The trap that costs marks here: you know the acetanilide reaction, but NEET tests whether you can pick the correct indicator for a related titration, identify which analytical group a cation belongs to when it appears in two groups, or recognise the difference between a transient colour flash and a true end-point. These are practical-chemistry recall and application questions — not synthesis problems.
Acetanilide preparation involves acetylation of aniline using acetic anhydride (or glacial acetic acid). The reaction:
C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
The product is recrystallised from hot water. NCERT's Class 12 Chemistry Laboratory Manual sets it out as Experiment 10.1 (NCERT Chemistry Lab Manual Class 12, Unit 10, pages 104–105).
Why NEET cares: The exam bundles acetanilide preparation with broader practical-chemistry questions — titrimetric analysis (indicator selection, end-point recognition) and qualitative analysis (cation group assignment). You'll see 1–2 questions per paper drawn from this practical bundle (observed in 2024 and 2025 papers).
Key concepts tested:
Indicator–pH matching: Phenolphthalein works when equivalence-point pH > 7 (weak acid + strong base). Methyl orange works when equivalence-point pH < 7 (strong acid + weak base). Mixing these up is a common distractor.
End-point recognition: The first persistent colour change (lasting ≥30 seconds) marks the end-point. A transient flash that fades on swirling is NOT the end-point.
Cation group overlap: Pb²⁺ precipitates in both Group I (as PbCl₂, white, soluble in hot water) and Group II (as PbS, black). Questions exploit this dual membership.
Watch-out: When a stem describes "colour faded after swirling," the answer is always "end-point not yet reached" — do not pick the option that says titration is complete.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
In the preparation of acetanilide, aniline reacts with acetic anhydride. What is the byproduct of this reaction?
Answer: A. The acetylation of aniline with acetic anhydride produces acetanilide and acetic acid (CH₃COOH) as the byproduct. Acylation of amines by anhydrides is in NCERT Class 12 Chemistry Chapter 9, page 270.
Why B is wrong: Water is the byproduct when glacial acetic acid (not acetic anhydride) is used as the acetylating agent. With anhydride, the leaving group is acetate, yielding acetic acid. (trap: confusing the two acetylating agents)
Why C is wrong: Ethanol is not produced in any acetylation of amines — it would require an ester hydrolysis mechanism, which is irrelevant here. (trap: random organic byproduct confusion)
Why D is wrong: Acetone is a ketone and has no mechanistic connection to the nucleophilic acyl substitution occurring here. (trap: similar 'acet-' prefix misleads)
Acetanilide is purified after preparation by:
Answer: B. Acetanilide has moderate solubility in hot water and low solubility in cold water, making recrystallisation from hot water the standard purification method (NCERT Chemistry Lab Manual Class 12, Unit 10, page 105).
Why A is wrong: Steam distillation is used for volatile water-immiscible compounds like aniline itself, not for the solid product acetanilide. (trap: confusing purification of reactant vs product)
Why C is wrong: Sublimation is reserved for solids with high vapour pressure (e.g., naphthalene, camphor). Acetanilide does not sublime readily under normal lab conditions. (trap: overgeneralising sublimation to all organic solids)
Why D is wrong: Solvent extraction with ether is used for liquid-liquid separations, not for purifying a solid that dissolves well in hot water. (trap: mixing up separation techniques)
In qualitative analysis, which cation is known to appear in both Group I and Group II?
Answer: D. Pb²⁺ precipitates as PbCl₂ (white, soluble in hot water) in Group I with dilute HCl, and also as PbS (black) in Group II with H₂S in acidic medium. This dual membership is a standard NEET practical-chemistry fact.
Why A is wrong: Cu²⁺ belongs exclusively to Group II (precipitates as CuS, black). It does not form an insoluble chloride with dilute HCl, so it never appears in Group I. (trap: both Cu²⁺ and Pb²⁺ are Group II, but only Pb²⁺ crosses into Group I)
Why B is wrong: Zn²⁺ belongs to Group IV (precipitates as ZnS with H₂S in basic medium). No overlap with earlier groups. (trap: confusing group reagent conditions)
Why C is wrong: Fe³⁺ belongs to Group III (precipitates as Fe(OH)₃ with NH₄OH). It has no overlap with Groups I or II. (trap: confusing analytical groups)
For the titration of oxalic acid (weak acid) against NaOH (strong base), the suitable indicator is:
Answer: B. Weak acid + strong base titration has an equivalence-point pH > 7 (in the range 8–10). Phenolphthalein changes colour between pH 8.2–10, which matches this equivalence-point pH. This is the standard indicator-selection principle for acid-base titrations.
Why A is wrong: Methyl orange (pH 3.1–4.4) is suitable for strong acid + weak base titrations where the equivalence-point pH < 7. Using it here would give an end-point before the actual equivalence point. (trap: indicator pH-range mismatch — the most common NEET distractor in this topic)
Why C is wrong: Methyl red (pH 4.4–6.2) also has its range below neutral pH. For weak acid + strong base, the equivalence point is well above pH 6.2, making methyl red unsuitable. (trap: 'red' sounds similar to 'orange' — students pick either without checking the pH range)
Why D is wrong: Litmus has a very gradual colour change around pH 5–8 with no sharp transition, making it unreliable for precise end-point detection in any titration. (trap: litmus is a general indicator, not a titration indicator)
In a titration, a student observes a pink colour that fades back to colourless after swirling. The correct interpretation is:
Answer: A. The end-point is defined as the first persistent colour change (lasting ≥30 seconds). A colour that fades on swirling is a transient local excess that disappears upon mixing — the bulk solution has not yet reached equivalence.
Why B is wrong: Claiming the end-point has been reached contradicts the definition: end-point requires persistence. A transient flash means unreacted acid still remains in solution. (trap: students confuse 'any colour change' with 'end-point' — the overshoot trap)
Why C is wrong: Indicator decomposition would cause permanent loss of colour throughout the solution, not a brief flash that fades. Phenolphthalein is stable under normal titration conditions. (trap: inventing a mechanism that doesn't apply here)
Why D is wrong: Excess base would produce a permanent deep pink (with phenolphthalein), not a fading flash. If it fades, the base was locally in excess but globally insufficient. (trap: confusing local vs bulk concentration)
25 mL of 0.1 N oxalic acid is titrated against NaOH. If the end-point is reached at 12.5 mL of NaOH, what is the normality of NaOH?
Answer: C. Using N₁V₁ = N₂V₂: 0.1 × 25 = N₂ × 12.5, so N₂ = 2.5 / 12.5 = 0.2 N. The normality equation applies directly at the equivalence point.
Why A is wrong: 0.05 N results from dividing 0.1 by 2 without applying the titration formula — a careless arithmetic shortcut. (trap: halving the acid normality instead of computing V-ratio)
Why B is wrong: 0.1 N assumes equal normalities because you forgot that V_acid ≠ V_base. The formula requires the volume ratio to determine the unknown normality. (trap: assuming equal normality when volumes differ)
Why D is wrong: 0.4 N results from doubling the correct 0.2 N because oxalic acid is diprotic, but normality already includes the n-factor. (trap: counting the n-factor twice)
A student performing qualitative analysis adds dilute HCl to a solution and obtains a white precipitate. The precipitate dissolves when the solution is heated. This confirms the presence of:
Answer: D. PbCl₂ is white and notably soluble in hot water — this hot-water solubility is the confirmatory test distinguishing Pb²⁺ from other Group I cations. AgCl and Hg₂Cl₂ are also white but insoluble in hot water.
Why A is wrong: AgCl is white and insoluble in hot water (it requires NH₃ to dissolve). If the precipitate dissolved on heating, it cannot be Ag⁺. (trap: both give white precipitate with HCl, but solubility behaviour differs)
Why B is wrong: Cu²⁺ does not precipitate with dilute HCl at all (CuCl₂ is soluble). It belongs to Group II, not Group I. (trap: Cu²⁺ gives coloured compounds, but students may confuse group reagent conditions)
Why C is wrong: Hg₂Cl₂ is white but insoluble in hot water (and turns black with NH₃). Hot-water solubility rules it out. (trap: all three Group I chlorides are white — must differentiate by confirmatory tests)
20 mL of 0.05 M H₂SO₄ is titrated against NaOH using phenolphthalein indicator. Using the molarity-stoichiometry relationship (balanced equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O), what volume of 0.1 M NaOH is required to reach the end-point?
Answer: C. Using M_a V_a / n_a = M_b V_b / n_b: (0.05 × 20) / 1 = (0.1 × V_b) / 2, so 1.0 = 0.05 V_b, giving V_b = 20 mL. Here n_a = 1 (coefficient of H₂SO₄) and n_b = 2 (coefficient of NaOH) from the balanced equation. Phenolphthalein is suitable because its colour change (about pH 8–10) falls inside the steep pH jump at the equivalence point of a strong acid–strong base titration.
Why A is wrong: 10 mL results from ignoring the stoichiometric coefficient of NaOH (using n_b = 1 instead of 2): 0.05 × 20 / 1 = 0.1 × V / 1 → V = 10 mL. (trap: forgetting the 1:2 mole ratio between H₂SO₄ and NaOH)
Why B is wrong: 40 mL counts the factor of 2 twice: converting H₂SO₄ to normality (0.10 N, which already includes its two H⁺) and then also applying the 1:2 mole ratio: 0.10 × 20 × 2 = 0.1 × V → V = 40 mL. (trap: applying the stoichiometric factor twice)
Why D is wrong: 5 mL results from swapping the coefficients (n_a = 2, n_b = 1): 0.05 × 20 / 2 = 0.1 × V / 1 → V = 5 mL. (trap: confusing which coefficient goes with which species)
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Pattern: Practical chemistry — titrimetric calculation with indicator selection (from P.CHE.U20.PRACTICAL_BUNDLE, observed 2024–2025).
Given
• 25.0 mL of Na₂CO₃ solution (weak base), unknown normality• Titrant: 0.1 N HCl• End-point volume of HCl: 20.0 mL• Indicator used: methyl orange
Required
Normality of Na₂CO₃ solution and verification that the indicator choice is correct.
Concept
At the equivalence point, equivalents of acid = equivalents of base (normality equation). For strong acid + weak base titration, the equivalence-point pH < 7, so methyl orange (pH 3.1–4.4) is the correct indicator.
Formula
N₁V₁ = N₂V₂
Substitution
N(Na₂CO₃) × 25.0 = 0.1 × 20.0
Calculation
N(Na₂CO₃) = (0.1 × 20.0) / 25.0 = 2.0 / 25.0 = 0.08 N
Note: The stoichiometric coefficients (1, 2 in the balanced equation Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂) are already embedded in the normality (equivalent weight = M/2 for Na₂CO₃). These are exact integers and do not affect significant-figure count.
Final answer
Normality of Na₂CO₃ = 0.08 N (2 significant figures, matching the precision of the given data).
Indicator verification: Strong acid (HCl) + weak base (Na₂CO₃) → equivalence-point pH < 7 → methyl orange is correct. Phenolphthalein would give a premature end-point (it would change colour before true equivalence).
Common trap
Students pick phenolphthalein by default ("phenolphthalein is always used in titrations") without checking the equivalence-point pH. For this combination, phenolphthalein would indicate the half-neutralisation point (conversion to NaHCO₃), not the full equivalence point.
Similar NEET-style question
"30 mL of 0.05 N NaOH is titrated against HCl. If the end-point is reached at 15 mL of HCl, calculate the normality of HCl and state which indicator is appropriate." (Answer: N(HCl) = 0.1 N; strong acid + strong base → any indicator works, but phenolphthalein is conventional.)
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Aniline + acetic anhydride (or CH3COCl) in pyridine → N-phenylacetamide (acetanilide) + CH3COOH. Crystals are recrystallised from hot water (acetanilide is sparingly soluble in cold water). Pyridine traps HCl; for acetic anhydride, no catalyst needed. White flaky crystals, m.p. 114 °C.
-- NCERT Chemistry Lab Manual Class 12, Unit 10, p. 104More in Practical and Analytical Chemistry: 6 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
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