Write the major product(s) of the following reactions: (i) C₆H₅Cl (chlorobenzene) xrightarrow[Δ]Conc. H₂SO₄ (i

Question

Q. Write the major product(s) of the following reactions:

(i)

C_6H_5Cl

(chlorobenzene)

\xrightarrow[\Delta]{\text{Conc. } H_2SO_4}

(ii) 4-hydroxybenzyl alcohol,

HOCH_2\text{-}C_6H_4\text{-}OH

(para-substituted benzene)

\xrightarrow[\Delta]{HCl}

(iii)

(CH_3)_3C\text{-}OH \xrightarrow{Cu,\ 573\ K}

CBSECBSE Class XII Board 2020Subjective· 3mImportance★★★★★

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Concept understanding — Electrophilic Aromatic Substitution

Electrophilic Aromatic Substitution – The First Meeting

Imagine you have a benzene ring — that perfect, flat hexagon of six carbons with alternating double bonds. It's stable, almost stubbornly so. You want to attach something new to it, say a bromine atom or a nitro group. But benzene doesn't react like an alkene. It doesn't just add across a double bond. Instead, it does something more elegant: it kicks out a hydrogen and keeps its aromatic ring intact.

That's the heart of Electrophilic Aromatic Substitution (EAS).

The Intuition: Why "Substitution" and Not "Addition"?

Benzene's stability comes from its delocalised $\pi$ electrons — a cloud above and below the ring. This cloud is electron-rich, so it attracts electrophiles (electron-loving species). But if an electrophile simply added to a double bond, the ring would break its aromaticity, losing that huge stabilisation. That would be energetically costly.

So benzene does something smarter: it lets the electrophile attack, temporarily breaks aromaticity to form a high-energy intermediate (the arenium ion), and then loses a proton to restore the aromatic ring. The net result? A hydrogen is replaced by the electrophile. The ring is back to its stable, aromatic self.

Note

The key trade-off: temporary loss of aromaticity is acceptable because the final product regains it. Addition reactions would permanently destroy aromaticity — benzene avoids that.

The Precise Statement

Electrophilic Aromatic Substitution is a reaction in which an electrophile ( $E^+$ ) replaces a hydrogen atom on an aromatic ring, proceeding through a sigma complex (arenium ion) intermediate, and restoring aromaticity after deprotonation.

The general equation:

\ce{Ar-H + E+ -> Ar-E + H+}

where Ar represents an aromatic ring.

The Mechanism in Three Steps

Generation of the electrophile – Many EAS reactions need a catalyst to create a strong enough $E^+$ . For example, bromination uses $\ce{FeBr3}$ to polarise $\ce{Br2}$ into $\ce{Br^+}$ .
Attack by the aromatic ring – The $\pi$ electrons of benzene attack the electrophile, forming a sigma complex (also called the arenium ion or Wheland intermediate). This intermediate is non-aromatic — it has four $\pi$ electrons delocalised over five carbons, and one $sp^3$ carbon bearing the electrophile and a hydrogen.

\text{Benzene} + E^+ \longrightarrow \text{Sigma complex (non-aromatic)}

Deprotonation – A base (often the counterion of the catalyst, like $\ce{FeBr4-}$ ) removes the proton from the $sp^3$ carbon. The pair of electrons from the C–H bond flows back into the ring, restoring the aromatic sextet.

\text{Sigma complex} + \text{Base} \longrightarrow \text{Product} + \ce{H+}

Watch out

The sigma complex is not aromatic. It's a high-energy intermediate. Students often mistakenly think it's still aromatic — it isn't. That's why the step is fast and the complex is short-lived.

Why This Matters for Exams

EAS is the gateway to understanding how to put groups onto benzene rings. The rate-determining step is usually the formation of the sigma complex (step 2). The regiochemistry (where the electrophile goes) depends on whether the ring already has a substituent — that's the topic of activating/deactivating groups and ortho/para vs. meta directors.

But for now, remember this: …

Why this formula?

Electrophilic Aromatic Substitution: Why the Mechanism Holds

The Core Puzzle: Why Benzene Doesn't Just Add

Benzene ( $C_6H_6$ ) has three double bonds — so why doesn't it undergo addition reactions like alkenes?

The answer lies in aromatic stabilisation: benzene's delocalised $\pi$ -electron cloud (the "aromatic sextet") is about 150 kJ/mol more stable than a hypothetical cyclohexatriene with localised double bonds.

If benzene simply added an electrophile (like $Br_2$ ), it would lose this stabilisation — a huge energy penalty.

So, nature chooses a different path: substitution instead of addition, preserving the aromatic ring.

The Key Formula: The Reaction Profile

The rate-determining step in electrophilic aromatic substitution (EAS) is the formation of the arenium ion (σ-complex):

\text{Ar-H} + E^+ \xrightarrow{\text{slow}} \text{Ar-E}^+ \text{H} \quad (\text{σ-complex})

Then, fast deprotonation restores aromaticity:

\text{Ar-E}^+ \text{H} + B^- \xrightarrow{\text{fast}} \text{Ar-E} + B\text{H}

Why This Holds: The Energy Barrier Logic

First step (slow): The electrophile $E^+$ attacks the electron-rich ring. The σ-complex is non-aromatic — it has only 4 π-electrons delocalised over 5 carbons (the sixth carbon is $sp^3$ hybridised). This intermediate is higher in energy than the starting benzene.
Second step (fast): A base removes the proton, restoring the aromatic sextet. This step is strongly exothermic — the system regains ~150 kJ/mol of stabilisation.

The overall reaction is exothermic, but the activation energy is dominated by the destabilisation of the σ-complex.

The Rate Law: Why It's First Order in Both

From the mechanism:

\text{Rate} = k[\text{Arene}][E^+]

Reasoning:

The slow step involves one molecule of arene and one molecule of electrophile.
No other species appear before the rate-determining step.
Therefore, the rate law is bimolecular — first order in each reactant.

This is not derived from the overall stoichiometry — it comes directly from the molecularity of the slow step.

The Hammett Equation: Quantifying Substituent Effects

For substituted benzenes, the rate constant $k$ relative to benzene ( $k_0$ ) follows:

\log \frac{k}{k_0} = \sigma \rho

Why This Holds

$\sigma$ (sigma constant): Measures the electronic effect of a substituent (electron-donating or withdrawing) relative to hydrogen. It is derived from the ionisation constants of benzoic acids — a purely empirical scale.
$\rho$ (rho constant): Measures the sensitivity of the reaction to substituent effects. A positive $\rho$ means the reaction is favoured by electron-withdrawing groups (rare in EAS); a negative $\rho$ means electron-donating groups accelerate the reaction.

Why it works:

The σ-complex has a positive charge delocalised over the ring. Substituents that stabilise this positive charge (electron-donating groups like $-\text{OH}$ , $-\text{NH}_2$ ) lower the activation energy — hence $\sigma$ is negative for such groups. Electron-withdrawing groups ( $-\text{NO}_2$ , $-\text{CN}$ ) destabilise the σ-complex — $\sigma$ is positive.

The linear free-energy relationship holds because the transition state resembles the σ-complex in charge distribution.

The Directing Effect: Why Ortho/Para vs Meta

The position of substitution is governed by the stability of the σ-complex for each possible attack site. …

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The key idea is to apply Markovnikov’s rule for electrophilic addition to alkenes and to recognise that carbocation rearrangements can occur. For (i) chlorobenzene with conc. $H_2SO_4$ , the major product is 4-chlorobenzenesulfonic acid (para-substitution). For (ii) 4-hydroxybenzyl alcohol with HCl, the major product is 4-hydroxybenzyl chloride (SN1 via a resonance-stabilised benzylic carbocation). For (iii) $(CH_3)_3C\text{-}OH$ with Cu at 573 K, the major product is 2-methylpropene (dehydration to an alkene).

Let’s work through each reaction step by step, focusing on the why behind each choice.

(i) Chlorobenzene + Conc. $H_2SO_4$ (heat)

What’s happening?

Concentrated sulfuric acid is a strong electrophile. It generates the sulfonium ion ( $SO_3H^+$ ) or simply $SO_3$ (depending on conditions), which acts as an electrophile in aromatic electrophilic substitution. Chlorobenzene has a chlorine atom attached to the benzene ring.
Directing effect of chlorine

Chlorine is an ortho-para director but also a deactivator (due to its strong -I effect). However, under forcing conditions (conc. $H_2SO_4$ , heat), sulfonation occurs. The major product is the para isomer because the ortho position is sterically hindered by the bulky chlorine atom and the incoming sulfonic acid group.
Why not meta?

The chlorine lone pairs can donate into the ring (+M effect), stabilising the arenium ion intermediate at ortho and para positions. Meta attack would not benefit from this resonance.
Final product

The reaction yields 4-chlorobenzenesulfonic acid (para-chlorobenzenesulfonic acid) as the major product. …

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

Showing the 12 most recent of 30 on this concept.

CBSE 2026Set ANNUAL1 mark
$-NH_2$
›Reveal solutionSolution

The $-NH_2$ group is a powerful activating, ortho/para-directing group, so aniline reacts instantly with bromine water at all three activated ring positions to give 2,4,6-tribromoaniline as a white precipitate.

The lone pair on the amino nitrogen delocalises into the aromatic ring by resonance, strongly raising electron density especially at the ortho ( $2,6$ ) and para ( $4$ ) positions. This makes those three positions so reactive toward electrophiles that no Lewis-acid catalyst is required (unlike ordinary benzene bromination, which needs $FeBr_3$ ), and substitution does not stop after one bromination — it proceeds at all three activated sites simultaneously: …

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CBSE 2026Set SEM31 markMCQ
$Q. The reagent which can be used for the following transformation is: phenol (C6H5OH) -> salicylaldehyde (2-hydroxybenzaldehyde, OH and CHO on adjacent ring carbons)$
(a) i) CHCl3, NaOH, 60-80 C ii) dil. HCl
(b) i) CO2, NaOH, 120-140 C ii) dil. HCl
(c) i) CCl4, NaOH, 60-80 C ii) dil. HCl
(d) i) HCHO, NaOH ii) dil. HCl
›Reveal solutionSolution
Phenol + CHCl3 + NaOH (60-80 C) then acidification gives 2-hydroxybenzaldehyde (salicylaldehyde) by the Reimer-Tiemann reaction. Correct option (a).

In the Reimer-Tiemann reaction, chloroform (CHCl3) with aqueous NaOH generates dichlorocarbene (:CCl2), the electrophile. It attacks the phenoxide ring, chiefly at the ortho position; subsequent hydrolysis on acidification (dil. HCl) converts the -CHCl2 group into -CHO, introducing an aldehyde group ortho to -OH.

Product: salicylaldehyde (2-hydroxybenzaldehyde).

CO2/NaOH (option b) is the Kolbe reaction, giving salicylic acid (-COOH), not the aldehyde. …
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CBSE 2025Set ANNUAL1 mark
$Q. Aniline does not undergo Friedel-Crafts reaction. Give reason.$
›Reveal solutionSolution

The catalyst itself reacts with aniline's basic amino group, deactivating the ring before any substitution can occur.

Friedel–Crafts reactions (alkylation/acylation) require the Lewis acid catalyst $AlCl_3$ . Aniline's $-NH_2$ group is strongly basic (it has a lone pair on nitrogen), so it readily reacts with $AlCl_3$ to form a salt/complex ( $C_6H_5\overset{+}{N}H_2-AlCl_3^-$ ).

…

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$Q. Fill in the blank: When phenol is reacted with concentrated nitric acid, the product formed is ________.$
›Reveal solutionSolution

Phenol reacts with concentrated nitric acid to give 2,4,6-trinitrophenol (picric acid) via nitration at all three activated ortho/para positions.

The -OH group of phenol is a strong activating, ortho/para-directing group. With concentrated HNO3 (a strong nitrating agent), phenol undergoes exhaustive electrophilic nitration at both ortho pos …

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CBSE 2025Set ANNUAL1 mark
$Q. What happens when aniline is treated with bromine water?$
›Reveal solutionSolution

The -NH2 group strongly activates the benzene ring, so aniline reacts with bromine water even without a catalyst, substituting at all three positions ortho/para to -NH2 at once.

Aniline's -NH2 group is a powerful electron-donating, ring-activating group (o,p-director). It makes the ring so reactive that bromine water reacts directly, without needing a Lewis-acid catalyst, substituting simultaneously at both ortho positions and the para position:

$\text{C}_6\text{H}_5\text{NH}_2 + 3\text{Br}_2 (\text{aq}) \rightarrow 2,4,6\text{-tribromoaniline}\downarrow (\text{white ppt}) + 3\text{HBr}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Reaction of bromine water with phenol gives:$
(a) 2, 4, 6-Tribromophenol
(b) o-Bromophenol and p-Bromophenol
(c) o-Bromophenol
(d) p-Bromophenol
›Reveal solutionSolution

Phenol's -OH group strongly activates the ring at all three of the ortho/ortho/para positions, so with excess aqueous bromine (bromine water) all three positions get substituted at once, giving 2,4,6-tribromophenol as a white precipitate — no catalyst needed.

The -OH group donates electron density into the ring by resonance, making the ortho and para positions highly electron-rich. Bromine water (dilute aqueous Br2) is reactive enough on its own (unlike with benzene, which needs a Lewis-acid catalyst like FeBr3) to brominate all three activated positions (2, …

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$Q. Identify the structure of the missing component in the given reaction sequence : Toluene --(conc. HNO3 + conc. H2SO4)--> ? --(Fe/HCl)--> 4-aminotoluene$
›Reveal solutionSolution

Nitration of toluene (methyl = o,p-director) followed by reduction of the nitro group gives the target amine — the missing intermediate is the nitro compound before reduction.

Toluene, treated with a nitrating mixture (conc. HNO3 + conc. H2SO4), undergoes electrophilic aromatic substitution. The methyl group is an ortho/para-directing, ring-activating substituent, so nitration occurs mainly at the para (and ortho) position, giving predominantly 4-nitrotoluene (p-nitrotoluene) as the major product. This nitro compound, on reduction with F …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. In the chlorination of benzene, the reactive species is$
(a) Cl+
(b) Cl-
(c) Cl2
(d) Cl2-
›Reveal solutionSolution

Chlorination of benzene proceeds via electrophilic attack by Cl+.

In the presence of a Lewis acid catalyst such as anhydrous FeCl3 or AlCl3, Cl2 is polarised and heterolysed to generate an electrophilic chlorine species, Cl+ (as part of a complex with the catalyst, e.g. [FeCl4]- Cl+). This Cl+ then attacks the electron-rich benzene ring …

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CBSE 2024Set 56/3/11 markMCQ
$Q. For the following question, two statements are given - one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Aniline does not undergo Friedel-Crafts reaction. Reason (R) : Friedel-Crafts reaction is an electrophilic substitution reaction.$
›Reveal solutionSolution
Aniline fails in Friedel-Crafts alkylation/acylation because the amino group forms a complex with the Lewis acid catalyst (AlCl₃), making the ring strongly deactivated. The Reason is true but does not explain this specific failure — it only states a general fact about the reaction type.

Concept first: Electrophilic Aromatic Substitution (EAS) and why aniline is special

Friedel-Crafts reactions are classic EAS reactions. In EAS, an electrophile attacks the electron-rich benzene ring. The more electron-rich the ring, the faster the reaction. Activating groups (like –NH₂, –OH, –OCH₃) donate electrons to the ring, making it more reactive toward electrophiles. So at first glance, aniline (C₆H₅NH₂) should be highly reactive in Friedel-Crafts reactions — the –NH₂ group is a strong activator.

But real chemistry is not that simple. The catalyst in Friedel-Crafts reactions is a Lewis acid, typically anhydrous AlCl₃. AlCl₃ is a strong electron-pair acceptor. The lone pair on the nitrogen of aniline is basic — it readily coordinates to AlCl₃, forming a salt-like complex. This complex changes everything.

Let’s walk through the reasoning step by step.

What the Assertion says: Aniline does not undergo Friedel-Crafts reaction. This is a well-known experimental fact. If you try to alkylate or acylate aniline using AlCl₃ and an alkyl halide or acyl halide, you get either no reaction or a messy tar. The desired product is not formed.

Why the Assertion is true: When aniline is mixed with AlCl₃, the nitrogen’s lone pair donates to the aluminium, forming C₆H₅NH₂·AlCl₃. This complex has a positive charge on nitrogen (or at least a strongly polarised N–Al bond). The –NH₂ group is no longer an electron-donating group — it becomes a strong electron-withdrawing group (–NH₂⁺AlCl₃⁻). This deactivates the ring so severely that even a powerful electrophile like the acylium ion cannot attack it. The ring becomes less reactive than nitrobenzene. So the reaction simply does not proceed.

What the Reason says: Friedel-Crafts reaction is an electrophilic substitution reaction. This is a true statement — it is the textbook definition. Both alkylation and acylation proceed via an electrophilic attack on the aromatic ring. …
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CBSE 2024Set A11 markMCQ
$\text{CH}_3\text{Cl}$
(a) Toluene
(b) O – chloroanisole
(c) Ortho and para-methylanisoles
(d) p – chloroanisole
›Reveal solutionSolution

Friedel–Crafts alkylation of anisole gives o- and p-methylanisole — option (c).

Anisole ( $\text{C}_6\text{H}_5\text{OCH}_3$ ) reacting with $\text{CH}_3\text{Cl}$ in the presence of anhydrous $\text{AlCl}_3$ is a Friedel–Crafts alkylation: a methyl group is introduced onto the ring. The methoxy group $-\text{OCH}_3$ is activating and ortho/para-directing, so the new methyl enters mainly at the ortho and para positions, giving a mixture of …
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CBSE 2024Set ANNUAL1 mark
$Q. Write directive influence of -OCH3 group present in anisole for electrophilic substitution reaction.$
›Reveal solutionSolution

The -OCH3 group in anisole donates electron density into the ring by resonance, activating the ring and directing incoming electrophiles preferentially to the ortho and para positions.

The oxygen of -OCH3 has lone pairs that can be delocalised into the benzene ring by resonance (+M/+R effect), increasing electron density specifically at the ortho and para positions relative to the -OCH3 group.

Although oxygen's electronegativity also exerts a small electron-withdrawing inductive (-I) effect, the stronger resonance donation dominates overall, making the ring more reactive than benzene itself (activating) towards electrophiles. …

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CBSE 2024Set ANNUAL1 mark
$Q. Write chemical name of white precipitate obtained on the reaction of phenol with bromine water.$
›Reveal solutionSolution

Phenol is highly reactive towards electrophilic bromination because the -OH group strongly activates the ring; even with dilute bromine water (no catalyst needed) it substitutes at all three available ortho/para positions at once.

The -OH group of phenol is a powerful activating, ortho/para-directing group (via resonance donation of its oxygen lone pair into the ring), making the ring far more reactive than benzene towards electrophiles. …

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Q.Write the major product(s) of the following reactions:

(i) Chlorobenzene + Conc. H2SO4H_2SO_4H2​SO4​ (heat)

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(i) Chlorobenzene + Conc. $H_2SO_4$ (heat)

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