Hello, aspiring chemists! Today, we're diving deep into the fascinating world of amines – those wonderful organic compounds containing nitrogen. We'll explore two super important aspects: their basicity and how we can cleverly separate different types of amines. This topic is a cornerstone for both your board exams and the challenging JEE, so let's build a strong foundation, starting from the very basics!

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### Understanding Amines: The Nitrogen Connection

First things first, what exactly are amines? Imagine the simplest nitrogen-containing inorganic compound you know: Ammonia (NH₃). It has one nitrogen atom bonded to three hydrogen atoms, and crucially, it has a lone pair of electrons on that nitrogen.

Now, if you replace one or more of these hydrogen atoms with alkyl groups (like -CH₃, -C₂H₅) or aryl groups (like -C₆H₅, a benzene ring), you get an amine! So, amines are essentially organic derivatives of ammonia.

We classify amines into three types based on how many hydrogen atoms on the nitrogen have been replaced by organic groups:

1. Primary Amines (1° Amines): Here, only one hydrogen atom of ammonia is replaced by an alkyl or aryl group.
* Example: Methylamine (CH₃NH₂), Aniline (C₆H₅NH₂)
* Think of it as having one "R" group and two "H" atoms on the nitrogen. (R-NH₂)

2. Secondary Amines (2° Amines): In this case, two hydrogen atoms of ammonia are replaced by alkyl or aryl groups. These two groups can be the same or different.
* Example: Dimethylamine ((CH₃)₂NH), N-Methylaniline (C₆H₅NHCH₃)
* It's like having two "R" groups and one "H" atom on the nitrogen. (R₂NH)

3. Tertiary Amines (3° Amines): All three hydrogen atoms of ammonia are replaced by alkyl or aryl groups. Again, these groups can be the same or different.
* Example: Trimethylamine ((CH₃)₃N), N,N-Dimethylaniline (C₆H₅N(CH₃)₂)
* Here, you have three "R" groups and no "H" atoms on the nitrogen. (R₃N)

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### Basicity of Amines: The Power of the Lone Pair

Why are amines considered bases? It all boils down to that lone pair of electrons on the nitrogen atom!

Remember your Brønsted-Lowry definition of a base? A base is a proton acceptor. Or, if you recall Lewis theory, a base is an electron pair donor. Nitrogen in amines has a readily available lone pair, making it an excellent candidate for both!

When an amine acts as a base, it accepts a proton (H⁺) to form an ammonium ion (or a substituted ammonium ion):

R-NH₂  +  H⁺   ⇌   R-NH₃⁺  (Primary ammonium ion)

(Amine)      (Proton)      (Conjugate Acid)

The strength of an amine's basicity depends on how readily that lone pair of electrons is available for donation or protonation. The more available the lone pair, the stronger the base.

We often compare basicity using pKb values.
* A smaller pKb value means a stronger base (and a larger Kb value).
* A larger pKb value means a weaker base (and a smaller Kb value).

#### Factors Affecting Basicity

Several factors influence the availability of the lone pair on nitrogen:

1. Inductive Effect (+I Effect):
* Alkyl groups (like -CH₃, -C₂H₅) are known as electron-donating groups because they push electron density through sigma bonds. This is called the positive inductive effect (+I effect).
* When an alkyl group is attached to the nitrogen atom of an amine, it pushes electron density towards the nitrogen. This makes the lone pair on the nitrogen more concentrated and thus, more available for donation or protonation.
* Analogy: Imagine the nitrogen's lone pair is a small lamp. Alkyl groups are like little power boosters, making the lamp shine brighter (i.e., making the lone pair more available).
* Impact: More alkyl groups generally mean a stronger +I effect, leading to increased basicity.

2. Steric Hindrance:
* "Steric hindrance" simply means bulkiness. If the groups attached to nitrogen are very large, they can physically block the approach of a proton (H⁺) to the lone pair.
* Analogy: If your lamp is in a crowded room with big furniture all around it, it's harder for someone to reach it and turn it on, even if it's shining brightly.
* Impact: Steric hindrance can reduce basicity, especially in crowded tertiary amines.

3. Solvation Effects (in Aqueous Solution):
* When an amine accepts a proton, it forms a positively charged ammonium ion (R-NH₃⁺). This ion can be stabilized by hydrogen bonding with water molecules.
* Water molecules, being polar, surround and "solvate" the positive charge, distributing it and making the ion more stable.
* The more stable the conjugate acid (the ammonium ion), the stronger the parent base.
* The extent of hydrogen bonding depends on the number of H atoms attached to the positively charged nitrogen.
* Primary ammonium ions (R-NH₃⁺) can form three H-bonds.
* Secondary ammonium ions (R₂NH₂⁺) can form two H-bonds.
* Tertiary ammonium ions (R₃NH⁺) can form only one H-bond.
* Impact: Greater solvation stabilizes the conjugate acid more, increasing basicity. This effect is strongest for 1° amines and weakest for 3° amines.

4. Resonance Effect (-R Effect):
* If the lone pair on nitrogen can participate in resonance with an adjacent pi system (like a benzene ring), its availability for protonation decreases significantly.
* Example: In aniline (C₆H₅NH₂), the lone pair on the nitrogen is delocalized into the benzene ring. This means it's not "sitting" exclusively on the nitrogen, making it less available.
* Analogy: Imagine you have a toy (the lone pair) that you usually keep in your pocket (on the nitrogen). If you start sharing it with your friends (the benzene ring), it's less available for you to give away instantly to a new person.
* Impact: Resonance dramatically *decreases* basicity.

#### Comparing Basicity: Different Scenarios

Let's put these factors together and compare the basicity of various amines:

##### A. Alkyl Amines vs. Ammonia (Gas Phase vs. Aqueous Phase)

* In the Gas Phase: Only the inductive effect matters.
* More alkyl groups = stronger +I effect = more available lone pair.
* So, the order is: 3° amine > 2° amine > 1° amine > Ammonia (NH₃).
* Example: (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃

* In Aqueous Solution (The Tricky Part - JEE Important!): Here, we have a competition between the +I effect (which favors 3° > 2° > 1°) and solvation effect (which favors 1° > 2° > 3°). Steric hindrance also plays a role, especially for bulkier groups.
* For methyl amines (R = CH₃): The order observed is (CH₃)₂NH (2°) > CH₃NH₂ (1°) > (CH₃)₃N (3°) > NH₃.
* Why 2° first? It gets a good balance of +I effect (two methyl groups) and good solvation (two H-bonds possible).
* Why 1° second? It has less +I effect but excellent solvation (three H-bonds).
* Why 3° third? Although it has the strongest +I effect, its solvation is very poor (only one H-bond) and it experiences significant steric hindrance around the nitrogen, making protonation difficult.
* For ethyl amines (R = C₂H₅): The order changes slightly to (C₂H₅)₂NH (2°) > (C₂H₅)₃N (3°) > (C₂H₅)NH₂ (1°) > NH₃.
* Here, ethyl groups are bulkier than methyl groups. The increased steric hindrance in the 1° amine's conjugate acid starts to outweigh its solvation advantage compared to the 3° amine. The 2° amine still maintains the best balance.


JEE Focus: Memorizing these specific orders for methyl and ethyl amines in aqueous solution is crucial. Understand *why* the order changes – it's the subtle interplay of inductive effect, solvation, and steric hindrance.

##### B. Aryl Amines vs. Alkyl Amines/Ammonia

* Aryl amines (like aniline, C₆H₅NH₂) are significantly weaker bases than ammonia and alkyl amines.
* Reason: The lone pair of electrons on the nitrogen in aniline is delocalized into the benzene ring through resonance.

       NH₂                  NH₂⁺

       |                      |

      C₆H₅  <--->  [Resonance Structures where lone pair is in ring]

    

This delocalization makes the lone pair less available for donation to a proton.
* Comparison: A typical alkyl amine might have a pKb around 3-4, ammonia around 4.75, while aniline has a pKb around 9.4. This huge difference clearly shows aniline is a much weaker base.

##### C. Effect of Substituents on Aniline Basicity

* Electron-donating groups (EDGs) on the benzene ring (e.g., -CH₃, -OCH₃) *increase* the electron density on the nitrogen, making the lone pair more available, and thus *increase* basicity.
* Electron-withdrawing groups (EWGs) on the benzene ring (e.g., -NO₂, -Cl, -COOH) *decrease* the electron density on the nitrogen (by pulling electrons away from the ring, and subsequently from the nitrogen), making the lone pair *less* available, and thus *decrease* basicity.
* Example: p-Nitroaniline is a much weaker base than aniline because the strong electron-withdrawing -NO₂ group further delocalizes the nitrogen's lone pair and stabilizes the unprotonated amine.

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### Separation of Primary, Secondary, and Tertiary Amines: Hinsberg's Method

Imagine you have a bottle containing a mixture of 1°, 2°, and 3° amines. How do you tell them apart, let alone separate them? This is where the brilliant Hinsberg's Method comes into play!

The core idea of Hinsberg's method is to exploit the different reactivity of 1°, 2°, and 3° amines with a specific reagent, followed by treatment with a base and then an acid.

The reagent used is Benzenesulfonyl Chloride (C₆H₅SO₂Cl), often called Hinsberg's reagent.

Let's break down how each type of amine reacts:

#### 1. Primary (1°) Amines:

* Step 1: Reaction with Benzenesulfonyl Chloride (Hinsberg's Reagent)
* A 1° amine (R-NH₂) reacts with C₆H₅SO₂Cl in the presence of a base (like NaOH or pyridine) to form an N-alkylbenzenesulfonamide. This is a solid, insoluble compound.
* Key: This sulfonamide has an acidic hydrogen atom directly attached to the nitrogen.

      R-NH₂  +  C₆H₅SO₂Cl  →  C₆H₅SO₂-NH-R  +  HCl

      (1° Amine)               (N-alkylbenzenesulfonamide - insoluble solid)

    

* Step 2: Treatment with a Strong Base (e.g., KOH or NaOH)
* Since the N-alkylbenzenesulfonamide has an acidic hydrogen (the H attached to the N, which is next to the strong electron-withdrawing -SO₂- group), it reacts with the strong base (KOH or NaOH) to form a potassium or sodium salt.
* This salt is soluble in alkali (water-soluble).

      C₆H₅SO₂-NH-R  +  KOH  →  C₆H₅SO₂-N⁻K⁺-R  +  H₂O

      (Insoluble sulfonamide)    (Soluble potassium salt)

    

* Step 3: Acidification
* Upon acidification of the soluble salt, the original insoluble N-alkylbenzenesulfonamide is regenerated.

      C₆H₅SO₂-N⁻K⁺-R  +  HCl  →  C₆H₅SO₂-NH-R  +  KCl

      (Soluble salt)           (Insoluble sulfonamide regenerated)

    

#### 2. Secondary (2°) Amines:

* Step 1: Reaction with Benzenesulfonyl Chloride (Hinsberg's Reagent)
* A 2° amine (R₂NH) reacts with C₆H₅SO₂Cl to form an N,N-dialkylbenzenesulfonamide. This is also a solid, insoluble compound.
* Key: This sulfonamide does NOT have any acidic hydrogen atom attached to the nitrogen (because both hydrogens have been replaced by alkyl groups).

      R₂NH  +  C₆H₅SO₂Cl  →  C₆H₅SO₂-N(R)₂  +  HCl

      (2° Amine)               (N,N-dialkylbenzenesulfonamide - insoluble solid)

    

* Step 2: Treatment with a Strong Base (e.g., KOH or NaOH)
* Because there is no acidic hydrogen on the nitrogen, the N,N-dialkylbenzenesulfonamide does NOT react with KOH/NaOH and thus remains insoluble in the alkaline solution.

#### 3. Tertiary (3°) Amines:

* Step 1: Reaction with Benzenesulfonyl Chloride (Hinsberg's Reagent)
* A 3° amine (R₃N) does NOT have any hydrogen atom attached to the nitrogen. Therefore, it does NOT react with benzenesulfonyl chloride.
* It remains as the original 3° amine, which is typically insoluble in water but soluble in acid.

* Step 2: Treatment with a Strong Base (e.g., KOH or NaOH)
* Since it didn't react with the Hinsberg's reagent, the 3° amine will remain in the organic layer. When you add base, the 3° amine is also insoluble in the alkaline solution.

#### Summary of Hinsberg's Test:

Amine Type	Reaction with C₆H₅SO₂Cl	Behavior in KOH/NaOH Solution	Behavior on Acidification
Primary (1°) Amine	Forms N-alkylbenzenesulfonamide (insoluble solid)	Dissolves to form a clear solution (due to acidic H)	Insoluble sulfonamide precipitates out
Secondary (2°) Amine	Forms N,N-dialkylbenzenesulfonamide (insoluble solid)	Remains insoluble (no acidic H)	No change (still insoluble)
Tertiary (3°) Amine	No reaction	Remains as an insoluble layer/precipitate (or sometimes forms a soluble salt if the amine is significantly basic, but this is the primary distinction for Hinsberg)	Dissolves to form a soluble salt (as it's a base)

JEE Focus: The key distinction in Hinsberg's method is the solubility of the sulfonamide product in aqueous alkali. This relies entirely on the presence or absence of an acidic hydrogen on the nitrogen of the sulfonamide. Also, understand that the initial amine starting material, if unreacted (like 3° amine), will be insoluble in the basic solution.

#### Practical Separation Steps using Hinsberg's Method:

Let's say you have a mixture of a 1°, 2°, and 3° amine. Here's how you'd separate them:

1. Add Hinsberg's Reagent (C₆H₅SO₂Cl) and Excess KOH/NaOH:
* The 1° amine will react to form an N-alkylbenzenesulfonamide, which then immediately reacts with the excess KOH to form a soluble potassium salt. This will be in the aqueous layer.
* The 2° amine will react to form an N,N-dialkylbenzenesulfonamide, which is insoluble in KOH and will be a solid precipitate.
* The 3° amine will not react with Hinsberg's reagent and will remain as the unreacted amine, which will typically form a separate organic layer (or be suspended if it's a solid).

2. Separate the Layers/Solids:
* You can filter off the solid N,N-dialkylbenzenesulfonamide (from 2° amine).
* Then, separate the remaining aqueous layer (containing the 1° amine salt) from the organic layer (containing the 3° amine).

3. Recover the Amines:
* From the filtered solid (2° amine product): Treat the N,N-dialkylbenzenesulfonamide with strong acid (e.g., conc. HCl) and then neutralize with a base to regenerate the original 2° amine.
* From the aqueous layer (1° amine salt): Acidify this solution (e.g., with HCl) to regenerate the insoluble N-alkylbenzenesulfonamide. Filter this solid, and then hydrolyze it (e.g., by boiling with concentrated HCl or NaOH) to get back the original 1° amine.
* From the organic layer (3° amine): Acidify the organic layer (e.g., with HCl). The 3° amine, being basic, will react to form a water-soluble ammonium salt. Separate the aqueous acidic layer, then make it alkaline with NaOH to regenerate the original 3° amine, which can then be extracted.

This methodical approach allows chemists to effectively distinguish and separate these closely related organic compounds! Understanding the reactivity differences is key, and Hinsberg's method is a classic example of using chemical properties for separation.

Alright, aspiring chemists, buckle up! We're about to embark on a deep dive into the fascinating world of amines – specifically, their basicity and the clever ways we can distinguish and separate primary, secondary, and tertiary amines. This topic is absolutely crucial for JEE, as it tests your fundamental understanding of organic reactivity and practical application.

Let's begin!

## Amines: Basicity and the Dance of Electrons

Amines are organic compounds derived from ammonia (NH3) where one or more hydrogen atoms are replaced by alkyl or aryl groups. The defining characteristic of an amine is the presence of a lone pair of electrons on the nitrogen atom. This lone pair is the key to their basicity.

### What is Basicity? The Proton Acceptor Role

Recall your general chemistry basics:
* A Brønsted-Lowry base is a proton (H+) acceptor.
* A Lewis base is an electron pair donor.

Amines fit both definitions perfectly! The lone pair on the nitrogen atom can readily accept a proton from an acid, or donate itself to an electron-deficient species. The stronger the amine's ability to donate its lone pair (or accept a proton), the stronger its basicity.

We quantify basicity using the base dissociation constant (Kb) or its negative logarithm, pKb.
* Larger Kb value (meaning the equilibrium lies more towards the formation of the conjugate acid) indicates a stronger base.
* Smaller pKb value (since pKb = -log Kb) indicates a stronger base.

Consider the general reaction of an amine with water:
(Imagine a reaction image here: R-NH2 + H2O <=> R-NH3+ + OH-)

$K_b = frac{[R-NH_3^+][OH^-]}{[R-NH_2]}$

### Factors Influencing the Basicity of Amines

The availability of the lone pair on the nitrogen atom determines an amine's basicity. Several factors influence this availability:

#### 1. Inductive Effect (+I Effect)

Alkyl groups (like methyl, ethyl, etc.) are electron-donating groups (+I effect). When alkyl groups are attached to the nitrogen atom:
* They push electron density towards the nitrogen.
* This increases the electron density on the nitrogen, making the lone pair more available for donation.
* The more alkyl groups, the stronger this effect.

In the Gas Phase:
In the absence of a solvent, only the inductive effect plays a significant role.
* Tertiary amines (R3N) have three alkyl groups, providing the maximum +I effect.
* Secondary amines (R2NH) have two alkyl groups.
* Primary amines (RNH2) have one alkyl group.
* Ammonia (NH3) has no alkyl groups.

Therefore, the order of basicity in the gas phase is:
Tertiary Amine > Secondary Amine > Primary Amine > Ammonia
(e.g., (CH3)3N > (CH3)2NH > CH3NH2 > NH3)

#### 2. Solvation Effect (Aqueous Solution Basicity)

This is where things get interesting and slightly counter-intuitive, especially for JEE. In aqueous solutions, the stability of the conjugate acid (the ammonium ion, R-NH3+) formed after protonation, becomes crucial. This stability is primarily due to hydrogen bonding with water molecules.

* When an amine accepts a proton, it forms a positively charged ammonium ion.
* This positive charge can be stabilized by surrounding water molecules through hydrogen bonding with the N-H protons.
* Greater stabilization of the conjugate acid means a more favorable equilibrium for protonation, hence a stronger base.

Let's look at the hydrogen bonding possibilities for different ammonium ions:
* Primary ammonium ion (R-NH3+): Has three N-H bonds available for hydrogen bonding. It is extensively solvated and thus highly stabilized.
* Secondary ammonium ion (R2NH2+): Has two N-H bonds available for hydrogen bonding. Moderately solvated.
* Tertiary ammonium ion (R3NH+): Has only one N-H bond available for hydrogen bonding. Least solvated among the three.

The Combined Effect in Aqueous Solution:
The basicity in aqueous solution is a complex interplay of:
1. Inductive effect: Favors tertiary > secondary > primary.
2. Solvation effect: Favors primary > secondary > tertiary (due to maximum H-bonding).
3. Steric hindrance: Bulky alkyl groups can hinder the approach of the proton and the water molecules for solvation, slightly reducing basicity, especially for tertiary amines.

JEE Focus: This combined effect leads to a different basicity order in aqueous solution compared to the gas phase! You must know this distinction.

For methyl-substituted amines (R = CH3):
The order of basicity is:
Secondary ((CH3)2NH) > Primary (CH3NH2) > Tertiary ((CH3)3N) > Ammonia (NH3)
(pKb values: ~3.27 for secondary, ~3.36 for primary, ~4.2 for tertiary, ~4.75 for ammonia)

For ethyl-substituted amines (R = C2H5):
The order of basicity is slightly different due to bulkier ethyl groups and a slightly greater inductive effect:
Secondary ((C2H5)2NH) > Tertiary ((C2H5)3N) > Primary (C2H5NH2) > Ammonia (NH3)
(pKb values: ~3.06 for secondary, ~3.25 for tertiary, ~3.35 for primary, ~4.75 for ammonia)

Why the difference between methyl and ethyl? Ethyl groups are bulkier and have a stronger +I effect. For ethyl amines, the increased +I effect of the third ethyl group in tertiary amine slightly outweighs the reduced solvation, placing tertiary *before* primary, but still *after* secondary (which benefits maximally from both +I and good solvation).

#### 3. Resonance Effect (-R Effect)

This effect dramatically impacts the basicity of aromatic amines (e.g., aniline).
* In aniline (C6H5-NH2), the lone pair of electrons on the nitrogen atom is in conjugation with the $pi$-electron system of the benzene ring.
* This lone pair participates in resonance, delocalizing into the ring.

(Imagine resonance structures of aniline showing lone pair delocalization into the ring)

Due to resonance:
* The lone pair is less available on the nitrogen atom for protonation.
* The nitrogen atom acquires a partial positive charge in some resonance structures, making it even less inclined to accept another positive charge (a proton).

Consequently, aniline is a much weaker base than aliphatic amines and even ammonia.
(Aniline pKb ~ 9.38, Ammonia pKb ~ 4.75).

Effect of Substituents on Aromatic Amine Basicity:
* Electron-Donating Groups (EDGs) like -CH3, -OCH3, -NH2, -OH (via +I or +R effects) on the benzene ring will increase the electron density on the nitrogen, making the lone pair more available, and thus increase basicity. This effect is stronger at ortho and para positions than meta.
* Electron-Withdrawing Groups (EWGs) like -NO2, -CN, -COOH, -SO3H, halogens (via -I or -R effects) on the benzene ring will decrease the electron density on the nitrogen (by pulling electrons from the ring and further from the nitrogen), making the lone pair less available, and thus decrease basicity. This effect is stronger at ortho and para positions than meta.

Example: Compare basicity of Aniline, p-Nitroaniline, and p-Methoxyaniline.
* Aniline: Baseline.
* p-Nitroaniline: -NO2 is a strong EWG (-M and -I). It pulls electron density away from the nitrogen, severely decreasing its basicity. (pKb ~ 13)
* p-Methoxyaniline: -OCH3 is an EDG (+M effect, despite -I). Its +M effect dominates, pushing electron density towards the ring and thus towards nitrogen, increasing basicity relative to aniline. (pKb ~ 8.7)

Anomalous Ortho Effect: Sometimes, substituents at the ortho position can cause steric hindrance or other complex interactions that can either increase or decrease basicity in unexpected ways, making ortho-substituted anilines often weaker bases than their meta or para counterparts, irrespective of the electronic nature of the substituent. For JEE, focus on the electronic effects predominantly unless explicitly mentioned.

#### 4. Hybridization Effect

The hybridization state of the nitrogen atom also affects basicity.
* sp3 hybridized nitrogen (e.g., in amines): The lone pair is in an sp3 orbital.
* sp2 hybridized nitrogen (e.g., in imines, pyridine): The lone pair is in an sp2 orbital.
* sp hybridized nitrogen (e.g., in nitriles): The lone pair is in an sp orbital.

As the s-character in the hybrid orbital increases (sp3 < sp2 < sp):
* The orbital becomes smaller and closer to the nucleus.
* The electrons in that orbital are held more tightly by the nucleus.
* This makes the lone pair less available for donation.

Therefore, the order of basicity is:
sp3 N (Amines) > sp2 N (Imines/Pyridine) > sp N (Nitriles)
(Example: CH3CH2NH2 > CH3CH=NH > CH3C≡N (very weakly basic))

### Summary of Basicity Factors:
| Factor | Effect | Example/Order |
| :---------------- | :---------------------------------------------------------------------- | :------------------------------------------------------------------------------------------------------------ |
| Inductive (+I) | Electron donation to N, increases lone pair availability | Gas Phase: 3° > 2° > 1° > NH3 |
| Solvation | Stabilization of conjugate acid by H-bonding with water | Aqueous Phase (Methyl): 2° > 1° > 3° > NH3
Aqueous Phase (Ethyl): 2° > 3° > 1° > NH3 |
| Resonance (-R)| Delocalization of lone pair into ring/conjugated system, decreases availability | Aromatic Amines (Aniline) << Aliphatic Amines
EWGs decrease basicity, EDGs increase basicity (on ring) |
| Hybridization | Higher s-character holds lone pair tighter | sp3 N > sp2 N > sp N |

## Separation of Primary, Secondary, and Tertiary Amines: The Hinsberg Method

Now that we understand the chemical nature of amines, let's explore a classical method used to separate a mixture of primary, secondary, and tertiary amines – the Hinsberg Test. This method exploits the differential reactivity of these amines with benzene sulfonyl chloride (C6H5SO2Cl).

### Hinsberg's Reagent: Benzene Sulfonyl Chloride (C6H5SO2Cl)

This reagent has an acidic chlorine atom attached to a sulfonyl group. The reaction is a type of nucleophilic substitution at the sulfur atom, where the amine acts as the nucleophile.

(Imagine structure of Benzene Sulfonyl Chloride)

### The Reactions:

Let's examine how each type of amine reacts:

#### 1. Primary Amine (R-NH2)

* Reaction with Hinsberg's Reagent: A primary amine reacts with benzene sulfonyl chloride in the presence of a strong base (like KOH or NaOH) to form an N-alkylbenzenesulfonamide. This reaction involves the loss of a proton and a chloride ion.

R-NH2 + C6H5SO2Cl $xrightarrow{KOH}$ C6H5SO2-NH-R + HCl
(The HCl produced is neutralized by KOH)

* Behavior in NaOH Solution: The resulting N-alkylbenzenesulfonamide has a hydrogen atom directly attached to the nitrogen. This hydrogen is acidic because it is bonded to a highly electronegative nitrogen which is further bonded to a strong electron-withdrawing sulfonyl group.
Therefore, this sulfonamide reacts with the strong base (NaOH or KOH) to form a water-soluble salt.

C6H5SO2-NH-R + NaOH $
ightarrow$ C6H5SO2-N(-)R Na(+) (Water Soluble) + H2O

Key Observation: Primary amines give a clear solution with Hinsberg's reagent in NaOH, which on acidification (with dil. HCl) regenerates the insoluble sulfonamide.

#### 2. Secondary Amine (R2NH)

* Reaction with Hinsberg's Reagent: A secondary amine reacts with benzene sulfonyl chloride to form an N,N-dialkylbenzenesulfonamide.

R2NH + C6H5SO2Cl $xrightarrow{KOH}$ C6H5SO2-N-R2 + HCl
(The HCl produced is neutralized by KOH)

* Behavior in NaOH Solution: The resulting N,N-dialkylbenzenesulfonamide has no hydrogen atom attached to the nitrogen. Thus, it is not acidic and does not react with NaOH. It remains as an insoluble precipitate or oily layer in the alkaline solution.

C6H5SO2-N-R2 + NaOH $
ightarrow$ No reaction (Insoluble)

Key Observation: Secondary amines give an insoluble precipitate or oily layer with Hinsberg's reagent in NaOH, which remains insoluble even after acidification.

#### 3. Tertiary Amine (R3N)

* Reaction with Hinsberg's Reagent: A tertiary amine has no hydrogen atoms on the nitrogen. Therefore, it does not react with benzene sulfonyl chloride under these conditions.

R3N + C6H5SO2Cl $
ightarrow$ No reaction

* Behavior in NaOH Solution: The tertiary amine itself is basic. If the amine is soluble in the aqueous medium, it will remain dissolved. When the reaction mixture is then made acidic (e.g., with HCl), the tertiary amine, being basic, will react to form a water-soluble ammonium salt.

R3N + HCl $
ightarrow$ R3NH(+)Cl(-) (Water Soluble)

Key Observation: Tertiary amines do not react with Hinsberg's reagent. They remain unaffected and dissolve in dilute acid (like HCl) as their ammonium salt.

### Step-by-Step Separation Procedure:

Imagine you have a mixture of primary, secondary, and tertiary amines. Here’s how you would separate them:

1. Step 1: Reaction with Hinsberg's Reagent + NaOH.
* Add benzene sulfonyl chloride to the mixture of amines.
* Then, add excess aqueous NaOH solution and shake vigorously.
* Observation:
* The primary amine forms a water-soluble sulfonamide salt (a clear solution).
* The secondary amine forms an insoluble N,N-dialkylbenzenesulfonamide (a precipitate or oily layer).
* The tertiary amine remains unreacted and, if water-soluble, stays in the aqueous layer; if not, it forms a separate organic layer.

2. Step 2: Separation of Layers.
* Filter the mixture or separate using a separatory funnel.
* The insoluble precipitate (or oily layer) collected is the secondary amine derivative.

3. Step 3: Acidification of the Aqueous Layer.
* Take the aqueous layer (which contains the primary amine salt and potentially the unreacted tertiary amine) and acidify it with dilute HCl.
* Observation:
* The water-soluble salt of the primary amine derivative will decompose upon acidification to regenerate the original insoluble N-alkylbenzenesulfonamide. This will precipitate out.
* If any tertiary amine was present in the aqueous layer, it will react with HCl to form a water-soluble ammonium salt.

4. Step 4: Separation and Recovery.
* Filter the acidified solution. The insoluble precipitate obtained is the primary amine derivative.
* The tertiary amine (as its soluble ammonium salt) remains in the aqueous filtrate.
* To recover the primary amine, treat its insoluble sulfonamide derivative with conc. HCl and heat, followed by basification.
* To recover the tertiary amine, make the aqueous filtrate (containing the tertiary ammonium salt) strongly basic with NaOH, which converts the salt back to the free tertiary amine. The tertiary amine can then be extracted with an organic solvent (like ether) and distilled.

Amine Type	Reaction with C6H5SO2Cl in NaOH	Observation in NaOH	Action on Acidification of Aqueous Layer	Final State (after separation/recovery)
Primary (RNH2)	Forms C6H5SO2NHR, which then reacts with NaOH to form a soluble salt.	Clear Solution	Salt converts back to insoluble C6H5SO2NHR.	Recovered as insoluble sulfonamide, then hydrolyze to get RNH2.
Secondary (R2NH)	Forms C6H5SO2NR2, which has no acidic H.	Insoluble Precipitate/Oily Layer	No change (remains insoluble).	Recovered as insoluble sulfonamide.
Tertiary (R3N)	No reaction with C6H5SO2Cl.	Dissolves (if water soluble) or forms separate layer	Reacts with acid to form soluble R3NH+Cl- salt.	Recovered as soluble ammonium salt, then free R3N by basification.

JEE Advanced Tip: While benzene sulfonyl chloride is the classical reagent, sometimes *p*-toluenesulfonyl chloride is also used. The principle remains the same. The Hinsberg test is a cornerstone for distinguishing and separating amines, so understanding the reactivity differences is vital.

This detailed explanation should equip you with a strong conceptual understanding of amine basicity and the practical application of the Hinsberg method for their separation. Keep practicing with examples and you'll master this!

This section provides effective mnemonics and shortcuts to quickly recall the basicity order of amines and the mechanism of their separation, particularly focusing on the Hinsberg test.

### Mnemonics & Shortcuts for Amines: Basicity and Separation

#### 1. Basicity of Amines

Amines are basic due to the lone pair of electrons on the nitrogen atom. Electron-donating groups (+I effect) increase basicity, while electron-withdrawing groups (-I effect) decrease basicity. The order of basicity depends on the phase (gas or aqueous).

* General Basicity Rule:
* Mnemonic: "EDG = Enhanced Basicity". (Electron Donating Groups increase Basicity).

* Gas Phase Basicity: In the gas phase, only the inductive effect (+I effect) of alkyl groups plays a role. More alkyl groups mean stronger +I effect, higher electron density on nitrogen, and thus higher basicity.
* Mnemonic: "3-2-1 Go!"
* This means: 3° amine > 2° amine > 1° amine > NH₃. (Simply count the alkyl groups, higher count = higher basicity in gas phase).

* Aqueous Phase Basicity: In aqueous solution, three factors influence basicity:
1. +I Effect: Electron-donating alkyl groups increase basicity.
2. Solvation Effect: Stability of the conjugate acid (alkylammonium ion) by hydrogen bonding with water molecules. More N-H bonds allow for greater solvation (1° > 2° > 3°).
3. Steric Hindrance: Bulky alkyl groups hinder solvation and protonation of the amine.
* Mnemonic for Alkyl Amines (Aqueous Phase):
* For Methyl Amines (R = CH₃): Think "M213"
* Meaning: 2° > 1° > 3° > NH₃
* For Ethyl Amines (R = C₂H₅): Think "E231"
* Meaning: 2° > 3° > 1° > NH₃
* Why different? For methyl groups, solvation effect is significant, balancing the +I effect effectively. For ethyl groups, the +I effect is stronger, pushing the 3° amine's basicity higher than 1° due to better electron donation, despite increased steric hindrance compared to methyl. However, 2° remains the strongest due to an optimal balance of +I and solvation.

#### 2. Separation of Primary, Secondary, and Tertiary Amines (Hinsberg Test)

The Hinsberg test uses Benzene Sulfonyl Chloride (PhSO₂Cl) as the reagent to distinguish and separate 1°, 2°, and 3° amines based on their reactivity and the solubility of their products in alkali.

* Reagent Mnemonic: "Hinsberg uses BS (Benzene Sulfonyl) Chloride."

* Reactions & Products:
* 1° Amine (R-NH₂):
* Reacts with PhSO₂Cl to form an N-alkylbenzenesulfonamide.
* Key Feature: This product has an acidic hydrogen attached to the nitrogen atom (R-NH-SO₂Ph).
* Mnemonic: "1° SAS" (One-Acidic-Soluble).
* Meaning: The product from 1° amine is Acidic and therefore Soluble in aqueous KOH/NaOH (forms a salt).
* 2° Amine (R₂NH):
* Reacts with PhSO₂Cl to form an N,N-dialkylbenzenesulfonamide.
* Key Feature: This product lacks an acidic hydrogen on the nitrogen atom (R₂N-SO₂Ph).
* Mnemonic: "2° INS" (Two-Insoluble).
* Meaning: The product from 2° amine is Insoluble in aqueous KOH/NaOH.
* 3° Amine (R₃N):
* Does not react with PhSO₂Cl (it has no hydrogen attached to nitrogen).
* Key Feature: It remains as the unreacted 3° amine, which is basic.
* Mnemonic: "3° NRR" (Three-No Reaction-Recovered).
* Meaning: The 3° amine shows No Reaction and can be Recovered by acid extraction (as it is basic, it dissolves in dilute HCl).

Amine Type	Hinsberg Reagent (PhSO₂Cl)	Product in NaOH/KOH	Outcome / Mnemonic
Primary (1°)	Reacts	Soluble (acidic H)	1° SAS (One-Acidic-Soluble)
Secondary (2°)	Reacts	Insoluble (no acidic H)	2° INS (Two-Insoluble)
Tertiary (3°)	No Reaction	Insoluble (but dissolves in HCl)	3° NRR (Three-No Reaction-Recovered)

JEE Tip: While the Hinsberg test is classic, newer methods like mass spectrometry are used in labs. However, for JEE, understanding Hinsberg is crucial. Remember the specific solubility patterns to answer questions quickly.

Understanding the basicity and separation of amines intuitively involves grasping why they act as bases and how their structural differences lead to distinct reactivities.

Intuitive Understanding of Amine Basicity

Amines are organic derivatives of ammonia (NH₃), where one or more hydrogen atoms are replaced by alkyl or aryl groups. Their basic nature stems from the presence of a lone pair of electrons on the nitrogen atom, which can be donated to an electron-deficient species (Lewis base) or accept a proton (Brønsted-Lowry base).

• The Lone Pair is Key: Imagine the nitrogen's lone pair as an available "hand" ready to grab a proton (H⁺). The more available and eager this hand is, the stronger the base.


• Electron-Donating Groups (Alkyl Groups): Alkyl groups (like -CH₃, -C₂H₅) are electron-donating by the inductive effect (+I effect). They push electron density towards the nitrogen, making its lone pair more concentrated and thus more available for donation.
  
  
  
  • Primary Amine (R-NH₂): One alkyl group pushing electrons.
  
  
  • Secondary Amine (R₂-NH): Two alkyl groups pushing electrons – even more basic than primary in gas phase.
  
  
  • Tertiary Amine (R₃-N): Three alkyl groups pushing electrons – appears most basic in gas phase due to maximum +I effect.
  
  
  
  


• Aqueous vs. Gas Phase Basicity (JEE Important!): This is a common point of confusion.
  
  
  
  • Gas Phase: Basicity follows the order: Tertiary > Secondary > Primary > Ammonia. This is because only the +I effect of alkyl groups is at play.
  
  
  • Aqueous Phase: Here, two additional factors come into play:
    
    
    
    1. Solvation Effect: The stability of the conjugate acid (protonated amine, RNH₃⁺) formed after accepting a proton. This conjugate acid is stabilized by hydrogen bonding with water molecules.
       
       
       
       • Primary amine's conjugate acid (RNH₃⁺) can form 3 H-bonds.
       
       
       • Secondary amine's conjugate acid (R₂NH₂⁺) can form 2 H-bonds.
       
       
       • Tertiary amine's conjugate acid (R₃NH⁺) can form 1 H-bond.
       
       
       • More H-bonds = more stable conjugate acid = stronger base.
       
       
       
       
    
    
    2. Steric Hindrance: Bulky alkyl groups around the nitrogen can physically block the approach of the proton and solvent molecules, hindering both protonation and solvation.
    
    
    
    
  
  
  
  


• The Compromise: In aqueous solution, the actual basicity is a balance between the electron-donating effect (+I), solvation effect, and steric hindrance. This often results in secondary amines being the strongest bases, followed by primary or tertiary, depending on the size of the alkyl group.
  
  
  
  • For methylamines: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ (2° > 1° > 3°).
  
  
  • For ethylamines: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃ (2° > 3° > 1°).
  
  
  
  


• Aromatic Amines (e.g., Aniline): These are much weaker bases than aliphatic amines. The lone pair on nitrogen is delocalized (involved in resonance) with the benzene ring, making it less available for protonation. Imagine the lone pair being "busy" within the ring structure.

Intuitive Understanding of Amine Separation (Hinsberg's Test)

The separation of primary (1°), secondary (2°), and tertiary (3°) amines relies on their differing reactivities with Hinsberg's Reagent (Benzene sulfonyl chloride, C₆H₅SO₂Cl). This difference is primarily due to the number of acidic hydrogens directly attached to the nitrogen atom in the amine.

Amine Type	Reaction with Hinsberg's Reagent	Product with NaOH/KOH	Solubility in Aqueous KOH	Intuition
Primary Amine (RNH₂)	Reacts to form N-alkylbenzenesulfonamide.	Forms a water-soluble potassium salt.	Soluble	Has 2 H's on N. One reacts with PhSO₂Cl. The remaining H on N becomes highly acidic due to the electron-withdrawing -SO₂ group, so it reacts with KOH to form a salt and dissolves.
Secondary Amine (R₂NH)	Reacts to form N,N-dialkylbenzenesulfonamide.	No reaction with NaOH/KOH.	Insoluble	Has 1 H on N. This H reacts with PhSO₂Cl. The product has no H on N, so it cannot react with KOH to form a salt and remains insoluble.
Tertiary Amine (R₃N)	Does not react.	Remains unreacted amine. Dissolves in dilute acid (e.g., HCl) to form a soluble salt.	Insoluble (but reacts with acid)	Has no H's on N. No site for reaction with PhSO₂Cl. It's an unreacted base, so it can be distinguished by dissolving in dilute acid.

Practical Tip for JEE: Remember the Hinsberg test by focusing on the 'acidic H' on the nitrogen *after* the initial reaction with sulfonyl chloride. If an acidic H remains, it's soluble in base. If not, it's insoluble. If no reaction occurs, it's tertiary.

Analogies help simplify complex concepts by relating them to familiar scenarios. For amines, understanding their basicity and separation methods can be made easier with the right comparisons.

Basicity of Amines: The "Generous Giver" Analogy

Imagine the lone pair of electrons on the nitrogen atom of an amine as a valuable item (like money or a rare possession) that the nitrogen atom is willing to donate. The basicity of an amine depends on how readily this item can be donated.

• Nitrogen Atom: The person who owns the valuable item.


• Lone Pair of Electrons: The valuable item (e.g., money) that can be donated.


• Alkyl Groups (Electron-Donating Groups): These are like "hype men" or "boosters" who give the nitrogen atom more confidence and resources (electron density). More alkyl groups (up to a point in aqueous solution) make the nitrogen more electron-rich and thus more willing to donate its lone pair, increasing basicity.


• Steric Hindrance (in gas phase, or less solvating solvents): This is like too many "hype men" crowding around the nitrogen atom, making it difficult for the "receiver" (a proton, H+) to physically reach the lone pair. Even if the nitrogen is very rich in electrons, if access is blocked, donation is hindered.


• Solvation Effect (in aqueous solution): When the nitrogen donates its lone pair and becomes a positively charged ammonium ion (the conjugate acid), it needs stabilization. Solvation by water molecules is like having "bodyguards" or a "support system" that stabilizes the positively charged ion.
  
  
  
  • More H-atoms on the N-H bond (as in primary amines) mean more points for water molecules to form H-bonds, leading to greater stabilization of the conjugate acid. This makes the primary amine more willing to donate its lone pair in the first place, thus increasing its effective basicity in water.
  
  
  • Tertiary amines, having fewer H-atoms on nitrogen, experience less effective solvation, which can reduce their basicity in aqueous solution despite having more alkyl groups.
  
  
  
  

JEE Note: The interplay of inductive effect, steric hindrance, and solvation effect is crucial for understanding the anomalous basicity order (secondary > primary > tertiary > ammonia for methyl amines, and secondary > tertiary > primary > ammonia for ethyl amines) in aqueous solutions. The gas-phase basicity order is straightforward: tertiary > secondary > primary > ammonia, primarily due to the inductive effect.

Separation of Primary, Secondary, and Tertiary Amines: The "Special Coat" Analogy (Hinsberg's Method)

Imagine you have a group of three friends: Primary (P), Secondary (S), and Tertiary (T) amines. You want to identify and separate them based on how they interact with a special "tailor" and a "special liquid."

• The Amines: Three friends, P, S, and T.


• Hinsberg's Reagent (Benzenesulfonyl chloride): A "special tailor" who offers a unique "coat" (sulfonyl amide).


• Primary Amine (P): When P encounters the tailor, it readily accepts and wears the coat. This coat has a special "button" (an acidic hydrogen atom). Because of this button, when P (wearing the coat) is put into a "special liquid" (aqueous NaOH solution), it can react and become soluble.


• Secondary Amine (S): S also accepts and wears the coat from the tailor. However, S's coat lacks the special "button" (no acidic hydrogen atom). So, when S (wearing its coat) is put into the "special liquid" (aqueous NaOH), it cannot react and remains insoluble (precipitates as a solid).


• Tertiary Amine (T): T is very stubborn and refuses to wear the coat altogether. It doesn't react with the tailor. So, when T (no coat) is put into the "special liquid," it remains as an unreacted, insoluble organic liquid layer.

Separation Process (Analogous Steps):

1. You add the "tailor" (Hinsberg's reagent) to the mixture of P, S, and T. P and S get their coats, T does not react.


2. You then add the "special liquid" (aqueous NaOH).
   
   
   
   • P (with coat and button) dissolves, forming a clear solution.
   
   
   • S (with coat but no button) remains as an insoluble solid precipitate.
   
   
   • T (no coat) remains as an insoluble organic liquid layer.
   
   
   
   


3. Now, you can easily separate the dissolved P, the solid S, and the liquid T by simple physical methods (e.g., decantation, filtration, or using a separating funnel), allowing you to collect each friend individually.

CBSE vs JEE: Hinsberg's method is a classical qualitative test and separation technique for P, S, T amines, important for both CBSE boards and JEE. Understanding the underlying chemistry of acidic hydrogen and salt formation is key.

Keep practicing and relating these concepts to everyday scenarios for better retention!

Prerequisites for Amines: Basicity and Separation

To effectively grasp the concepts of basicity and separation of primary, secondary, and tertiary amines, a strong foundation in several fundamental organic chemistry principles is essential. Revisiting these topics will ensure a smooth learning curve for this advanced section, crucial for both JEE and board exams.

Key Concepts to Review:

• Nomenclature and Structure of Amines:
  
  
  
  • Ability to correctly identify and name primary (1°), secondary (2°), and tertiary (3°) amines based on the number of alkyl/aryl groups attached to the nitrogen atom.
  
  
  • Understanding the hybridization (sp³) and geometry (pyramidal) of the nitrogen atom in amines, and the presence of a lone pair of electrons.
  
  
  
  


• General Organic Acid-Base Chemistry:
  
  
  
  • Lewis Acid-Base Theory: Comprehension of amines acting as Lewis bases due to the presence of a lone pair on nitrogen, capable of donating electrons.
  
  
  • Brønsted-Lowry Acid-Base Theory: Understanding amines as Brønsted-Lowry bases, capable of accepting a proton (H⁺).
  
  
  • Acidity and Basicity Scales (pKa and pKb): Knowledge of how pK_a and pK_b values relate to the strength of an acid or a base. Remember that a stronger base has a higher K_b and a lower pK_b. For a conjugate acid, a lower pK_a means a stronger acid, implying a weaker conjugate base.
  
  
  • Factors Affecting Acid/Base Strength: General understanding of how electron-donating and electron-withdrawing groups influence acidity and basicity.
  
  
  
  


• Electronic Effects:
  
  
  
  • Inductive Effect (+I and -I): Understanding how alkyl groups exhibit a +I effect (electron-donating) and how this can increase electron density on nitrogen, potentially increasing basicity. Conversely, electron-withdrawing groups exert a -I effect.
  
  
  • Resonance Effect (Mesomeric Effect): Knowledge of how delocalization of the lone pair on nitrogen into an aromatic ring (e.g., in anilines) affects its availability for protonation, thereby influencing basicity.
  
  
  
  


• Steric Effects:
  
  
  
  • Basic awareness of how the bulkiness of groups around a reactive site can hinder access for reagents (steric hindrance). This plays a role in solvation and protonation of amines.
  
  
  
  


• Solvent Effects and Hydrogen Bonding:
  
  
  
  • Understanding how solvent molecules (especially water) can stabilize charged species (like protonated amines) through solvation and hydrogen bonding. This is crucial for explaining basicity trends in aqueous solutions versus the gas phase.
  
  
  • Recall the concept of intermolecular hydrogen bonding and its impact on physical properties.
  
  
  
  


• Chemical Equilibrium:
  
  
  
  • Fundamental understanding of chemical equilibrium, equilibrium constants (K_a, K_b), and Le Chatelier's principle.
  
  
  
  


• Basic Separation Techniques:
  
  
  
  • Familiarity with acid-base extraction principles, where compounds can be separated based on their acidic or basic properties by converting them into water-soluble salts. This is directly relevant to the separation of amine mixtures.
  
  
  
  

Having a solid grasp of these prerequisites will make the intricate details of amine basicity and their practical separation techniques much clearer and easier to master for your exams.

Related Topics: Amines Basicity & Separation

Understanding the basicity and separation techniques for primary (1°), secondary (2°), and tertiary (3°) amines is foundational and connects to several other crucial topics in organic chemistry. Mastery of these related concepts enhances your overall understanding and problem-solving abilities, especially for JEE Main and Advanced.

Here are the key related topics:

1.

General Organic Chemistry (GOC) – Acid-Base Principles

This is the most fundamental prerequisite. A strong grasp of GOC concepts is essential to explain and predict amine basicity.

• Factors Affecting Basicity: Review the influence of:
  
  
  
  • Inductive Effect (+I and -I): Electron-donating groups (alkyl groups) increase basicity by stabilizing the conjugate acid, while electron-withdrawing groups decrease it.
  
  
  • Resonance Effect (+R and -R): Delocalization of the lone pair of electrons (e.g., in anilines) reduces basicity.
  
  
  • Hybridization: Basicity decreases as the s-character of the orbital holding the lone pair increases (e.g., sp³ > sp² > sp).
  
  
  • Steric Hindrance (Solvation Effect): In aqueous solutions, the stability of the conjugate acid is also affected by solvation. Steric hindrance can reduce the number of hydrogen bonds formed, thus affecting basicity (particularly relevant for methyl amines in water).
  
  
  
  


• Lewis and Brønsted-Lowry Definitions: Amines act as Lewis bases (electron pair donors) and Brønsted-Lowry bases (proton acceptors).

2.

Aromatic Amines (Anilines) – Basicity

The basicity of aromatic amines is a direct application of GOC principles, particularly the resonance effect.

• Reduced Basicity: Understand why anilines are significantly less basic than aliphatic amines and ammonia due to the resonance delocalization of the nitrogen lone pair into the benzene ring.


• Substituent Effects: Analyze how electron-donating and electron-withdrawing groups on the benzene ring affect the basicity of anilines (e.g., nitroaniline vs. methoxyaniline).

3.

Reactions of Amines (Especially Differentiating Reactions)

Several reactions are specifically used to differentiate between 1°, 2°, and 3° amines, and thus are directly related to their separation.

• Hinsberg's Test (Reaction with Benzenesulfonyl Chloride):
  
  
  
  • This is a cornerstone for distinguishing and separating amines. Understand the different reactions of 1°, 2°, and 3° amines with benzenesulfonyl chloride and the solubility properties of the products formed.
  
  
  • 1° amine: Forms an N-alkylbenzenesulfonamide, which is acidic and soluble in KOH/NaOH.
  
  
  • 2° amine: Forms an N,N-dialkylbenzenesulfonamide, which is neutral and insoluble in KOH/NaOH.
  
  
  • 3° amine: Does not react but forms a salt with the reagent, which is soluble in acid.
  
  
  
  


• Reaction with Nitrous Acid (NaNO₂/HCl):
  
  
  
  • A classic test: 1° aliphatic amines form diazonium salts (which decompose to alcohols with N₂ evolution). 1° aromatic amines form stable arenediazonium salts.
  
  
  • 2° amines (aliphatic or aromatic) form N-nitrosamines (yellow oily compounds).
  
  
  • 3° aliphatic amines form soluble salts, while 3° aromatic amines undergo para-nitrosation (electrophilic substitution).
  
  
  
  


• Carbylamine Reaction (Isocyanide Test): A specific test for 1° primary amines (aliphatic and aromatic) producing foul-smelling isocyanides when heated with chloroform and alcoholic KOH.


• Acylation and Alkylation: Understand how amines react as nucleophiles in these reactions, which are relevant to forming derivatives and understanding reactivity.

4.

Separation Techniques

Beyond Hinsberg's method, general organic chemistry separation techniques are applicable.

• Acid-Base Extraction: Amines, being basic, form water-soluble salts with dilute acids. This property is exploited to separate them from non-basic or acidic compounds.


• Fractional Distillation: For amines with sufficiently different boiling points.

JEE Tip: Questions often combine basicity comparisons with identification tests. Be prepared to explain basicity orders and apply the results of Hinsberg's or nitrous acid tests to identify unknown amines.

Navigating the topic of amines requires a sharp understanding, as it's rife with conceptual traps designed to test your clarity. Be vigilant of the following common exam pitfalls related to basicity and separation of amines.

Common Exam Traps in Basicity of Amines

• 
  Aqueous Basicity Order vs. Gas Phase Basicity Order (JEE Specific):
  
  
  
  • Trap: Memorizing a single basicity order for aliphatic amines.
  
  
  • Reality: The order of basicity for primary (1°), secondary (2°), and tertiary (3°) aliphatic amines changes depending on the solvent.
    
    
    
    • Gas Phase: 3° > 2° > 1° > NH₃ (Purely inductive effect)
    
    
    • Aqueous Phase (e.g., Methylamines): 2° > 1° > 3° > NH₃ (Inductive effect + Solvation effect + Steric hindrance to solvation)
    
    
    • Aqueous Phase (e.g., Ethylamines): 2° > 3° > 1° > NH₃ (Slightly different due to greater steric hindrance of ethyl groups)
    
    
    
    
  
  
  • Key to avoid: Always consider both the inductive effect of alkyl groups and the stabilizing effect of hydrogen bonding with solvent molecules (solvation effect) on the conjugate acid, along with steric hindrance, especially for JEE.
  
  
  
  


• 
  Aromatic Amines Basicity (Aniline and its derivatives):
  
  
  
  • Trap: Assuming aromatic amines are as basic as aliphatic amines.
  
  
  • Reality: Aniline is significantly less basic than aliphatic amines (and ammonia) because the lone pair on nitrogen is delocalized into the benzene ring via resonance. This makes the lone pair less available for protonation.
  
  
  • Trap: Misjudging the effect of substituents on aromatic amine basicity.
  
  
  • Reality:
    
    
    
    • Electron-Donating Groups (EDG) (e.g., -OCH₃, -CH₃) increase basicity by stabilizing the conjugate acid/destabilizing the lone pair delocalization.
    
    
    • Electron-Withdrawing Groups (EWG) (e.g., -NO₂, -COOH) decrease basicity by further destabilizing the lone pair or increasing its delocalization into the ring.
    
    
    • Ortho effect: Ortho substituents, regardless of their electronic nature, often decrease the basicity of aniline due to a combination of steric and electronic factors (sometimes referred to as Steric Inhibition of Resonance or SIR, though it's more complex).
    
    
    
    
  
  
  
  


• 
  Amides vs. Amines Basicity:
  
  
  
  • Trap: Treating amides (RCONHR') as basic like amines.
  
  
  • Reality: Amides are extremely weak bases (almost neutral). The lone pair on nitrogen is strongly delocalized into the highly electronegative carbonyl oxygen via resonance, making it unavailable for protonation.
  
  
  
  


• 
  pKb vs. Kb Values:
  
  
  
  • Trap: Confusing the interpretation of pKb and Kb.
  
  
  • Reality:
    
    
    
    • A higher Kb value indicates a stronger base.
    
    
    • A lower pKb value indicates a stronger base (pKb = -log Kb).
    
    
    
    Always pay attention to whether Kb or pKb is given.
    
  
  
  
  

Common Exam Traps in Separation (Hinsberg Test)

• 
  Incorrect Definition of 1°, 2°, 3° Amines:
  
  
  
  • Trap: Defining primary, secondary, and tertiary amines based on the number of carbons attached to the alpha carbon, or simply the degree of branching.
  
  
  • Reality: The classification (1°, 2°, 3°) for amines is based on the number of alkyl or aryl groups directly attached to the nitrogen atom.
    
    
    
    • Primary (1°): One alkyl/aryl group, two H atoms (R-NH₂)
    
    
    • Secondary (2°): Two alkyl/aryl groups, one H atom (R-NH-R')
    
    
    • Tertiary (3°): Three alkyl/aryl groups, no H atoms (R-N(R')-R'')
    
    
    
    
  
  
  
  


• 
  Hinsberg Test Product Solubility (JEE & CBSE):
  
  
  
  • Trap: Misinterpreting the solubility of the sulfonamide products in aqueous KOH/NaOH.
  
  
  • Reality:
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    

    Amine Type	Reaction with Hinsberg Reagent (C₆H₅SO₂Cl)	Product Solubility in Aq. NaOH/KOH
    Primary (1°)	Forms N-alkylbenzenesulfonamide (R-NH-SO₂C₆H₅).	Soluble. The product has an acidic N-H proton, which reacts with NaOH to form a soluble salt.
    Secondary (2°)	Forms N,N-dialkylbenzenesulfonamide (R-N(R')-SO₂C₆H₅).	Insoluble. The product lacks an acidic proton on nitrogen, so it does not react with NaOH.
    Tertiary (3°)	Does not react.	Insoluble in NaOH, but soluble in dilute HCl (as it's a base and forms a salt). This distinction is crucial for differentiation.

    
    
  
  
  • Key to avoid: Remember that the acidity of the N-H proton in primary sulfonamides is due to the strong electron-withdrawing nature of the sulfonyl group. This proton is absent in secondary and tertiary derivatives.
  
  
  
  

By understanding these common traps, you can approach questions on amine basicity and separation with greater confidence and accuracy.

🔑 Key Takeaways: Amines – Basicity and Separation

Understanding the basicity of amines and their separation methods is crucial for both JEE and CBSE exams. This section summarizes the most important points you need to remember.

1. Basicity of Amines

Amines are basic due to the presence of a lone pair of electrons on the nitrogen atom, which can accept a proton (Brønsted base) or donate electrons (Lewis base).

• 
  Factors Affecting Basicity:
  
  
  
  • Inductive Effect (+I): Alkyl groups are electron-donating (+I effect). They increase electron density on nitrogen, making the lone pair more available for protonation. Thus, alkyl amines are generally stronger bases than ammonia.
  
  
  • Steric Hindrance: Bulky alkyl groups can hinder the approach of a proton, slightly decreasing basicity, especially in the gas phase for tertiary amines.
  
  
  • Solvation Effects (Aqueous Phase): Stability of the conjugate acid (ammonium ion) by hydrogen bonding with water molecules. More H-bonds mean greater stability, enhancing basicity. Primary amines form more H-bonds than secondary, and secondary more than tertiary.
  
  
  • Resonance Effect (-R): Aromatic amines (like aniline) are weaker bases than aliphatic amines because the lone pair on nitrogen is delocalized into the benzene ring through resonance, making it less available for protonation. Electron-withdrawing groups on the ring further decrease basicity, while electron-donating groups increase it.
  
  
  
  


• 
  Order of Basicity:
  
  
  
  • 
    In Gaseous Phase (based purely on +I effect and some steric hindrance):
    

    3° amine > 2° amine > 1° amine > NH₃

    
    

    Example: (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃

    
    
  
  
  • 
    In Aqueous Phase (combining +I effect, steric hindrance, and solvation):
    

    This order is critical and depends on the alkyl group.

    
    
    
    
    • For Methyl Amines (CH₃): 2° > 1° > 3° > NH₃
      (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
    
    
    • For Ethyl Amines (C₂H₅): 2° > 3° > 1° > NH₃
      (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃
    
    
    
    

    JEE Tip: Remember these specific orders for methyl and ethyl amines. If the alkyl group isn't specified, assume ethyl or methyl and apply the respective order.

    
    
  
  
  • 
    Aromatic vs. Aliphatic Amines: Aliphatic amines are significantly stronger bases than aromatic amines.
    

    Aliphatic amine > NH₃ > Aromatic amine

    
    
  
  
  
  

2. Separation of Primary, Secondary, and Tertiary Amines (Hinsberg Test)

The Hinsberg Test is a classic method to distinguish and separate 1°, 2°, and 3° amines using Benzene Sulphonyl Chloride (C₆H₅SO₂Cl) as the reagent.

Amine Type	Reaction with C₆H₅SO₂Cl	Product Nature	Solubility in NaOH	Observation & Separation
Primary (1°) Amine (RNH₂)	Reacts to form N-alkylbenzenesulfonamide.	Acidic (H attached to N)	Soluble (forms a salt with NaOH)	Clear solution in NaOH. Upon acidification, a solid precipitates.
Secondary (2°) Amine (R₂NH)	Reacts to form N,N-dialkylbenzenesulfonamide.	Neutral (no H attached to N)	Insoluble	Insoluble solid/oil layer in NaOH. Remains insoluble upon acidification.
Tertiary (3°) Amine (R₃N)	No reaction (no H attached to N to be replaced by sulfonyl group).	Basic amine itself.	Insoluble in NaOH, but soluble in dilute HCl (forms a salt).	Remains as an insoluble layer in NaOH. Dissolves in dilute HCl.

JEE & CBSE Focus: Be able to identify the products of the Hinsberg reaction for each amine type and understand the solubility differences that enable separation. This is a frequently asked concept.

Remember these key points for quick revision and to tackle related problems effectively! Keep practicing!

Problem-Solving Approach: Basicity and Separation of Amines

Understanding the basicity and separation methods for primary (1°), secondary (2°), and tertiary (3°) amines is crucial for both theoretical questions and practical organic chemistry problems in JEE and board exams. A systematic approach is key.

1. Approach for Comparing Basicity of Amines

Amines are basic due to the presence of a lone pair of electrons on the nitrogen atom, which can accept a proton. The greater the availability of this lone pair, the stronger the base.

1. Identify the Type of Amine: Determine if the amine is aliphatic or aromatic, and whether it's 1°, 2°, or 3°.


2. Analyze Electronic Effects:
   
   
   
   • 
     Inductive Effect (+I): Alkyl groups are electron-donating (+I effect). They increase electron density on the nitrogen, making the lone pair more available for protonation.
     
     
     
     • In the gas phase, the order of basicity for aliphatic amines is generally 3° > 2° > 1° > NH₃ due to the cumulative +I effect.
     
     
     
     
   
   
   • 
     Resonance Effect (-R): Aromatic amines (e.g., aniline) are weaker bases than aliphatic amines because the lone pair on nitrogen is delocalized into the benzene ring through resonance, making it less available.
     
     
     
     • Substituents on the aromatic ring significantly influence basicity:
       
       
       
       • Electron-Donating Groups (EDGs) like -OCH₃, -CH₃ (+M or +I) increase electron density on nitrogen, enhancing basicity.
       
       
       • Electron-Withdrawing Groups (EWGs) like -NO₂, -COOH, -X (-M or -I) decrease electron density on nitrogen, reducing basicity.
       
       
       
       
     
     
     
     
   
   
   
   


3. Consider Solvation Effects (Aqueous Phase):
   
   
   
   • In aqueous solution, the basicity order of aliphatic amines is influenced by the stability of the conjugate acid (ammonium ion) formed. This stability depends on the extent of solvation (hydrogen bonding with water molecules).
     
   • The conjugate acid of 1° amines forms more H-bonds than 2° or 3° amines (1° > 2° > 3°).
     
   • JEE Tip: For aliphatic amines in aqueous solution, the basicity order is typically 2° > 1° > 3° for methylamines and 2° > 3° > 1° for ethylamines. This is a crucial distinction from gas-phase basicity and a common JEE question. The exact order depends on the balance between +I effect, steric hindrance, and solvation.
     
   
   
   
   


4. Steric Hindrance: Large alkyl groups can sterically hinder the approach of a proton to the nitrogen, slightly reducing basicity, especially in 3° amines in aqueous solution.


5. Compare pKb values: If given, a lower pKb value indicates a stronger base.

General Rule for Comparison: Resonance effects are usually more dominant than inductive effects. Solvation effects are critical in aqueous media.

2. Approach for Separation and Distinction of Amines

The primary chemical method for distinguishing and separating 1°, 2°, and 3° amines is the Hinsberg Test, complemented by the Nitrous Acid Test.

Amine Type	Hinsberg Test (Reagent: Benzenesulfonyl Chloride in aq. KOH/NaOH)	Nitrous Acid Test (Reagent: NaNO₂ + HCl, 0-5°C)
1° Amine (R-NH₂, Ar-NH₂)	Forms N-substituted benzenesulfonamide (R-NH-SO₂-C₆H₅). This amide has an acidic H-atom attached to N, making it soluble in aqueous KOH/NaOH.	Aliphatic 1°: Forms unstable diazonium salt, which decomposes immediately to alcohol with evolution of N₂ gas. (R-OH + N₂↑) Aromatic 1°: Forms stable arenediazonium salt, which gives an orange dye with β-naphthol (coupling reaction).
2° Amine (R₂NH, Ar₂NH, Ar-NH-R)	Forms N,N-disubstituted benzenesulfonamide (R₂N-SO₂-C₆H₅). This amide has no acidic H-atom attached to N, making it insoluble in aqueous KOH/NaOH.	Forms N-nitrosamine (R₂N-NO), a yellow oily compound. Gives a positive Liebermann's nitroso test (blue/green to red, then blue solution upon warming with phenol and conc. H₂SO₄).
3° Amine (R₃N, Ar-NR₂)	Does not react with benzenesulfonyl chloride to form an amide. Reacts with the reagent to form an insoluble salt of the amine with HCl. It remains insoluble in aqueous KOH/NaOH.	Aliphatic 3°: Forms a soluble trialkylammonium nitrite salt. Aromatic 3°: Undergoes electrophilic substitution at the para position to form a p-nitroso compound (e.g., p-nitroso-N,N-dimethylaniline), often observed as a green precipitate.

Strategy for Separation:

1. Treat the mixture of amines with Hinsberg's reagent (benzenesulfonyl chloride) in the presence of an alkali (like NaOH).


2. Acidify the resulting solution.


3. Separate the components based on their solubility:
   
   
   
   • The 1° amine derivative will be soluble in alkali, but upon acidification, the N-substituted sulfonamide (acidic) will precipitate.
   
   
   • The 2° amine derivative will be insoluble in alkali and will remain as an oily layer or precipitate.
   
   
   • The 3° amine will remain unreacted and can be separated by steam distillation or solvent extraction, as it will be basic and soluble in acid to form a salt.
   
   
   
   

CBSE vs JEE: Both Hinsberg and Nitrous Acid tests are important for CBSE. For JEE, understanding the precise products and their solubility/physical appearance for each class of amine in both tests is critical.

For CBSE board exams, a thorough understanding of the basicity of amines and their separation methods is crucial. Expect direct questions on comparing basicity orders and mechanisms, as well as descriptive questions on chemical tests like Hinsberg's method.

1. Basicity of Amines

Amines are basic due to the presence of a lone pair of electrons on the nitrogen atom, which can be donated to an acid (Lewis base). The strength of an amine as a base depends on the availability of this lone pair.

• Factors Affecting Basicity:
  
  
  
  • Inductive Effect: Electron-donating groups (+I effect, e.g., alkyl groups) increase electron density on nitrogen, enhancing basicity. Electron-withdrawing groups (-I effect) decrease electron density, reducing basicity.
  
  
  • Resonance Effect: In aromatic amines (e.g., aniline), the lone pair on nitrogen is delocalized into the benzene ring, making it less available for protonation. This significantly reduces their basicity compared to aliphatic amines.
  
  
  • Solvation Effect (in Aqueous Phase): The stability of the conjugate acid (ammonium ion) formed after protonation also affects basicity. Greater solvation (hydrogen bonding with water) stabilizes the conjugate acid, thereby increasing basicity. This effect is particularly important for aliphatic amines.
  
  
  • Steric Hindrance: Bulky alkyl groups can hinder the approach of a proton to the nitrogen atom, reducing basicity, especially in the gaseous phase or in highly substituted amines.
  
  
  
  


• CBSE Focus: Comparative Basicity Orders
  
  
  
  • Aliphatic Amines (in Gaseous Phase): Tertiary > Secondary > Primary > NH₃ (Purely based on +I effect and steric factors).
  
  
  • Aliphatic Amines (in Aqueous Phase): This order is a balance of +I effect, solvation effect, and steric hindrance. For simple alkyl groups (e.g., methyl, ethyl), the order varies:
    
    
    
    • Methyl substituted amines: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ (2° > 1° > 3° > NH₃)
    
    
    • Ethyl substituted amines: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃ (2° > 3° > 1° > NH₃)
    
    
    
    

    Important: You must remember these specific orders for CBSE.

    
    
  
  
  • Aromatic Amines vs. Aliphatic Amines: Aliphatic amines are significantly more basic than aromatic amines and ammonia. Aniline is much less basic than ethylamine due to resonance stabilization of the lone pair on nitrogen in aniline.
  
  
  • Effect of Substituents on Aromatic Amines:
    
    
    
    • Electron-donating groups (+R, +I): (e.g., -OCH₃, -CH₃) at ortho and para positions increase basicity.
    
    
    • Electron-withdrawing groups (-R, -I): (e.g., -NO₂, -COOH, -SO₃H, -X) at ortho and para positions decrease basicity.
    
    
    
    
  
  
  
  

2. Separation of Primary, Secondary, and Tertiary Amines

The most common and CBSE-favored method for distinguishing and separating 1°, 2°, and 3° amines is the Hinsberg Test.

• Reagent: Benzenesulfonyl chloride (C₆H₅SO₂Cl) in the presence of an alkali (e.g., KOH or NaOH).


• Principle: The reactivity of primary, secondary, and tertiary amines with benzenesulfonyl chloride differs, leading to distinct observations when treated with an alkali.

Type of Amine	Reaction with C6H5SO2Cl	Addition of KOH/NaOH	Observation & Inference
Primary Amine (R-NH2)	Forms N-alkylbenzenesulfonamide (R-NH-SO2-C6H5). This product has an acidic hydrogen attached to nitrogen.	The sulfonamide dissolves, forming a clear solution.	Clear solution: Indicates a primary amine. The acidic H on N allows it to react with KOH/NaOH to form a soluble salt.
Secondary Amine (R2NH)	Forms N,N-dialkylbenzenesulfonamide (R2N-SO2-C6H5). This product does NOT have an acidic hydrogen on nitrogen.	The sulfonamide remains insoluble/precipitates.	Insoluble precipitate: Indicates a secondary amine. No acidic H, so it does not react with alkali to form a soluble salt.
Tertiary Amine (R3N)	Does NOT react with benzenesulfonyl chloride (lacks an acidic H on nitrogen necessary for reaction with sulfonyl chloride).	The tertiary amine itself is generally insoluble in water but soluble in acidic solution. In a basic medium, it remains as an insoluble oil/layer.	No reaction/Insoluble layer (amine itself): Indicates a tertiary amine. The unreacted amine can be dissolved by adding dilute HCl.

Tip: Practice writing the full reactions for the Hinsberg test, including the formation of the soluble salt for primary amines and the non-reactive nature of tertiary amines. This is a common question in CBSE.

Welcome, future engineers! This section highlights the crucial aspects of amine basicity and separation methods, frequently tested in JEE. Mastering these concepts is key to scoring well in organic chemistry.

JEE Focus Area 1: Basicity of Amines

Understanding the factors influencing amine basicity and the relative order in different phases is paramount for JEE. The stability of the conjugate acid formed is directly proportional to the basicity of the amine.

• Factors Affecting Basicity:
  
  
  
  • +I Effect (Inductive Effect): Alkyl groups are electron-donating, increasing electron density on nitrogen and enhancing basicity. More alkyl groups generally mean more basicity (e.g., 3° > 2° > 1° in the gas phase).
  
  
  • Solvation Effect: In aqueous solution, the conjugate acid (ammonium ion) is stabilized by hydrogen bonding with water molecules. Greater number of H-atoms on N means better solvation and stabilization. This effect is dominant for smaller alkyl groups.
  
  
  • Resonance Effect: Aromatic amines (e.g., aniline) are less basic than aliphatic amines because the lone pair on nitrogen is delocalized into the benzene ring, making it less available for protonation.
  
  
  • Hybridization Effect: Lone pair in sp³ hybridized N (amines) is more available than in sp² (imines, amides) or sp (nitriles) due to less s-character and lower electronegativity.
  
  
  • Steric Hindrance: Bulky alkyl groups can hinder the approach of a proton and also reduce the extent of solvation, thus decreasing basicity in solution.
  
  
  
  


• Order of Basicity (Crucial for JEE):
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  

  Environment	Alkyl Group	Basicity Order	Reason
  Gaseous Phase	Any alkyl group	3° > 2° > 1° > NH3	Purely governed by +I effect. Maximum electron donation by three alkyl groups to N.
  Aqueous Phase	Methyl (-CH3)	2° > 1° > 3° > NH3	Balance of +I effect, steric hindrance, and solvation effect. (Order is approx. 2° (477) > 1° (473) > 3° (422) > NH3 (398) in pKa values). Common JEE trap!
  Aqueous Phase	Ethyl (-CH2CH3)	2° > 3° > 1° > NH3	For bulkier alkyl groups, steric hindrance to solvation becomes more significant for 3° amines.
  General Rule: Aliphatic Amines > NH3 > Aromatic Amines

  
  


• Substituent Effects on Aromatic Amine Basicity:
  
  
  
  • Electron-Donating Groups (EDG) (e.g., -OCH₃, -CH₃) on the benzene ring increase basicity by stabilizing the conjugate acid or increasing electron density on N.
  
  
  • Electron-Withdrawing Groups (EWG) (e.g., -NO₂, -CN, -COOH, -SO₃H) decrease basicity by destabilizing the conjugate acid or pulling electron density away from N.
  
  
  • Ortho Effect: Substituents at the ortho position can cause anomalous basicity due to steric inhibition of resonance, steric hindrance to protonation, or intramolecular H-bonding. This is a subtle but important point for advanced JEE problems.
  
  
  
  

JEE Focus Area 2: Separation of Primary, Secondary, and Tertiary Amines

The Hinsberg test is the classic and most important method for distinguishing and separating these amines for JEE. You must know the reagents, reactions, and solubility patterns.

• Hinsberg's Method:
  
  
  
  • Reagent: Benzene sulfonyl chloride (C₆H₅SO₂Cl) or p-Toluenesulfonyl chloride.
  
  
  • Procedure & Observations:
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    

    Amine Type	Reaction with C6H5SO2Cl + NaOH	Product	Solubility in NaOH/KOH
    Primary (1°)	Reacts to form N-alkylbenzenesulfonamide (R-NH-SO2C6H5).	White precipitate (if insoluble) or solution.	Soluble (due to acidic H attached to N, forming a salt with NaOH).
    Secondary (2°)	Reacts to form N,N-dialkylbenzenesulfonamide (R2N-SO2C6H5).	White precipitate.	Insoluble (no acidic H on N to react with NaOH).
    Tertiary (3°)	Does not react with sulfonyl chloride in the presence of base (forms an insoluble oil/layer if amine itself is insoluble, or remains dissolved).	Unreacted amine.	Insoluble in NaOH (unless it forms a salt with the acid, then reverts to amine layer upon basification).

    
    
  
  
  • Separation Process: Treat the mixture with Hinsberg's reagent and aqueous KOH. The primary amine derivative dissolves, the secondary amine derivative forms a precipitate, and the tertiary amine remains unreacted (as a separate layer or dissolved depending on its solubility). Subsequent acidification/extraction separates them further.
  
  
  
  


• Other Methods (less common for JEE separation, but underlying principles important):
  
  
  
  • Fractional Distillation: Based on boiling point differences (not effective if boiling points are close).
  
  
  • Selective Salt Formation: Tertiary amines can be separated from primary and secondary amines by forming an insoluble salt with specific reagents (e.g., picric acid), but Hinsberg's is more general for all three.
  
  
  
  

JEE Tip: Always remember the specific basicity order in aqueous solutions for methyl and ethyl amines – it’s a classic JEE differentiator. For Hinsberg's, focus on the solubility of the sulfonamide derivatives in base.