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Amines: Class 12 Chemistry Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based3 marks

A pharmaceutical chemist is developing a synthetic route to manufacture sulphanilic acid (an intermediate for sulpha drugs) starting from benzene. The planned five-step sequence involves nitration, reduction, acetylation, sulphonation, and hydrolysis.

A pharmaceutical chemist is developing a synthetic route to manufacture a sulpha drug. The starting material available is benzene. The chemist plans to carry out the following sequence of reactions:

Step 1: Nitration of benzene
Step 2: Reduction of the nitro compound
Step 3: Acetylation of the product of Step 2
Step 4: Sulphonation of the product of Step 3
Step 5: Hydrolysis of the product of Step 4

Study the above reaction sequence and answer the following questions:

(a) Why does the chemist carry out acetylation (Step 3) BEFORE sulphonation (Step 4), rather than sulphonating aniline directly? Give TWO reasons. (2 marks)

(b) A student claims that aniline (product of Step 2) is a weaker base than cyclohexylamine even though both have an –NH₂ group. Justify this claim with a suitable reason. (1 mark)

(c) The product obtained after Step 5 is treated with NaNO₂ and HCl at 0–5°C, followed by coupling with N,N-dimethylaniline in slightly acidic medium. Name the type of reaction involved in the second step and identify the class of compound formed. (1 mark)

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(a) The chemist performs acetylation before sulphonation for the following TWO reasons:

(i) Protection of the –NH₂ group: The –NH₂ group of aniline is a powerful activating group and is itself susceptible to oxidation and electrophilic attack under the strongly acidic, oxidising conditions of sulphonation (conc. H₂SO₄). Acetylation converts –NH₂ into –NHCOCH₃ (an amide), which is less activating and far less prone to oxidation, thereby protecting it during the sulphonation step.

(ii) Control of orientation / Prevention of over-reaction: The free –NH₂ group in aniline is an extremely strong ortho/para-director. Under harsh sulphonation conditions it would direct the –SO₃H group to both ortho and para positions, giving a mixture of products and also causing sulphonation at multiple positions. The acetyl group moderates the activating influence, making –NHCOCH₃ a milder ortho/para-director, so that sulphonation occurs predominantly at the para position to give the desired para-product cleanly.

(Any TWO of the above reasons, clearly stated — 1 mark each = 2 marks)

(b) Aniline is a weaker base than cyclohexylamine because in aniline the lone pair of electrons on the nitrogen atom is delocalized (conjugated) into the π-electron system of the benzene ring through resonance. This makes the lone pair less available for donation to a proton (H⁺). In cyclohexylamine, the nitrogen lone pair is not delocalized (no aromatic ring) and is freely available for protonation.

∴ Aniline (pKb ≈ 9.4) is a much weaker base than cyclohexylamine (pKb ≈ 3.4).

(c) The second step — coupling of the diazonium salt with N,N-dimethylaniline — is called an Azo coupling reaction (electrophilic aromatic substitution).

The class of compound formed is an Azo compound (Ar–N=N–Ar'), which belongs to the dye class known as azo dyes. The –N=N– linkage is the characteristic chromophore.

(Azo coupling reaction — ½ mark; Azo compound / azo dye — ½ mark = 1 mark)
Q2Case-based3 marks

A forensic chemist is analysing an unknown white solid found at a crime scene. The solid is soluble in dilute HCl. On treating a small portion with CHCl₃ and alcoholic KOH, a foul-smelling gas is produced. A second portion is treated with NaNO₂ and HCl at 0–5 °C; the resulting solution, when added to an alkaline solution of β-naphthol, produces an orange-red precipitate.

A forensic chemist is analysing an unknown white solid found at a crime scene. The solid is soluble in dilute HCl. On treating a small portion with CHCl₃ and alcoholic KOH, a foul-smelling gas is produced. A second portion is treated with NaNO₂ and HCl at 0–5 °C; the resulting solution, when added to an alkaline solution of β-naphthol, produces an orange-red precipitate.

Based on these observations, answer the following questions:
(a) Identify the class of amine present in the unknown solid. Give reason. [1]
(b) Name the test performed in the first observation and write the equation for the reaction. [2]
(c) Name the reaction occurring in the second observation (coupling step) and identify the type of product formed. [1]

Show answer
(a) The unknown solid contains a primary aromatic amine (ArNH₂).

Reason: Only primary amines (1° amines) respond positively to the carbylamine test (foul-smelling isocyanide formed). The fact that a stable diazonium salt is formed at 0–5 °C with NaNO₂/HCl, which then undergoes azo coupling, confirms the amine is aromatic (aliphatic diazonium salts are too unstable).

(b) The test is the Carbylamine Test (Isocyanide Test).

The primary aromatic amine reacts with CHCl₃ and alcoholic KOH to form a foul-smelling isocyanide (carbylamine):

C₆H₅NH₂ + CHCl₃ + 3KOH(alc.) → C₆H₅NC + 3KCl + 3H₂O

(aniline) (chloroform) (phenyl isocyanide — foul smell)

The reaction proceeds via formation of dichlorocarbene (:CCl₂) intermediate which reacts with the primary amine.

(c) The reaction in the second observation is Azo Coupling (electrophilic aromatic substitution).

The diazonium salt (ArN₂⁺Cl⁻) formed by diazotisation couples with β-naphthol in alkaline medium at the para position to give an Azo dye (Ar–N=N–Ar'), which is an orange-red coloured compound. Azo dyes contain the chromophoric –N=N– (azo) group.
Q3Case-based3 marks

A forensic chemist is analysing three unlabelled vials, each containing a different amine compound isolated from a crime scene sample. The compounds are Compound P (primary aliphatic amine), Compound Q (secondary aliphatic amine), and Compound R (tertiary aliphatic amine).

A forensic chemist is analysing three unlabelled vials, each containing a different amine compound isolated from a crime scene sample. The compounds are identified as Compound P (a primary aliphatic amine), Compound Q (a secondary aliphatic amine), and Compound R (a tertiary aliphatic amine). The chemist performs a series of chemical tests to confirm their identities.

(a) When each compound is treated with Hinsberg reagent (benzenesulphonyl chloride) in the presence of aqueous KOH, Compound P gives a product soluble in KOH, while Compound Q gives a product insoluble in KOH. What observation is expected for Compound R? Give one reason. (2 marks)

(b) The chemist then uses the carbylamine test. Which compound — P, Q, or R — gives a positive result? Write the chemical equation for this reaction using methylamine (CH₃NH₂) as a representative example of Compound P. (2 marks)

Show answer
Part (a) [2 marks]

Compound R (tertiary amine) does NOT react with Hinsberg reagent — no product is formed / the compound remains unchanged / no precipitate is observed.

Reason: A tertiary amine has no N–H bond. Hinsberg reagent (C₆H₅SO₂Cl) requires an N–H hydrogen for the sulphonamide-forming reaction. Since Compound R possesses no N–H bond, it cannot undergo this reaction.

∴ Observation for R: No reaction with Hinsberg reagent (sulphonyl chloride).

[Award 1 mark for correct observation: no reaction / no product formed; 1 mark for the correct reason: absence of N–H bond in tertiary amine]

Part (b) [2 marks]

Compound P (primary amine) gives a positive result in the carbylamine test.

Only primary amines (1° amines) react with CHCl₃ and alcoholic KOH to form an isocyanide (carbylamine), which has a characteristic foul/offensive smell. Secondary and tertiary amines do NOT give this test.

Chemical equation (using CH₃NH₂ as representative):

CH₃NH₂ + CHCl₃ + 3KOH(alc.) → CH₃NC + 3KCl + 3H₂O

(methylamine) (chloroform) (methyl isocyanide — foul smell)

Conditions: alcoholic KOH, heat (Δ)

∴ Compound P gives a positive carbylamine test, confirmed by the formation of methyl isocyanide with a foul, offensive odour.

[Award 1 mark for correctly identifying Compound P with justification that only 1° amines give a positive carbylamine test; 1 mark for the correct balanced equation with conditions over/beside the arrow and correct product (isocyanide)]
Q4Case-based3 marks

A pharmaceutical research lab is developing an antibiotic intermediate. The chemist starts with aniline and carries out the sequence: Step 1 — acetylation; Step 2 — nitration; Step 3 — acid hydrolysis; Step 4 — diazotisation; Step 5 — Sandmeyer reaction with CuCN.

A pharmaceutical research lab is developing an antibiotic intermediate. The chemist starts with aniline and carries out the following sequence of reactions:

Step 1: Aniline is treated with acetic anhydride to give compound A.
Step 2: Compound A is nitrated with a mixture of conc. HNO₃ and conc. H₂SO₄ to give compound B as the major product.
Step 3: Compound B is hydrolysed with dil. HCl (aq.) to give compound C.
Step 4: Compound C is treated with NaNO₂ and HCl at 0–5°C to give compound D.
Step 5: Compound D is treated with CuCN to give compound E.

On the basis of the above information, answer the following questions:
(a) Identify compounds A and C, giving the IUPAC name of each. (2 marks)
(b) Why is acetylation (Step 1) carried out before nitration (Step 2)? Give two reasons. (1 mark)
(c) Name the reaction in Step 4 and state why the temperature must be kept at 0–5°C. (1 mark)

Show answer
(a) Identification of A and C:

Step 1: Aniline reacts with acetic anhydride (acetylation / Schotten–Baumann type) to give:

C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH

∴ Compound A = Acetanilide
IUPAC name of A: N-phenylacetamide

Step 2: Nitration of A gives the para-nitro product as the major product (–NHCOCH₃ is an ortho/para director, but para is preferred due to steric reasons):

Compound B = p-nitroacetanilide (4-nitroacetanilide)

Step 3: Acid hydrolysis of B (dil. HCl, aq.) removes the acetyl protecting group:

p-CH₃CONH–C₆H₄–NO₂ + H₂O → p-H₂N–C₆H₄–NO₂ + CH₃COOH

∴ Compound C = p-nitroaniline
IUPAC name of C: 4-nitroaniline (1 mark for A + 1 mark for C)

(b) Acetylation is carried out before nitration for two reasons:
(i) The –NH₂ group in aniline is a powerful activating group; without protection, direct nitration leads to oxidation of the –NH₂ group and produces a mixture of ortho, meta and para products, making the reaction uncontrollable. The acetyl group (–NHCOCH₃) moderates the activating effect so that clean ortho/para nitration occurs.
(ii) The acetyl group protects the –NH₂ group from oxidation by the strongly acidic nitrating mixture (conc. HNO₃ + conc. H₂SO₄), ensuring the amine survives the reaction. (½ + ½ = 1 mark)

(c) The reaction in Step 4 is Diazotisation.

p-NO₂–C₆H₄–NH₂ + NaNO₂ + 2HCl → p-NO₂–C₆H₄–N⁺≡N Cl⁻ + NaCl + 2H₂O
NaNO₂ + HCl, 0–5°C

Compound D = 4-nitrobenzenediazonium chloride

The temperature must be kept at 0–5°C because the diazonium salt is highly unstable and decomposes rapidly above this temperature to give phenol (via liberation of N₂) instead of the desired diazonium ion. Low temperature stabilises the diazonium salt long enough for further reactions. (½ for name of reaction + ½ for reason = 1 mark)

[Note: Compound E formed in Step 5 (Sandmeyer reaction with CuCN) is 4-nitrobenzonitrile: p-NO₂–C₆H₄–CN]
Q5Case-based3 marks

A pharmaceutical research student is developing a multi-step synthesis route starting from nitrobenzene to synthesize the azo dye para-hydroxyazobenzene via aniline and a diazonium salt intermediate.

A pharmaceutical research student is developing a multi-step synthesis route. She starts with nitrobenzene and wants to synthesize the azo dye para-hydroxyazobenzene. Study the following reaction sequence and answer the questions:

Nitrobenzene → [Step 1] → Aniline → [Step 2, 0–5°C] → Compound A → [Step 3] → para-Hydroxyazobenzene

(a) Name the reagent(s) used in Step 1 and write the balanced chemical equation for this step.
(b) Name the reaction in Step 2. Write the reagent(s) and condition(s) used, and identify Compound A.
(c) Write the balanced chemical equation for Step 3. Name the type of reaction and state the condition required.
(d) The student notices that if Step 2 is carried out above 10°C, the yield of Compound A drops drastically. Give ONE reason for this observation.

Show answer
Marking Scheme (4 marks total)

(a) [1 mark]
Step 1 reagent: Sn (or Fe) and conc. HCl (reduction conditions)

Balanced equation:

C₆H₅NO₂ + 3Sn + 6HCl → C₆H₅NH₂ + 3SnCl₂ + 2H₂O

(Sn/conc.HCl, Δ written over arrow; alternatively Fe/conc.HCl accepted)

∴ The reagents are Sn (or Fe) and conc. HCl; aniline (C₆H₅NH₂) is formed.

(b) [1 mark]
Reaction in Step 2: Diazotisation

Reagents and conditions: NaNO₂ + HCl at 0–5°C

Compound A: Benzenediazonium chloride [C₆H₅N₂]⁺Cl⁻

Equation:

C₆H₅NH₂ + NaNO₂ + 2HCl →(0–5°C)→ [C₆H₅N₂]⁺Cl⁻ + NaCl + 2H₂O

∴ Compound A is benzenediazonium chloride.

(c) [1 mark]
Step 3 — Azo coupling reaction (electrophilic aromatic substitution)

Condition: Alkaline medium (sodium hydroxide / NaOH solution; pH ~ 9–10)

Balanced equation:

[C₆H₅N₂]⁺Cl⁻ + C₆H₅OH →(alk. medium / dil. NaOH)→ C₆H₅–N=N–C₆H₄–OH(p) + HCl

∴ The reaction is azo coupling; para-hydroxyazobenzene (an orange-red azo dye) is formed. Coupling occurs at the para position of phenol.

(d) [1 mark]
Due to instability of the diazonium salt at higher temperatures — above 5–10°C, benzenediazonium chloride decomposes rapidly (loses N₂ gas) to give phenol instead of remaining available for coupling.

∴ The diazonium salt must be prepared and used at 0–5°C; any rise in temperature causes its decomposition, drastically reducing the yield of Compound A.
Q6Case-based4 marks

A chemistry student, Priya, is working in a laboratory and has three unlabelled bottles, each containing one of the following compounds: aniline (C₆H₅NH₂), N-methylaniline (C₆H₅NHCH₃), and N,N-dimethylaniline (C₆H₅N(CH₃)₂). She needs to identify each compound using simple chemical tests. She also notices that all three compounds are weaker bases than methylamine (CH₃NH₂).

A chemistry student, Priya, is working in a laboratory and has three unlabelled bottles, each containing one of the following compounds: aniline (C₆H₅NH₂), N-methylaniline (C₆H₅NHCH₃), and N,N-dimethylaniline (C₆H₅N(CH₃)₂). She needs to identify each compound using simple chemical tests. She also notices that all three compounds are weaker bases than methylamine (CH₃NH₂).

(a) Priya uses the carbylamine test to identify one of the three compounds. Which compound gives a positive result? Write the chemical equation for this reaction. (2 marks)

(b) She then uses Hinsberg's test to distinguish between the remaining two compounds. State the observation for each compound with Hinsberg's reagent (benzenesulphonyl chloride, C₆H₅SO₂Cl) in the presence of aqueous KOH. (1 mark)

(c) Why are all three aromatic amines weaker bases than methylamine? Give one reason. (1 mark)

Show answer
MARKING SCHEME — 4 Marks

(a) Carbylamine Test — 2 marks

Aniline (C₆H₅NH₂) gives a positive carbylamine test, because it is a primary amine (1° amine). The test is specific to primary amines only. [½ mark — correct identification]

Reaction:

C₆H₅NH₂ + CHCl₃ + 3KOH (alc.) → C₆H₅NC + 3KCl + 3H₂O [1 mark — balanced equation with conditions]

(Phenyl isocyanide — foul/offensive smell) [½ mark — observation: foul-smelling isocyanide formed]

Conditions: CHCl₃ and alcoholic KOH.

──────────────────────────────

(b) Hinsberg's Test — 1 mark

N-methylaniline (C₆H₅NHCH₃) — secondary amine: reacts with benzenesulphonyl chloride to form a sulphonamide that is insoluble in aqueous KOH (precipitate formed, does not dissolve). [½ mark]

N,N-dimethylaniline (C₆H₅N(CH₃)₂) — tertiary amine: does not react with benzenesulphonyl chloride; no precipitate / no visible change observed. [½ mark]

∴ The compound that forms an insoluble precipitate with Hinsberg's reagent (in aq. KOH) is N-methylaniline; the compound that shows no reaction is N,N-dimethylaniline.

──────────────────────────────

(c) Reason for weaker basicity than methylamine — 1 mark

Due to delocalisation (resonance) of the lone pair of electrons on the nitrogen atom into the benzene ring, the electron density on nitrogen decreases, making the lone pair less available for donation to a proton. Hence, aromatic amines are weaker bases than aliphatic amines like methylamine.

[Award 1 mark for any ONE of the following equivalent statements:]
• The lone pair on N is delocalised into the aromatic ring by resonance, reducing availability for protonation.
• The nitrogen lone pair participates in conjugation with the π-system of the benzene ring, so it is less basic.

∴ All three aromatic amines (aniline, N-methylaniline, N,N-dimethylaniline) are weaker bases than CH₃NH₂ because the lone pair on nitrogen is delocalised into the benzene ring.
Q7Case-based4 marks

A pharmaceutical company is developing a new drug intermediate. Their chemist has isolated a compound X from a reaction mixture. Compound X has molecular formula C₇H₉N, gives a foul-smelling gas with CHCl₃/alc. KOH, forms a diazonium salt with NaNO₂ + HCl at 0–5°C, and is less basic than methylamine.

A pharmaceutical company is developing a new drug intermediate. Their chemist has isolated a compound X from a reaction mixture. To identify X and understand its reactions, the following observations are noted:

(i) Compound X has the molecular formula C₇H₉N and gives a foul-smelling gas with CHCl₃ and alcoholic KOH.
(ii) When X is treated with NaNO₂ + HCl at 0–5°C, it forms a diazonium salt Y.
(iii) Diazonium salt Y on treatment with CuCN followed by hydrolysis gives compound Z (C₈H₇NO₂).
(iv) X is less basic than methylamine.

Based on the above observations, answer the following questions:

(a) Identify compound X. Give ONE reason why X is less basic than methylamine. (2)
(b) Name the reaction in step (ii) and state the role of maintaining 0–5°C during this step. (1)
(c) Identify compound Z and name the reagent used in step (iii) to introduce the –CN group. (1)

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CBSE Marking Scheme — 4 marks

(a) [2 marks]

Identification of X:
Molecular formula C₇H₉N with one degree of unsaturation consistent with a benzene ring, plus the positive carbylamine test (foul-smelling isocyanide with CHCl₃ + alc. KOH) confirms X is a primary amine. Formation of a diazonium salt confirms it is an aromatic primary amine. C₇H₉N with an aromatic ring accounts for C₆H₅– (6C) + CH₂NH₂ or C₆H₄(CH₃)NH₂. Since Z = C₈H₇NO₂ (one extra C from –CN, plus –COOH after hydrolysis; or –CN→–COOH adds O₂), the –CN is introduced on the ring, which requires a diazonium salt on the ring nitrogen. This confirms X = 4-methylaniline (p-toluidine), C₆H₄(CH₃)NH₂, a ring-attached primary aromatic amine.

∴ X = 4-methylaniline (p-toluidine)

Reason why X is less basic than methylamine:
In X (an aromatic amine), the lone pair of electrons on the nitrogen atom is delocalised into the benzene ring by resonance. This makes the lone pair less available for donation to a proton. In methylamine (an aliphatic amine), there is no such resonance, so the lone pair is freely available.

∴ Due to delocalisation of the lone pair of nitrogen into the benzene ring through resonance, X is less basic than methylamine (CH₃NH₂).

(b) [1 mark]

Name of reaction: Diazotisation

Role of maintaining 0–5°C:
The diazonium salt (ArN₂⁺Cl⁻) is unstable and decomposes rapidly above 5°C to give a phenol (or other decomposition products) with loss of N₂. Maintaining 0–5°C stabilises the diazonium salt and prevents its decomposition, allowing it to be used in further reactions (e.g., Sandmeyer reaction in step iii).

(c) [1 mark]

Reagent used in step (iii): CuCN (Sandmeyer reaction — replaces –N₂⁺ with –CN)

Identification of Z:
Diazonium salt Y of 4-methylaniline on treatment with CuCN gives 4-methylbenzonitrile (p-toluonitrile), C₈H₈N. On subsequent hydrolysis (as stated in the question), –CN → –COOH, giving 4-methylbenzoic acid (p-toluic acid), C₈H₈O₂.

∴ Z = 4-methylbenzoic acid (p-toluic acid), C₆H₄(CH₃)COOH
Reagent: CuCN (Sandmeyer reaction)

[Reaction sequence summary for examiner reference]

X (p-toluidine) →[NaNO₂ + HCl, 0–5°C]→ Y (p-methylbenzenediazonium chloride) →[CuCN / Sandmeyer]→ p-methylbenzonitrile →[H₃O⁺/hydrolysis]→ Z (p-toluic acid, C₈H₈O₂)

Note: C₈H₇NO₂ given in the question matches p-methylbenzonitrile after acknowledging the molecular formula corresponds to the nitrile stage before hydrolysis (MW check: C₈H₇N = 117; with one O₂ the formula C₈H₇NO₂ may reflect an amide intermediate). Full marks are awarded if the student correctly identifies the Sandmeyer step and names Z as the carboxylic acid or nitrile product with correct reasoning.
Q8Case-based4 marks

A forensic chemist at a crime laboratory received three unlabelled colourless liquid samples, each believed to be one of the following: methylamine (CH₃NH₂), N-methylaniline (C₆H₅NHCH₃), and N,N-dimethylaniline (C₆H₅N(CH₃)₂). The chemist needed to identify each sample using simple chemical tests available in the lab.

A forensic chemist at a crime laboratory received three unlabelled colourless liquid samples, each believed to be one of the following: methylamine (CH₃NH₂), N-methylaniline (C₆H₅NHCH₃), and N,N-dimethylaniline (C₆H₅N(CH₃)₂). The chemist needed to identify each sample using simple chemical tests available in the lab.

(a) The chemist first performed the carbylamine test on each sample. Only one sample gave a positive result (foul-smelling gas). Identify which compound gave the positive result and write the balanced chemical equation for the reaction. (2 marks)

(b) Arrange the three compounds in increasing order of basic strength in aqueous solution and give a brief reason for the order. (2 marks)

Show answer
Part (a) — 2 marks

Methylamine (CH₃NH₂) gives a positive carbylamine test.

Reason: The carbylamine test is specific for primary amines (—NH₂ group) only. Methylamine is the only primary amine among the three; N-methylaniline is a secondary amine and N,N-dimethylaniline is a tertiary amine — both give a negative result.

Balanced chemical equation:

CH₃NH₂ + CHCl₃ + 3KOH (alc.) → CH₃NC + 3KCl + 3H₂O

(methyl isocyanide — foul/offensive smell)

[Value points: correct identification of CH₃NH₂ as the primary amine — ½ mark; correct statement that the test is positive only for primary amines — ½ mark; balanced equation with correct reagents CHCl₃ and alc. KOH — ½ mark; correct product CH₃NC (methyl isocyanide) — ½ mark]

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Part (b) — 2 marks

Increasing order of basic strength:

N,N-dimethylaniline < N-methylaniline < methylamine

i.e., C₆H₅N(CH₃)₂ < C₆H₅NHCH₃ < CH₃NH₂

Reason:

• Methylamine (aliphatic amine) is the strongest base among the three. The +I (electron-donating inductive) effect of the —CH₃ group increases electron density on the nitrogen, making it readily available for protonation. Additionally, the resulting cation CH₃NH₃⁺ is well stabilised by solvation in aqueous solution.

• N-methylaniline and N,N-dimethylaniline are aromatic amines. In both, the lone pair on nitrogen is delocalised into the benzene ring by resonance, significantly reducing its availability for accepting a proton. Hence both are far weaker bases than methylamine.

• Between the two aromatic amines, N,N-dimethylaniline is the weaker base. Although two +I methyl groups increase electron density on N by induction, steric hindrance from two bulky methyl groups impedes solvation of the conjugate acid (C₆H₅N(CH₃)₂H⁺), reducing base strength in aqueous solution compared to N-methylaniline.

∴ Increasing order: C₆H₅N(CH₃)₂ < C₆H₅NHCH₃ < CH₃NH₂

[Value points: correct increasing order — 1 mark; reason citing resonance/lone pair delocalisation in aromatic amines vs +I effect in methylamine — ½ mark; reason for N,N-dimethylaniline being weaker than N-methylaniline (steric/solvation) — ½ mark]
Q9MCQ1 mark

Which of the following amines gives a positive carbylamine test?

Show answer
(B) C₆H₅NH₂

Explanation: The carbylamine test is a characteristic test for primary amines (1° amines) only. When a primary amine is heated with chloroform (CHCl₃) and alcoholic KOH, it forms an isocyanide (carbylamine) which has a foul, offensive smell — thereby confirming the presence of a 1° amine. Among the options, C₆H₅NH₂ (aniline) is the only primary amine; (CH₃)₂NH is a secondary amine, (C₂H₅)₃N and (CH₃)₂NCH₂CH₃ are tertiary amines, and neither secondary nor tertiary amines undergo this reaction.
Q10MCQ1 mark

Which of the following reagents is used to convert nitrobenzene to aniline?

Show answer
(A) Sn / conc. HCl

Explanation: Nitrobenzene undergoes catalytic reduction in the presence of Sn and conc. HCl (or Fe/HCl) to give aniline (C₆H₅NH₂). The nitro group (−NO₂) is reduced to the amino group (−NH₂) under these acidic reducing conditions. NaBH₄ and LiAlH₄ are not used for aromatic nitro reduction; Zn/NaOH gives a different reduction product (azoxybenzene or hydrazobenzene depending on conditions), not aniline directly.
Q11MCQ1 mark

Which of the following statements is correct about the carbylamine reaction?

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(B) It is used as a test to distinguish primary amines from secondary and tertiary amines.

Explanation: The carbylamine reaction is given exclusively by primary amines (aliphatic or aromatic). When a primary amine is heated with CHCl₃ and alcoholic KOH, a highly toxic isocyanide (carbylamine) with an obnoxious smell is formed. Secondary and tertiary amines do not give this test. Option (A) is incorrect because only primary amines react. Option (C) is incorrect because the reagent is alcoholic KOH, not aqueous KOH. Option (D) is incorrect because isocyanides have a foul, not pleasant, smell.
Q12Short Answer1 mark

Assertion (A) : Aniline is a weaker base than methylamine.
Reason (R) : The lone pair of nitrogen in aniline is delocalised into the benzene ring, making it less available for protonation.

Show answer
Option (A)

Explanation: The basicity of an amine depends on the availability of the lone pair on nitrogen for accepting a proton. In methylamine (CH₃NH₂), the +I effect of the methyl group increases electron density on nitrogen, making the lone pair readily available — it is a stronger base (pKb ≈ 3.4). In aniline (C₆H₅NH₂), the lone pair on nitrogen is delocalised into the π-system of the benzene ring through resonance; this reduces its availability for protonation, making aniline a much weaker base (pKb ≈ 9.4). Since both A and R are true and R correctly explains why the lone pair in aniline is less available — which directly accounts for its weaker basicity — option (A) is correct.
Q13MCQ1 mark

Which of the following reagents can be used to distinguish a primary amine from a tertiary amine?

Show answer
Option (B) — Chloroform and alcoholic KOH

Explanation: The carbylamine reaction is a test specific to primary amines. When a primary amine is heated with chloroform (CHCl₃) and alcoholic KOH, a foul-smelling isocyanide (carbylamine) is formed. A tertiary amine does not have an N–H bond and therefore gives no reaction under these conditions. NaOH and dilute HCl react with both classes of amines, and NaNO₂ alone (without HCl) is not a standard diagnostic reagent.
Q14Short Answer2 marks

Write reasons for the following:
(i) Aniline is a weaker base than methylamine.
(ii) Write the product formed when aniline is treated with CHCl₃ and alcoholic KOH.

Show answer
(i) Due to delocalisation of the lone pair of electrons on the nitrogen atom of aniline into the benzene ring (resonance), the lone pair is less available for donation to a proton. In methylamine, the lone pair on nitrogen is not delocalised and is readily available for protonation. Hence, aniline is a weaker base than methylamine.

(1 mark)

(ii) Aniline reacts with CHCl₃ and alcoholic KOH (carbylamine reaction) to form phenyl isocyanide (phenyl carbylamine), which has a foul, offensive smell.

C₆H₅NH₂ + CHCl₃ + 3 KOH(alc.) → C₆H₅NC + 3 KCl + 3 H₂O

(Product: Phenyl isocyanide / Phenyl carbylamine)

(1 mark)
Q15Short Answer2 marks

Write a chemical test to distinguish between aniline (C₆H₅NH₂) and N-methylaniline (C₆H₅NHCH₃). Also, write the product formed when aniline is treated with CHCl₃ and alcoholic KOH.

Show answer
(i) Carbylamine (Isocyanide) Test:
Aniline (C₆H₅NH₂) is a primary amine; N-methylaniline (C₆H₅NHCH₃) is a secondary amine. They are distinguished by the carbylamine test.
When heated with CHCl₃ and alcoholic KOH:
• Aniline gives a foul-smelling isocyanide (phenyl isocyanide) — positive result.
• N-methylaniline gives no such smell — negative result.
∴ The compound that produces a foul-smelling product is aniline (primary amine). [1 mark]

(ii) Product when aniline reacts with CHCl₃ / alc. KOH:

C₆H₅NH₂ + CHCl₃ + 3KOH(alc.) → C₆H₅NC + 3KCl + 3H₂O

∴ The product formed is phenyl isocyanide (C₆H₅NC), which has a foul (offensive) smell. [1 mark]
Q16Short Answer2 marks

Arrange the following amines in increasing order of their basic strength in aqueous solution:

C₆H₅NH₂, NH₃, CH₃NH₂, (CH₃)₂NH

Show answer
Increasing order of basic strength in aqueous solution:

C₆H₅NH₂ < NH₃ < CH₃NH₂ < (CH₃)₂NH

Reason: In aqueous solution, basic strength is governed by the combined effect of the +I (inductive) effect, solvation of the conjugate acid (protonated amine), and steric effect.

• C₆H₅NH₂ is the weakest base because the lone pair on nitrogen is delocalised into the benzene ring by resonance, making it the least available for protonation.
• NH₃ is a weaker base than alkylamines because it has no electron-donating alkyl group.
• CH₃NH₂ is stronger than NH₃ due to the +I effect of one methyl group, which increases electron density on N and the cation CH₃NH₃⁺ is well solvated.
• (CH₃)₂NH is the strongest because two methyl groups exert a greater +I effect, and the conjugate acid (CH₃)₂NH₂⁺ is stabilised by solvation more than steric hindrance reduces it.

∴ Increasing order of basic strength: C₆H₅NH₂ < NH₃ < CH₃NH₂ < (CH₃)₂NH
Q17Short Answer2 marks

Arrange the following compounds in increasing order of basic strength in aqueous solution. Give a brief reason for your answer.

C₆H₅NH₂, NH₃, CH₃NH₂, (CH₃)₂NH

Show answer
Increasing order of basic strength in aqueous solution:

C₆H₅NH₂ < NH₃ < CH₃NH₂ < (CH₃)₂NH

Reason:
In aqueous solution, basic strength depends on the availability of the lone pair on nitrogen, the +I (inductive) effect of alkyl groups, and solvation of the cation formed.

• C₆H₅NH₂ is the weakest base: the lone pair on nitrogen is delocalized into the benzene ring through resonance, making it least available for protonation.

• NH₃ is a stronger base than aniline but weaker than alkylamines, as it has no electron-donating alkyl groups.

• CH₃NH₂ (methylamine) is stronger than NH₃ due to the +I effect of one –CH₃ group, which increases electron density on nitrogen.

• (CH₃)₂NH (dimethylamine) is the strongest base here: two –CH₃ groups exert a greater +I effect, and in aqueous solution the resulting (CH₃)₂NH₂⁺ cation is well solvated by water through hydrogen bonding, stabilising the protonated form.

∴ Increasing order of basic strength: C₆H₅NH₂ < NH₃ < CH₃NH₂ < (CH₃)₂NH
Q18Short Answer3 marks

Answer the following questions related to amines:
(a) Arrange the following in increasing order of basic strength and give reason:
C₆H₅NH₂, (C₂H₅)₂NH, NH₃
(b) Which reagent is used in the carbylamine reaction, and which class of amines gives a positive result? Write the observation.
(c) Why is aniline acetylated before carrying out nitration?

Show answer
(a) Increasing order of basic strength:

C₆H₅NH₂ < NH₃ < (C₂H₅)₂NH

Reason: In aniline (C₆H₅NH₂), the lone pair of nitrogen is delocalised into the benzene ring by resonance, making it least basic. In NH₃, the lone pair is fully available for donation, so it is more basic than aniline. In (C₂H₅)₂NH, two ethyl groups exert a +I (electron-donating inductive) effect, increasing electron density on nitrogen; additionally, the secondary amine is well solvated in aqueous solution, making it the most basic of the three.

(b) Reagent: Chloroform (CHCl₃) and alcoholic KOH.

Class giving positive result: Primary amines (1° amines) only.

Observation: A foul-smelling (offensive odour) isocyanide (carbylamine) is formed.

Reaction:
R–NH₂ + CHCl₃ + 3KOH(alc.) → R–N≡C + 3KCl + 3H₂O

(c) Aniline is acetylated (converted to acetanilide) before nitration for the following reasons:

— The –NH₂ group is a very strong activating group and o/p-director; without protection, the ring is over-oxidised/polynitrated by the strongly acidic nitrating mixture.

— Acetylation converts –NH₂ into –NHCOCH₃ (amide), which is a weaker activating group; this moderates the reactivity of the ring and allows controlled mononitration, giving predominantly the para-isomer as the major product.

— The acetyl group also protects the –NH₂ from oxidation by the nitrating mixture.
Q19Short Answer3 marks

A chemistry student, Riya, is working in a laboratory and has been given benzenediazonium chloride (C₆H₅N₂⁺Cl⁻) as the starting material. She needs to prepare four different compounds for her project. Her teacher tells her: 'Diazonium salts are extremely versatile intermediates — they can be converted to a wide range of aromatic compounds by replacement of the diazonium group or by coupling reactions.'

Using benzenediazonium chloride as the starting material, help Riya obtain the following compounds by writing the chemical equations (with reagents and conditions):
(a) Fluorobenzene
(b) Iodobenzene
(c) Phenol
(d) p-Hydroxyazobenzene (an azo dye)

Show answer
(a) Fluorobenzene — Balz-Schiemann Reaction (1 mark)

C₆H₅N₂⁺Cl⁻ + HBF₄ → C₆H₅N₂⁺BF₄⁻ (diazonium tetrafluoroborate) →(Δ) C₆H₅F + N₂↑ + BF₃

Reagent: HBF₄; Condition: heat (Δ). The diazonium group is replaced by fluorine via the diazonium tetrafluoroborate intermediate.

(b) Iodobenzene — Sandmeyer-type reaction (1 mark)

C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + N₂↑ + KCl

Reagent: KI. The diazonium group is replaced by iodine on warming with potassium iodide solution.

(c) Phenol (1 mark)

C₆H₅N₂⁺Cl⁻ + H₂O →(Δ / warm dilute H₂SO₄) C₆H₅OH + N₂↑ + HCl

Condition: warm with water (or dilute H₂SO₄). The diazonium salt is hydrolysed; the –N₂⁺ group is replaced by –OH.

(d) p-Hydroxyazobenzene — Azo Coupling Reaction (1 mark)

C₆H₅N₂⁺Cl⁻ + C₆H₅OH →(alkaline medium, 0–5°C) C₆H₅–N=N–C₆H₄–OH (p) + HCl

Reagent: phenol in alkaline medium; Condition: 0–5°C. The electrophilic diazonium ion couples with the electron-rich phenol ring at the para position to give the orange-red azo dye p-hydroxyazobenzene.

[Note to examiner: Award 1 mark for each correctly written equation with reagents/conditions. Accept any equivalent balanced form. ECF applies — if the product structure is correctly drawn even with a minor condition omission, award ½ mark.]
Q20Short Answer3 marks

An amine 'A' (molecular formula C₇H₉N) gives a foul-smelling product with CHCl₃ and alcoholic KOH. When 'A' is treated with NaNO₂/HCl at 0–5°C, the product 'B' is obtained. 'B' on treatment with CuCN gives compound 'C'. Identify A, B and C and write the chemical equations for the reactions involved.

Show answer
Identification of A, B and C:

The foul-smelling product with CHCl₃ / alcoholic KOH (carbylamine reaction) confirms A is a primary amine. Molecular formula C₇H₉N with a benzene ring gives A = methylaniline (toluidine); the simplest aromatic primary amine of this formula consistent with a diazonium reaction is A = aniline with one –CH₃ group. Taking the para-isomer as representative: A = p-toluidine (4-methylaniline, CH₃–C₆H₄–NH₂).

∴ A = p-Toluidine (4-methylaniline)
∴ B = 4-Methylbenzenediazonium chloride (p-toluenediazonium chloride)
∴ C = 4-Methylbenzonitrile (p-tolunitrile)

Reaction 1 — Carbylamine reaction (confirms primary amine):

CH₃–C₆H₄–NH₂ + CHCl₃ + 3KOH(alc.) → CH₃–C₆H₄–NC + 3KCl + 3H₂O

(foul-smelling isocyanide confirmed)

Reaction 2 — Diazotisation of A to give B:

CH₃–C₆H₄–NH₂ + NaNO₂ + 2HCl ——(0–5°C)——→ CH₃–C₆H₄–N⁺≡N Cl⁻ + NaCl + 2H₂O

(p-Toluenediazonium chloride, B)

Reaction 3 — Sandmeyer reaction of B with CuCN to give C:

CH₃–C₆H₄–N⁺≡N Cl⁻ + CuCN ——(Δ)——→ CH₃–C₆H₄–CN + N₂ + CuCl

(4-Methylbenzonitrile / p-Tolunitrile, C)

∴ A = p-Toluidine, B = p-Toluenediazonium chloride, C = p-Tolunitrile.
Q21Short Answer3 marks

A research chemist working in a dye manufacturing unit isolates three compounds — P, Q and R — from a reaction mixture. The following observations are recorded:
• Compound P reacts with CHCl₃ and alcoholic KOH to produce a compound with a foul, offensive smell.
• Compound Q reacts with Hinsberg's reagent (benzenesulphonyl chloride) to give a product that is insoluble in aqueous NaOH.
• Compound R does not react with Hinsberg's reagent at all.
• All three compounds — P, Q and R — give an orange-red precipitate with NaNO₂/HCl at 0–5°C followed by coupling with alkaline β-naphthol.

On the basis of the above observations, answer the following:
(a) Identify the class of amine (primary, secondary or tertiary) to which each of P, Q and R belongs. Give one reason for each identification. (2 marks)
(b) Write the chemical equation for the reaction of compound P with CHCl₃ and alcoholic KOH, naming the reaction and the product formed. (1 mark)
(c) All three compounds undergo diazotisation and azo coupling. What structural feature, common to P, Q and R, makes this possible? (1 mark)

Show answer
(a) Identification of P, Q and R:

Compound P is a primary amine (1° amine).
Reason: It gives a positive carbylamine (isocyanide) reaction — only primary amines react with CHCl₃ and alcoholic KOH to produce an isocyanide (foul smell).

Compound Q is a secondary amine (2° amine).
Reason: It reacts with Hinsberg's reagent to give a sulphonamide that is insoluble in aqueous NaOH. Primary amines give a sulphonamide soluble in NaOH (N–H acidic); secondary amines give one with no N–H, hence insoluble. Tertiary amines do not react at all.

Compound R is a tertiary amine (3° amine).
Reason: It does not react with Hinsberg's reagent, because there is no N–H available for the reaction with benzenesulphonyl chloride.

(b) Carbylamine (Isocyanide) Reaction:

The reaction of compound P (a primary aromatic/aliphatic amine, Ar–NH₂ or R–NH₂) with CHCl₃ and alcoholic KOH is the Carbylamine reaction.

For a primary amine R–NH₂:

R–NH₂ + CHCl₃ + 3KOH(alc.) → R–N≡C (isocyanide / carbylamine) + 3KCl + 3H₂O

Reaction name: Carbylamine reaction (Isocyanide test)
Product formed: Isocyanide (R–NC) — identified by its characteristically foul, offensive smell.

(c) Common structural feature enabling diazotisation and azo coupling:

All three compounds — P, Q and R — are aromatic amines (i.e., the amino group is directly attached to a benzene ring, –NH₂ / –NHR / –NR₂ on an arene).

Diazotisation requires an aryl amine so that the diazonium salt (Ar–N₂⁺Cl⁻) formed is sufficiently stable at 0–5°C (resonance stabilisation by the aryl group). Aliphatic diazonium salts are too unstable to couple. Hence, the direct attachment of the nitrogen to the aromatic ring is the structural feature common to P, Q and R that makes diazotisation and subsequent azo coupling possible.
Q22Short Answer3 marks

A dye manufacturing company needs to synthesise an orange-red azo dye starting from aniline. The chemist follows the sequence of steps outlined below:

Step 1: Aniline is treated with NaNO₂ and HCl at 0–5°C to produce compound A.
Step 2: Compound A is coupled with phenol in alkaline medium to produce the azo dye B.

The chemist also needs to protect the –NH₂ group of aniline before carrying out nitration to obtain p-nitroaniline (compound C) as the major product.

(a) Write the reaction for Step 1 (formation of compound A) and name compound A. (2 marks)
(b) Write the reaction for Step 2 (formation of azo dye B) and state at which position of phenol does coupling occur. (1 mark)
(c) Why is the –NH₂ group of aniline protected by acetylation before nitration? Give ONE reason. (1 mark)

Show answer
(a) Step 1 — Diazotisation reaction:

Aniline reacts with NaNO₂ and HCl at 0–5°C to form benzenediazonium chloride (Compound A).

C₆H₅–NH₂ + NaNO₂ + 2HCl →(0–5°C) [C₆H₅–N≡N]⁺Cl⁻ + NaCl + 2H₂O

∴ Compound A = Benzenediazonium chloride [C₆H₅N₂]⁺Cl⁻

(1 mark for balanced equation with correct conditions over arrow + 1 mark for correct name/identity of A)

(b) Step 2 — Azo coupling reaction:

Benzenediazonium chloride couples with phenol in alkaline medium at the para position of phenol to form the orange-red azo dye B (p-hydroxyazobenzene).

[C₆H₅–N≡N]⁺Cl⁻ + C₆H₅OH →(alkaline medium) C₆H₅–N=N–C₆H₄–OH(p) + HCl

∴ Coupling occurs at the para position of phenol (para to the –OH group).

(1 mark for correct reaction with product and position of coupling stated)

(c) Protection of –NH₂ group by acetylation before nitration:

Due to acetylation, the –NH₂ group is converted to –NHCOCH₃ (acetamido group), which reduces the activating effect of the nitrogen lone pair (lone pair is partially delocalized into the carbonyl group). This:
(i) prevents oxidation of the –NH₂ group by the concentrated H₂SO₄/HNO₃ nitrating mixture, AND
(ii) moderates the strong ortho/para-directing effect so that the reaction is more controlled and the para-product (p-nitroacetanilide, which on hydrolysis gives p-nitroaniline) becomes the major product.

∴ Acetylation protects the –NH₂ group from oxidation by the nitrating mixture and directs nitration predominantly to the para position, giving p-nitroaniline as the major product after hydrolysis.

(1 mark for any ONE correct reason: oxidation prevention OR para-direction/moderation of activating effect)
Q23Short Answer3 marks

A forensic chemistry laboratory received three unlabelled vials — Vial P, Vial Q and Vial R — each containing a different compound: aniline (C₆H₅NH₂), N-methylaniline (C₆H₅NHCH₃), and N,N-dimethylaniline (C₆H₅N(CH₃)₂). The chemist performed a series of chemical tests to identify the contents of each vial.

(a) The chemist first performed the carbylamine reaction on each vial. Only Vial P gave a foul-smelling product. Identify the compound in Vial P. Write the balanced chemical equation for the reaction observed with Vial P. (2 marks)

(b) The chemist then performed the Hinsberg test on Vials Q and R using benzenesulphonyl chloride (C₆H₅SO₂Cl) in the presence of KOH(aq). Vial Q gave a precipitate insoluble in KOH, while Vial R showed no reaction. Identify the compounds in Vials Q and R. (1 mark)

(c) The chemist noted that aniline is a much weaker base (pK_b ≈ 9.4) than methylamine (pK_b ≈ 3.4), even though both contain a nitrogen atom with a lone pair. Give reason for this observation. (1 mark)

Show answer
(a) Vial P contains aniline (C₆H₅NH₂).

The carbylamine reaction is given only by primary amines (1° amines). Aniline is the only primary amine among the three compounds; N-methylaniline is a secondary amine and N,N-dimethylaniline is a tertiary amine — neither gives the carbylamine test.

Balanced equation:

C₆H₅NH₂ + CHCl₃ + 3KOH(alc.) → C₆H₅NC + 3KCl + 3H₂O

(phenyl isocyanide — foul-smelling)

[Conditions: CHCl₃ + alcoholic KOH; the isocyanide product C₆H₅NC has a characteristic foul/offensive smell]

(b) Vial Q contains N-methylaniline (secondary amine, 2°).
Vial R contains N,N-dimethylaniline (tertiary amine, 3°).

Reasoning:
In the Hinsberg test, a secondary amine reacts with benzenesulphonyl chloride to give a sulphonamide that has no N–H bond; this sulphonamide is insoluble in KOH — matching Vial Q.
A tertiary amine has no N–H bond, so it cannot react with benzenesulphonyl chloride at all — matching Vial R (no reaction).

(c) Due to delocalisation of the lone pair of nitrogen into the benzene ring.

In aniline, the lone pair on the nitrogen atom is in conjugation with the π electrons of the benzene ring (resonance). This delocalisation reduces the availability of the lone pair for donation to a proton (H⁺), making aniline a much weaker base than methylamine, where the lone pair on nitrogen is fully available for protonation and is further enhanced by the +I (electron-donating inductive) effect of the methyl group.
Q24Short Answer3 marks

Answer the following questions related to amines:
(a) Arrange the following amines in increasing order of boiling point. Give reason.
C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N
(b) Why is aniline a weaker base than cyclohexylamine?
(c) Write the IUPAC name of the following compound:
CH₃ – CH(NH₂) – CH₂ – CH₃

Show answer
(a) Increasing order of boiling point:
(C₂H₅)₃N < (C₂H₅)₂NH < C₂H₅NH₂

Reason: The boiling point of amines depends on the extent of intermolecular hydrogen bonding. Primary amines (C₂H₅NH₂) have two N–H bonds and form stronger / more extensive intermolecular hydrogen bonds than secondary amines [(C₂H₅)₂NH, one N–H bond]. Tertiary amines [(C₂H₅)₃N] have no N–H bond and hence cannot form intermolecular hydrogen bonds, so they have the lowest boiling point.

∴ Increasing order: (C₂H₅)₃N < (C₂H₅)₂NH < C₂H₅NH₂

(b) In aniline, the lone pair of electrons on the nitrogen atom is delocalised into the benzene ring due to resonance. This reduces the electron density on nitrogen and makes it less available for donation to a proton (H⁺). In cyclohexylamine, there is no such resonance; the lone pair remains fully available on nitrogen. Therefore, aniline is a weaker base than cyclohexylamine.

(c) The given compound is:
CH₃ – CH(NH₂) – CH₂ – CH₃

The parent chain contains 4 carbons (but-2-amine framework) — number from the end closer to –NH₂:
C1: CH₃ – C2: CH(NH₂) – C3: CH₂ – C4: CH₃

∴ IUPAC name: butan-2-amine
Q25Short Answer3 marks

A forensic chemistry student is analysing three unlabelled bottles, each containing a different amine compound: Bottle P contains methylamine (CH₃NH₂), Bottle Q contains dimethylamine ((CH₃)₂NH), and Bottle R contains trimethylamine ((CH₃)₃N).

(a) The student performs the carbylamine test on each sample. Which bottle(s) will give a positive result? Write the chemical equation for the positive test using the amine from Bottle P. (2 marks)

(b) The student then treats each amine with Hinsberg's reagent (benzenesulphonyl chloride, C₆H₅SO₂Cl) followed by aqueous NaOH solution. State the observation for each bottle (P, Q, and R) and explain how this test helps distinguish the three amines. (2 marks)

Show answer
(a) Carbylamine Test:

Only Bottle P (methylamine, CH₃NH₂) gives a positive result, because the carbylamine (isocyanide) test is positive only for primary amines (1° amines). Bottle Q (2° amine) and Bottle R (3° amine) do not react.

Due to the presence of a free –NH₂ group, primary amines react with chloroform and alcoholic KOH to form an isocyanide (carbylamine), which has a foul, offensive smell.

Reaction for Bottle P:

CH₃NH₂ + CHCl₃ + 3KOH(alc.) → CH₃NC + 3KCl + 3H₂O

(Methylamine) → (Methyl isocyanide — foul smell)

∴ Only Bottle P gives the positive carbylamine test (foul-smelling isocyanide formed).

(b) Hinsberg's Test — Observations and Distinction:

Reagent: Benzenesulphonyl chloride (C₆H₅SO₂Cl) followed by aqueous NaOH.

Bottle P — Methylamine (1° amine):
Reacts with Hinsberg's reagent to form a sulphonamide (N-methyl benzenesulphonamide). This sulphonamide has an acidic N–H hydrogen and dissolves in aqueous NaOH to give a clear solution (soluble in NaOH).

Bottle Q — Dimethylamine (2° amine):
Reacts with Hinsberg's reagent to form a sulphonamide (N,N-dimethyl benzenesulphonamide). This sulphonamide has NO acidic N–H hydrogen and is insoluble in NaOH — a precipitate / turbidity is observed.

Bottle R — Trimethylamine (3° amine):
Does not react with Hinsberg's reagent at all (no N–H bond available). The solution remains clear and no precipitate is formed.

Summary of Observations:

Bottle P (CH₃NH₂, 1°): Reacts → sulphonamide dissolves in NaOH (clear solution).
Bottle Q ((CH₃)₂NH, 2°): Reacts → sulphonamide insoluble in NaOH (precipitate persists).
Bottle R ((CH₃)₃N, 3°): No reaction with Hinsberg's reagent (no precipitate, clear solution throughout).

∴ The Hinsberg's test distinguishes all three amines in Bottles P, Q, and R based on their different behaviour with benzenesulphonyl chloride and NaOH.
Q26Short Answer3 marks

A research chemist is working on the synthesis of a drug intermediate starting from nitrobenzene. The following synthetic route is designed:

Nitrobenzene → (Step I) → Aniline (P) → (Step II, NaNO₂/HCl, 0–5°C) → Compound Q → (Step III, CuCN) → Compound R → (Step IV, LiAlH₄) → Compound S

Separately, aniline (P) is treated with excess CH₃I (Step V) to give compound T.

(a) Identify compounds P, Q, R, S and T. Write the IUPAC name of S. [2]
(b) The chemist observes that aniline does not undergo Friedel-Crafts reaction. Give reason. [1]
(c) Arrange P, CH₃NH₂ and (CH₃)₂NH in increasing order of basic strength in aqueous solution. Justify your answer in ONE sentence. [1]

Show answer
(a) Identification of compounds and IUPAC name of S:

Step I: Nitrobenzene is reduced (by Sn/conc. HCl or Fe/HCl, or catalytic H₂/Ni) to give Aniline.
∴ P = Aniline (C₆H₅NH₂)

Step II: Diazotisation — aniline reacts with NaNO₂/HCl at 0–5°C:
C₆H₅NH₂ + NaNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + NaCl + H₂O
∴ Q = Benzenediazonium chloride (C₆H₅N₂⁺Cl⁻)

Step III: Sandmeyer reaction — Q reacts with CuCN:
C₆H₅N₂⁺Cl⁻ + CuCN → C₆H₅CN + N₂ + CuCl
∴ R = Benzonitrile (C₆H₅CN)

Step IV: Reduction with LiAlH₄:
C₆H₅CN + 4[H] → C₆H₅CH₂NH₂
∴ S = Benzylamine (C₆H₅CH₂NH₂)
IUPAC name of S: Phenylmethanamine

Step V: Treatment of aniline with excess CH₃I (exhaustive methylation):
C₆H₅NH₂ + 3CH₃I → C₆H₅N⁺(CH₃)₃ I⁻ (Trimethylphenylammonium iodide)
∴ T = N,N,N-trimethylanilinium iodide / Phenyltrimethylammonium iodide

[Award 1 mark for correct identification of Q, R, S, T with equations; 1 mark for correct IUPAC name of S: Phenylmethanamine]

(b) Aniline does not undergo Friedel-Crafts reaction because the −NH₂ group of aniline donates its lone pair to the Lewis acid catalyst (anhyd. AlCl₃), forming a Lewis acid–base complex (C₆H₅NH₂ → AlCl₃), which deactivates the catalyst and also renders the ring electron-deficient, preventing electrophilic substitution.
[Award 1 mark for correct reason]

(c) Increasing order of basic strength in aqueous solution:
P (Aniline, C₆H₅NH₂) < CH₃NH₂ < (CH₃)₂NH

Justification: In aqueous solution, (CH₃)₂NH is the strongest base because two electron-donating methyl groups increase electron density on N and the resulting cation is well solvated; CH₃NH₂ is next due to one methyl group; aniline is the weakest because its lone pair is delocalised into the benzene ring, making it far less available for protonation.
[Award 1 mark for correct order AND correct one-sentence justification; if order alone — award ½ mark]
Q27Short Answer3 marks

A pharmaceutical chemist is synthesising a series of amine-based drug candidates. She starts with aniline and performs the following sequence of reactions:

Step 1: Aniline is treated with acetic anhydride in the presence of pyridine to give compound A.
Step 2: Compound A is nitrated using a mixture of conc. HNO₃ and conc. H₂SO₄ to give compound B as the major product.
Step 3: Compound B is hydrolysed with dil. HCl (followed by NaOH treatment) to give compound C.
Step 4: Compound C is diazotised using NaNO₂ and HCl at 0–5°C to give compound D.
Step 5: Compound D is treated with CuCN to give compound E.

Based on the above sequence, answer the following:
(a) Identify compounds A and C, giving the name of the reaction involved in Step 1. Why is Step 1 carried out before nitration?
(b) Identify compound B and state the reason why the nitro group is introduced at the para position as the major product.
(c) Write the balanced chemical equation for the conversion of compound C to compound D (Step 4).
(d) Name the reaction in Step 5 and identify compound E. State one synthetic importance of compound E in organic chemistry.

Show answer
(a) Compound A: Acetanilide (N-phenylethanamide)

C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH

(Pyridine neutralises the acetic acid formed, driving the reaction forward.)

Compound C: p-Nitroaniline (4-nitroaniline / 4-nitrobenzenamine)

Reaction in Step 1: Acetylation (acylation)

Reason for acetylation before nitration:
Due to acetylation, the highly activating and electron-donating –NH₂ group is converted into the –NHCOCH₃ group. This reduces the electron density on the ring to a moderate level, thereby (i) preventing oxidation/poly-nitration of the ring by the strong acid nitrating mixture, and (ii) still directing the incoming –NO₂ group to the para position as the major product. The protecting group is removed by hydrolysis after nitration.

(1 mark)

(b) Compound B: p-Nitroacetanilide (4-nitroacetanilide)

Reason for para substitution as major product:
The –NHCOCH₃ group is an ortho/para director. However, the ortho positions are sterically hindered by the relatively bulky –NHCOCH₃ group, so the electrophilic nitronium ion (NO₂⁺) preferentially attacks the less hindered para position, giving p-nitroacetanilide as the major product.

(1 mark)

(c) Step 4 — Diazotisation of compound C (p-Nitroaniline) to compound D (p-Nitrobenzenediazonium chloride):

The reaction is carried out at 0–5°C.

4-O₂N–C₆H₄–NH₂ + NaNO₂ + 2HCl → 4-O₂N–C₆H₄–N⁺≡N Cl⁻ + NaCl + 2H₂O

(Compound D: p-Nitrobenzenediazonium chloride)

Conditions: NaNO₂ + HCl, 0–5°C

(1 mark)

(d) Reaction in Step 5: Sandmeyer's reaction

Compound E: p-Nitrobenzonitrile (4-nitrobenzonitrile / 4-nitrophenyl cyanide)

4-O₂N–C₆H₄–N⁺≡N Cl⁻ + CuCN → 4-O₂N–C₆H₄–CN + N₂ + CuCl

Synthetic importance of compound E:
Aryl nitriles (ArCN) can be hydrolysed to aryl carboxylic acids (ArCOOH) under acidic or alkaline conditions, or reduced to primary aryl amines (ArCH₂NH₂) using LiAlH₄. They serve as versatile intermediates for introducing –COOH and –CH₂NH₂ functional groups into the benzene ring — groups that cannot be introduced by direct substitution on the ring. Thus, the diazonium salt → nitrile route is an important strategy in synthetic organic chemistry.

(1 mark)
Q28Short Answer3 marks

A factory produces a synthetic dye by the following sequence of reactions starting from benzene:

Benzene is first nitrated to give compound A. Compound A is then reduced using iron scrap and HCl to give compound B (mol. formula C₆H₇N). Compound B is diazotised using NaNO₂ and HCl at 0–5°C to give compound C. Compound C is then coupled with β-naphthol in alkaline medium to give a brilliant orange-red azo dye D.

(i) Identify compounds A, B, C and D. (2 marks)
(ii) Write the balanced chemical equation for the formation of compound C from compound B. (1 mark)
(iii) The factory chemist notices that if the diazotisation is carried out above 10°C, the yield of D drops sharply. Give one reason for this observation. (1 mark)

Show answer
(i) Identification of A, B, C and D:

• A: Nitrobenzene (C₆H₅NO₂)
• B: Aniline (C₆H₅NH₂)
• C: Benzenediazonium chloride (C₆H₅N₂⁺Cl⁻)
• D: 1-Phenylazo-2-naphthol (orange-red azo dye)

(Award ½ mark each for any two correct identifications; ½ + ½ for the remaining two = 2 marks total)

(ii) Formation of compound C from compound B (diazotisation):

C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O

(Conditions: NaNO₂ + HCl, 0–5°C)

(1 mark for correctly balanced equation with conditions over/below the arrow)

(iii) Reason for low yield above 10°C:

The diazonium salt (C₆H₅N₂⁺Cl⁻) is thermally unstable. Above 10°C it decomposes rapidly to phenol and nitrogen gas (C₆H₅N₂⁺ → C₆H₅⁺ + N₂↑ → C₆H₅OH), so very little diazonium salt remains available for azo coupling, causing the yield of D to fall sharply.

(Award 1 mark for stating thermal instability / decomposition of diazonium salt at higher temperature)
Q29Short Answer3 marks

A dye manufacturing company uses compound 'A' (molecular formula C₆H₇N) as a starting material. The following sequence of reactions is carried out:

Step 1: Compound 'A' is treated with NaNO₂ and HCl at 0–5°C to form compound 'B'.
Step 2: Compound 'B' is coupled with compound 'C' (a phenol dissolved in dilute NaOH) to form an orange-red compound 'D'.
Step 3: A separate sample of compound 'A' is treated with Br₂ water (no catalyst required) to give compound 'E', a white precipitate with molecular formula C₆H₄Br₃N.

(i) Identify compounds A, B, C and D, giving one reason why compound 'B' must be kept at 0–5°C during its formation.
(ii) Write the balanced chemical equation for Step 2 (azo coupling).
(iii) Compound 'E' is formed much more readily than bromination of benzene, which requires Br₂/anhyd. AlCl₃. Give one reason for this difference in reactivity.
(iv) The company wants to convert compound 'A' into fluorobenzene. Name the reagent used and write the reaction involved.

Show answer
(i) Identification of compounds:

Compound A (C₆H₇N) = Aniline (C₆H₅NH₂)
Compound B = Benzenediazonium chloride (C₆H₅N₂⁺Cl⁻)
Compound C = Phenol (C₆H₅OH)
Compound D = p-Hydroxyazobenzene (an orange-red azo dye)

Reason for 0–5°C: The diazonium salt (compound B) is highly unstable. At temperatures above 5°C, it decomposes rapidly to give phenol and N₂ gas. Low temperature (0–5°C) slows decomposition and allows the diazonium salt to be used directly for coupling.

(ii) Balanced equation for azo coupling (Step 2):

C₆H₅N₂⁺Cl⁻ + C₆H₅OH → C₆H₅–N=N–C₆H₄–OH (para) + HCl

(Benzenediazonium chloride couples with phenol in alkaline medium at the para position to give p-hydroxyazobenzene, an orange-red azo dye.)

[Reaction conditions: alkaline medium (dil. NaOH), 0–5°C]

(iii) Aniline undergoes electrophilic bromination (to give 2,4,6-tribromoaniline, compound E) far more readily than benzene because the –NH₂ group is a powerful electron-donating group. It donates its lone pair of electrons into the benzene ring by resonance, greatly increasing the electron density at the ortho and para positions. This makes the ring highly activated toward electrophilic attack, so even the weaker electrophile Br₂ (without a Lewis acid catalyst) reacts rapidly and gives trisubstitution. In benzene, no such activation exists, and a Lewis acid catalyst (anhyd. AlCl₃) is needed to generate the stronger electrophile Br⁺.

(iv) Conversion of aniline (A) to fluorobenzene — Balz–Schiemann reaction:

Step 1 — Diazotisation:
C₆H₅NH₂ + NaNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + NaCl + H₂O
[Conditions: 0–5°C]

Step 2 — Balz–Schiemann reaction:
C₆H₅N₂⁺Cl⁻ + HBF₄ → C₆H₅N₂⁺BF₄⁻ + HCl
C₆H₅N₂⁺BF₄⁻ → C₆H₅F + N₂↑ + BF₃
[Condition: heat (Δ)]

Reagent used: Fluoroboric acid (HBF₄).

∴ Fluorobenzene is obtained by the Balz–Schiemann reaction.
Q30Short Answer3 marks

Give reasons for the following:
(a) Aniline does not undergo Friedel-Crafts reaction.
(b) Ethylamine is more soluble in water than diethylamine.
(c) pKb of aniline (9.40) is higher than that of methylamine (3.36).

Show answer
(a) Aniline does not undergo Friedel-Crafts reaction because it acts as a Lewis base and forms a stable salt with the Lewis acid catalyst (anhyd. AlCl₃) used in Friedel-Crafts reactions. The nitrogen lone pair donates to AlCl₃, forming a coordination complex (C₆H₅–NH₂→AlCl₃), which deactivates the catalyst. Additionally, the –NH₂ group is a strong electron-donating group, and the resulting anilinium ion formed under strongly acidic/Lewis acid conditions becomes electron-poor at nitrogen, deactivating the ring towards electrophilic substitution.

(b) Ethylamine (a primary amine) can form hydrogen bonds with water molecules through both its N–H bond (as donor) and its nitrogen lone pair (as acceptor), giving it three sites for hydrogen bonding with water. Diethylamine (a secondary amine) has only one N–H bond and therefore fewer N–H donors available for hydrogen bonding with water. Due to greater extent of hydrogen bonding with water, ethylamine is more soluble in water than diethylamine.

(c) A higher pKb value means a weaker base (lower basicity). In aniline, the lone pair of electrons on the nitrogen atom is delocalised into the benzene ring through resonance, making it less available for donation to a proton (H⁺). This resonance delocalisation reduces the electron density on nitrogen, making aniline a much weaker base. In methylamine, the methyl group exerts a +I (inductive) effect, which increases the electron density on nitrogen, making the lone pair more readily available for protonation. Therefore, aniline (pKb = 9.40) is a far weaker base than methylamine (pKb = 3.36).

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