When phenol is treated with bromine water, white precipitate is obtained. Give the structure and the name of t

NCERT Exemplar · Q28

Q. When phenol is treated with bromine water, white precipitate is obtained. Give the structure and the name of the product.

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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 reacts with bromine water via electrophilic substitution at all three activated positions, yielding 2,4,6-tribromophenol as a white precipitate.

The key to this reaction lies in understanding how the –OH group activates the benzene ring. The oxygen’s lone pairs donate electron density into the ring through resonance, making the ortho and para positions strongly nucleophilic. Bromine water ( $\text{Br}_2$ in water) provides a source of electrophilic bromine ( $\text{Br}^+$ ), which attacks these activated sites.

Unlike bromination of benzene (which requires a Lewis acid catalyst like $\text{FeBr}_3$ ), phenol is so reactive that it undergoes triple substitution even in mild conditions — no catalyst needed. The water in bromine water also helps by polarising the $\text{Br}_2$ molecule, generating $\text{Br}^+$ more readily.

Electrophile generation: In bromine water, $\text{Br}_2$ is polarised by water molecules, producing a small concentration of $\text{Br}^+$ ions:

\text{Br}_2 + \text{H}_2\text{O} \rightleftharpoons \text{Br}^+ + \text{Br}^- + \text{H}_2\text{O}

First substitution: The $\text{Br}^+$ attacks the para position (most activated, least sterically hindered) of phenol. A $\sigma$ -complex forms, then loses $\text{H}^+$ to restore aromaticity, giving 4-bromophenol.
Second substitution: The remaining ortho positions are still activated by the –OH group. Another $\text{Br}^+$ attacks one ortho position, yielding 2,4-dibromophenol.
Third substitution: The last ortho position (position 6) is still sufficiently activated. A third $\text{Br}^+$ attacks here, producing 2,4,6-tribromophenol.
Precipitation: 2,4,6-Tribromophenol is sparingly soluble in water due to its large nonpolar bromine atoms and the heavy molecular mass. It forms a white precipitate — often described as a "creamy white" or "pale yellow" solid in practice, but exam answers specify white. …

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

Mistake 1: Writing the wrong product structure (mono-brominated instead of tri-brominated)

Why it happens:

Students often think bromination of phenol is similar to benzene — one substitution at a time. They forget that the –OH group is a powerful activating group, making the ring highly reactive toward electrophilic substitution.

How to avoid:

Remember: Phenol + bromine water (excess bromine) gives 2,4,6-tribromophenol, not just monobromophenol.
The –OH group directs bromine to the ortho and para positions. With excess bromine, all three positions get substituted.

Correct structure:

\ce{OH} \quad \text{attached to a benzene ring with Br at positions 2, 4, and 6}

(Image: A benzene ring with –OH at C1, Br at C2, C4, C6)

Mistake 2: Writing the wrong IUPAC name

Why it happens:

Students confuse common names with IUPAC names, or they number the ring incorrectly.

How to avoid:

The IUPAC name is 2,4,6-tribromophenol.
Numbering starts from the carbon attached to –OH (C1). Bromines go to positions 2, 4, and 6.
Common name: Tribromophenol (acceptable in many exam contexts, but IUPAC is safer).

Correct name:

2,4,6-Tribromophenol

Mistake 3: Forgetting the white precipitate observation

Why it happens:

Students focus only on the product structure and miss the physical state — which is often asked in exams.

How to avoid:

Always note: 2,4,6-tribromophenol is a white precipitate (insoluble in water).
This is a test for phenol — the formation of a white precipitate with bromine water confirms the presence of a phenolic –OH group.

Key point:

White precipitate = 2,4,6-tribromophenol

Mistake 4: Writing the reaction without balancing or showing bromine water correctly

Why it happens:

Students write “Br₂” without specifying “bromine water” or forget that 3 moles of Br₂ are needed.

How to avoid:

Write the balanced equation:

\ce{C6H5OH + 3Br2 -> C6H2Br3OH + 3HBr}

Bromine water is Br₂ in water — it provides excess bromine for complete substitution.

Mistake 5: Confusing this with Friedel-Crafts bromination or catalytic bromination

Why it happens: …

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

TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
$\mathrm{p-CH_3C_6H_4COCl}$
›Reveal solutionSolution

A fails on temperature (diazonium salts need 273–278 K, not 285 K); C fails on identity (Hinsberg's reagent is benzenesulphonyl chloride, not an acyl chloride); only B — acetanilide being less reactive than aniline towards nitration — is correct. Option (C), "B only".

The concept first

All three statements orbit one theme: how the nitrogen lone pair of an aromatic amine behaves, and how we control it.

Diazonium chemistry. Aniline $+$ $\mathrm{NaNO_2/HCl}$ gives $\mathrm{C_6H_5N_2^+Cl^-}$ . The ion is stabilised by resonance with the ring — which is precisely why the aromatic diazonium salt can be isolated at all, whereas aliphatic ones decompose instantly. But even so, it is only kinetically stable in ice-cold conditions. Above about $278$ K, the excellent leaving group $\mathrm{N_2}$ departs and water attacks, giving phenol.

Protecting the amino group. Aniline is so strongly activated that direct nitration is a mess: the strongly acidic nitrating mixture protonates the $-\mathrm{NH_2}$ to $-\mathrm{NH_3^+}$ (a meta-directing deactivator), and oxidation side-reactions occur. The classical fix is to acetylate aniline to acetanilide (N-phenylethanamide). Now the nitrogen lone pair is shared with the acetyl carbonyl, so the ring is still activated but much less so, the amine can't be protonated, and nitration gives clean para-nitroacetanilide, which is then hydrolysed back. Understanding why we bother acetylating is exactly what makes statement B obviously true.

Hinsberg's reagent. Benzenesulphonyl chloride, $\mathrm{C_6H_5SO_2Cl}$ , distinguishes the three classes of amine: $1^\circ$ amines give a sulphonamide with an acidic N–H, soluble in alkali; $2^\circ$ amines give a sulphonamide with no N–H, insoluble in alkali; $3^\circ$ amines do not react. Note the $\mathrm{SO_2}$ — that is the fingerprint.

Step-by-step

Statement A — "Aniline forms a stable benzene diazonium chloride at 285 K".

Diazotisation is carried out at $273$ – $278$ K ( $0$ – $5\,^\circ\mathrm{C}$ ). $285$ K is $12\,^\circ\mathrm{C}$ — above the safe window. At that temperature the salt decomposes:
$\mathrm{C_6H_5N_2^+Cl^- + H_2O \longrightarrow C_6H_5OH + N_2\uparrow + HCl}$
So it is not stable at 285 K. A is incorrect. ✗ …
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TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
$\text{C}_6\text{H}_6$
›Reveal solutionSolution
Friedel–Crafts alkylation needs an electron-rich ring. The methyl group of toluene is activating, while $-$ NO $_2$ and $-$ COCH $_3$ are strongly deactivating (rings carrying them do not undergo Friedel–Crafts at all). Toluene — option (D).

The concept first

In Friedel–Crafts alkylation, a Lewis acid generates a carbocation:
$\text{R}\!-\!\text{Cl} + \text{AlCl}_3 \longrightarrow \text{R}^+ + [\text{AlCl}_4]^-$
That carbocation is an electrophile; it is attacked by the aromatic ring's $\pi$ electrons. So the reaction rate is governed by one thing: how much electron density the ring can offer.

Substituents fall into two camps:

Activating (ring-enriching): $-$ CH $_3$ and other alkyls ( $+I$ , hyperconjugation), $-$ OH, $-$ OR, $-$ NH $_2$ ( $+R$ ). They raise the rate.

Deactivating (ring-depleting): $-$ NO $_2$ , $-$ CHO, $-$ COR, $-$ COOH, $-$ CN, $-$ SO $_3$ H ( $-I$ and $-R$ ). They lower the rate — and in the special case of Friedel–Crafts they lower it so much that the reaction fails entirely. (Nitrobenzene is even used as a solvent for Friedel–Crafts reactions, precisely because it will not react.)

Step-by-step

Step 1 — Rank the substituents by their electronic effect.

(A) Benzene — no substituent; the baseline.

(B) Nitrobenzene, $-$ NO $_2$ — one of the strongest deactivators known ( $-I$ and $-R$ ). Ring badly electron-poor.

(C) Acetophenone, $-$ COCH $_3$ — a carbonyl; $-I$ and $-R$ , strongly deactivating. Also, its lone pair can complex with AlCl $_3$ , poisoning the catalyst.

(D) Toluene, $-$ CH $_3$ — electron-donating by $+I$ and hyperconjugation. Ring electron-rich.

Step 2 — Apply the rule. Rate of electrophilic substitution increases with ring electron density: …
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TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ

\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{Cu_2Br_2/HBr}} \mathrm{C_6H_5Br}

›Reveal solution

The Sandmeyer reaction replaces $-\mathrm{N_2^+}$ using a cuprous salt: $\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{Cu_2Br_2/HBr}} \mathrm{C_6H_5Br}$ . That is option (A); the others are the Gattermann, hydrolysis, and Balz–Schiemann reactions respectively.

The concept first: one diazonium salt, four different name reactions

Aryl diazonium salts are a crossroads. From the same $\mathrm{C_6H_5N_2Cl}$ you can go in many directions, and each direction has its own name and its own reagent. Examiners test whether you can tell them apart, so learn them as a set, not one at a time:

Reagent	Product	Name of reaction
$\mathrm{Cu_2Cl_2/HCl}$ , $\mathrm{Cu_2Br_2/HBr}$ , $\mathrm{CuCN/KCN}$	$\mathrm{ArCl}$ , $\mathrm{ArBr}$ , $\mathrm{ArCN}$	Sandmeyer
Cu powder + $\mathrm{HCl/HBr}$	$\mathrm{ArCl}$ , $\mathrm{ArBr}$	Gattermann
$\mathrm{H_2O}$ , warm	$\mathrm{ArOH}$ (phenol)	hydrolysis
$\mathrm{HBF_4}$ , then $\Delta$	$\mathrm{ArF}$	Balz–Schiemann
$\mathrm{KI}$	$\mathrm{ArI}$	(direct, no Cu needed)
$\mathrm{H_3PO_2}/\mathrm{H_2O}$ or $\mathrm{C_2H_5OH}$	$\mathrm{ArH}$	deamination

The distinguishing mark of Sandmeyer is the cuprous (Cu(I)) salt, $\mathrm{Cu_2X_2}$ , in the corresponding halogen acid.

Step-by-step: test each option

$\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{Cu_2Br_2/HBr}} \mathrm{C_6H_5Br}$ — cuprous bromide in HBr, giving bromobenzene. This is exactly the Sandmeyer prescription. ✓
$\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{Cu/HBr}} \mathrm{C_6H_5Br}$ — the product is the same, but the reagent is copper powder, not the cuprous salt. This is the Gattermann reaction — a close cousin, deliberately placed here as the trap. ✗
$\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{H_2O}} \mathrm{C_6H_5OH}$ — this is simple hydrolysis to phenol (which is why diazonium salts must be kept cold). Not Sandmeyer. ✗ …

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TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
$Q. The major product formed in the following reactions is$
i) NaOH, C $_2$ H $_5$ I
ii) CH $_3$ COCl, anhyd. AlCl $_3$
iii) Zn-Hg, conc. HCl (A) [Image of benzene ring with OC $_2$ H $_5$ and C $_2$ H $_5$ substituents at 1,2 positions] (B) [Image of benzene ring with OC $_2$ H $_5$ and C $_2$ H $_5$ substituents at 1,3 positions] (C) [Image of benzene ring with OH and C $_2$ H $_5$ substituents at 1,2 positions] (D) [Image of benzene ring with OC $_2$ H $_5$ and C $_2$ H $_5$ substituents at 1,4 positions]
›Reveal solutionSolution
The sequence involves Williamson ether synthesis, Friedel–Crafts acylation, and Clemmensen reduction; the final product is 1-ethoxy-4-ethylbenzene, which corresponds to the 1,4‑disubstituted pattern — option (D).

Concept & Intuition

This problem tests your ability to predict the outcome of a multi‑step synthesis on an aromatic ring. The three steps are classic transformations:

Williamson ether synthesis – converts a phenol into an aryl alkyl ether.

Friedel–Crafts acylation – introduces an acyl group (here, acetyl) ortho/para to the activating ether group.

Clemmensen reduction – reduces the carbonyl (C=O) to a methylene (CH₂), giving an alkyl side chain.

The key insight: the ether group (–OC₂H₅) is a strong activating and ortho/para‑directing substituent. So the acylation will occur at the para position (less steric hindrance than ortho). The final reduction then turns the –COCH₃ into –CH₂CH₃, leaving the ether untouched. The product is a 1,4‑disubstituted benzene with –OC₂H₅ and –C₂H₅.

Step‑by‑Step Reasoning

Step (i): NaOH, C₂H₅I The starting material is phenol (C₆H₅OH). NaOH deprotonates it to phenoxide (C₆H₅O⁻), which then attacks ethyl iodide (C₂H₅I) in an Sₙ2 reaction. This is a Williamson ether synthesis, giving ethoxybenzene (phenetole):

$\text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5\text{O}^- \xrightarrow{\text{C}_2\text{H}_5\text{I}} \text{C}_6\text{H}_5\text{O}\text{C}_2\text{H}_5$

Step (ii): CH₃COCl, anhyd. AlCl₃ This is a Friedel–Crafts acylation. The –OC₂H₅ group is a strong ortho/para‑director because the oxygen lone pairs donate electron density into the ring.

The ortho positions are sterically hindered by the ethoxy group.

The para position is more accessible. Therefore, the acetyl group (–COCH₃) attaches para to the ethoxy group, forming 4‑ethoxyacetophenone:

$\text{C}_6\text{H}_5\text{O}\text{C}_2\text{H}_5 + \text{CH}_3\text{COCl} \xrightarrow{\text{AlCl}_3} p\text{-}\text{C}_2\text{H}_5\text{O}\text{-}\text{C}_6\text{H}_4\text{-}\text{COCH}_3$

Step (iii): Zn‑Hg, conc. HCl …
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TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
$_3$
›Reveal solutionSolution
Friedel–Crafts alkylation of benzene with n-propyl chloride proceeds via a carbocation rearrangement, so the major product is isopropyl benzene, not n-propyl benzene.

The reaction described is a classic Friedel–Crafts alkylation. In this reaction, an alkyl halide reacts with benzene in the presence of a Lewis acid catalyst (here, anhydrous AlCl₃) to form an alkylbenzene. The key mechanistic step is the generation of a carbocation from the alkyl halide. That carbocation then attacks the benzene ring.

The trap in this specific case is that the n-propyl carbocation (CH₃CH₂CH₂⁺) is primary and therefore relatively unstable. Under the reaction conditions, it rapidly rearranges to the more stable secondary carbocation (CH₃CH⁺CH₃) via a 1,2-hydride shift. This rearranged carbocation is what actually attacks benzene, giving isopropyl benzene as the major product.

Let’s walk through the reasoning step by step.

Generation of the electrophile. Anhydrous AlCl₃ coordinates with the chlorine atom of n-propyl chloride, weakening the C–Cl bond. This leads to the formation of a carbocation and an AlCl₄⁻ complex:

$\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{AlCl}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2^+ + \text{AlCl}_4^-$

Carbocation rearrangement. The primary carbocation (CH₃CH₂CH₂⁺) is not very stable. A hydrogen atom from the second carbon (along with its bonding pair) shifts to the positively charged carbon. This 1,2-hydride shift converts the primary carbocation into a more stable secondary carbocation:

$\text{CH}_3\text{CH}_2\text{CH}_2^+ \xrightarrow{\text{1,2-hydride shift}} \text{CH}_3\text{CH}^+\text{CH}_3$

Electrophilic attack on benzene. The rearranged secondary carbocation acts as the electrophile. It attacks the electron-rich benzene ring, forming a sigma complex (arenium ion): $\text{C}_6\text{H}_6 + \text{CH}_3\text{CH}^+\text{CH}_3 \rightarrow \text{C}_6\text{H}_6^+(\text{CH}(\text{CH}_3)_2)$ …
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TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
$0\,^\circ\mathrm{C}$
›Reveal solutionSolution
The strongly activating $-\mathrm{OH}$ controls the substitution, so two equivalents of bromine enter the positions ortho and para to it — giving 3,5-dibromo-2-hydroxybenzaldehyde, option (B).

The concept first

When a ring carries two substituents, the more powerfully activating one decides where the electrophile goes.

$\mathrm{-OH}$ : strongly activating via its lone pair ( $+R \gg -I$ ), ortho/para directing.

$\mathrm{-CHO}$ : deactivating ( $-R, -I$ ), meta directing. The phenol group is far stronger, so it wins. Conveniently, the positions ortho/para to the OH of salicylaldehyde are also meta to the CHO — the two groups' preferences agree, which is why the reaction is so clean.

Step-by-step

Number the ring. Salicylaldehyde is 2-hydroxybenzaldehyde: $\mathrm{CHO}$ at C-1, $\mathrm{OH}$ at C-2.

Find the positions activated by the OH (at C-2).

ortho to C-2 → C-1 (blocked by the CHO) and C-3 ✓

para to C-2 → C-5 ✓ So the two available activated sites are C-3 and C-5.

Cross-check against the CHO (at C-1). Meta to C-1 → C-3 and C-5 — exactly the same two carbons. Both directing effects therefore point to C-3 and C-5. …
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Q.When phenol is treated with bromine water, white precipitate is obtained. Give the structure and the name of the product.

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

Method: Electrophilic Aromatic Substitution (Bromination of Phenol)

Why This Happens (Concept First)

Steps of the Method

Mistake 1: Writing the wrong product structure (mono-brominated instead of tri-brominated)

Mistake 2: Writing the wrong IUPAC name

Mistake 3: Forgetting the white precipitate observation

Mistake 4: Writing the reaction without balancing or showing bromine water correctly

Mistake 5: Confusing this with Friedel-Crafts bromination or catalytic bromination

The concept first

Step-by-step

The concept first

Step-by-step

The concept first: one diazonium salt, four different name reactions

Step-by-step: test each option

Concept & Intuition

Step‑by‑Step Reasoning

The concept first

Step-by-step

Q.When phenol is treated with bromine water, white precipitate is obtained. Give the structure and the name of the product.

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