- to a CH3-CO- (methyl ketone) group, which then reacts. So really everything must become a methyl ketone before the yellow iodoform appears.
Why does ethanol give the test but methanol does not?
Ethanol is CH3-CH2-OH. This is CH3-CH(OH)-H, so it has the CH3-CH(OH)- pattern (R = H). I2/NaOH oxidises it to acetaldehyde CH3-CHO, which is a methyl 'ketone' type carbonyl (CH3-CO-H), so it gives iodoform. Methanol is just CH3-OH with no second carbon, so it can never form a CH3-CO- group. No methyl-next-to-carbonyl means no iodoform.
Why does acetaldehyde give iodoform but higher aldehydes like propanal do not?
Acetaldehyde is CH3-CHO. The CH3 is directly attached to the C=O, so it fits the CH3-CO- rule and gives iodoform. Propanal is CH3-CH2-CHO. Here the carbon attached to the C=O is a CH2, not a CH3. No methyl sits on the carbonyl carbon, so it fails. Acetaldehyde is the ONLY aldehyde that gives a positive iodoform test.
What is the yellow solid and why does it smell?
The yellow precipitate is iodoform, CHI3 (triiodomethane). It has a sharp, medicine-like or antiseptic smell. In the exam, 'yellow precipitate with characteristic smell' is code for 'iodoform test is positive.' If chlorine or bromine is used instead of iodine you get chloroform (CHCl3) or bromoform (CHBr3), which are liquids, not yellow solids; only iodine gives the yellow test you can see.
What is the difference between the haloform reaction and the iodoform test?
They are the same reaction, just used for two purposes. The haloform reaction is the general reaction: a methyl ketone + NaOX (X = Cl, Br, I) gives a carboxylate salt (RCOO-) plus a haloform CHX3. The iodoform test is the special case using iodine (NaOI). Because iodoform is a visible yellow solid, iodine is chosen to DETECT the CH3CO- or CH3CH(OH)- group. So haloform = the chemistry; iodoform test = using it as a lab detector.
How many carbons does the product acid lose?
The carboxylic acid you get always has ONE carbon less than the starting carbonyl compound. That lost carbon is the methyl group, which leaves as CHX3 (the haloform). Example: acetophenone C6H5-CO-CH3 (8 carbons) gives sodium benzoate C6H5-COO- (7 carbons) plus CHI3. Remember: the CH3 walks away as the haloform.
⚠️ The NEET trap ✗ Students see any aldehyde or any -OH compound and mark 'gives iodoform test.' They tick propanal, methanol, or a normal secondary alcohol like CH3CH2CH(OH)CH3. ✓ Only CH3-CO-R (methyl ketone), CH3-CHO (acetaldehyde), CH3-CH2-OH (ethanol), and CH3-CH(OH)-R (methyl carbinol) give a positive test. The methyl must sit right next to the C=O or right next to the CH-OH. 🧠 Check ONE thing: is there a CH3 stuck onto the carbonyl carbon, or onto the carbon holding -OH? If yes, yellow. If no, no test.
Real NEET questions
NEET 2018
Compound A (C8H10O) reacts with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell. A and Y are respectively:
A · C6H5CH(OH)CH3 and I2 ✓
B · C6H5CH2CH2OH and I2
C · H3C-C6H4-CH2OH and I2
D · (CH3)2C6H3OH and I2
Solution: A yellow precipitate with a characteristic smell is iodoform (CHI3), so this is a positive iodoform test. NaOI is made from I2 + 2NaOH, so Y = I2. For the test to pass, compound A must have a CH3CO- or CH3CH(OH)- group. 1-Phenylethanol C6H5-CH(OH)-CH3 has the CH3-CH(OH)- pattern; NaOI first oxidises it to acetophenone C6H5-CO-CH3 (a methyl ketone), which then gives iodoform plus sodium benzoate. Option (A) is correct.
ReNEET 2026
For the sequence: Benzene --(i) CH3COCl, AlCl3--> --(ii) NaOCl--> P + Q. Choose the correct option.
A · If P is the sodium salt of a carboxylic acid, Q is a primary alcohol
B · P and Q are aromatic compounds
C · If P gives a carboxylic acid on acidification, Q gives a poisonous gas on exposure to air and light ✓
D · Both P and Q are carbonyl compounds
Solution: Step (i) Friedel-Crafts acylation gives acetophenone C6H5-CO-CH3, a methyl ketone. Step (ii) NaOCl runs the haloform reaction: the CH3CO- group gives sodium benzoate (P) and chloroform CHCl3 (Q). Acidifying P gives benzoic acid. Chloroform, on standing in air and light, is oxidised to phosgene COCl2, a poisonous gas. So option (C) is correct. Note Q is chloroform (not a carbonyl, not aromatic, not an alcohol), which rules out the others.
NEET 2021
Match the reaction (b) R-CO-CH3 --NaOX--> R-COO- + CHX3 with its correct name.
A · Hell-Volhard-Zelinsky reaction
B · Gattermann-Koch reaction
C · Haloform reaction ✓
D · Esterification
Solution: A methyl ketone R-CO-CH3 treated with sodium hypohalite NaOX loses its methyl group as the haloform CHX3 and forms the carboxylate salt R-COO-. This is exactly the definition of the haloform reaction. When X = iodine it becomes the iodoform test. So the answer is (C) Haloform reaction. (In the full NEET 2021 match question this pairs as (b)-(iii).)
Solved Aldehydes, Ketones And Carboxylic Acid NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. Acetone is CH3-CO-CH3. It has a CH3 attached to the carbonyl carbon (it is a methyl ketone), so it gives a positive iodoform test, forming CHI3 and sodium acetate.
Does the haloform reaction affect a C=C double bond in the molecule?
No. NCERT states clearly that this oxidation does not affect a carbon-carbon double bond if one is present. Only the methyl group next to the carbonyl reacts.
Why is iodine used for the test and not chlorine or bromine?
Iodoform (CHI3) is a yellow solid you can see and it has a sharp smell, so it is a clear signal. Chloroform and bromoform are colourless liquids, so they are not useful for a visible detection test.
Is the iodoform test the same as Tollens' or Fehling's test?
No. Tollens' and Fehling's tests detect aldehydes (they are oxidation tests for -CHO). The iodoform test detects a CH3CO- or CH3CH(OH)- group. A methyl ketone gives iodoform but does NOT give Tollens' or Fehling's, which is a common NEET distinguishing point.
Does the product carboxylic acid have the same number of carbons as the reactant?
No, it has one carbon fewer. The methyl group leaves as the haloform (CHX3), so the acid formed is always one carbon shorter than the starting carbonyl compound.
