Nitration is an example of aromatic electrophilic substitution and its rate depends upon the group already pre

NCERT Exemplar · Q42

Q. Nitration is an example of aromatic electrophilic substitution and its rate depends upon the group already present in the benzene ring. Out of benzene and phenol, which one is more easily nitrated and why?

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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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Phenol is nitrated far more easily than benzene because the –OH group is a strong activating group that donates electron density into the ring via resonance, making the aromatic ring much more reactive toward electrophilic attack.

The core idea: what makes a ring “easy” to nitrate?

Nitration is an electrophilic aromatic substitution reaction. The attacking species is the nitronium ion ( $\text{NO}_2^+$ ), which is an electron-seeking (electrophilic) reagent. For the reaction to happen quickly, the benzene ring must be electron-rich — it must have a high electron density that can attract and stabilise the incoming $\text{NO}_2^+$ .

So the question reduces to: Which ring — benzene or phenol — has higher electron density in the aromatic π-system?

1. Benzene: the neutral baseline

Benzene has six π-electrons perfectly delocalised over six carbons. There is no substituent to push or pull electron density. Its electron density is uniform and moderate. The nitronium ion finds it “okay” to attack, but the reaction requires fairly harsh conditions — concentrated nitric and sulphuric acids, often with heat.

2. Phenol: the –OH group changes everything

In phenol, the –OH group is attached directly to the ring. Oxygen is highly electronegative, so you might guess it withdraws electrons. But that’s not the full story — and this is where the classic exam trap lies.

Watch out

A common mistake is to think that because oxygen is electronegative, the –OH group withdraws electrons by induction. In fact, the resonance effect dominates here, and it is strongly electron-donating.

The lone pairs on oxygen can delocalise into the π-system of the benzene ring. Draw the resonance structures:

One lone pair from oxygen forms a π-bond to the ipso carbon (the carbon attached to –OH).
This pushes negative charge into the ortho and para positions of the ring.
The result: the ring becomes significantly more electron-rich than benzene, especially at the ortho and para carbons.

\text{–OH} \xrightarrow{\text{resonance}} \text{increases electron density at ortho and para positions}

3. Why this makes nitration faster

The rate-determining step in nitration is the attack of $\text{NO}_2^+$ on the ring to form a σ-complex (arenium ion). This intermediate is positively charged and unstable. A ring that already has high electron density can better stabilise this positive charge — it “gives” electrons more readily. …

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Here are the common mistakes students make when answering this question, along with how to avoid each one.

1. Mistake: Saying "Phenol is more reactive because it has an –OH group" without explaining why

What students write: "Phenol is more reactive because of the –OH group."
Why it’s wrong: This is incomplete. The examiner expects the mechanism — how the –OH group affects the ring.

How to avoid: Always link the –OH group to electron donation and activation of the ring.

Correct logic: The –OH group donates electrons via resonance (lone pair on oxygen delocalises into the ring). This increases electron density at ortho and para positions, making the ring more nucleophilic and thus more reactive toward electrophilic substitution (nitration).

2. Mistake: Confusing activating vs deactivating groups

What students write: "Phenol is more reactive because –OH is a deactivating group."
Why it’s wrong: –OH is a strong activating group, not deactivating.

How to avoid: Memorise the shortlist:

Activating (electron-donating): –OH, –NH₂, –OCH₃, –CH₃ (alkyl groups)
Deactivating (electron-withdrawing): –NO₂, –CN, –COOH, –CHO, –SO₃H

Tip: For exams, remember: All ortho/para directors (except halogens) are activating.

3. Mistake: Forgetting to mention resonance vs inductive effect

What students write: "–OH group donates electrons by inductive effect."
Why it’s wrong: The inductive effect of –OH is actually electron-withdrawing (due to high electronegativity of oxygen). The net activation comes from resonance, which dominates.

How to avoid: Clearly state:

"The –OH group shows a +R effect (resonance donation) that outweighs its –I effect (inductive withdrawal). This makes the ring electron-rich."

4. Mistake: Not comparing with benzene explicitly

What students write: "Phenol is easily nitrated."
Why it’s wrong: The question asks for a comparison — you must state why phenol is more reactive than benzene.

How to avoid: Always write a direct comparison:

"Benzene has no activating group, so its electron density is uniform and lower. Phenol, due to the –OH group, has higher electron density on the ring, making it more reactive toward electrophiles like NO₂⁺."

5. Mistake: Writing the wrong reaction conditions

What students write: "Phenol undergoes nitration with conc. HNO₃ and conc. H₂SO₄ at 50–60°C."
Why it’s wrong: Phenol is so reactive that it undergoes nitration even with dilute HNO₃ at room temperature — no need for conc. acid or heat.

How to avoid: Know the conditions:

Benzene: Conc. HNO₃ + conc. H₂SO₄, 50–60°C
Phenol: Dilute HNO₃, room temperature (or even with nitrating mixture, but much milder) …

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

COMEDK 2026Set 2026-A1 markMCQ
$\begin{aligned} &\text { What is the major product }\left[\mathrm{P}_3\right] \text { formed when } n \text { - Hexane undergoes the given series of reactions: }\\ &\begin{aligned} \text { n-Hexane } & \frac{\mathrm{V}_2 \mathrm{O}_5}{20 \mathrm{~atm}, 773 \mathrm{k}}\left[\mathrm{P}_1\right] \\ {\left[\mathrm{P}_1\right] } & \xrightarrow[\triangle]{\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}, \text { Anhyd.AlCl }}\left[\mathrm{P}_2\right] \\ {\left[\mathrm{P}_2\right] } & \xrightarrow[333 \mathrm{~K}]{\text { Conc. } \mathrm{HNO}_3+\mathrm{H}_2 \mathrm{SO}_4}\left[\mathrm{P}_3\right] \text { Major product } \end{aligned} \end{aligned}$
›Reveal solutionSolution
The reaction sequence converts n-hexane to benzene (via aromatization), then ethylates it to ethylbenzene, and finally nitrates it to give para-ethylnitrobenzene as the major product. The correct option is (C).

The key to this problem is recognizing that each step in the sequence is a classic organic transformation, and the final product’s structure is determined by the directing effects of substituents already on the ring. Let’s walk through it step by step.

First step: Aromatization of n-hexane

n-Hexane ( $\mathrm{C_6H_{14}}$ ) is treated with $\mathrm{V_2O_5}$ at 20 atm and 773 K. This is a typical catalytic reforming or dehydrogenation condition. Vanadium pentoxide acts as a catalyst to cyclize and dehydrogenate the straight-chain alkane into an aromatic ring. The product $\mathrm{[P_1]}$ is benzene ( $\mathrm{C_6H_6}$ ).

Why? Under high temperature and pressure, alkanes with six or more carbons can undergo dehydrocyclization — losing hydrogen atoms and forming a stable aromatic ring. No other functional groups are introduced here.

Second step: Friedel–Crafts alkylation

Benzene ( $\mathrm{[P_1]}$ ) reacts with ethyl chloride ( $\mathrm{C_2H_5Cl}$ ) in the presence of anhydrous $\mathrm{AlCl_3}$ under heat ( $\triangle$ ). This is a classic Friedel–Crafts alkylation: the $\mathrm{AlCl_3}$ generates an ethyl carbocation (or an ethyl– $\mathrm{AlCl_3}$ complex) that attacks the benzene ring.

The product $\mathrm{[P_2]}$ is ethylbenzene ( $\mathrm{C_6H_5CH_2CH_3}$ ).

Note: Alkylation gives a mono-substituted product because the ethyl group is an activating, ortho/para-directing group, but under these conditions further alkylation is minimal (or we assume mono-substitution as the intended step).

Third step: Nitration of ethylbenzene

Ethylbenzene ( $\mathrm{[P_2]}$ ) is treated with concentrated nitric acid and sulfuric acid (a nitrating mixture) at 333 K. This is an electrophilic aromatic substitution. The nitronium ion ( $\mathrm{NO_2^+}$ ) is the attacking species.

The ethyl group is an ortho/para-directing group (it donates electron density via hyperconjugation and inductive effect). Therefore, the nitro group will preferentially attach at the ortho or para positions relative to the ethyl group. …
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COMEDK 2026Set 2026-M1 markMCQ
$\text { The decreasing order of reactivity towards electrophilic substitutions is: }$
›Reveal solutionSolution
The reactivity of a benzene ring toward electrophilic substitution is governed by the electron-donating or electron-withdrawing nature of its substituent. The correct decreasing order is III (anisole) > I (toluene) > II (benzene) > IV (trifluoromethylbenzene), which corresponds to option (B).

Concept & Intuition

Electrophilic substitution on benzene proceeds via a positively charged intermediate (the arenium ion). Substituents that donate electrons (by resonance or induction) stabilize this intermediate and activate the ring, making it more reactive than benzene itself. Substituents that withdraw electrons destabilize the intermediate and deactivate the ring. The key is to compare the net electronic effect of each group:

–OCH₃ (methoxy): Strongly activating. The oxygen lone pairs donate into the ring by resonance, overwhelming the weak inductive withdrawal.

–CH₃ (methyl): Mildly activating. It donates electrons by hyperconjugation and the +I effect.

–H (benzene): The reference point — no substituent effect.

–CF₃ (trifluoromethyl): Strongly deactivating. The three fluorine atoms pull electrons inductively, and there is no resonance donation to compensate.

Thus the order is: anisole > toluene > benzene > trifluoromethylbenzene.

Step-by-step reasoning

Identify the substituent effects

III (–OCH₃): The oxygen lone pairs are conjugated with the π-system, creating strong resonance donation (the methoxy group is ortho/para-directing and strongly activating).

I (–CH₃): The methyl group donates via hyperconjugation (C–H σ bonds overlap with the ring π-orbitals) and a small +I effect. This is a mild activator.

II (–H): No substituent — reactivity is the baseline.

IV (–CF₃): The highly electronegative fluorines pull electron density through σ-bonds (strong –I effect). No resonance donation is possible, so the ring becomes electron-poor and deactivated.

Rank the activating power

Strongest activator → weakest activator → deactivator:

–OCH₃ > –CH₃ > –H > –CF₃.

Translate to reactivity order …
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COMEDK 2026Set 2026-M1 markMCQ
$\text { Match reactions in Column I with the corresponding products formed as given in Column II. }$
›Reveal solutionSolution
The key is to match each reaction with its characteristic product by recalling the specific conditions: diazonium hydrolysis gives phenol, bromination in CS₂ at low temperature gives p-bromophenol, Reimer–Tiemann reaction gives salicylaldehyde, and aqueous bromination gives 2,4,6-tribromophenol. The correct matching is A→R, B→P, C→S, D→Q, which corresponds to option (A).

Concept and Intuition

This problem tests your knowledge of four classic organic reactions of benzene derivatives. Each reaction is distinguished by the reagent and conditions, which dictate the product. Instead of memorizing blindly, think about why each condition leads to a specific outcome:

Diazonium salts are excellent leaving groups; in warm water, they simply hydrolyze to give phenol.

Bromination of phenol can be controlled: in a non-polar solvent (CS₂) at low temperature, only one bromine substitutes at the para position (due to steric and electronic control). In aqueous solution, the reaction is much more vigorous, leading to trisubstitution.

Reimer–Tiemann reaction uses chloroform and strong base to install a formyl group ortho to the phenol – a classic example of a carbene intermediate.

Let’s go through each match step by step.

Step-by-Step Matching

1. Reaction A: Diazonium salt with warm water

Benzenediazonium chloride ( $\mathrm{C_6H_5N_2^+Cl^-}$ ) is unstable in water. The $\mathrm{N_2^+}$ group is a very good leaving group, and upon heating, it is replaced by a hydroxyl group from water.

Product: Phenol ( $\mathrm{C_6H_5OH}$ ).

So A → R (Phenol).

2. Reaction B: Phenol with $\mathrm{Br_2}$ in $\mathrm{CS_2}$ at 273 K

In a non-polar solvent like carbon disulfide and at low temperature, bromination of phenol is monosubstitution. The –OH group is strongly activating and ortho/para-directing. However, the ortho positions are somewhat sterically hindered; the major product is the para isomer.

Product: p-Bromophenol.

So B → P (p-Bromophenol).

3. Reaction C: Phenol with $\mathrm{CHCl_3}$ and aqueous $\mathrm{NaOH}$ , then $\mathrm{HCl}$

This is the Reimer–Tiemann reaction. Chloroform in the presence of strong base generates dichlorocarbene ( $:\mathrm{CCl_2}$ ), which attacks the electron-rich ortho position of the phenoxide ion. After hydrolysis with HCl, the product is an ortho-hydroxybenzaldehyde.

Product: 2-Hydroxybenzaldehyde (salicylaldehyde). …
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KCET 2026Set D31 markMCQ
$Q. Nitration of aniline in strong acidic medium gives significant amount of m-nitroaniline because (A) In electrophilic substitution reaction, amino group is meta directing (B) In strong acidic medium, aniline is present as anilinium ion (C) -NH2 group always directs to meta position (D) m-nitroaniline has higher molar mass than o&p nitroanilines$
›Reveal solutionSolution

Protonation of aniline's amino group under strongly acidic nitrating conditions removes the usual ortho/para-directing effect of $-NH_2$ and replaces it with the meta-directing effect of the positively charged $-NH_3^+$ group.

Step 1 — Normal behaviour of aniline

Free aniline's $-NH_2$ group has a lone pair that donates electron density into the ring by resonance, strongly activating the ring and directing electrophiles to the ortho and para positions.

Step 2 — What happens in strongly acidic medium

Nitration uses a mixture of concentrated $HNO_3$ and $H_2SO_4$ , a highly acidic environment. Under these conditions, the nitrogen lone pair of aniline is protonated before nitration can occur, converting $-NH_2$ into $-NH_3^+$ (the anilinium ion).

Step 3 — Why anilinium directs meta …
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KCET 2025Set D-41 markMCQ
$\mathrm{Br_2}$
›Reveal solutionSolution
The two compounds differ only in being a primary versus a secondary amine, so we need a test that is exclusive to primary amines — the carbylamine test ( $\mathrm{CHCl_3}$ + alc. KOH).

Step 1 — Classify the two amines.

Aniline, $\mathrm{C_6H_5-NH_2}$ — the nitrogen carries two hydrogens and one carbon → primary (1°) aromatic amine.

N-methylaniline, $\mathrm{C_6H_5-NH-CH_3}$ — the nitrogen carries one hydrogen and two carbons → secondary (2°) amine.

So any reagent that distinguishes them must respond differently to 1° versus 2° amines. Anything that only probes the aromatic ring or the basicity of $-\mathrm{NH}-$ will react with both.

Step 2 — Test each reagent.

(A) Acetic anhydride. Acetylation needs an N–H bond. Aniline has two, N-methylaniline has one — both react:
$\mathrm{C_6H_5NH_2 + (CH_3CO)_2O \longrightarrow C_6H_5NHCOCH_3 + CH_3COOH}$ $\mathrm{C_6H_5NH(CH_3) + (CH_3CO)_2O \longrightarrow C_6H_5N(CH_3)COCH_3 + CH_3COOH}$
Both give a substituted amide, so no distinction.

(B) $\mathrm{Br_2}$ water. The $-\mathrm{NH_2}$ and $-\mathrm{NHCH_3}$ groups are both strongly activating, ortho/para-directing groups. Both anilines therefore undergo rapid ring bromination to give a white precipitate of the 2,4,6-tribromo derivative. No distinction.

(C) Conc. HCl + anhydrous $\mathrm{ZnCl_2}$ . This is Lucas reagent — it is a test for alcohols (1°/2°/3°), distinguishing them by how fast a turbidity of the alkyl chloride appears. It has nothing to do with telling two amines apart; both amines simply form soluble ammonium salts with the acid. Not applicable. …
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COMEDK 2025Set 2025-M1 markMCQ
$[\mathrm{Z}]$
›Reveal solutionSolution
The reaction sequence converts phenol to benzene (via reduction), then nitrates it to nitrobenzene, and finally brominates it at the meta position relative to the nitro group, giving 3-bromonitrobenzene. The correct option is (D).

Concept & Intuition

This problem tests your understanding of how functional groups direct electrophilic aromatic substitution, and how a reaction sequence can completely change the directing effect. The key is to track the activating/deactivating and directing nature of the substituent at each step. Phenol’s –OH group is strongly activating and ortho/para-directing. But the first step (Zn dust, heat) removes that –OH entirely, giving benzene. Then nitration adds a meta-directing nitro group. Finally, bromination under conditions that prevent side reactions (cold & dark, FeBr₃) will place bromine meta to the nitro group. A common pitfall is to forget that the –OH is gone after step 1, and to mistakenly apply phenol’s directing effects to later steps.

Step-by-step reasoning

Step 1: Reduction of phenol to benzene

Phenol ( $\ce{C6H5OH}$ ) is heated with zinc dust. This is a classic reduction that removes the hydroxyl group, replacing it with hydrogen. The product [X] is benzene ( $\ce{C6H6}$ ).

Why? Zinc dust at high temperature acts as a reducing agent, cleaving the C–O bond and adding H (from trace moisture or the Zn itself). No substituent remains to direct further reactions — we now have a plain benzene ring.

Step 2: Nitration of benzene to nitrobenzene

Benzene [X] is treated with concentrated nitric acid and heat. This is standard electrophilic nitration. The product [Y] is nitrobenzene ( $\ce{C6H5NO2}$ ).

Why? The nitronium ion ( $\ce{NO2^+}$ ) attacks the benzene ring. Since there is no activating or deactivating group yet, the first substitution occurs at any position equally, but the product is simply nitrobenzene. The nitro group is strongly deactivating and meta-directing for further substitutions.

Step 3: Bromination of nitrobenzene …
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COMEDK 2024Set 2024-A1 markMCQ
$\text { IV }>\text { III }>\text { I }>\text { II }$
›Reveal solutionSolution
Electrophilic aromatic substitution is fastest on the ring bearing an activating group and slowest on the ring bearing a strong deactivator: isopropyl (activating) > benzene (reference) > chloro (weak deactivator) > nitro (strong deactivator), i.e. IV > III > I > II.

Rate of electrophilic substitution tracks the electron density of the ring:

Isopropylbenzene (IV): the isopropyl group is electron-donating (+I, hyperconjugation) → activated, most reactive.

Benzene (III): the unsubstituted reference. …
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COMEDK 2024Set 2024-E1 markMCQ
$_3$
›Reveal solutionSolution
Reactivity in Friedel–Crafts alkylation follows the ring's electron density: nitro (strong deactivator) $<$ bromo (weak deactivator) $<$ benzene $<$ methyl (activator). The increasing order is D < B < A < C, option (C).

Concept

Friedel–Crafts alkylation is an electrophilic aromatic substitution, and its rate is set by how electron-rich the ring is. Electron-donating groups raise the electron density and activate the ring; electron-withdrawing groups lower it and deactivate the ring. The stronger the effect, the larger the change in reactivity.

Solution

D (nitrobenzene): $-\text{NO}_2$ withdraws electrons strongly by both resonance and induction — the ring is the most deactivated, so D is least reactive.

B (bromobenzene): $-\text{Br}$ is net deactivating (inductive withdrawal dominates), so B is less reactive than plain benzene.

A (benzene): the reference, with no substituent. …
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KCET 2023Set D-21 markMCQ
$Q. Aniline does not undergo (A) Nitration (B) Sulphonation (C) Friedel-Craft reaction (D) Bromination$
›Reveal solutionSolution
Aniline is highly activated and basic, so it reacts with the Lewis acid in Friedel-Crafts conditions to form a salt that deactivates the ring — making the reaction fail. The correct option is (C).

The key here is to understand how the amino group ( $-\text{NH}_2$ ) in aniline behaves. It is a strongly activating and ortho/para-directing group because the lone pair on nitrogen can donate into the benzene ring through resonance. That makes aniline extremely reactive toward electrophilic substitution — too reactive, in fact, for some conditions.

But the same lone pair also makes aniline basic. In the presence of a strong Lewis acid (like $\text{AlCl}_3$ ), the nitrogen donates its lone pair to the acid, forming a salt. That salt has a positive charge on nitrogen, which is a powerful electron-withdrawing group. The ring becomes deactivated, and the Friedel-Crafts reaction fails.

Let’s go through each option.

Nitration — Aniline undergoes nitration readily. In fact, it is so reactive that direct nitration with $\text{HNO}_3/\text{H}_2\text{SO}_4$ gives a mixture including meta-product due to oxidation of the amino group. To avoid this, the amino group is first protected (e.g., by acetylation to acetanilide), then nitrated. But the point is: nitration does occur, so this is not the answer.

Sulphonation — Aniline reacts with concentrated $\text{H}_2\text{SO}_4$ to form anilinium hydrogen sulphate at low temperature. On heating, it undergoes sulphonation to give ortho- and para-aminobenzenesulphonic acids. So sulphonation is possible.

Friedel-Crafts reaction — This is the one that fails. In a Friedel-Crafts alkylation or acylation, the catalyst is a Lewis acid like $\text{AlCl}_3$ . Aniline, being a strong base, forms a complex with $\text{AlCl}_3$ : …
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COMEDK 2021Set 20211 markMCQ
$Q. Coupling reaction is an example of (A) nucleophilic addition reaction. (B) nucleophilic substitution reaction. (C) electrophilic substitution reaction. (D) electrophilic addition reaction.$
›Reveal solutionSolution

That mechanism - electrophile attacks the aromatic ring, H+ is lost, aromaticity restored - is precisely electrophilic aromatic SUBSTITUTION (the ring's H is replaced by the -N=N-Ar group).

Concept: Diazo coupling.

In a coupling reaction the benzenediazonium ion, C6H5-N2+, acts as a weak ELECTROPHILE and attacks a highly activated aromatic ring (phenol in mild alkali, or aniline in mildly acidic medium). The activated ring's pi cloud attacks the terminal N of the diazonium ion, a sigma (arenium) complex forms, and loss of H+ restores aromaticity, giving the azo dye Ar-N=N-Ar'. …
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COMEDK 2021Set 20211 markMCQ
$_3$
›Reveal solutionSolution
Benzene undergoes a Gattermann–Koch formylation with CO and HCl in the presence of anhydrous AlCl₃, giving benzaldehyde as the product.

The reaction you’re asking about is the Gattermann–Koch reaction — a classic method to introduce an aldehyde group directly onto an aromatic ring. The key idea is that carbon monoxide (CO) and hydrogen chloride (HCl) combine, under the Lewis acid catalysis of anhydrous AlCl₃, to generate a reactive electrophile: the formyl cation ( $\mathrm{HCO}^+$ ). This electrophile then attacks the electron-rich benzene ring in an electrophilic aromatic substitution.

Why does this work? AlCl₃ coordinates with CO, polarising it, and then HCl provides the proton, forming a complex that effectively acts as $\mathrm{HCO}^+$ . This is a strong electrophile, capable of substituting a hydrogen on benzene. The product is benzaldehyde — no further substitution occurs because the aldehyde group is deactivating and meta-directing, so the reaction stops at mono-formylation.

Let’s walk through the mechanism step by step.

Generation of the electrophile Anhydrous AlCl₃ (a strong Lewis acid) coordinates with the oxygen of carbon monoxide, pulling electron density away from the carbon. This makes the carbon highly electrophilic. HCl then adds, forming a complex that can be thought of as $\mathrm{HCO}^+$ (formyl cation) stabilised by $\mathrm{AlCl_4^-}$ :

$\mathrm{CO + HCl + AlCl_3 \rightarrow HCO^+ \, AlCl_4^-}$

Electrophilic attack on benzene The formyl cation attacks the benzene ring, forming a sigma complex (arenium ion). The positive charge is delocalised over the ring:

$\mathrm{C_6H_6 + HCO^+ \rightarrow C_6H_6\!-\!CHO^+}$

Deprotonation to restore aromaticity The $\mathrm{AlCl_4^-}$ ion (or a chloride ion) abstracts a proton from the sigma complex, regenerating the aromatic ring and yielding benzaldehyde: …
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Q.Nitration is an example of aromatic electrophilic substitution and its rate depends upon the group already present in the benzene ring. Out of benzene and phenol, which one is more easily nitrated and why?

Concept understanding — Electrophilic Aromatic Substitution

Electrophilic Aromatic Substitution – The First Meeting

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

The Precise Statement

The Mechanism in Three Steps

Why This Matters for Exams

Why this formula?

Electrophilic Aromatic Substitution: Why the Mechanism Holds

The Core Puzzle: Why Benzene Doesn't Just Add

The Key Formula: The Reaction Profile

Why This Holds: The Energy Barrier Logic

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

The Hammett Equation: Quantifying Substituent Effects

Why This Holds

The Directing Effect: Why Ortho/Para vs Meta

The core idea: what makes a ring “easy” to nitrate?

1. Benzene: the neutral baseline

2. Phenol: the –OH group changes everything

3. Why this makes nitration faster

Concept: Acidity of Phenol — Activating Effect of –OH Group

Method: Resonance Effect Analysis for Electrophilic Substitution

Steps

1. Mistake: Saying "Phenol is more reactive because it has an –OH group" without explaining why

2. Mistake: Confusing activating vs deactivating groups

3. Mistake: Forgetting to mention resonance vs inductive effect

4. Mistake: Not comparing with benzene explicitly

5. Mistake: Writing the wrong reaction conditions

Concept and Intuition

Step-by-Step Matching

Step 1 — Normal behaviour of aniline

Step 2 — What happens in strongly acidic medium

Step 3 — Why anilinium directs meta …

Concept

Solution

Q.Nitration is an example of aromatic electrophilic substitution and its rate depends upon the group already present in the benzene ring. Out of benzene and phenol, which one is more easily nitrated and why?

The core idea: what makes a ring “easy” to nitrate?

1. Benzene: the neutral baseline

2. Phenol: the –OH group changes everything

3. Why this makes nitration faster

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