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?
The iodoform reaction of acetone with I₂/NaOH produces CHI₃ and which organic by-product?
During the preparation of aniline yellow, the diazotisation of aniline must be carried out at:
Isopropyl alcohol (CH₃CH(OH)CH₃) gives a positive iodoform test because:
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?
Which of the following alcohols will give a positive iodoform test?
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?
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:
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').] ---
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, p. 220Use when normality is awkward (e.g., diprotic acids). Stoichiometric coefficients from balanced equation.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | molarity | mol/L |
| V | volume | L |
| n | coefficient | - |
Equivalents of acid = equivalents of base at end-point. Or for redox: equivalents of oxidant = equivalents of reductant.
| Symbol | Quantity | SI Unit |
|---|---|---|
| N | normality | eq/L |
| V | volume | mL or L |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Cations like Pb²⁺ precipitate in BOTH Group I (with HCl) and Group II (with H2S) — assigning to only one group misses the redundancy.
Cation that appears in two analytical groups, e.g. Pb²⁺ (Group I + Group II) or Hg²⁺ vs Hg2²⁺.
Apply confirmatory tests for each candidate group; do not assume mutual exclusivity.
Category: Overthinking
Continuing to add titrant past the first persistent colour change because the colour seemed to fade after a swirl.
Question describes 'colour faded after swirling' or 'persistent colour' — distinguishes transient vs end-point.
End-point = first PERSISTENT colour change (lasts ≥30 s). Transient fades back to original on swirling.
Category: Similar Terms
Phenolphthalein (pH 8.2–10) and methyl orange (pH 3.1–4.4) only mark equivalence when the eq-pt pH falls within their range; using the wrong indicator gives an end-point that disagrees with the actual equivalence point.
Titration prompt mentions a specific weak/strong combination but asks which indicator is suitable.
Match the indicator's pH-change range to the equivalence-point pH: phenolphthalein for eq-pt > 7, methyl orange for eq-pt < 7.
Root cause: concept gap
Pb²⁺ appears in both Group I (PbCl2 white ppt, soluble in hot water) and Group II (PbS black ppt). Confirm with hot-water solubility test.
Root cause: concept gap
Strong-acid + strong-base: any indicator (eq pt = 7). Weak-acid + strong-base: phenolphthalein (eq pt > 7). Strong-acid + weak-base: methyl orange (eq pt < 7).
Root cause: rushed under time pressure
Slow drop-wise addition near the end-point; first persistent colour change is the end-point. Re-do if overshot.
3 questions from NEET 2024, 2025. Answers verified against NTA official keys.
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
wrong zero error direction
Sign convention is easily flipped.
Test yourself on this topic with real past-paper questions:
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