Write short notes on the following: (i) Carbylamine reaction (ii) Diazotisation (iii) Hofmann's bromamide reac

Exercises · 9.7

Q. Write short notes on the following:

(i) Carbylamine reaction

(ii) Diazotisation

(iii) Hofmann's bromamide reaction

(iv) Coupling reaction

(v) Ammonolysis

(vi) Acetylation

(vii) Gabriel phthalimide synthesis.

Punjab PsebTextbookSubjective· 5mImportance★★★★★

19% · 21/108 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

These seven reactions are the backbone of amine chemistry in organic synthesis. Carbylamine is a test for primary amines; diazotisation converts primary aromatic amines to diazonium salts; Hofmann's bromamide degrades amides to amines; coupling joins diazonium salts with phenols/amines; ammonolysis replaces halogens with amino groups; acetylation protects amines; Gabriel phthalimide makes pure primary amines from alkyl halides.

(i) Carbylamine Reaction

This is a test for primary amines — both aliphatic and aromatic. When a primary amine is heated with chloroform ( $CHCl_3$ ) and alcoholic potassium hydroxide ( $KOH$ ), a foul-smelling isocyanide (carbylamine) is formed.

Why it works: The reaction involves the formation of a dichlorocarbene intermediate ( $:CCl_2$ ) from chloroform under basic conditions. This highly reactive species attacks the nitrogen of the primary amine, leading to the isocyanide.

Reaction:

$RNH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} RNC + 3KCl + 3H_2O$

Watch out

Secondary and tertiary amines do not give this test. Also, the isocyanides are extremely toxic and have a penetrating, nauseating odour — perform this test only in a fume hood.

(ii) Diazotisation

This reaction converts primary aromatic amines into diazonium salts at low temperatures (0–5°C). The amine is treated with nitrous acid ( $HNO_2$ ), generated in situ from sodium nitrite ( $NaNO_2$ ) and a mineral acid ( $HCl$ or $H_2SO_4$ ).

Why it works: Nitrous acid acts as a nitrosating agent. It attacks the amino group, forming an N-nitroso intermediate, which then tautomerises and loses water to give the diazonium ion.

Reaction:

$ArNH_2 + NaNO_2 + 2HCl \xrightarrow{0-5^\circ C} ArN_2^+Cl^- + NaCl + 2H_2O$

Tip

The temperature must be kept low because diazonium salts are unstable and decompose to phenols at higher temperatures. This is a classic example of kinetic control in organic reactions.

(iii) Hofmann's Bromamide Reaction

This is a degradation reaction that converts an amide into a primary amine with one fewer carbon atom. The amide is treated with bromine ( $Br_2$ ) and aqueous sodium hydroxide ( $NaOH$ ).

Why it works: The reaction proceeds through a rearrangement — the alkyl group migrates from the carbonyl carbon to the nitrogen, expelling a molecule of $CO_2$ . This is an example of a rearrangement reaction (Hofmann rearrangement).

Reaction:

$RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + 2NaBr + Na_2CO_3 + 2H_2O$

RCONH_2 \xrightarrow{Br_2/NaOH} RNH_2 + CO_2

(iv) Coupling Reaction

This is a reaction between a diazonium salt and an activated aromatic compound (like phenol or an aromatic amine) to form an azo compound ( $-N=N-$ linkage). The product is brightly coloured and used as a dye.

Why it works: The diazonium ion is a weak electrophile. It attacks the electron-rich aromatic ring of phenol (at the para position, or ortho if para is blocked) under mild conditions (pH 8–10 for phenols, pH 4–5 for amines).

Reaction (with phenol):

$C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{NaOH} C_6H_5-N=N-C_6H_4-OH(p) + HCl$

Watch out

The coupling does not occur with deactivated aromatic rings (e.g., nitrobenzene). The attacking species is the diazonium cation, which is why the reaction is called electrophilic substitution.

(v) Ammonolysis

This is the nucleophilic substitution of an alkyl halide (or other leaving group) by ammonia ( $NH_3$ ) to form an amine. It is a classic $S_N2$ reaction.

Why it works: Ammonia acts as a nucleophile, attacking the electrophilic carbon bearing the halogen. The product is a primary amine, but it can further react to give secondary, tertiary amines, and even quaternary ammonium salts — so the reaction is not selective.

Reaction:

$RX + NH_3 \xrightarrow{excess} RNH_2 + HX$

(Then: $RNH_2 + RX \rightarrow R_2NH + HX$ , and so on.)

Tip

To maximise the yield of the primary amine, use a large excess of ammonia. This is a common trick in exam problems.

(vi) Acetylation

This is the process of introducing an acetyl group ( $-COCH_3$ ) into a compound, typically by reacting an amine or alcohol with acetyl chloride ( $CH_3COCl$ ) or acetic anhydride ( $(CH_3CO)_2O$ ).

Why it works: The reaction is a nucleophilic acyl substitution. The lone pair on the nitrogen (or oxygen) attacks the electrophilic carbonyl carbon of the acetylating agent, displacing the leaving group ( $Cl^-$ or $CH_3COO^-$ ).

Reaction (with an amine): …

Start your 14-day free trial to unlock the full solution →

Here is the solution method for each of these classic reactions, focusing on the why before the what. No single mechanism covers all seven — each reaction below is labelled by its own governing mechanism (nucleophilic substitution, electrophilic substitution, rearrangement, or nucleophilic acyl substitution).

(i) Carbylamine Reaction

Method: Nucleophilic Addition-Elimination (on a carbene intermediate)

This is a specific test for primary amines ( $RNH_2$ ). It does not work with secondary or tertiary amines.

Steps:

Formation of Dichlorocarbene (:CCl $_2$ ): Chloroform ( $CHCl_3$ ) reacts with a strong base (like $KOH$ ) to generate a highly reactive, electron-deficient species called dichlorocarbene.

CHCl_3 + KOH \rightarrow :CCl_2 + KCl + H_2O

Nucleophilic Attack: The lone pair on the nitrogen of the primary amine ( $RNH_2$ ) acts as a nucleophile and attacks the electron-deficient carbon of the carbene.

RNH_2 + :CCl_2 \rightarrow [RNH_2-CCl_2]

Elimination: The intermediate undergoes dehydrohalogenation (loss of $HCl$ ) to form an isocyanide (also called carbylamine). This product has a foul, penetrating smell.

[RNH_2-CCl_2] \xrightarrow{-2HCl} R-N \equiv C

Key Result: Formation of a foul-smelling isocyanide confirms a primary amine.

(ii) Diazotisation

Method: Electrophilic Substitution (on the amine nitrogen)

This reaction converts a primary aromatic amine (like aniline, $C_6H_5NH_2$ ) into a diazonium salt at low temperatures (0-5°C).

Steps:

Generation of Nitrosonium Ion ( $NO^+$ ): Nitrous acid ( $HNO_2$ ), generated in situ from $NaNO_2$ and $HCl$ , forms the powerful electrophile, the nitrosonium ion.

NaNO_2 + HCl \rightarrow HNO_2 + NaCl

HNO_2 + H^+ \rightarrow H_2O + \overset{+}{N}=O

Electrophilic Attack: The $NO^+$ ion attacks the lone pair on the nitrogen of the aromatic amine.
Tautomerization and Elimination: The intermediate loses water to form the stable diazonium salt ( $C_6H_5N_2^+Cl^-$ ).

C_6H_5NH_2 + NO^+ \rightarrow C_6H_5N_2^+ + H_2O

Key Result: A highly versatile intermediate for substitution reactions (replacing $N_2^+$ with $Cl$ , $Br$ , $CN$ , $OH$ , etc.).

(iii) Hofmann's Bromamide Reaction

Method: Rearrangement (followed by hydrolysis)

This reaction converts a primary amide ( $RCONH_2$ ) into a primary amine ( $RNH_2$ ) with one less carbon atom.

Steps:

Formation of N-Bromoamide: The amide reacts with bromine ( $Br_2$ ) in the presence of a base ( $NaOH$ ). The amide nitrogen is deprotonated, and then brominated.

RCONH_2 + Br_2 + NaOH \rightarrow RCONHBr + NaBr + H_2O

Rearrangement (The Key Step): The N-bromoamide is treated with more base. The alkyl group ( $R$ ) from the carbonyl carbon migrates to the nitrogen atom, displacing the bromine atom. This forms an isocyanate intermediate ( $R-N=C=O$ ).

RCONHBr + NaOH \rightarrow R-N=C=O + NaBr + H_2O

Hydrolysis: The isocyanate is unstable in water and undergoes hydrolysis to form a primary amine and carbon dioxide.

R-N=C=O + 2NaOH \rightarrow RNH_2 + Na_2CO_3

Key Result: A primary amine with one less carbon than the starting amide.

(iv) Coupling Reaction

Method: Electrophilic Aromatic Substitution

This reaction joins a diazonium salt (the electrophile) with a highly activated aromatic ring (like phenol or an aromatic amine) to form an azo compound ( $-N=N-$ ).

Steps:

Formation of the Electrophile: The diazonium salt ( $ArN_2^+$ ) is a weak electrophile.
Attack on Activated Ring: The diazonium ion attacks the para position (or ortho if para is blocked) of a phenol ( $C_6H_5OH$ ) or an aromatic amine ( $C_6H_5NH_2$ ). The ring must have strong electron-donating groups ( $-OH$ , $-NH_2$ ) to be sufficiently activated.
Rearomatization: The intermediate loses a proton ( $H^+$ ) to restore the aromatic ring, forming the brightly colored azo dye.

ArN_2^+Cl^- + C_6H_5OH \xrightarrow{pH \ 9-10} Ar-N=N-C_6H_4OH + HCl

Key Result: Formation of a colored azo compound, used extensively in dye chemistry.

(v) Ammonolysis

Method: Nucleophilic Substitution ( $S_N2$ )

This is the reaction of an alkyl halide (or other substrate with a leaving group) with ammonia ( $NH_3$ ) to form an amine.

Steps:

Nucleophilic Attack: Ammonia ( $NH_3$ ) acts as a nucleophile. Its lone pair attacks the electrophilic carbon bearing the halogen ( $X$ ).

R-X + :NH_3 \rightarrow [R-NH_3]^+X^-

Deprotonation: The intermediate salt is deprotonated by another molecule of ammonia (or base) to give the primary amine. $[R-NH_3]^+X^- + NH_3 \rightarrow R-NH_2 + NH_4X$ …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make when writing short notes on these named reactions, along with precise corrections to ensure exam-ready answers.

General Mistake: Writing "Short Notes" as Full Essays

The Mistake: Students write paragraphs instead of bullet points or structured points, losing marks for presentation.

How to Avoid: For "short notes," always use a format:

Definition/Type of reaction
Key Reagent(s)
Mechanism (1-2 steps only)
Product(s)
Key condition/limitation

(i) Carbylamine Reaction

Common Mistake: Thinking it works for all amines.

The Truth: It is a test for primary amines only (aliphatic and aromatic). Secondary and tertiary amines do not give this reaction.

How to Avoid: Memorise: Only 1° amines + CHCl₃ + alc. KOH → foul-smelling isocyanide (carbylamine).

Correct Note:

Reagents: Primary amine ( $R-NH_2$ ), chloroform ( $CHCl_3$ ), alcoholic KOH.
Product: Isocyanide ( $R-NC$ ) — foul-smelling.
Mechanism: Formation of dichlorocarbene (:CCl₂) which attacks the amine.
Limitation: Not given by 2° or 3° amines.

(ii) Diazotisation

Common Mistake: Writing that the reaction works at room temperature or with any acid.

The Truth: Diazotisation requires ice-cold temperature (0–5°C) and nitrous acid (HNO₂) generated in situ from NaNO₂ + HCl/H₂SO₄. Above 5°C, the diazonium salt decomposes.

How to Avoid: Always write: NaNO₂ + HCl, 0–5°C.

Correct Note:

Reaction: Primary aromatic amine + HNO₂ (from NaNO₂ + HCl) → Diazonium salt.
Condition: Temperature 0–5°C (to prevent decomposition).
Product: Benzenediazonium chloride ( $C_6H_5N_2^+Cl^-$ ).
Use: Intermediate for azo dyes, substitution reactions.

(iii) Hofmann's Bromamide Reaction

Common Mistake: Confusing it with Hofmann elimination (which gives alkenes).

The Truth: This is a rearrangement of an amide to a primary amine with one less carbon.

How to Avoid: Remember the carbon loss: $R-CONH_2 \rightarrow R-NH_2$ (one C removed).

Correct Note:

Reagents: Amide ( $RCONH_2$ ) + Bromine ( $Br_2$ ) + NaOH (aq).
Product: Primary amine ( $R-NH_2$ ) with one carbon less than the amide.
Mechanism: Hofmann rearrangement (migration of R from C to N).
Key point: Used to prepare 1° amines from amides.

(iv) Coupling Reaction

Common Mistake: Thinking it works with any aromatic compound.

The Truth: The diazonium salt couples only with highly activated aromatic rings (phenols, anilines) — not with benzene or nitrobenzene.

How to Avoid: Remember the coupling partner must have strong electron-donating groups ( $-OH$ , $-NH_2$ , $-NHR$ ).

Correct Note:

Reaction: Diazonium salt + phenol/aniline → Azo dye ( $-N=N-$ ).
Condition: pH 9–10 for phenols (alkaline), pH 4–5 for anilines (weakly acidic).
Product: Coloured azo compound (e.g., Orange I, Yellow).
Mechanism: Electrophilic substitution at the para position (or ortho if para blocked).

(v) Ammonolysis

Common Mistake: Writing that it gives only primary amines.

The Truth: Ammonolysis of alkyl halides gives a mixture of primary, secondary, tertiary amines and quaternary ammonium salts (due to further alkylation).

How to Avoid: Always mention mixture formation and the need for excess NH₃ to favour primary amine.

Correct Note:

Reaction: Alkyl halide ( $R-X$ ) + $NH_3$ → $R-NH_2$ + $R_2NH$ + $R_3N$ + $R_4N^+X^-$ .
Condition: Excess $NH_3$ (to minimise further substitution).
Limitation: Not a clean method for pure primary amines.
Better alternative: Gabriel phthalimide synthesis (for pure 1° amine).

(vi) Acetylation

Common Mistake: Using the wrong reagent (e.g., acetic acid instead of acetyl chloride/acetic anhydride).

The Truth: Acetylation of amines requires acetyl chloride ( $CH_3COCl$ ) or acetic anhydride ( $(CH_3CO)_2O$ ) — not acetic acid (which is too weak).

How to Avoid: Write: $CH_3COCl$ or $(CH_3CO)_2O$ in presence of pyridine (to neutralise HCl). …

Start your 14-day free trial to unlock the full solution →

←Previous question9.6 — Describe a method for the identification of primary, secondary and ter…→Next question9.8 — Accomplish the following conversions: (i) Nitrobenzene to benzoic acid…

Q.Write short notes on the following:

(i) Carbylamine Reaction

(ii) Diazotisation

(iii) Hofmann's Bromamide Reaction

(iv) Coupling Reaction

(v) Ammonolysis

(vi) Acetylation

🎓Unlock everything free for 14 days

Unlock everything free for 14 days