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^-

Jammu Kashmir JkboseMCQ· 1mImportance★★★★★

41% · 44/108 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 — 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. …

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

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. …

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

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.

--- …

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

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

JKBOSE Class 12 Annual Regular Examination 2019Set KD1 mark
$Q. What is the directive influence of phenolic group?$
›Reveal solutionSolution
-OH on the phenol ring is a strong electron-donating, ortho/para-directing group in electrophilic substitution, due to resonance donation of a lone pair into the ring.

Directive influence of the phenolic -OH group

In phenol, the oxygen of the -OH group has lone pairs that can conjugate (delocalise) into the attached benzene ring by resonance (the +R/+M mesomeric effect). This pushes extra electron density specifically onto the ortho and para carbons of the ring (as can be seen by drawing the resonance structures of phenol, where negative charge appears at the ortho and para positions).

Consequences:

Activating group: because the ring is electron-richer than benzene, phenol undergoes electrophilic aromatic substitution (nitration, halogenation, sulphonation, Friedel-Crafts reactions) much faster and under much milder conditions than benzene itself — e.g. phenol reacts instantly with bromine water at room temperature (no catalyst needed) to give 2,4,6-tribromophenol, whereas benzene needs a Lewis-acid catalyst. …
Start your 14-day free trial to see the full solution →
JKBOSE Class 12 Annual Regular Examination 2018Set SZ1 mark
$Q. Give the reaction of Phenol with Bromine water.$
›Reveal solutionSolution

Phenol's –OH group activates the benzene ring so strongly (by resonance) that it reacts instantly with bromine water — no catalyst required, unlike benzene — brominating at all three positions open to it (both ortho and the one para) at once.

Reaction: when bromine water is added to an aqueous solution of phenol at room temperature, a white precipitate forms immediately: C₆H₅OH + 3Br₂(aq) → C₆H₂Br₃OH (2,4,6-tribromophenol, white precipitate) + 3HBr.

Why it is so much more reactive than benzene: the lone pair of electrons on the oxygen of the –OH group is delocalised into the aromatic ring by resonance, which strongly increases the electron density specifically at the ortho and para positions. This makes phenol enormously more reactive than benzene towards electrophilic aromatic substitution — so much so that bromination occurs readily at ordinary temperature without needing a Lewis-acid catalyst (such as FeBr₃, which benzene requires), and substitution occurs simultaneously at all three available o/p positions, giving the fully tribrominated product directly.

…

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

←Previous questionQ15 — The gas evolved when methylamine reacts with nitrous acid is __. (i) N…→Next questionQ17 — Reduction of aromatic nitro compounds using Fe and HCl gives __. (i) a…

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 ____.

🎓Unlock everything free for 14 days

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 ____.

Unlock everything free for 14 days