Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylam

Exercises · 9.2

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

(i) Methylamine and dimethylamine

(ii) Secondary and tertiary amines

(iii) Ethylamine and aniline

(iv) Aniline and benzylamine

(v) Aniline and N-methylaniline.

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Concept understanding — Nucleophilic Substitution Reactions

Nucleophilic Addition – From Intuition to Precision

Imagine you have a molecule with a carbon–oxygen double bond — a carbonyl group ( $\ce{C=O}$ ). That oxygen is greedy for electrons; it pulls them away from carbon, leaving the carbon slightly positive ( $\delta^+$ ) and the oxygen slightly negative ( $\delta^-$ ). Now, if you bring a species that is rich in electrons (a nucleophile, meaning "nucleus-loving"), it will naturally be attracted to that electron-deficient carbon. The nucleophile attacks the carbon, the $\pi$ bond breaks, and the oxygen picks up a proton (or some other electrophile) to become stable. That, in a nutshell, is nucleophilic addition.

Note

The key idea: a nucleophile adds across a polar multiple bond (usually $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds.

The Precise Statement

Nucleophilic addition is a reaction in which a nucleophile (an electron-rich species) forms a sigma bond with an electrophilic carbon atom of a polar multiple bond (typically a carbonyl group, $\ce{C=O}$ , or a nitrile, $\ce{C≡N}$ ), while the $\pi$ bond breaks. The resulting intermediate then captures a proton (or another electrophile) to give a neutral product.

In general form:

\ce{R2C=O + Nu^- ->[H^+] R2C(OH)Nu}

The nucleophile ( $\ce{Nu^-}$ ) attacks the carbonyl carbon; the oxygen becomes negatively charged; then a proton ( $\ce{H^+}$ ) from the medium attaches to the oxygen, yielding an alcohol.

Why It Happens – The Driving Force

The carbonyl carbon is electrophilic because:

Oxygen is more electronegative than carbon, so the $\ce{C=O}$ bond is polarised: $\ce{C^{\delta+}=O^{\delta-}}$ .
The $\pi$ bond is weaker than a $\sigma$ bond, so it can break relatively easily.

A nucleophile (like $\ce{OH^-}$ , $\ce{CN^-}$ , or $\ce{NH3}$ ) has a lone pair or a negative charge. It seeks positive centres. The attack forms a new $\sigma$ bond, and the $\pi$ electrons move entirely to oxygen, creating an alkoxide ion ( $\ce{R2C-O^-}$ ). This intermediate is then quenched by a proton.

Watch out

A common mistake: thinking the nucleophile attacks the oxygen. No — oxygen is already electron-rich; the nucleophile goes to the carbon because it is electron-deficient.

A Concrete Example – Addition of HCN to a Ketone

Take acetone ( $\ce{CH3COCH3}$ ) and hydrogen cyanide ( $\ce{HCN}$ ). In the presence of a base, $\ce{CN^-}$ (the nucleophile) attacks the carbonyl carbon:

\ce{CH3COCH3 + CN^- -> CH3C(O^-)(CN)CH3}

The alkoxide intermediate then picks up a proton from $\ce{HCN}$ (or from water) to give a cyanohydrin:

\ce{CH3C(O^-)(CN)CH3 + H^+ -> CH3C(OH)(CN)CH3}

The product is acetone cyanohydrin. Notice: two new sigma bonds formed ( $\ce{C-CN}$ and $\ce{O-H}$ ), and the $\pi$ bond is gone.

What Makes a Good Nucleophile?

Strong nucleophiles are usually negatively charged or have lone pairs:

$\ce{OH^-}$ , $\ce{CN^-}$ , $\ce{NH2^-}$ , $\ce{CH3O^-}$ , $\ce{H^-}$ (from hydride reagents like $\ce{NaBH4}$ or $\ce{LiAlH4}$ )
Neutral but polarisable: $\ce{NH3}$ , $\ce{H2O}$ (weaker, but can add under acidic conditions)

Tip

In exams, remember: hydride ion ( $\ce{H^-}$ ) is a nucleophile in reductions (e.g., $\ce{NaBH4}$ reduces aldehydes/ketones to alcohols via nucleophilic addition).

The Big Picture – Why This Matters

Nucleophilic addition is the fundamental reaction of carbonyl compounds. It is how:

Aldehydes and ketones form alcohols (with $\ce{NaBH4}$ or $\ce{LiAlH4}$ )
Cyanohydrins are made (important in organic synthesis)
Grignard reagents ( $\ce{RMgX}$ ) add to carbonyls to form new carbon–carbon bonds
Hemiacetals and acetals form (key in carbohydrate chemistry)

Every time you see a $\ce{C=O}$ group, think: this carbon is a target for nucleophiles.

Final Answer

Nucleophilic addition is a reaction where an electron-rich nucleophile attacks the electrophilic carbon of a polar multiple bond (typically $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds — one to the nucleophile and one to a proton (or other electrophile). The driving force is the polarity of the $\ce{C=O}$ bond and the stability gained by forming stronger sigma bonds.

Nucleophilic addition reactions of carbonyl compounds form a central part of the NCERT Class 12 Chemistry chapter on aldehydes and ketones, and questions on cyanohydrin formation or the role of hydride nucleophiles like NaBH4 are common in CBSE boards and JEE Main. Students searching "nucleophilic addition mechanism class 12 chemistry important questions" will find this carbonyl-carbon-attack explanation is the standard NCERT approach.

Why this formula?

Nucleophilic Substitution Reactions: Why the Key Formulas Hold

Nucleophilic substitution reactions are a cornerstone of organic chemistry. Instead of just memorizing the rate laws, let's understand why they arise from the molecular events.

1. The Two Main Mechanisms: A Tale of Timing

The key formulas (rate laws) for nucleophilic substitution come directly from how many molecules are involved in the rate-determining step (RDS) — the slowest step that controls the overall reaction speed.

SN1: Unimolecular — The Leaving Group Goes First

The Rate Law:

\text{Rate} = k[\text{RX}]

Why?

The reaction happens in two steps:

Slow step (RDS): The C–X bond breaks spontaneously, forming a carbocation intermediate. Only the substrate ( $\text{RX}$ ) is involved.

\text{RX} \xrightarrow{\text{slow}} \text{R}^+ + \text{X}^-

Fast step: The nucleophile ( $\text{Nu}^-$ ) attacks the carbocation.

\text{R}^+ + \text{Nu}^- \xrightarrow{\text{fast}} \text{RNu}

Since the slow step depends only on the concentration of $\text{RX}$ , the rate law has no dependence on $[\text{Nu}^-]$ . The nucleophile arrives after the carbocation is formed — it cannot affect the speed of the first step.

Key insight: The rate is determined by how easily the leaving group leaves, not by how fast the nucleophile attacks.

SN2: Bimolecular — Simultaneous Attack and Departure

The Rate Law:

\text{Rate} = k[\text{RX}][\text{Nu}^-]

Why?

The reaction occurs in one concerted step:

The nucleophile attacks the carbon from one side at the same time as the leaving group departs from the opposite side.
Both $\text{RX}$ and $\text{Nu}^-$ must collide with the correct orientation and sufficient energy.

The rate depends on the frequency of productive collisions between the two molecules. This is directly proportional to the product of their concentrations:

\text{Rate} \propto [\text{RX}] \times [\text{Nu}^-]

Key insight: Both partners are involved in the transition state simultaneously — if either is missing, the reaction cannot proceed.

2. The Transition State: Why the Formulas Are Not Just "Given"

For SN2, the transition state has a pentavalent carbon (five bonds partially formed/broken). The energy barrier depends on steric hindrance — bulkier groups around carbon make it harder for the nucleophile to approach, which is why SN2 is favored at primary carbons.

For SN1, the transition state of the slow step involves only bond breaking (no bond formation yet). The carbocation stability (tertiary > secondary > primary) determines the activation energy — this is why SN1 is favored at tertiary carbons.

3. The Temperature Dependence: The Arrhenius Equation

Both rate constants $k$ follow the Arrhenius equation:

k = A e^{-E_a / RT}

$A$ = frequency factor (how often collisions occur with correct orientation)
$E_a$ = activation energy (the energy barrier for the RDS)
$R$ = gas constant
$T$ = temperature

This is not a separate formula — it explains why the rate constants change with temperature. Higher $T$ increases the fraction of molecules with energy $\geq E_a$ , speeding up the reaction.

4. Summary: The "Why" in One Table

Mechanism	Rate Law	Why? (Molecular Reason)
SN1	$\text{Rate} = k[\text{RX}]$	Slow step involves only the substrate breaking apart. Nucleophile waits.
SN2	$\text{Rate} = k[\text{RX}][\text{Nu}^-]$	Both molecules must collide in the single, concerted step.

Final takeaway: The formulas are not arbitrary — they are direct consequences of which molecules are present in the slowest step. Always ask: "What is happening in the rate-determining step?" The answer gives you the rate law.

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

Reasoning:

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.
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).
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).
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.
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.

✓Final answer

Hinsberg test: methylamine's sulfonamide dissolves in KOH; dimethylamine's does not. Hinsberg test: the secondary amine's sulfonamide is insoluble in alkali; the tertiary amine does not react with benzenesulfonyl chloride. Azo dye test: aniline gives an orange-red dye on coupling with beta-naphthol; ethylamine gives none. Bromine-water test: aniline gives an instant white precipitate (2,4,6-tribromoaniline); benzylamine does not. Hinsberg test: aniline's sulfonamide dissolves in KOH; N-methylaniline's does not.

The key idea is that primary, secondary, and tertiary amines react differently with Hinsberg's reagent (benzenesulfonyl chloride) and with nitrous acid - these differences form the basis of simple chemical tests. For each pair, a single reagent gives a distinct observable change (like solubility, gas evolution, or colour) that identifies one compound from the other.

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).

Watch out

A common mistake is to think that the $2^\circ$ amine also dissolves - it does not, because its sulfonamide lacks the acidic proton needed to form a water-soluble salt.

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).

Tip

The key distinction: the $2^\circ$ amine's sulfonamide is insoluble in both alkali and acid, while the $3^\circ$ amine itself is insoluble in alkali but dissolves in dilute HCl.

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.

\ce{C6H5NH2 ->[NaNO2/HCl][0-5°C] C6H5N2+Cl- ->[β-naphthol/OH-] Orange-red azo 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.

Watch out

The carbylamine test is positive for all primary amines, both aliphatic and aromatic. So it cannot separate aniline from benzylamine.

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).

Tip

For aromatic amines, the Hinsberg test is often quicker and more dramatic - the difference in solubility is immediately visible.

Test: Add benzenesulfonyl chloride and KOH.

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

✓Final answer

Hinsberg test: methylamine gives a clear solution; dimethylamine gives an insoluble oil. Hinsberg test: secondary amine gives an insoluble sulfonamide; tertiary amine does not react (dissolves in acid). Azo dye test: aniline gives an orange-red dye; ethylamine gives no dye. Bromine water test: aniline gives a white precipitate; benzylamine does not. Hinsberg test: aniline gives a clear solution; N-methylaniline gives an insoluble oil.

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:

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.
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.
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:

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

Observation & Inference:

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

C_6H_5SO_2Cl + CH_3NH_2 \xrightarrow{KOH} C_6H_5SO_2NHCH_3 \text{ (soluble)}

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

C_6H_5SO_2Cl + (CH_3)_2NH \xrightarrow{KOH} C_6H_5SO_2N(CH_3)_2 \text{ (precipitate)}

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

(ii) Secondary vs. Tertiary Amines

Method: Hinsberg Test (same as above)

Steps:

Treat each amine with $C_6H_5SO_2Cl$ and aqueous KOH.
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:

Dissolve each compound in dilute HCl and cool to 0–5°C.
Add a cold solution of sodium nitrite ( $NaNO_2$ ) to form a diazonium salt.
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.

C_6H_5NH_2 \xrightarrow{NaNO_2/HCl} C_6H_5N_2^+Cl^- \xrightarrow{\beta\text{-naphthol}} \text{Orange-red 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:

Take each compound in a separate test tube.
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.

C_6H_5NH_2 + 3Br_2 \rightarrow 2,4,6\text{-}Br_3C_6H_2NH_2 \downarrow + 3HBr

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:

Treat each compound with $C_6H_5SO_2Cl$ and aqueous KOH.
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

Pair	Test	Observation for First	Observation for Second
(i) Methylamine vs. Dimethylamine	Hinsberg	Clear solution	Precipitate
(ii) Secondary vs. Tertiary	Hinsberg	Precipitate	No reaction
(iii) Ethylamine vs. Aniline	Azo dye test	No dye	Orange-red dye
(iv) Aniline vs. Benzylamine	Bromine water	White precipitate	No precipitate
(v) Aniline vs. N-Methylaniline	Hinsberg	Clear solution	Precipitate

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 ( $C_6H_5SO_2Cl$ $C_{6} H_{5} S O_{2} C l$ + 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

Pair	Best Test	Key Observation
Methylamine vs. Dimethylamine	Hinsberg test	1° → clear solution; 2° → insoluble sulfonamide
2° vs. 3° amine	Hinsberg test	2° → insoluble sulfonamide; 3° → no reaction
Ethylamine vs. Aniline	Azo dye test	Aniline → orange dye; Ethylamine → no dye
Aniline vs. Benzylamine	Bromine water	Aniline → white precipitate; Benzylamine → none
Aniline vs. N-Methylaniline	Hinsberg test	Aniline → soluble; N-Methylaniline → insoluble

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.

←Previous question9.1 — Write IUPAC names of the following compounds and classify them into pr…→Next question9.3 — Account for the following: (i) pKb of aniline is more than that of met…

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

1. Methylamine and dimethylamine

2. Secondary and tertiary amines

3. Ethylamine and aniline

4. Aniline and benzylamine

5. Aniline and N-methylaniline

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