Q.Which arenium ion(s) (sigma complex) is/are involved in the bromination of aniline? (Two or more options may be correct.)
(i) the arenium ion from ortho attack, drawn in its most stable resonance form with the positive charge on nitrogen as an iminium (+NH2 doubly bonded to the ring) and the H and Br on the adjacent (ortho) sp3 carbon
(ii) the arenium ion from ortho attack with the positive charge on a ring carbon (the –NH2 kept neutral) and the H and Br on the ortho sp3 carbon
(iii) the arenium ion from para attack, drawn in its most stable resonance form with the positive charge on nitrogen as an iminium (+NH2 doubly bonded to the ring) and the H and Br on the para sp3 carbon
(iv) the arenium ion from para attack with the positive charge on a ring carbon (the –NH2 kept neutral) and the H and Br on the para sp3 carbon
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 π 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:
Ar−H+EX+Ar−E+HX+
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 FeBrX3 to polarise BrX2 into BrX+.
Attack by the aromatic ring – The π 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 π electrons delocalised over five carbons, and one sp3 carbon bearing the electrophile and a hydrogen.
Benzene+E+⟶Sigma complex (non-aromatic)
Deprotonation – A base (often the counterion of the catalyst, like FeBrX4X−) removes the proton from the sp3 carbon. The pair of electrons from the C–H bond flows back into the ring, restoring the aromatic sextet.
Sigma complex+Base⟶Product+HX+
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 (C6H6) has three double bonds — so why doesn't it undergo addition reactions like alkenes?
The answer lies in aromatic stabilisation: benzene's delocalised π-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 Br2), 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):
Ar-H+E+slowAr-E+H(σ-complex)
Then, fast deprotonation restores aromaticity:
Ar-E+H+B−fastAr-E+BH
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 sp3 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:
Rate=k[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 (k0) follows:
logk0k=σρ
Why This Holds
σ (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.
ρ (rho constant): Measures the sensitivity of the reaction to substituent effects. A positive ρ means the reaction is favoured by electron-withdrawing groups (rare in EAS); a negative ρ 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 −OH, −NH2) lower the activation energy — hence σ is negative for such groups. Electron-withdrawing groups (−NO2, −CN) destabilise the σ-complex — σ 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. …
The –NH2 group makes aniline an ortho/para director because the nitrogen lone pair can help delocalise the positive charge of the arenium ion — giving an especially stable iminium (+NH2=) resonance form for ortho and para attack. …
Aniline brominates at the ortho and para positions because the nitrogen lone pair can be donated into the ring, delocalising the arenium-ion positive charge onto nitrogen (an iminium). The correct arenium ions are those for ortho and para attack shown with this +NH2= form.
Concept – the arenium (sigma) ion
In electrophilic aromatic substitution, the electrophile (Br+) adds to a ring carbon to give a resonance-stabilised carbocation (arenium ion / sigma complex) before the proton is lost. The more stable this intermediate, the more favoured that position.
Why ortho/para for aniline
For ortho and para attack, one resonance structure places the positive charge on the carbon bearing –NH2. The nitrogen lone pair can then form a double bond to that carbon, giving an iminium ion (+NH2=) in which every atom has a full octet. This is an especially stable contributor and is unavailable for meta attack. That extra stabilisation is exactly why –NH2 is a strong ortho/para director.
Choosing the options
(i) ortho attack shown as the iminium (+NH2=) resonance form — correct, the key stabilised arenium ion. …
To explain WHY a substituent directs electrophilic aromatic substitution to particular ring positions, draw every resonance structure of the arenium ion (sigma complex) formed by attack at each position and identify which position produces an extra, unusually stable resonance contributor.
Steps
For the substrate under attack, draw the arenium ion resulting from electrophilic addition at each candidate position (ortho, meta, para relative to the directing group).
For attack ortho or para to an electron-donating group with a lone pair (like -NH2), draw the resonance structure in which the positive charge sits on the ring carbon bearing that group.
Show that the group's lone pair can then form a new pi bond to that carbon (e.g. nitrogen becoming part of a C=NH2+ iminium unit), pushing the positive charge onto the heteroatom and giving every atom a complete octet - an extra, especially stable resonance contributor.
Confirm that this specific resonance form is NOT available for attack at the meta position (the positive charge there can never reach a carbon adjacent to the directing group), so the meta arenium ion lacks this extra stabilisation.
Conclude that the lower-energy (more stabilised) arenium ion at ortho/para attack is why the group is an ortho/para director and ring-activator. …
Q.[benzene ring]-NH2 --HNO3/H2SO4, 288K--> ____ is a major product.
(a) para-Nitroaniline (NH2, NO2 para)
(b) ortho-Nitroaniline (NH2, NO2 ortho)
(c) meta-Nitroaniline (NH2, NO2 meta)
(d) 1,3-Dinitrobenzene (two NO2 groups meta, no NH2 shown)
›Reveal solutionSolution
Nitrating aniline with HNO3/H2SO4 partly protonates the -NH2 to -NH3+ (a meta director), so the reaction gives a mixture of ortho, meta, and para nitroanilines; the major SINGLE product is still para-nitroaniline.
Aniline's –NH2 group is normally a strong ortho/para director. But in the strongly acidic HNO3/H2SO4 medium, a large fraction of aniline is protonated to the anilinium ion (–NH3+), which is a deactivating, META-directing group. So nitration of aniline actually gives a MIXTURE of all three isomers (ortho, meta, and para nitroaniline) — unusual for an activating substituent, and a well-known exception highlighted in NCERT.
Q.Which reagent is used to distinguish aniline and benzylamine?
(a) Br2/H2O
(b) C6H5SO2Cl
(c) CHCl3 + KOH
(d) CH3COCl/pyridine
›Reveal solutionSolution
Bromine water gives a white ppt (2,4,6-tribromoaniline) with aniline but not with benzylamine.
In aniline (C6H5NH2), the -NH2 is directly on the ring and strongly activates it, so aniline reacts instantly with bromine water to give a white precipitate of 2,4,6-tribromoaniline:
Q.Aniline + HNO3/H2SO4 at 288 K -> In this reaction, which product is obtained in greater proportion?
(a) o-nitroaniline
(b) m-nitroaniline
(c) p-nitroaniline
(d) a dinitrobenzene (no -NH2 group)
›Reveal solutionSolution
Nitration of aniline under strongly acidic conditions is complicated because much of the aniline is protonated to the anilinium ion, but the overall product mixture is still dominated by ortho and, most of all, para substitution.
In concentrated H2SO4, most aniline exists as the anilinium ion (C6H5NH3+), which is weakly meta-directing/deactivating; however, a small fraction of free -NH2 (a powerful ortho/para director) still directs nitration, and because the -NH2 group is a much stronger activator than the deactivated anilinium ring, the observed product distribut …
Q.Which product is obtained by nitration of aniline?
(A) o-nitroaniline
(B) m-nitroaniline
(C) p-nitroaniline
(D) All above
›Reveal solutionSolution
Protonation of aniline in acid makes the ring less selective → o, m and p nitroanilines all form.
Concept:−NH2 is normally o/p-directing, but in strong acid aniline becomes anilinium (−NH3+), a deactivating m-director. The competition between the free amine and its cation gives a mixture: substantial para (~51%), signif …
Q.Phenol --(X, 273K)--> parabromophenol
In the above reaction reagent 'X' is ______
(a) Bromine water
(b) Br2/FeBr3
(c) Br2/CH3COOH
(d) Br2/CS2
›Reveal solutionSolution
Phenol is so strongly activated toward electrophilic substitution that even mild bromine (dissolved in a non-polar solvent, at low temperature) brominates it; using a non-polar solvent at low temperature favours controlled monosubstitution at the less hindered para position.
Phenol reacts readily with molecular bromine even without a Lewis acid catalyst (unlike benzene) because the ring is strongly activated by the -OH group. With aqueous bromine (bromine water), the reaction proceeds all the way to 2,4,6-tribromophenol (an instant white precipitate, used as a qualitative test for phenol). To obtain a controlled MONO-bromination product, phenol is instead treated with Br2 dissol …
Nitration of aniline gives para (about 51%) > meta (about 47%) > ortho (about 2%).
The -NH2 group is strongly ortho/para directing. But nitration is done in a strongly acidic medium (HNO3/H2SO4), where aniline is largely protonated to the anilinium ion (C6H5NH3+). The -NH3+ group is deactivating and meta-directing.
Q.Which reagent is used for bromination of methyl phenyl ether?
(A) Br2 / CH3COOH
(B) Br2 / Red P
(C) Br2 / FeBr3
(D) HBr / Δ
›Reveal solutionSolution
[!TLDR]
Anisole's ring is activated by –OCH3, so plain Br2 in acetic acid brominates it (no FeBr3 needed) — option (A).
Concept
The methoxy group is an electron-donating, ortho/para-directing activator. It raises the ring's electron density enough that electrophilic bromination occurs readily with molecular bromine; a Lewis-acid catalyst (needed for deactivated/benzene rings) is unnecessary and would only be used for less reactive arenes.
Solution
(A) Br2/CH3COOH — acetic acid is a suitable polar solvent; the activated ring brominates directly to give mainly p-bromoanisole. Correct. …