Key Fact
Haloform reaction
Methyl ketones (or 2° alcohols oxidisable to methyl ketones, ethanol) + X₂/NaOH → CHX₃ (haloform) + RCOO⁻. Iodoform test: distinguishes methyl ketones; yellow precipitate of CHI₃.
-- NCERT Class 12 Chemistry, Ch. 8, p. 240Haloform
8 MCQs9-step worked example
Source: NCERT Alcohols, Phenols, Ethers and Carbonyl CompoundsPYQ coverage: NEET 2026Official key: NTA-verifiedLast updated: 24 Sep 2026
Haloform, explained for NEET
The haloform reaction is one of the cleanest "yes/no" tests in organic chemistry — and the trap that costs marks is treating it as if every ketone qualifies.
What the reaction is. When a compound containing a methyl carbonyl group (CH₃–CO–) or a structure oxidisable to one is treated with excess NaOH/I₂ (or NaOH/Cl₂, NaOH/Br₂), the three α-hydrogens on the methyl group are successively replaced by halogen atoms. The resulting trihalomethyl group is then cleaved by base, yielding a haloform (CHI₃, CHCl₃, or CHBr₃) and the sodium salt of a carboxylic acid with one fewer carbon.
The iodoform variant uses I₂/NaOH. A yellow precipitate of iodoform (CHI₃, mp 119 °C) is the positive test result, as noted in NCERT Class 12 Chemistry Chapter 8, Part 2, page 244.
Which substrates give a positive iodoform test — the complete list:
- Methyl ketones: CH₃–CO–R (acetone, acetophenone, etc.)
- Ethanal (acetaldehyde): CH₃–CHO
- Ethanol: CH₃–CH₂–OH (oxidised in situ to ethanal)
- Secondary alcohols of the form CH₃–CH(OH)–R (oxidised in situ to methyl ketones)
The high-frequency trap: aspirants mark non-methyl ketones (e.g., diethyl ketone, benzophenone) or primary alcohols longer than ethanol as positive. They are not. The structural requirement is strict — a CH₃CO– unit must be present or generatable by mild oxidation. Pentan-3-one (CH₃CH₂COCH₂CH₃) has no methyl group adjacent to the carbonyl and does NOT give the test.
NEET context. This topic appears as a direct recall or single-step application question, typically asking "which of the following gives a positive iodoform test?" with four structures. Speed matters: if you can spot the CH₃CO– motif in seconds, the mark is free. If you guess, the –1 negative marking penalty hits.
Can you answer these Haloform MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which one of the following compounds gives a positive iodoform test?
Show answer and why every option is right or wrong
Answer: B. Acetophenone contains the CH₃–CO– group directly bonded to the carbonyl, satisfying the structural requirement for the iodoform reaction. NCERT Class 12 Chemistry Chapter 8, Part 2, page 244 lists methyl ketones as positive substrates.
Why A is wrong: A is wrong because benzophenone has two phenyl groups on the carbonyl (C₆H₅–CO–C₆H₅) — no CH₃–CO– unit is present. It is not a methyl ketone.
Why C is wrong: C is wrong because propanal is an aldehyde whose carbonyl carbon bears –CH₂CH₃ and –H, not –CH₃. NCERT requires a CH₃CO– or CH₃CH(OH)– group, and propanal has neither. Among the aldehydes only ethanal (CH₃CHO) carries CH₃CO– (trap: assuming every small aldehyde behaves like acetaldehyde).
Why D is wrong: D is wrong because pentan-3-one is identical to diethyl ketone. The carbonyl carbon bears –CH₂CH₃ groups, not –CH₃. No methyl carbonyl unit exists (trap: confusing the CH₃ at the chain end with a CH₃ on the carbonyl).
MCQ 2Easy RecallPractice
The iodoform test uses I₂/NaOH. What is the characteristic observation for a positive test?
Show answer and why every option is right or wrong
Answer: A. A positive iodoform test is indicated by the formation of a yellow precipitate of CHI₃ (iodoform, mp 119 °C), as stated in NCERT Class 12 Chemistry Chapter 8, Part 2, page 244.
Why B is wrong: B is wrong because the haloform reaction does not produce a gas as the diagnostic observation. Iodoform is a solid precipitate, not a volatile product under these conditions.
Why C is wrong: C is wrong because the precipitate is yellow, not white, and iodoform is insoluble in aqueous NaOH. Confusing iodoform with a metal hydroxide precipitate leads to this error.
Why D is wrong: D is wrong because a violet colouration is associated with FeCl₃ tests for phenols or enols, not with the iodoform reaction. This confuses two different qualitative tests.
MCQ 3Easy RecallPractice
Which of the following alcohols gives a positive iodoform test?
Show answer and why every option is right or wrong
Answer: D. Ethanol is oxidised in situ by I₂/NaOH to ethanal (CH₃CHO), which contains the CH₃–CO– unit and undergoes the haloform reaction. NCERT Class 12 Chemistry Chapter 8, Part 2, page 244.
Why A is wrong: A is wrong because methanol (CH₃OH) is oxidised to formaldehyde (HCHO), which lacks the CH₃–CO– group. Formaldehyde has no α-methyl hydrogens to be halogenated (trap: assuming any short-chain alcohol qualifies).
Why B is wrong: B is wrong because 2-methylpropan-2-ol is a tertiary alcohol — it resists oxidation by mild reagents like I₂/NaOH. No carbonyl is generated, so no haloform reaction occurs.
Why C is wrong: C is wrong because propan-1-ol is oxidised to propanal (CH₃CH₂CHO). The carbonyl carbon bears a –CH₂CH₃ group, not –CH₃. The methyl group is at the far end of the chain, not adjacent to C=O.
MCQ 4Direct ApplicationPractice
Among the following, how many compounds give a positive iodoform test?
(i) Acetone (ii) Pentan-2-one (iii) Pentan-3-one (iv) Acetaldehyde (v) Benzaldehyde (vi) Propan-2-ol
Show answer and why every option is right or wrong
Answer: C. Four compounds qualify. Acetone (CH₃COCH₃) — methyl ketone ✓. Pentan-2-one (CH₃COCH₂CH₂CH₃) — methyl ketone ✓. Acetaldehyde (CH₃CHO) — contains CH₃CO– unit ✓. Propan-2-ol (CH₃CH(OH)CH₃) — oxidised in situ to acetone ✓. Pentan-3-one lacks a CH₃CO– group. Benzaldehyde (C₆H₅CHO) has no methyl on the carbonyl. Total = 4.
Why A is wrong: A is wrong — this count of 3 likely excludes propan-2-ol. Propan-2-ol is a secondary alcohol of the form CH₃–CH(OH)–R, which is oxidised in situ to acetone (a methyl ketone). It qualifies (trap: forgetting that certain alcohols are oxidised to methyl ketones under test conditions).
Why B is wrong: B is wrong — a count of 5 likely includes either pentan-3-one or benzaldehyde. Pentan-3-one has ethyl groups on both sides of C=O (no CH₃CO–), and benzaldehyde has a phenyl group, not a methyl, on the carbonyl carbon. Neither qualifies.
Why D is wrong: D is wrong — all six cannot be positive. Both pentan-3-one and benzaldehyde fail the structural requirement of having a CH₃CO– unit or a precursor oxidisable to one.
MCQ 5Direct ApplicationPractice
Propan-2-ol gives a positive iodoform test, but propan-1-ol does not. The correct reason is:
Show answer and why every option is right or wrong
Answer: C. Under I₂/NaOH conditions, propan-2-ol (a secondary alcohol) is oxidised to acetone (CH₃COCH₃), which has the CH₃CO– unit required for the haloform reaction. Propan-1-ol is oxidised to propanal (CH₃CH₂CHO), where the carbonyl bears –CH₂CH₃, not –CH₃. NCERT Class 12 Chemistry Chapter 8, Part 2, page 244.
Why A is wrong: A is wrong because acidity is irrelevant to the iodoform test. The test depends on the presence of a CH₃CO– unit (or its in-situ generation), not on alcohol acidity.
Why B is wrong: B is wrong because propan-1-ol CAN be oxidised — it yields propanal. The issue is not oxidisability but the structure of the oxidation product: propanal lacks the CH₃CO– group.
Why D is wrong: D is wrong because the statement is factually correct. Propanal does not have a methyl group attached to the carbonyl carbon, so no haloform reaction occurs from propan-1-ol (trap: assuming all oxidisable alcohols qualify).
MCQ 6Direct ApplicationPractice
Compound X gives a positive iodoform test and on treatment with I₂/NaOH produces sodium benzoate as one of the products. Compound X is:
Show answer and why every option is right or wrong
Answer: D. Acetophenone (C₆H₅COCH₃) is a methyl ketone. The haloform reaction replaces the three methyl hydrogens with iodine, then base cleaves the CI₃ group, yielding iodoform (CHI₃) and sodium benzoate (C₆H₅COONa). This matches both clues: positive iodoform test and sodium benzoate as a product.
Why A is wrong: A is wrong because benzaldehyde (C₆H₅CHO) lacks a CH₃CO– group — the carbonyl bears H, not CH₃. It would undergo Cannizzaro reaction with NaOH, not the haloform reaction (trap: confusing the aldehyde H with a methyl group).
Why B is wrong: B is wrong because benzophenone (C₆H₅COC₆H₅) has two phenyl groups on the carbonyl — no CH₃CO– unit. It does not undergo the haloform reaction.
Why C is wrong: C is wrong because phenylacetic acid is a carboxylic acid, not a ketone or aldehyde. It has no carbonyl group susceptible to the haloform reaction conditions.
MCQ 7Concept TrapPractice
A student claims: "All ketones give a positive iodoform test." This claim is:
Show answer and why every option is right or wrong
Answer: A. The haloform reaction requires specifically a CH₃CO– group. Ketones such as diethyl ketone or benzophenone, which lack this motif, do not give the test. The claim overgeneralises from methyl ketones to all ketones. NCERT Class 12 Chemistry Chapter 8, Part 2, page 244.
Why B is wrong: B is wrong because having α-hydrogens is necessary but not sufficient. The reaction specifically requires three replaceable hydrogens on a methyl group directly bonded to the carbonyl. Ketones with –CH₂– groups adjacent to C=O do not form a trihalomethyl fragment that can be cleaved (trap: overgeneralising α-hydrogen halogenation to the haloform cleavage step).
Why C is wrong: C is wrong because the test is not restricted to aromatic ketones. Any methyl ketone — aliphatic (acetone) or aromatic (acetophenone) — gives a positive result. This answer is doubly wrong: it both excludes qualifying aliphatic methyl ketones and implicitly includes non-methyl aromatic ketones like benzophenone.
Why D is wrong: D is wrong because reagent strength does not overcome the structural requirement. Even with excess I₂/NaOH, a ketone without a CH₃CO– group cannot generate the trihalomethyl intermediate needed for cleavage.
MCQ 8CalculationPractice
An unknown liquid Y (molecular formula C₄H₈O) gives a positive iodoform test and does not reduce Tollens' reagent. The structure of Y is:
Show answer and why every option is right or wrong
Answer: B. Step 1: Positive iodoform test → CH₃CO– group must be present (or generatable). Step 2: Does NOT reduce Tollens' → not an aldehyde. Among the options, butan-2-one (CH₃COCH₂CH₃) is a methyl ketone (positive iodoform) and a ketone (negative Tollens'). Both clues match.
Why A is wrong: A is wrong because butanal is an aldehyde — it WOULD reduce Tollens' reagent (positive silver mirror). This contradicts the second clue. Additionally, butanal (CH₃CH₂CH₂CHO) has no CH₃CO– group, so it would also fail the iodoform test.
Why C is wrong: C is wrong because 2-methylpropanal is an aldehyde (positive Tollens'), contradicting the given observation. Its carbonyl bears an isopropyl group and H, not a methyl group, so the iodoform test would also be negative.
Why D is wrong: D is wrong because tetrahydrofuran is a cyclic ether with no carbonyl group. It gives neither a positive iodoform test nor a positive Tollens' test. While the molecular formula matches, the functional group is entirely different (trap: matching only the molecular formula without considering functional group tests).
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How do you solve a Haloform question? A worked example
- 1
Given
Five organic compounds to evaluate for the iodoform test (I₂/NaOH).
- 2
Required
Identify which compounds give a positive iodoform test (yellow precipitate of CHI₃).
- 3
Concept
The haloform reaction requires a CH₃–CO– group, either already present or generated by in-situ oxidation of the substrate. Methyl ketones, ethanal, ethanol, and secondary alcohols of the form CH₃–CH(OH)–R qualify.
- 4
Formula / Rule
Structural test: does the compound have or generate a CH₃CO– unit under I₂/NaOH conditions?
- 5
Substitution (compound-by-compound check)
| Compound | Structure | Oxidation product (if alcohol) | CH₃CO– present? | Result |
|---|---|---|---|---|
| Ethanol | CH₃CH₂OH | → CH₃CHO (ethanal) | Yes | ✓ |
| Propan-1-ol | CH₃CH₂CH₂OH | → CH₃CH₂CHO (propanal) | No (–CH₂CH₃ on C=O) | ✗ |
| Propan-2-ol | CH₃CH(OH)CH₃ | → CH₃COCH₃ (acetone) | Yes | ✓ |
| Pentan-3-one | CH₃CH₂COCH₂CH₃ | (already a ketone) | No (–CH₂CH₃ on both sides) | ✗ |
| Ethanal | CH₃CHO | (already an aldehyde) | Yes | ✓ | - 6
Calculation
Count of positive results = 3 (ethanol, propan-2-ol, ethanal).
- 7
Final answer
Ethanol, propan-2-ol, and ethanal give a positive iodoform test. Propan-1-ol and pentan-3-one do not.
- 8
Common trap
The most common error is including pentan-3-one because "it's a ketone" — but having a carbonyl is not enough. The CH₃ must be directly bonded to the C=O carbon. Pentan-3-one has –CH₂CH₃ groups on both sides. A second common error is excluding ethanol because "it's an alcohol, not a carbonyl compound" — but ethanol is oxidised in situ to ethanal.
- 9
Similar NEET-style question
"Among the following — acetophenone, benzophenone, butan-2-one, pentan-3-one, and methanol — how many give a positive iodoform test?" (Answer: 2 — acetophenone and butan-2-one are methyl ketones; the others lack the CH₃CO– unit.)
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What to remember before solving Haloform questions
Where do students lose marks on Haloform?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Root cause: concept gap
Correction
Iodoform test ONLY positive for: methyl ketones (CH₃-CO-R), ethanol, secondary alcohols of form CH₃-CH(OH)-R, ethanal.
More in Alcohols, Phenols, Ethers and Carbonyl Compounds: 2 exam traps and mistakes · 2 formulas · 2 question patterns from its other lessons.
Haloform questions from past NEET papers
1 question from NEET 2026. Answers verified against NTA official keys.
NEET 2026
1If P is the sodium salt of a carboxylic acid, Q is a primary alcohol.
2P and Q are aromatic compounds.
3If P gives a carboxylic acid on acidification, Q gives a poisonous gas on exposure to air and light.
4Both P and Q are carbonyl compounds.
NTA Answer: Option 3(final)
All 23 past-paper questions from Alcohols, Phenols, Ethers and Carbonyl Compounds →
How does NEET ask about Haloform?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Identify which compounds give iodoform test. Methyl ketones, ethanal, ethanol, secondary alcohols of form CH3-CH(OH)-R.
RecallEasy
Common distractors
includes non methyl ketones
Treats all ketones as positive
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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