Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equati

Exercises · 9.6

Q. Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

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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) …

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. …

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The Hinsberg test distinguishes primary, secondary, and tertiary amines based on their reaction with benzenesulfonyl chloride ( $\ce{C6H5SO2Cl}$ ) in the presence of aqueous alkali. Primary amines form a soluble sulfonamide salt; secondary amines form an insoluble sulfonamide; tertiary amines do not react and remain as an insoluble oil or dissolve in acid.

The Concept: Why the Hinsberg Test Works

The key to identifying amines lies in how many hydrogen atoms are attached to the nitrogen. Primary amines ( $\ce{RNH2}$ ) have two hydrogens, secondary amines ( $\ce{R2NH}$ ) have one, and tertiary amines ( $\ce{R3N}$ ) have none. The Hinsberg reagent — benzenesulfonyl chloride — is an electrophile that attacks the lone pair on nitrogen. But the fate of the product depends on whether there is a hydrogen left on the nitrogen after the attack.

If the nitrogen still has a hydrogen, the product is an acidic sulfonamide that can lose a proton in base, becoming water-soluble. If the nitrogen has no hydrogen (as in the tertiary case), no stable sulfonamide forms — the reagent simply gets hydrolysed or the amine remains unchanged.

This difference in solubility in alkali is the entire basis of the test. No fancy instruments needed — just a test tube, some base, and a bit of acid.

Step-by-Step Procedure

1. Prepare the sample.

Take a small amount of the amine (about 0.5 mL or a few crystals) in a clean test tube. Add 2–3 mL of 10% aqueous sodium hydroxide ( $\ce{NaOH}$ ) and a few drops of benzenesulfonyl chloride ( $\ce{C6H5SO2Cl}$ ). Shake the mixture well.

2. Observe the initial reaction.

If the amine is primary, the mixture becomes clear as the sulfonamide salt dissolves in the alkaline solution.

If the amine is secondary, a white or pale yellow precipitate (the sulfonamide) forms immediately.

If the amine is tertiary, no precipitate forms — the amine may remain as an oily layer or dissolve in the organic phase. The mixture stays cloudy or separates into two layers.

3. Confirm with acidification.

Add dilute hydrochloric acid ( $\ce{HCl}$ ) dropwise to the mixture.

For the primary amine case: the clear solution turns cloudy as the sulfonamide precipitates out (because the salt is converted back to the neutral sulfonamide).
For the secondary amine case: the precipitate remains unchanged (it was already the neutral sulfonamide).
For the tertiary amine case: the oily layer dissolves in the acid, forming a clear solution (the tertiary amine gets protonated to a water-soluble salt).

Watch out

A common mistake is to think that tertiary amines give no reaction at all. They do react — but not with the sulfonyl chloride. Instead, they simply get protonated by the acid in the confirmation step. Also, if the amine is very bulky, the secondary amine might react slowly, so shake thoroughly and wait a minute.

Chemical Equations

Primary amine (e.g., aniline, $\ce{C6H5NH2}$ )

Step 1: Formation of the sulfonamide salt (soluble in alkali):

\ce{C6H5NH2 + C6H5SO2Cl + NaOH -> C6H5SO2N^{-}Na^{+}C6H5 + H2O + HCl}

The sodium salt of N-phenylbenzenesulfonamide is water-soluble.

Step 2: On acidification, the free sulfonamide precipitates:

\ce{C6H5SO2N^{-}Na^{+}C6H5 + HCl -> C6H5SO2NHC6H5 + NaCl}

The neutral N-phenylbenzenesulfonamide is insoluble in water. …

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Method: Hinsberg Test (Benzenesulfonyl Chloride Test)

This is the standard board-exam method for distinguishing primary, secondary, and tertiary amines.

Principle

Amines react with benzenesulfonyl chloride ( $C_6H_5SO_2Cl$ ) in the presence of aqueous KOH. The solubility of the product in alkali differs for each class:

Primary amine → forms a sulfonamide soluble in alkali (due to acidic $-\text{NH}$ proton).
Secondary amine → forms a sulfonamide insoluble in alkali (no acidic proton).
Tertiary amine → no reaction (no $-\text{NH}$ group); the amine itself may be extracted by acid.

Step-by-Step Procedure

Take a small sample of the amine in a test tube.
Add benzenesulfonyl chloride ( $C_6H_5SO_2Cl$ ) and excess aqueous KOH.
Shake the mixture and warm gently.
Observe the result:

Observation	Inference
Clear solution forms (sulfonamide dissolves in alkali)	Primary amine
Precipitate forms (sulfonamide does not dissolve)	Secondary amine
No reaction; oily layer of amine remains (or dissolves in acid)	Tertiary amine

Chemical Equations

1. Primary amine (e.g., ethylamine)

C_2H_5NH_2 + C_6H_5SO_2Cl \xrightarrow{\text{KOH}} C_6H_5SO_2NHC_2H_5 + KCl + H_2O

The product N-ethylbenzenesulfonamide has an acidic $-\text{NH}$ proton and dissolves in KOH:

C_6H_5SO_2NHC_2H_5 + KOH \rightarrow C_6H_5SO_2N^-(C_2H_5)K^+ + H_2O

2. Secondary amine (e.g., diethylamine) …

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Here is a breakdown of the common mistakes students make when answering this specific question on Nucleophilic Substitution Reactions (specifically, the Hinsberg Test for amines), along with the correct method and how to avoid each error.

The Correct Method (Hinsberg Test)

The standard method for distinguishing between primary, secondary, and tertiary amines is the Hinsberg Test.

Reagent: Benzenesulfonyl chloride ( $C_6H_5SO_2Cl$ ) in the presence of aqueous KOH (or NaOH).

Procedure:

Shake the amine with benzenesulfonyl chloride and excess KOH solution.
Observe the result.

Chemical Equations:

Primary Amine ( $1^\circ$ ): Forms a sulfonamide which is soluble in alkali (because it has an acidic H attached to N).

C_6H_5SO_2Cl + RNH_2 \xrightarrow{KOH} C_6H_5SO_2NHR \text{ (soluble salt)} + KCl + H_2O

On acidification, the sulfonamide precipitates as a white solid.

C_6H_5SO_2NHR + HCl \rightarrow C_6H_5SO_2NHR \downarrow + KCl

Secondary Amine ( $2^\circ$ ): Forms a sulfonamide which is insoluble in alkali (no acidic H on N).

C_6H_5SO_2Cl + R_2NH \xrightarrow{KOH} C_6H_5SO_2NR_2 \text{ (insoluble precipitate)} + KCl + H_2O

Tertiary Amine ( $3^\circ$ ): Does not react with benzenesulfonyl chloride. It remains as an oily layer (or dissolves in acid if present, but in this basic medium it remains free).

Common Mistakes & How to Avoid Them

Mistake 1: Confusing the Solubility of the Products

The Error: Students often mix up which sulfonamide is soluble in KOH and which is not. They might say the secondary amine product is soluble, or that the primary amine product precipitates immediately.

Why it happens: They memorize the result without understanding the chemistry (the presence of an acidic hydrogen).

How to Avoid:

Focus on the N-H bond. A primary amine ( $RNH_2$ ) has two H atoms on N. After reaction, the product ( $R-NH-SO_2Ph$ ) still has one H on N. This H is acidic (due to the strong electron-withdrawing sulfonyl group). It reacts with KOH to form a water-soluble salt.
Secondary amine ( $R_2NH$ ) has one H on N. After reaction, the product ( $R_2N-SO_2Ph$ ) has zero H on N. No acidic H → no reaction with KOH → insoluble.
Trick to remember: "1° gives 1 H left → 1 soluble salt. 2° gives 2 R groups → 2 no H left → 2 insoluble."

Mistake 2: Forgetting to Acidify the Primary Amine Product

The Error: Students write that the primary amine gives a "clear solution" and stop there. They forget the crucial step of acidification to get the precipitate.

Why it happens: They only memorize the first step of the test.

How to Avoid:

Remember the purpose: The test is to identify the amine. A clear solution is not a visible solid. You need to regenerate the insoluble sulfonamide by adding dilute HCl to prove it was a primary amine.
Write the full sequence: Always write the two-step equation for primary amines: (1) Reaction with $C_6H_5SO_2Cl$ + KOH → soluble salt. (2) Add HCl → white precipitate.

Mistake 3: Thinking Tertiary Amines Give a Precipitate

The Error: Students write that tertiary amines react to form a solid or that they give a "no reaction" result but then incorrectly describe the physical state.

Why it happens: They confuse the Hinsberg test with other tests (like the carbylamine test).

How to Avoid:

Understand the mechanism: Tertiary amines have no H on the N atom. Nucleophilic substitution ( $S_N2$ ) at the sulfur atom requires the amine to attack and then lose a proton. If there is no H to lose, the reaction stops. The tertiary amine remains as an oily layer (or dissolves if the solution is acidic, but here it's basic).
Key phrase: "No reaction – remains as an oily layer."

Mistake 4: Writing Incorrect Reagents or Conditions

The Error: Students write the reagent as "benzene sulfonic acid" or forget to mention the aqueous KOH/NaOH medium. …

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Q.Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

The Concept: Why the Hinsberg Test Works

Step-by-Step Procedure

Chemical Equations

Primary amine (e.g., aniline, CX6HX5NHX2\ce{C6H5NH2}CX6​HX5​NHX2​)

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Primary amine (e.g., aniline, $\ce{C6H5NH2}$ )

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