In the nitration of benzene using a mixture of conc. $H_2SO_4$ and conc. $HNO_3$, the species which initiates the reaction is ____. (i) $NO_2$ (ii) $NO^+$ (iii) $NO_2^+$ (iv) $NO_2^-$

> [!TLDR] > The nitration of benzene proceeds via electrophilic aromatic substitution, where the attacking electrophile is the nitronium ion (NO2+), generated from the acid mixture - option (iii). Benzene's delocalised pi-system makes it a nucleophile; attacking such a stable aromatic ring needs a powerful electrophile. The mixture of concentrated nitric and sulphuric acids generates exactly that species. 1. Role of the acid mixture: HNO3 and H2SO4 are in equilibrium with H2NO3+ and HSO4-; the protonated nitric acid readily loses water to form NO2+ and H2O. Sulphuric acid also absorbs the water produced, driving the equilibrium forward. 2. Why NO2+ is the attacking species: it is a linear, positively charged, electron-deficient species that can accept a pair of electrons from benzene's pi-system. 3. The other options: (i) NO2 is a neutral free radical, not an electrophile in this mechanism; (ii) NO+ (nitrosonium) is generated under different (nitrosation) conditions and gives different products; (iv) NO2- (nitrite ion) is a nucleophile and would be repelled by benzene's electron cloud. …

In the nitration of benzene using a mixture of conc. H₂SO₄ and conc. HNO₃, the species which initiates the rea

NCERT Exemplar · Q16

H_2SO_4

(i)

NO_2

(ii)

NO^+

(iii)

NO_2^+

(iv)

NO_2^-

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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 nitration of benzene proceeds via electrophilic aromatic substitution, where the attacking electrophile is the nitronium ion (NO2+), generated from the acid mixture - option (iii).

Benzene's delocalised pi-system makes it a nucleophile; attacking such a stable aromatic ring needs a powerful electrophile. The mixture of concentrated nitric and sulphuric acids generates exactly that species.

Role of the acid mixture: HNO3 and H2SO4 are in equilibrium with H2NO3+ and HSO4-; the protonated nitric acid readily loses water to form NO2+ and H2O. Sulphuric acid also absorbs the water produced, driving the equilibrium forward.
Why NO2+ is the attacking species: it is a linear, positively charged, electron-deficient species that can accept a pair of electrons from benzene's pi-system.
The other options: (i) NO2 is a neutral free radical, not an electrophile in this mechanism; (ii) NO+ (nitrosonium) is generated under different (nitrosation) conditions and gives different products; (iv) NO2- (nitrite ion) is a nucleophile and would be repelled by benzene's electron cloud. …

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Common Mistakes in Identifying the Electrophile for Benzene Nitration

The Correct Answer

The species that initiates nitration of benzene is $NO_2^+$ (nitronium ion) — option (C).

Mistake #1: Choosing $NO_2$ (Neutral Nitrogen Dioxide)

Why students pick it:

They see "nitration" and think the nitro group ( $-NO_2$ ) comes from a neutral $NO_2$ molecule.

Why it's wrong:

Neutral $NO_2$ is a free radical, not an electrophile. Benzene requires a strong electrophile (electron-poor species) to attack its electron-rich ring. $NO_2$ is not sufficiently electron-deficient to initiate the reaction.

How to avoid:

Remember: Nitration uses a mixture of acids — the role of $H_2SO_4$ is to generate the active species. The neutral $NO_2$ is never the attacking agent in this classic reaction.

Mistake #2: Choosing $NO^+$ (Nitrosonium Ion)

Why students pick it:

The formula looks similar to $NO_2^+$ , and students confuse nitration ( $-NO_2$ addition) with nitrosation ( $-NO$ addition).

Why it's wrong:

$NO^+$ is the active species in nitrosation (e.g., diazotization of amines), not in nitration. It adds $-NO$ , not $-NO_2$ , to the ring.

How to avoid:

Nitration → adds $-NO_2$ → electrophile is $NO_2^+$
Nitrosation → adds $-NO$ → electrophile is $NO^+$
Count the oxygens: $NO_2^+$ has two oxygens (nitro group), $NO^+$ has one (nitroso group).

Mistake #3: Choosing $NO_2^-$ (Nitrite Ion)

Why students pick it:

They see a negative charge and think "reactive species" without checking charge compatibility.

Why it's wrong:

$NO_2^-$ is a nucleophile (electron-rich). Benzene needs an electrophile (electron-poor) to attack it. A negatively charged ion will be repelled by benzene's electron cloud, not attracted.

How to avoid:

Always ask: Is this species electron-rich or electron-poor?

Benzene is electron-rich → needs an electrophile (positive or neutral but electron-deficient).
$NO_2^-$ is negatively charged → it's a nucleophile → cannot initiate electrophilic substitution.

--- …

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

KEAM 2026Set eng-2026-04174 marksMCQ
$Br_2$
›Reveal solutionSolution

Using $Br_2$ in the non-polar solvent $CS_2$ at low temperature (273 K) suppresses polybromination and gives p-bromophenol as the major monobrominated product.

Phenol is strongly activated, so bromine water (polar, ionising) gives 2,4,6-tribromophenol.

To stop at monobromination, a low-polarity solvent and low temperature are used, which lowers the electrophilicity/availability of $Br^+$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$Q. When phenol is treated with excess of bromine water, it gives (A) o-bromophenol (B) o- and p-bromophenol (C) 1,3,5-tribromophenol (D) 2,4-dibromophenol (E) 2,4,6-tribromophenol$
›Reveal solutionSolution

Phenol with excess bromine water undergoes electrophilic substitution at the ortho and para positions, giving a white precipitate of 2,4,6-tribromophenol. …

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KEAM 2026Set pha-2026-0418F4 marksMCQ
$Br_2$
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Aniline $\rightarrow$ acetanilide (acetic anhydride/pyridine) $\rightarrow$ bromination gives mainly p-bromoacetanilide.

Aniline is acetylated to acetanilide $C_6H_5NHCOCH_3$ . The acetamido group is an activating ortho/para director, but the bulky $-NHCOCH_3$ hinders the ortho positions, so electrophilic bromination with $Br_2$ / $CH_3COOH$ occurs predominantly at the para po …
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KEAM 2026Set pha-2026-0420F4 marksMCQ
$Q. Which of the following reaction yieldstarry oxidation products? (A) Sulphonation of aniline (B) Nitration of aniline (C) Firedel-Crafts alkylation aniline (D) Firedel-Crafts alkylation of aniline (E) Bromination of aniline$
›Reveal solutionSolution

Aniline is readily oxidised; direct nitration with HNO $_3$ /H $_2$ SO $_4$ oxidises it to dark tarry products, so the amino group is protected (acetylated) first.

Aniline is very easily oxidised because the ring is electron-rich. When it is subjected to direct nitration with the strongly oxidising nitrating mixture ( $\mathrm{HNO_3/H_2SO_4}$ ), a large part of it is oxidised to dark, tarry products rather than cleanly nitrated. This is exactly why, in practice, aniline is first acetylated ( …
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KEAM 2025Set eng-2025-04264 marksMCQ
$H_2SO_4$
›Reveal solutionSolution

Sulphonation of phenol gives phenol-2,4-disulphonic acid; subsequent nitration replaces the $–SO_3H$ groups to yield picric acid (2,4,6-trinitrophenol).

Treating phenol with concentrated $H_2SO_4$ introduces sulphonic acid groups, giving phenol-2,4-disulphonic acid (X). On treatment with concentrated $HNO_3$ , the readily displaceable sulphonic groups are replaced by nitro groups and the ring is further nitrated, producing picric acid (2,4,6-trinitrophenol, Y). …
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KEAM 2025Set eng-2025-04284 marksMCQ
$C_6H_5NH_2 \xrightarrow[Pyridine]{(CH_3CO)_2O} A \xrightarrow[CH_3COOH]{Br_2} B$
›Reveal solutionSolution

Acetylation moderates aniline; the acetamido group is an o/p-director and bromination gives mainly the para product.

$\mathrm{C_6H_5NH_2}$ + $(\mathrm{CH_3CO})_2\mathrm{O}$ /pyridine $\to$ acetanilide (A), $\mathrm{C_6H_5NHCOCH_3}$ . The acetamido group is a strong ortho/para director; steric factors make the para product dominant. With $\mathrm{Br_2/CH_3COOH}$ the final product B is **p-bromoacetan …
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$+ Br_2 \xrightarrow{FeBr_3}$
›Reveal solutionSolution

$\text{-}OH$ activates and directs ortho/para. With the para position occupied by $\text{-}CH_3$ , electrophilic bromination goes ortho to the $\text{-}OH$ , yielding 2-bromo-4-methylphenol.

p-Cresol is 4-methylphenol, with $\text{-}OH$ and $\text{-}CH_3$ para to each other. Both substituents are ortho/para directors, but the $\text{-}OH$ group is a much stronger activator and controls the orientation. Its para position is already occupied by the methyl group, so electrophilic aromatic bromination (with $Br_2/FeBr_3$ ) occurs at the position ortho to the $\text{-}OH$ . The major product is **2-bromo-4-methylphen …
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$AlCl_3$ generates the acylium electrophile $CH_3CO^+$ , which substitutes a ring hydrogen (para to Cl). This is electrophilic aromatic substitution.

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$C_6H_6O$
›Reveal solutionSolution

X = phenol, Y = benzene, Z = acetophenone: phenol is reduced by Zn dust to benzene, which undergoes Friedel–Crafts acylation to give acetophenone ( $C_8H_8O$ ).

Identify X: $C_6H_6O$ is phenol. Heating phenol with zinc dust reduces it (removes the –OH):
$C_6H_5OH + Zn \longrightarrow C_6H_6 + ZnO$
so Y = benzene.

Benzene then undergoes Friedel–Crafts acylation with acetyl chloride and anhydrous $AlCl_3$ : …
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Q.In the nitration of benzene using a mixture of conc. $H_2SO_4$ and conc. $HNO_3$ , the species which initiates the reaction is ____.

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

Concept: Electrophilic Aromatic Substitution (Nitration)

Method: Identify the Active Electrophile Generated In Situ

Common Mistakes in Identifying the Electrophile for Benzene Nitration

The Correct Answer

Mistake #1: Choosing $NO_2$ (Neutral Nitrogen Dioxide)

Mistake #2: Choosing $NO^+$ (Nitrosonium Ion)

Mistake #3: Choosing $NO_2^-$ (Nitrite Ion)

Q.In the nitration of benzene using a mixture of conc. H2SO4H_2SO_4H2​SO4​ and conc. HNO3HNO_3HNO3​, the species which initiates the reaction is ____.

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Q.In the nitration of benzene using a mixture of conc. $H_2SO_4$ and conc. $HNO_3$ , the species which initiates the reaction is ____.

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