Q.Give one chemical test to distinguish between the following pairs of compounds.

Concept: Hinsberg test, the azo-dye (diazotisation/coupling) test, and the bromine-water test distinguish amines by degree and by ring activation.

Reasoning:

1. Methylamine vs. dimethylamine - Hinsberg test: treat with benzenesulfonyl chloride + KOH. Methylamine (1 degree) forms a sulfonamide with an acidic N-H that dissolves in excess KOH (clear solution); dimethylamine (2 degree) forms a sulfonamide with no N-H, which remains as an insoluble oil/solid.
2. Secondary vs. tertiary amines - Hinsberg test: the secondary amine gives an alkali-insoluble sulfonamide; the tertiary amine does not react with benzenesulfonyl chloride at all (it stays unreacted, but dissolves on acidification since the free amine itself is basic).
3. Ethylamine vs. aniline - Azo dye test: diazotise with NaNO2/HCl at 0-5 degC, then couple with alkaline beta-naphthol. Aniline's diazonium salt is stable and couples to give a bright orange-red dye; ethylamine's aliphatic diazonium salt decomposes immediately (no coupling, no dye).
4. Aniline vs. benzylamine - the carbylamine test cannot distinguish this pair (both are primary amines, so both give a positive isocyanide test). Use the bromine-water test instead: aniline's -NH2 is directly conjugated with the ring, so it reacts instantly with bromine water (no catalyst needed) to give a white precipitate of 2,4,6-tribromoaniline; benzylamine's -NH2 sits on a side-chain -CH2- group, not conjugated with the ring, so its ring behaves like an ordinary alkylbenzene and gives no such precipitate under the same conditions.
5. Aniline vs. N-methylaniline - Hinsberg test: aniline (1 degree) gives a sulfonamide that dissolves in KOH (clear solution); N-methylaniline (2 degree) gives a sulfonamide insoluble in KOH.

1. Methylamine and dimethylamine

Both are aliphatic amines, but methylamine is primary ($1^\circ$) and dimethylamine is secondary ($2^\circ$). The classic test uses Hinsberg's reagent (benzenesulfonyl chloride, $\ce{C6H5SO2Cl}$) in the presence of aqueous KOH.

• Methylamine ($1^\circ$): reacts to form a sulfonamide that has one acidic hydrogen on the N-atom. This dissolves in excess KOH to give a clear solution.
• Dimethylamine ($2^\circ$): forms a sulfonamide with no N-H hydrogen. It remains as an insoluble oil or solid (no dissolution in alkali).

Test: Add benzenesulfonyl chloride and a few drops of KOH solution. Shake well.

• If the mixture becomes clear (sulfonamide dissolves) -> methylamine.
• If an oily layer or precipitate remains -> dimethylamine.

2. Secondary and tertiary amines

Again, Hinsberg's test works beautifully. But here we also have the nitrous acid test as an alternative.

Using Hinsberg's reagent:

• Secondary amine ($2^\circ$): forms an insoluble sulfonamide (no N-H to dissolve in alkali).
• Tertiary amine ($3^\circ$): does not react with benzenesulfonyl chloride at all (no N-H to replace). It remains as an insoluble oil, but on acidification it dissolves (because the tertiary amine itself is basic and forms a salt).

Using nitrous acid ($\ce{NaNO2 + HCl}$):

• $2^\circ$ aliphatic amine -> forms a yellow oily N-nitrosamine (nitrosamine).
• $3^\circ$ aliphatic amine -> forms a water-soluble nitrite salt (no oil).

So either test separates them cleanly.

3. Ethylamine and aniline

Ethylamine is an aliphatic primary amine; aniline is an aromatic primary amine. The simplest test is the azo dye test (diazotisation followed by coupling).

• Aniline: on treatment with $\ce{NaNO2 + HCl}$ at 0-5°C gives a diazonium salt. This, when coupled with $\beta$-naphthol in alkaline medium, yields a brilliant orange-red azo dye.
• Ethylamine: forms a diazonium salt that is unstable and decomposes immediately at that temperature - no coupling, no dye.

Test: Diazotise at 0-5°C, then add alkaline $\beta$-naphthol.

• Orange-red precipitate -> aniline.
• No colour (or only nitrogen gas evolution) -> ethylamine.

4. Aniline and benzylamine

Both are primary amines, but aniline is aromatic (amino group directly on benzene ring) while benzylamine has the amino group on a side chain ($\ce{C6H5CH2NH2}$). The carbylamine test (isocyanide test) does not distinguish them, because it is positive for all primary amines, aliphatic and aromatic alike.

Correct test: Use the bromine water test instead - nitrous-acid/azo coupling isn't useful here either, since benzylamine's amino group is aliphatic (benzylic) and its diazonium salt just decomposes rather than giving a clean, comparable result.

• Aniline: reacts instantly with bromine water to give a white precipitate of 2,4,6-tribromoaniline (even without a catalyst), because the $-NH_2$ group is directly conjugated with the ring and strongly activates it.
• Benzylamine: does not give a precipitate with bromine water under the same conditions - its $-CH_2NH_2$ group is not conjugated with the ring, so the ring behaves like an ordinary (much less activated) alkylbenzene and bromination requires a catalyst.

Test: Add bromine water dropwise to the compound.

• White precipitate -> aniline.
• No precipitate (bromine colour persists or decolourises only slowly) -> benzylamine.

5. Aniline and N-methylaniline

Aniline is a primary aromatic amine; N-methylaniline is a secondary aromatic amine. The Hinsberg test works perfectly here.

• Aniline ($1^\circ$ aromatic): forms a sulfonamide that dissolves in KOH (clear solution).
• N-methylaniline ($2^\circ$ aromatic): forms a sulfonamide that is insoluble in KOH (remains as an oil or solid).

Alternatively, the nitrous acid test also works:

• Aniline -> diazonium salt (stable at 0-5°C) -> coupling gives azo dye.
• N-methylaniline -> forms a yellow oily N-nitrosoamine (no coupling possible).

Test: Add benzenesulfonyl chloride and KOH.

• Clear solution -> aniline.
• Insoluble oil/precipitate -> N-methylaniline.

Here is a clear, concept-first solution for distinguishing the given pairs of compounds using chemical tests.

Core Concepts: Sulfonylation, Diazotisation and Ring Activation

The tests below rely on three key ideas:

1. Hinsberg Test: Benzenesulfonyl chloride ($C_6H_5SO_2Cl$) reacts with primary and secondary amines. The product’s solubility in alkali (NaOH/KOH) depends on the presence of a free N–H hydrogen, so it separates 1°, 2° and 3° amines.
2. Azo Dye Test: Only aromatic primary amines form a diazonium salt stable at 0–5°C, which then couples with β-naphthol to give a bright dye. Aliphatic primary amines give unstable diazonium salts that decompose at once.
3. Bromine Water Test: An $-NH_2$ group attached directly to a benzene ring strongly activates it, so bromination happens instantly without any catalyst.

Note: The carbylamine reaction (CHCl₃ + alc. KOH) is positive for all primary amines — aliphatic and aromatic alike — so it confirms a 1° amine but cannot separate two primary amines from each other.

(i) Methylamine vs. Dimethylamine

Method: Hinsberg Test

Steps:

1. Take a small sample of each amine in separate test tubes.
2. Add benzenesulfonyl chloride ($C_6H_5SO_2Cl$) and a few drops of aqueous KOH (or NaOH).
3. Shake the mixture and observe.

Observation & Inference:

• Methylamine (1° amine): Forms a clear solution (sulfonamide salt is soluble in alkali).

• Dimethylamine (2° amine): Forms a precipitate (sulfonamide has no free N–H, so it is insoluble in alkali).

Result: Methylamine gives a clear solution; dimethylamine gives a precipitate.

(ii) Secondary vs. Tertiary Amines

Method: Hinsberg Test (same as above)

Steps:

1. Treat each amine with $C_6H_5SO_2Cl$ and aqueous KOH.
2. Observe solubility.

Observation & Inference:

• Secondary amine: Forms a precipitate (insoluble sulfonamide).
• Tertiary amine: Does not react (no N–H bond). It remains as an oily layer or dissolves in acid, but no precipitate forms.

Result: Secondary amine gives a precipitate; tertiary amine gives no reaction.

(iii) Ethylamine vs. Aniline

Method: Azo Dye Test (diazotisation followed by coupling)

Steps:

1. Dissolve each compound in dilute HCl and cool to 0–5°C.
2. Add a cold solution of sodium nitrite ($NaNO_2$) to form a diazonium salt.
3. Add an alkaline solution of β-naphthol.

Observation & Inference:

• Aniline (aromatic 1° amine): Forms a diazonium salt that is stable at 0–5°C and couples to give a bright orange-red azo dye.

• Ethylamine (aliphatic 1° amine): Its diazonium salt is unstable even at 0–5°C — it decomposes at once with brisk evolution of $N_2$; no dye is formed.

Result: Aniline gives an orange-red dye; ethylamine gives no dye.

⚠️ Do not use the carbylamine test for this pair — both ethylamine and aniline are primary amines, so both give a positive (foul-smelling isocyanide) result. It cannot distinguish them.

(iv) Aniline vs. Benzylamine

Method: Bromine Water Test

Steps:

1. Take each compound in a separate test tube.
2. Add bromine water dropwise at room temperature.

Observation & Inference:

• Aniline (aromatic 1° amine): The $-NH_2$ group is directly conjugated with the ring and strongly activates it, so bromine water reacts instantly (no catalyst) to give a white precipitate of 2,4,6-tribromoaniline.

• Benzylamine ($C_6H_5CH_2NH_2$): The $-NH_2$ sits on a side-chain $-CH_2-$ group, not on the ring, so the ring behaves like an ordinary alkylbenzene — no precipitate under the same conditions.

Result: Aniline gives an instant white precipitate; benzylamine gives none.

(v) Aniline vs. N-Methylaniline

Method: Hinsberg Test

Steps:

1. Treat each compound with $C_6H_5SO_2Cl$ and aqueous KOH.
2. Observe solubility.

Observation & Inference:

• Aniline (1° amine): Forms a clear solution (soluble sulfonamide salt).
• N-Methylaniline (2° amine): Forms a precipitate (insoluble sulfonamide).

Result: Aniline gives a clear solution; N-methylaniline gives a precipitate.

Quick Summary Table

Key takeaway: Match the amine class (1°, 2°, 3°) and aromatic vs. aliphatic nature to the correct test. The Hinsberg test separates 1°/2°/3° amines; the azo dye test picks out aromatic 1° amines; bromine water picks out a ring-activated (aryl) amine. Remember the carbylamine test is positive for all primary amines — aliphatic and aromatic — so it can never separate two primary amines from each other.

Here are the common mistakes students make when choosing distinction tests for amines, along with precise corrections.

General Mistake: Confusing the Test Reagent

Mistake: Using the wrong reagent (e.g., using AgNO₃ for all tests, or using Hinsberg reagent incorrectly).

Why it happens: Students memorize tests without understanding the chemical basis (basicity, solubility, or reaction type).

How to avoid: Always link the test to a specific property:

• Hinsberg test → distinguishes 1°, 2°, 3° amines based on sulfonamide solubility.
• Carbylamine test → positive for all primary amines (1°) — aliphatic and aromatic — so it identifies a 1° amine but cannot separate two primary amines from each other.
• Azo dye test → only for primary aromatic amines (like aniline).
• Bromine water test → only for amines whose $-NH_2$ is directly on the ring (instant white precipitate).

(i) Methylamine vs. Dimethylamine

Common Mistake: Misreading the Hinsberg test observations — expecting both sulfonamides to dissolve in alkali.

• Why wrong: Only the primary amine's sulfonamide has an acidic N–H left, so only it dissolves in KOH. The secondary amine's sulfonamide has no N–H and stays insoluble.
• Correct approach: Perform the Hinsberg test ($C_6H_5SO_2Cl$ + aqueous KOH):
  • Methylamine (1°) → sulfonamide dissolves in alkali → clear solution.
  • Dimethylamine (2°) → sulfonamide insoluble → precipitate/oily layer.
• Note: The carbylamine test also works for this particular pair (1° positive, 2° negative) — but the Hinsberg test is the standard choice because it cleanly separates all three classes.

How to avoid: Remember: “Free N–H on the sulfonamide = soluble in alkali.”

(ii) Secondary vs. Tertiary Amines

Common Mistake: Using the Hinsberg test but forgetting the solubility difference.

• Why wrong: Students often think both react, but:
  • Secondary amine (e.g., diethylamine) reacts with benzenesulfonyl chloride to form a solid sulfonamide that is insoluble in alkali.
  • Tertiary amine (e.g., triethylamine) does not react (no H on N) — it remains as an oily layer or dissolves in acid.
• Correct approach: Add benzenesulfonyl chloride (C₆H₅SO₂Cl) in presence of NaOH:
  • 2° amine → precipitate (sulfonamide) that does not dissolve in NaOH.
  • 3° amine → no precipitate; amine layer remains.

How to avoid: Draw the reaction mechanism: 2° amine has one H to lose, 3° has none.

(iii) Ethylamine vs. Aniline

Common Mistake: Using the Carbylamine test to separate them.

• Why wrong: Both are primary amines, so both give a positive carbylamine test (a foul-smelling isocyanide). The test confirms “primary” but does not distinguish this pair.
• Correct approach: Use the Azo dye test (diazotisation + coupling):
  • Aniline (aromatic 1°) → forms a diazonium salt stable at 0–5°C, then couples with β-naphthol to give a bright orange-red dye.
  • Ethylamine (aliphatic 1°) → its diazonium salt decomposes immediately (with $N_2$ evolution); no dye.

How to avoid: Remember: “Only aromatic primary amines give azo dyes.”

(iv) Aniline vs. Benzylamine

Common Mistake: Assuming both are aromatic amines and behave identically.

• Why wrong: Benzylamine has an aliphatic NH₂ attached to the ring via a CH₂ group. Aniline has NH₂ directly on the ring — only aniline's ring is strongly activated.
• Correct approach: Use the Bromine water test:
  • Aniline → instant white precipitate of 2,4,6-tribromoaniline (no catalyst needed).
  • Benzylamine → no precipitate; its ring behaves like an ordinary alkylbenzene.

Alternative to avoid: the Carbylamine test — both are 1°, so both are positive; it cannot separate them.

How to avoid: Check the nature of the amino group: directly attached to ring = aromatic (aryl) amine; on a side chain = aliphatic amine.

(v) Aniline vs. N-Methylaniline

Common Mistake: Forgetting that N-methylaniline is a secondary amine and treating both compounds as “aniline-like” primary amines.

• Why wrong: The two belong to different classes — aniline is 1°, N-methylaniline is 2° — and every class-based test hinges on exactly that difference.
• Correct approach: Use the Hinsberg test:
  • Aniline (1°) → sulfonamide soluble in NaOH → clear solution.
  • N-Methylaniline (2°) → sulfonamide insoluble in NaOH → precipitate.
• Note: The carbylamine test would also distinguish this pair (1° positive, 2° negative), but the Hinsberg test is the standard, cleaner choice.

How to avoid: Identify the degree of substitution on nitrogen before choosing the test.

Quick Reference Table

Final Tip for Exams

Always write the reaction equation (even a short one) to show the chemical change — examiners award marks for understanding, not just naming the test. For example:

“Aniline on diazotisation followed by coupling with β-naphthol gives an orange dye, while ethylamine does not.”

This proves you know the why behind the test.