Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and b

NCERT Exemplar · Q44

Q. Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidising agent?

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Concept understanding — Nucleophilic Addition

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 Addition: Why the Mechanism Works the Way It Does

Let's build this from first principles — understanding why nucleophilic addition happens, not just memorising the steps.

1. The Core Problem: Why Does Addition Happen at All?

A carbonyl group ( $\ce{C=O}$ ) has a polarised double bond:

Oxygen is more electronegative than carbon → it pulls electron density toward itself.
This creates a partial positive charge on carbon ( $\delta^+$ ) and a partial negative charge on oxygen ( $\delta^-$ ).

\ce{C^{\delta+}=O^{\delta-}}

Key insight: The carbon is electron-deficient — it wants electrons. A nucleophile (Nu⁻) is electron-rich — it wants to give electrons. This is a natural match.

2. The Two-Step Mechanism (Why Two Steps?)

Step 1: Nucleophilic Attack (Slow, Rate-Determining)

The nucleophile donates its lone pair to the electrophilic carbonyl carbon.

\ce{Nu^- + C=O -> [Nu-C-O]^-}

Why this happens:

The $\pi$ bond between C and O breaks — the electrons move entirely to oxygen.
Oxygen now has a full negative charge (alkoxide ion).
The carbon changes from $sp^2$ (trigonal planar) to $sp^3$ (tetrahedral).

This step is slow because the $\pi$ bond must break — it requires energy.

Step 2: Protonation (Fast)

The negatively charged oxygen picks up a proton ( $\ce{H^+}$ ) from the solvent or acid.

\ce{[Nu-C-O]^- + H^+ -> Nu-C-OH}

Why this happens:

The alkoxide ion is a strong base — it wants to neutralise its charge.
Protonation gives a stable neutral alcohol product.

3. The Key Formula: Rate Law Derivation

For a general nucleophilic addition:

\ce{Nu^- + R2C=O ->[k] products}

The rate law comes from the slow step (Step 1):

\text{Rate} = k[\text{Nu}^-][\ce{R2C=O}]

Why this form?

The reaction is bimolecular — two species must collide with correct orientation.
Doubling either concentration doubles the rate (first order in each).
This is second order overall.

Exam tip: This is why nucleophilic addition is often called addition-elimination when followed by loss of a leaving group (like in acyl substitution), but here it's just addition.

4. Why the Tetrahedral Intermediate Forms (And Why It's Unstable)

The intermediate is tetrahedral ( $sp^3$ hybridised carbon):

Bond angles: ~109.5°
Four groups around carbon: Nu, R, R', O⁻

Why it's unstable:

The negative charge on oxygen is high-energy.
The tetrahedral geometry is sterically crowded (especially with bulky R groups).
The intermediate collapses quickly — either back to starting materials or forward to product. …

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Iodination of an arene is reversible: $\text{ArH} + \text{I}_2 \rightleftharpoons \text{ArI} + \text{HI}$ . The HI formed is a good reducing agent — it converts the aryl iodide back to the arene — so on its own the reaction never accumulates product. An oxidising agent (HIO₄, or HNO₃) is added to oxidise the HI away, driving the equilibrium forward. That is why aryl iodides need an oxidising agent while aryl chlorides and bromides do not.

Why the usual method works for Cl₂ and Br₂

Chlorination and bromination of arenes proceed cleanly with just a Lewis acid catalyst (FeCl₃, AlCl₃), which polarises the halogen molecule into a strong electrophile:

\text{Cl}_2 + \text{FeCl}_3 \rightarrow \text{Cl}^{\delta+} \cdots \text{FeCl}_4^{\delta-}

The HCl or HBr released as byproduct does not attack the aryl halide product, so these reactions are effectively irreversible — no extra reagent is needed.

The problem with iodine: the reaction is reversible

Iodination is different in one decisive way. The reaction sits in an equilibrium:

\text{ArH} + \text{I}_2 \rightleftharpoons \text{ArI} + \text{HI}

The HI byproduct is a good reducing agent: it reduces the aryl iodide back to the parent arene (regenerating I₂), pulling the equilibrium backwards. It also doesn't help that I₂ is the weakest electrophile of the halogens, which makes the forward reaction sluggish to begin with — but the equilibrium is the core problem: even the product that does form is destroyed by the HI accumulating in the mixture.

The solution: oxidise away the HI

An oxidising agent — HIO₄ (periodic acid) is the one NCERT names; HNO₃ also works — is added to oxidise the HI as it forms, removing it from the equilibrium:

\text{ArH} + \text{I}_2 \rightleftharpoons \text{ArI} + \text{HI}

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Here’s a breakdown of the common mistakes students make on this concept, along with how to avoid each.

The Core Concept (Why the Question Exists)

The question tests your understanding of reactivity trends in electrophilic aromatic substitution (EAS) and the redox chemistry of halogens.

For Cl₂ and Br₂: The halogen molecule is already a strong enough electrophile (when activated by a Lewis acid like FeCl₃ or AlCl₃) to attack the benzene ring.
For I₂: Iodine is a much weaker electrophile than chlorine or bromine, and the reaction produces HI as a byproduct. HI is a good reducing agent that reduces the aryl iodide back to the arene (regenerating I₂), reversing the reaction.

The fix: An oxidizing agent (like HIO₄ or HNO₃) oxidizes the HI (the byproduct) back into I₂, preventing the reverse reaction and pushing the equilibrium forward.

Common Mistake #1: Confusing "Oxidizing Agent" with "Catalyst"

The Mistake:

Students say: "The oxidizing agent is needed because iodine is a weaker electrophile, so we need a stronger catalyst." They treat the oxidizing agent as if it’s just another Lewis acid catalyst.

Why it’s wrong:

A Lewis acid catalyst (like FeCl₃) activates the halogen by polarizing it (making it more electrophilic). An oxidizing agent does not activate iodine directly. Instead, it removes the byproduct (HI) that would otherwise destroy the aryl iodide product.

How to Avoid:

Remember the byproduct: For every I₂ that reacts, one HI is produced. HI is a strong reducing agent.
Trace the electron flow: HI + [O] → I₂ + H₂O. The oxidizing agent regenerates the starting material (I₂), not just activates it.
Exam tip: If a question asks "Why is an oxidizing agent needed?" your answer must mention preventing the reduction of the aryl iodide back to the arene by HI. Do not just say "to make iodine more reactive."

Common Mistake #2: Thinking the Oxidizing Agent Makes Iodine "More Electrophilic"

The Mistake:

Students write: "The oxidizing agent increases the electrophilicity of iodine."

Why it’s wrong:

Oxidizing agents do not directly increase the positive charge or polarity of I₂. They work indirectly by removing HI. The actual electrophile in the reaction is still I₂ (or I⁺ generated in situ, but that’s a separate mechanism). The oxidizing agent doesn’t touch the I₂ molecule itself.

How to Avoid:

Use precise language: Say "The oxidizing agent oxidizes the HI byproduct back to I₂, preventing the reverse reaction."
Draw the equilibrium: Show the reversible arrow: $\text{ArH} + \text{I}_2 \rightleftharpoons \text{ArI} + \text{HI}$ The oxidizing agent shifts the equilibrium to the right by removing HI.

Common Mistake #3: Forgetting the Role of HI as a Reducing Agent

The Mistake:

Students say: "HI is a strong acid, so it protonates the benzene ring and stops the reaction."

Why it’s wrong:

HI is indeed an acid, but the real problem is that it reduces the aryl iodide back to the arene. The reaction is reversible, and HI is the reducing agent that drives it backward. Protonation of the ring is not the main issue here (though it can happen, it’s secondary).

How to Avoid:

Remember the equilibrium: ArH + I₂ ⇌ ArI + HI. HI provides the electrons that reduce the ArI product back to ArH. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. The increasing order of reactivity towards nucleophilic addition is:$
(a) Acetophenone < Benzaldehyde < p-Tolualdehyde < p-Nitrobenzaldehyde
(b) Benzaldehyde < Acetophenone < p-Nitrobenzaldehyde < p-Tolualdehyde
(c) p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde < Acetophenone
(d) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde
›Reveal solutionSolution
Nucleophilic addition to a carbonyl is favoured by a more electrophilic (less hindered, less electron-rich) carbonyl carbon: aldehydes beat ketones, and electron-withdrawing ring substituents beat electron-donating ones.

Two factors govern reactivity toward nucleophilic addition here:

Aldehyde vs ketone: aldehydes are generally more reactive than ketones — fewer/smaller substituents mean less steric hindrance and less electron donation into the carbonyl carbon, keeping it more electrophilic.

Ring substituent on aromatic aldehydes: an electron-WITHDRAWING group (like –NO2) pulls electron density away from the carbonyl carbon, making it MORE electrophilic (more reactive); an electron-DONATING group (like –CH3) pushes electron density in, making the carbonyl LESS electrophilic (less reactive).

Ranking the four:

Acetophenone (a ketone) — least reactive (steric + extra alkyl/aryl electron donation). …
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GUJCET 2024Set 131 markMCQ
$'R' + CH_3-CO-CH_3 \xrightarrow{H^+}$
›Reveal solutionSolution

Schiff's base $=$ imine $\;C{=}N$ , formed by a carbonyl reacting with a primary amine.

Concept. A ketone/aldehyde condenses with a primary amine ( $RNH_2$ ) under acid catalysis to give a Schiff's base (imine). Hydrazine and phenylhydrazine give hydrazones, and hydroxylamine gives an o …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Salicylaldehyde on heating with zinc dust give ___ organic product.$
(a) Benzene
(b) Benzaldehyde
(c) Benzoic acid
(d) Benzyl alcohol
›Reveal solutionSolution

Zn dust distillation removes the -OH of salicylaldehyde, giving benzaldehyde.

Salicylaldehyde is 2-hydroxybenzaldehyde (a benzene ring bearing -CHO and -OH at adjacent positions). Heating a phenol with zinc dust replaces the aromatic -OH group with -H (reduction/removal of OH). Applying this to salicylaldehyde removes …

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GUJCET 2022Set 171 markMCQ
$C_6H_5-CHO + CH_3CHO \xrightarrow[\Delta]{OH^-}$
›Reveal solutionSolution

Base-catalysed cross-aldol + heat ⇒ dehydration ⇒ cinnamaldehyde $C_6H_5CH=CH-CHO$ .

Concept. Benzaldehyde has no α-H, so acetaldehyde supplies the enolate (nucleophile). The aldol adds, and on heating with base the β-hydroxy aldehyde loses water (condensation) to give the conjugated α,β-unsaturated aldehyde. …
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GUJCET 2019Set 131 markMCQ
$Q. Which of the major product obtained by hydrolysis of compound formed by reaction between formaldehyde and ethyl magnesium bromide? (A) 2 - Methyl - propan - 2 - ol (B) Propan - 1 - ol (C) Propan - 2 - ol (D) Ethan - 1 - ol$
›Reveal solutionSolution

Formaldehyde with a Grignard reagent gives a primary alcohol; ethyl MgBr adds C₂H₅ to give $C_2H_5CH_2OH$ = propan-1-ol.

Concept — Grignard on formaldehyde. $HCHO$ + $RMgX$ → $RCH_2OMgX$ → (hydrolysis) → $RCH_2OH$ , a primary alcohol with one extra carbon.

Steps. With $R = C_2H_5$ : …
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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. What is the final product 'C' in the following reaction? CH3CHO --HCN--> 'A' --H3O+--> 'B' --soda lime--> 'C'$
(a) Propanol
(b) Ethanol
(c) Propane
(d) Propanoic acid
›Reveal solutionSolution
CH3CHO -> cyanohydrin -> lactic acid -> (soda lime decarboxylation) -> ethanol.

Step by step:

CH3CHO + HCN -> CH3-CH(OH)-CN (A, acetaldehyde cyanohydrin).

A + H3O+ (hydrolysis of nitrile) -> CH3-CH(OH)-COOH (B, lactic acid). …
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←Previous questionQ43 — Alkyl fluorides are synthesised by heating an alkyl chloride/bromide i…→Next questionQ45 — Out of o- and p-dibromobenzene which one has higher melting point and …

Q.Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidising agent?

Concept understanding — Nucleophilic Addition

Nucleophilic Addition – From Intuition to Precision

The Precise Statement

Why It Happens – The Driving Force

A Concrete Example – Addition of HCN to a Ketone

What Makes a Good Nucleophile?

Why this formula?

Nucleophilic Addition: Why the Mechanism Works the Way It Does

1. The Core Problem: Why Does Addition Happen at All?

2. The Two-Step Mechanism (Why Two Steps?)

Step 1: Nucleophilic Attack (Slow, Rate-Determining)

Step 2: Protonation (Fast)

3. The Key Formula: Rate Law Derivation

4. Why the Tetrahedral Intermediate Forms (And Why It's Unstable)

Why the usual method works for Cl₂ and Br₂

The problem with iodine: the reaction is reversible

The solution: oxidise away the HI

Concept: Reversibility of Iodination and Removal of HI

Method: Equilibrium Analysis of Arene Halogenation

Step-by-Step Reasoning

The Core Concept (Why the Question Exists)

Common Mistake #1: Confusing "Oxidizing Agent" with "Catalyst"

Common Mistake #2: Thinking the Oxidizing Agent Makes Iodine "More Electrophilic"

Common Mistake #3: Forgetting the Role of HI as a Reducing Agent

Q.Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidising agent?

Why the usual method works for Cl₂ and Br₂

The problem with iodine: the reaction is reversible

The solution: oxidise away the HI

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