Iodoform and aniline yellow are two preparations that appear in the NEET practical-chemistry syllabus — one tests carbonyl/methyl-ketone chemistry, the other tests diazo-coupling. Both are recall-heavy but carry traps in reagent conditions and observation-based identification.
Iodoform (CHI₃) preparation. Acetone (or ethanol) is warmed with I₂ and NaOH. The haloform reaction cleaves the C–C bond adjacent to the carbonyl, producing a pale-yellow crystalline precipitate with a characteristic antiseptic odour. The net reaction with acetone:
CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3NaI + 3H₂O
Key observations: (i) yellow precipitate, (ii) melting point ~119 °C, (iii) characteristic smell. NEET questions commonly ask which substrate gives a positive iodoform test — the answer is any methyl ketone (RCOCH₃) or any alcohol oxidisable to a methyl ketone (CH₃CH(OH)R where R = H or alkyl, i.e., secondary alcohols with at least one methyl on the carbinol carbon, plus ethanol specifically).
Aniline yellow (p-aminoazobenzene) preparation. Aniline is diazotised at 0–5 °C with NaNO₂/HCl to form benzenediazonium chloride, which then couples with aniline in mildly acidic medium to give the yellow azo dye. Temperature control is critical — above 5 °C the diazonium salt decomposes, and the coupling fails.
Watch-out for NEET: Questions may present a substrate (e.g., 2-butanone, isopropyl alcohol, acetaldehyde) and ask whether it gives iodoform. The deciding test: does the molecule contain a –COCH₃ group, or can it be oxidised to one? Acetaldehyde (CH₃CHO) gives iodoform; benzaldehyde (C₆H₅CHO) does not — no adjacent methyl.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following compounds will NOT give a positive iodoform test with I₂/NaOH?
Answer: D. Benzaldehyde (C₆H₅CHO) lacks the CH₃CO– group and cannot be oxidised to form one. The iodoform reaction requires a methyl group adjacent to the carbonyl (NCERT Class 12 Chemistry, Aldehydes Ketones and Carboxylic Acids chapter).
Why A is wrong: Acetaldehyde (CH₃CHO) contains the CH₃CO– group and undergoes the haloform reaction readily.
Why B is wrong: Acetophenone (C₆H₅COCH₃) contains the CH₃CO– group directly, so it gives a positive iodoform test and is not the answer.
Why C is wrong: Ethanol is first oxidised to acetaldehyde by the I₂/NaOH system, which then undergoes the haloform reaction — so ethanol does give a positive test.
The iodoform reaction of acetone with I₂/NaOH produces CHI₃ and which organic by-product?
Answer: B. The haloform reaction cleaves the bond between the carbonyl carbon and the CI₃ group, giving the sodium salt of the acid with one fewer carbon. Acetone (3 C) → CHI₃ + CH₃COONa (2 C) (NCERT Class 12 Chemistry, Chapter 8 — Aldehydes, Ketones and Carboxylic Acids).
Why A is wrong: Sodium formate would require a 2-carbon substrate (acetaldehyde gives formate). Acetone has 3 carbons, so the by-product retains 2 carbons as acetate.
Why C is wrong: Sodium oxalate has two carboxylate groups and is not produced in a simple haloform cleavage of a methyl ketone.
Why D is wrong: Sodium benzoate would require a phenyl group adjacent to the carbonyl — acetophenone gives benzoate, not acetone.
During the preparation of aniline yellow, the diazotisation of aniline must be carried out at:
Answer: B. Benzenediazonium chloride is unstable and decomposes above 5 °C. Diazotisation is therefore performed at 0–5 °C to preserve the diazonium salt for the subsequent coupling step (NCERT Class 12 Chemistry, Amines chapter — diazonium salts section).
Why A is wrong: At 25–30 °C the diazonium salt decomposes to phenol + N₂, and coupling cannot proceed.
Why C is wrong: At 50–60 °C decomposition is rapid; diazonium salts are never prepared above 5 °C.
Why D is wrong: Reflux conditions would destroy the diazonium salt almost instantly — this temperature is used for reactions where diazo decomposition is the goal (e.g., Sandmeyer), not for coupling.
Isopropyl alcohol (CH₃CH(OH)CH₃) gives a positive iodoform test because:
Answer: A. Secondary alcohols of the form CH₃CH(OH)R are first oxidised by I₂/NaOH to the corresponding methyl ketone (here acetone), which then undergoes the haloform reaction (NCERT Class 12 Chemistry, Aldehydes Ketones and Carboxylic Acids chapter).
Why B is wrong: Isopropyl alcohol is an alcohol, not a ketone — it does not contain a carbonyl. The –COCH₃ group forms only after oxidation.
Why C is wrong: The hydroxyl group itself does not react with I₂ to directly form CHI₃ — the mechanism proceeds via oxidation to the ketone first.
Why D is wrong: Dehydration to propene is an acid-catalysed reaction irrelevant to alkaline I₂/NaOH conditions; propene does not give iodoform.
In the coupling step of aniline yellow preparation, benzenediazonium chloride reacts with aniline. The azo (–N=N–) linkage forms at which position of the coupling aniline ring?
Answer: C. The –NH₂ group is a strong activating group directing electrophilic attack to ortho and para positions. The para position is preferred due to less steric hindrance, giving p-aminoazobenzene (aniline yellow) as the major product (NCERT Class 12 Chemistry, Amines — coupling reactions of diazonium salts).
Why A is wrong: Ortho coupling is possible but is the minor product due to steric interference from the –NH₂ group; the major product is para-substituted.
Why B is wrong: Meta attack contradicts the directing effect of –NH₂, which is ortho/para directing. Electrophilic substitution at meta requires a deactivating group.
Why D is wrong: Ipso substitution would displace the –NH₂ group itself, which does not occur under these mild coupling conditions.
Which of the following alcohols will give a positive iodoform test?
Answer: A. 2-Butanol has the structure CH₃CH(OH)CH₂CH₃. Oxidation gives methyl ethyl ketone (CH₃COCH₂CH₃), which contains the –COCH₃ group and undergoes the iodoform reaction (NCERT Class 12 Chemistry, Aldehydes Ketones and Carboxylic Acids chapter).
Why B is wrong: 2-Methyl-2-propanol is a tertiary alcohol that resists oxidation under mild I₂/NaOH conditions. Even if forced, the product would be acetone + CO₂ decomposition — not a standard iodoform-positive substrate.
Why C is wrong: 1-Propanol oxidises to propanal (CH₃CH₂CHO), which lacks the CH₃CO– structural unit — no methyl attached to the carbonyl carbon.
Why D is wrong: Benzyl alcohol oxidises to benzaldehyde (C₆H₅CHO), which has no methyl on the carbonyl — does not give iodoform.
A student prepares iodoform from ethanol. The overall reaction consumes I₂ and NaOH. How many moles of NaOH are consumed per mole of ethanol in the complete iodoform reaction?
Answer: D. The overall reaction is: CH₃CH₂OH + 4I₂ + 6NaOH → CHI₃ + HCOONa + 5NaI + 5H₂O. Oxidation of ethanol to acetaldehyde consumes 1 I₂ + 2 NaOH; the subsequent haloform step on acetaldehyde consumes 3 I₂ + 4 NaOH (total = 4I₂, 6NaOH per mole of ethanol) (NCERT Class 12 Chemistry, Aldehydes Ketones and Carboxylic Acids — haloform reaction mechanism).
Why A is wrong: 4 NaOH accounts only for the haloform step on the aldehyde (3I₂ + 4NaOH), neglecting the initial oxidation step that converts ethanol to acetaldehyde (uses 1I₂ + 2NaOH).
Why B is wrong: 12 NaOH would correspond to twice the correct value and has no mechanistic basis.
Why C is wrong: 10 NaOH is an overcount — it conflates the iodoform stoichiometry with acetone's equation (which uses only 4 NaOH because no oxidation step is needed).
In the preparation of aniline yellow, if the temperature of the reaction mixture rises above 10 °C during diazotisation, the most likely product formed instead of benzenediazonium chloride is:
Answer: C. Above 5 °C, benzenediazonium chloride decomposes by loss of N₂, and the resulting phenyl cation reacts with water to form phenol (C₆H₅OH). This is the characteristic thermal decomposition pathway of aqueous diazonium salts (NCERT Class 12 Chemistry, Amines — reactions of diazonium salts).
Why A is wrong: Chlorobenzene formation requires Cu₂Cl₂ catalyst (Sandmeyer reaction), not mere thermal decomposition in aqueous HCl.
Why B is wrong: Nitrobenzene requires nitrating mixture (HNO₃ + H₂SO₄) and is unrelated to diazonium decomposition.
Why D is wrong: Biphenyl forms via Gomberg reaction with Cu (coupling two phenyl radicals) — not by simple thermal decomposition of diazonium salt in water.
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Pattern: Practical chemistry — organic preparation identification (aligned with NEET pattern: practical bundle)
Given
A student adds I₂ and NaOH to an unknown organic liquid and warms gently. A pale-yellow precipitate with a characteristic antiseptic smell forms. The precipitate melts at 119 °C. The unknown is one of: (A) diethyl ether, (B) acetone, (C) benzaldehyde, (D) acetic acid.
Required
Identify the unknown compound.
Concept
The iodoform test is positive for compounds containing the CH₃CO– group (methyl ketones) or compounds oxidisable to a methyl ketone (ethanol, secondary alcohols of pattern CH₃CHOH–R). The characteristic product is iodoform (CHI₃): pale-yellow crystals, m.p. 119 °C, antiseptic odour.
Formula
No quantitative formula needed — this is an identification problem. The diagnostic criterion is structural: presence of CH₃CO– or CH₃CHOH–.
Substitution / analysis
• (A) Diethyl ether (CH₃CH₂OCH₂CH₃): no carbonyl, no –OH. Cannot form CH₃CO–. Negative.• (B) Acetone (CH₃COCH₃): contains –COCH₃ directly. Positive.• (C) Benzaldehyde (C₆H₅CHO): carbonyl present but no methyl on the carbonyl carbon. Negative.• (D) Acetic acid (CH₃COOH): contains CH₃ adjacent to C=O, but the carboxylic acid does not undergo haloform cleavage under these conditions (the –OH of –COOH is not displaced by iodine in alkaline medium the way a ketone's α-H is). Negative.
Calculation
No numerical calculation. The logic is structural pattern-matching against the iodoform criterion.
Final answer
The unknown is (B) Acetone.
Common trap
Students sometimes pick acetic acid (D) reasoning that CH₃CO– is present in CH₃COOH. However, the carboxyl group's resonance stabilisation prevents α-iodination under mild I₂/NaOH conditions — carboxylic acids do not give the iodoform test.
Similar NEET-style question
"An organic compound gives a positive iodoform test and also reduces Tollens' reagent. The compound is: (A) acetone, (B) acetaldehyde, (C) formaldehyde, (D) acetophenone." [Answer: B — acetaldehyde has CH₃CHO (gives iodoform) and is an aldehyde (reduces Tollens').]
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Methyl ketones (or ethanol / 2-propanol) + I2 + NaOH → CHI3 (yellow ppt) + RCOO⁻Na⁺. The haloform test: pale-yellow iodoform crystals confirm presence of CH3-CO- or CH3-CH(OH)- group. Aniline yellow (p-aminoazobenzene): coupling of diazotised aniline with aniline (NaNO2 / HCl, 0–5 °C, then aniline).
-- NCERT Chemistry Lab Manual Class 12, Unit 8, p. 88More in Practical and Analytical Chemistry: 6 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
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