(a) Name the starting material used in the industrial preparation of phenol. (b) Write complete reaction for t

NCERT Exemplar · Q53

Q. (a) Name the starting material used in the industrial preparation of phenol.

(b) Write complete reaction for the bromination of phenol in aqueous and non-aqueous medium.

Gujarat GsebLong· 3mImportance★★★★★

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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 is so strongly activating that it undergoes electrophilic substitution without a Lewis acid catalyst — the oxygen’s lone pairs directly stabilise the arenium ion. In water, bromination gives 2,4,6-tribromophenol; in a non-polar solvent, it stops at monobromination. The industrial starting material is cumene (isopropylbenzene).

(a) Starting material for industrial phenol

The industrial route to phenol is the cumene process (also called the Hock process). The starting material is cumene — that’s isopropylbenzene, $C_6H_5CH(CH_3)_2$ . …

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Here is the clear, concept-first solution for your question on Electrophilic Aromatic Substitution (EAS) in phenol.

Method: Mechanism-Based Reasoning for Reactivity of Phenol

This method relies on understanding the activating effect of the -OH group. The lone pair on oxygen donates electron density into the benzene ring via resonance, making the ring highly nucleophilic (electron-rich). This explains all three parts of the question.

(a) Starting Material for Industrial Phenol

Method: Recall the Cumene Process (Hock process).

Steps:

Benzene is alkylated with propene to form cumene (isopropylbenzene).
Cumene is oxidized to cumene hydroperoxide.
This is cleaved with acid to yield phenol and acetone.

Answer: The starting material is cumene (isopropylbenzene).

Key point: This is the most common industrial route because it uses cheap feedstocks and produces two valuable products (phenol + acetone).

(b) Bromination of Phenol in Aqueous vs. Non-Aqueous Medium

Method: Compare the reactivity of the electrophile and the solvent effect.

1. In Aqueous Medium (Water)

Reagent: Bromine water ( $Br_2/H_2O$ ).
Observation: Immediate formation of a white precipitate of 2,4,6-tribromophenol.
Reaction:

\ce{C6H5OH + 3Br2 (aq) -> Br3C6H2OH (s) + 3HBr}

Why? Water polarizes $Br_2$ and the highly activated phenol undergoes tri-substitution at all three ortho/para positions.

2. In Non-Aqueous Medium (e.g., $CS_2$ , $CCl_4$ , or $CHCl_3$ )

Reagent: Bromine in a non-polar solvent.
Observation: Monobromination occurs, giving a mixture of ortho- and para-bromophenol.
Reaction:

\ce{C6H5OH + Br2 (in CS2) -> o-BrC6H4OH + p-BrC6H4OH + HBr}

Why? Without water to polarize $Br_2$ , the reaction is slower and controlled. The major product is para-bromophenol (due to less steric hindrance).

(c) Why Lewis Acid is Not Required for Bromination of Phenol …

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Here is a breakdown of the common mistakes students make regarding Electrophilic Aromatic Substitution (EAS) in phenol, specifically for the questions asked, along with precise corrections.

(a) Name the starting material used in the industrial preparation of phenol.

Common Mistake #1: Confusing the "Cumene Process" with older methods.

The Error: Students often write Benzene or Chlorobenzene as the starting material. While benzene is the ultimate source, the specific industrial starting material for the cumene process (which accounts for >90% of global phenol production) is Cumene (Isopropylbenzene).
Why it happens: Students memorize "phenol from benzene" but fail to distinguish between a laboratory method (Dow's process) and the modern industrial route.
How to Avoid: Memorize the Cumene Process as the standard industrial method. The sequence is:
- Benzene + Propene $\rightarrow$ Cumene (Isopropylbenzene)
- Cumene + $O_2$ $\rightarrow$ Cumene hydroperoxide
- Cumene hydroperoxide + $H^+$ $\rightarrow$ Phenol + Acetone
Correct Answer: Cumene (Isopropylbenzene).

(b) Write complete reaction for the bromination of phenol in aqueous and non-aqueous medium.

Common Mistake #2: Forgetting the solvent dictates the product.

The Error: Students write the same product (e.g., 2,4,6-tribromophenol) for both conditions, or they write the mono-brominated product for the aqueous medium.
Why it happens: They treat "bromination" as a single reaction, ignoring the massive activating effect of the -OH group and the role of the solvent.
How to Avoid: Remember the "Rule of Thumb" :
- Aqueous medium ( $Br_2/H_2O$ ): The -OH group activates the ring so strongly that polysubstitution occurs instantly. You get 2,4,6-tribromophenol (a white precipitate).
- Non-aqueous medium ( $Br_2/CS_2$ or $Br_2/CCl_4$ ): The reaction is controlled. You get mono-bromination at the ortho and para positions (major product is p-bromophenol).

Correct Reactions:

Aqueous medium (Polysubstitution):

\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2(aq) \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} \downarrow + 3\text{HBr}

*(Product: 2,4,6-Tribromophenol)*

Non-aqueous medium (Monosubstitution):

\text{C}_6\text{H}_5\text{OH} + \text{Br}_2 \xrightarrow{\text{CS}_2 \text{ or } \text{CCl}_4} \text{o-Bromophenol} + \text{p-Bromophenol} + \text{HBr}

(c) Explain why Lewis acid is not required in the bromination of phenol.

Common Mistake #3: Giving a vague or incorrect reason. …

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

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. [benzene ring]-NH2 --HNO3/H2SO4, 288K--> ____ is a major product.$
(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.

…

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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Which reagent is used to distinguish aniline and benzylamine?$
(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:

C6H5NH2 + 3 Br2 -> 2,4,6-Br3C6H2NH2 (white ppt) + 3 HBr. …

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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. Aniline + HNO3/H2SO4 at 288 K -> In this reaction, which product is obtained in greater proportion?$
(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 …

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GUJCET 2021Set 151 markMCQ
$Q. Which product is obtained by nitration of aniline? (A) o-nitroaniline (B) m-nitroaniline (C) p-nitroaniline (D) All above$
›Reveal solutionSolution

Protonation of aniline in acid makes the ring less selective → o, m and p nitroanilines all form.

Concept: $-NH_2$ is normally o/p-directing, but in strong acid aniline becomes anilinium $(-NH_3^+)$ , a deactivating m-director. The competition between the free amine and its cation gives a mixture: substantial para (~51%), signif …
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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. Phenol --(X, 273K)--> parabromophenol In the above reaction reagent 'X' is ______$
(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 …

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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. Give the correct order of proportion of products obtained by nitration of aniline.$
(a) o-nitroaniline > p-nitroaniline > m-nitroaniline
(b) m-nitroaniline > o-nitroaniline > p-nitroaniline
(c) m-nitroaniline > p-nitroaniline > o-nitroaniline
(d) p-nitroaniline > m-nitroaniline > o-nitroaniline
›Reveal solutionSolution

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.

…

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GUJCET 2015Set C1 markMCQ
$Br_2$
›Reveal solutionSolution
[!TLDR]

Anisole's ring is activated by –OCH $_3$ , so plain Br $_2$ in acetic acid brominates it (no FeBr $_3$ 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) Br $_2$ /CH $_3$ COOH — acetic acid is a suitable polar solvent; the activated ring brominates directly to give mainly p-bromoanisole. Correct. …
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←Previous questionQ52 — Write the mechanism of the reaction of HI with methoxybenzene.→Next questionQ54 — Explain the process (with reactions) by which phenol can be converted …

Q.(a) Name the starting material used in the industrial preparation of phenol.

(a) Starting material for industrial phenol

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