Q.Assertion: Like bromination of benzene, bromination of phenol is also carried out in the presence of a Lewis acid.
Reason: Lewis acid polarises the bromine molecule.
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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.
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
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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+.
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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+
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
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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.
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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. …
Bromination of benzene needs a Lewis acid (e.g. FeBr3) to generate a strong enough electrophile, because benzene's ring is not very reactive. Phenol's -OH group strongly activates the ring, so phenol reacts with Br2 in water directly, without any …
Phenol's ring is so strongly activated by the -OH group that it brominates instantly with Br2/water at room temperature, without needing a Lewis acid catalyst - unlike benzene, which does need one. So the assertion is wrong, though the reason (that Lewis acids polarise Br2) is a true, general statement about halogenation. The correct option is (iv).
Reasoning
Benzene's pi system is not nucleophilic enough to react with Br2 alone; a Lewis acid such as FeBr3 coordinates with Br2 and polarises it into a stronger electrophile, which benzene needs. Phenol, however, has its ring strongly activated by resonance donation from the -OH oxygen, making it far more electron-rich than benzene. Because of this, phenol reacts readily with Br2/water at room temperature with no catalyst, giving the white precipitate 2,4,6-trib …
Concept: Electrophilic Aromatic Substitution (EAS) in Phenol vs Benzene
Method: Compare Reactivity & Mechanism
Step 1 — Understand the Assertion
Bromination of benzene requires a Lewis acid (like FeBr3) to generate the electrophile Br+.
Bromination of phenol, however, occurs without a Lewis acid — the -OH group is strongly activating, so phenol reacts directly with Br2 at room temperature (even in water).
Result: Assertion is wrong.
Step 2 — Understand the Reason …
Here are the common mistakes students make with this Assertion-Reason question, along with how to avoid each.
Mistake 1: Assuming the Assertion is Correct
- The Error: Students often memorize that "bromination of benzene needs a Lewis acid (like FeBr3)" and incorrectly extend this rule to phenol. They assume the Assertion is true.
- Why it's Wrong: Phenol is highly activated due to the resonance donation from the –OH group. This makes the ring so electron-rich that it reacts with bromine without a catalyst. In fact, bromination of phenol occurs so readily that it gives 2,4,6-tribromophenol (a white precipitate) even with bromine water at room temperature. A Lewis acid is not required.
- How to Avoid: Always check the activating power of the substituent.
- Benzene: Needs a Lewis acid to polarize Br2.
- Phenol (and aniline): So reactive that they undergo electrophilic substitution without a catalyst. The –OH group itself polarizes the bromine molecule.
Mistake 2: Confusing the Role of the Lewis Acid
- The Error: Students think the Reason is wrong because they believe a Lewis acid's job is to generate an electrophile (like Br+), not just to "polarise" the molecule.
- Why it's Wrong: The Reason is actually correct. A Lewis acid (like FeBr3) does polarise the bromine molecule: Br2+FeBr3→Brδ+—Brδ−—FeBr3 This polarization creates a partial positive charge on one bromine, making it a strong enough electrophile to attack benzene.
- How to Avoid: Understand the mechanism. The Lewis acid doesn't create a free Br+ ion in most cases; it simply polarises the Br-Br bond, making one end electrophilic. The Reason is a valid general statement.
Mistake 3: Choosing the Wrong Option (A vs. B)
- The Error: Students who realize the Assertion is false often panic and pick (D) without checking if the Reason is correct. Others who think the Assertion is true pick (A) or (B) incorrectly.
- Why it's Wrong: The correct logic flow is:
- Assertion: "Bromination of phenol is carried out in the presence of a Lewis acid." → False. (Phenol doesn't need it).
- Reason: "Lewis acid polarises the bromine molecule." → True. (This is a correct general fact).
- Since Assertion is false and Reason is true, the answer is (D). …
- JKBOSE Class 12 Annual Regular Examination 2019Set KD1 markQ.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. …
- JKBOSE Class 12 Annual Regular Examination 2018Set SZ1 markQ.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.
…
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