You are given benzene, conc. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these rea

Exercises · 7.12

H_2SO_4

Odisha ChseTextbookSubjective· 3mImportance★★★★★

23% · 31/135 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 →

Phenol is prepared from benzene via a two-step process: first, sulfonation of benzene with conc. $H_2SO_4$ gives benzenesulfonic acid; second, fusion of its sodium salt with NaOH (followed by acidification) yields phenol. The overall transformation replaces a hydrogen on the benzene ring with an –OH group.

The Concept: Electrophilic Aromatic Substitution (EAS) as a Route to Phenol

Benzene is an electron-rich aromatic ring, but it does not easily undergo direct hydroxylation. The –OH group is a strong activating and ortho/para-directing group, but introducing it directly (e.g., with $OH^+$ ) is impractical because the conditions would also oxidize the product. Instead, we use a clever indirect route: introduce a good leaving group via EAS, then replace it by nucleophilic aromatic substitution under harsh conditions.

Sulfonation is a reversible EAS reaction. The sulfonic acid group ( $-SO_3H$ ) is a moderately good leaving group under strongly basic conditions. When benzenesulfonic acid (as its sodium salt) is fused with NaOH at high temperature, the hydroxide ion attacks the carbon bearing the sulfonate group, displacing it and forming the phenoxide ion. Acidification then gives phenol.

Watch out

A common mistake is to think that direct hydroxylation of benzene (e.g., with $H_2O_2$ or $OH^-$ ) works. It does not — benzene is too stable. The sulfonation–fusion route is the classic laboratory method.

Step-by-Step Preparation

1. Sulfonation of Benzene

Benzene reacts with concentrated sulfuric acid at moderate temperatures (around 80°C) to form benzenesulfonic acid. The electrophile is sulfur trioxide ( $SO_3$ ), which is present in conc. $H_2SO_4$ due to the equilibrium:

2H_2SO_4 \rightleftharpoons SO_3 + H_3O^+ + HSO_4^-

The mechanism is EAS: $SO_3$ is attacked by the $\pi$ electrons of benzene, forming a sigma complex (arenium ion), which then loses a proton to regenerate the aromatic ring.

The reaction equation is:

C_6H_6 + H_2SO_4 \xrightarrow{\Delta} C_6H_5SO_3H + H_2O

Note

This reaction is reversible. To drive it forward, we use excess conc. $H_2SO_4$ and remove water (or use fuming sulfuric acid). The product, benzenesulfonic acid, is a strong acid and is water-soluble.

2. Alkaline Fusion (with NaOH)

Benzenesulfonic acid is first neutralized by NaOH to its sodium salt, then fused with solid NaOH at a high temperature (around 300°C). This is a nucleophilic aromatic substitution (SNAr) reaction, but it follows an addition-elimination mechanism (not the typical SN2, because the aromatic ring is planar and the carbon is $sp^2$ hybridized).

The sulfonate group ( $-SO_3^-$ ) is a good leaving group under these conditions. The hydroxide ion attacks the carbon bearing the sulfonate group, forming a Meisenheimer complex (a negatively charged intermediate). The sulfonate group is then expelled, giving sodium phenoxide.

The equation for this step is:

C_6H_5SO_3Na + 2NaOH \xrightarrow{\text{fusion}} C_6H_5ONa + Na_2SO_3 + H_2O

Note: The benzenesulfonic acid is first neutralized to its sodium salt (by NaOH) before fusion. The reaction consumes 2 moles of NaOH per mole of sulfonate: one to form the salt, and one to provide the nucleophile.

Tip

The fusion temperature is critical. Too low, and the reaction is too slow; too high, and side reactions (like decomposition) occur. The phenoxide formed is stable at these temperatures.

3. Acidification to Phenol

The sodium phenoxide from the fusion is dissolved in water and then treated with a mineral acid (like dilute $H_2SO_4$ or HCl) to liberate phenol. Phenol is a weak acid ( $pK_a \approx 10$ ), so it is protonated by the strong acid.

The equation is:

C_6H_5ONa + H_2SO_4 \rightarrow C_6H_5OH + NaHSO_4

Or, more simply: …

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

Here are the common mistakes students make when preparing phenol from benzene using conc. $\text{H}_2\text{SO}_4$ and NaOH, along with how to avoid each.

Mistake 1: Forgetting the First Step (Sulfonation)

Students often try to jump directly from benzene to phenol, missing the crucial intermediate.

The Mistake: Writing a direct reaction like:

\text{C}_6\text{H}_6 + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{NaH}

This is chemically impossible under normal conditions. Benzene does not react directly with NaOH.

Why it’s wrong: The benzene ring is stabilized by resonance and is not nucleophilic enough to be attacked by the $\text{OH}^-$ ion. You need to first make the ring more reactive (or introduce a leaving group).
How to Avoid: Remember the two-step logic:
1. Activate the ring by introducing a good leaving group (sulfonic acid group, $-\text{SO}_3\text{H}$ ).
2. Displace the leaving group using the strong nucleophile ( $\text{OH}^-$ from NaOH).

Mistake 2: Writing the Wrong Reagent for Sulfonation

Students confuse the sulfonating agent.

The Mistake: Using dilute $\text{H}_2\text{SO}_4$ or writing the reagent as just " $\text{H}_2\text{SO}_4$ " without specifying "conc." (concentrated).
Why it’s wrong: Sulfonation requires the electrophile $\text{SO}_3$ (sulfur trioxide), which is generated in high concentration only from conc. $\text{H}_2\text{SO}_4$ (via the equilibrium: $2\text{H}_2\text{SO}_4 \rightleftharpoons \text{SO}_3 + \text{H}_3\text{O}^+ + \text{HSO}_4^-$ ). Dilute acid does not produce enough $\text{SO}_3$ .
How to Avoid: Always write conc. $\text{H}_2\text{SO}_4$ in the first step. The correct equation is:

\text{C}_6\text{H}_6 + \text{H}_2\text{SO}_4 (\text{conc.}) \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{SO}_3\text{H} + \text{H}_2\text{O}

Mistake 3: Forgetting the Fusion Step (High Temperature)

Students write the second step as a simple acid-base reaction at room temperature.

The Mistake: Writing:

\text{C}_6\text{H}_5\text{SO}_3\text{H} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{NaHSO}_3

This is incorrect. At room temperature, NaOH simply neutralizes the sulfonic acid to form the sodium salt (sodium benzenesulfonate).

Why it’s wrong: The $\text{SO}_3^-$ group is a poor leaving group at low temperatures. You need high temperature (fusion) to force the nucleophilic aromatic substitution ( $\text{S}_N\text{Ar}$ ) mechanism.
How to Avoid: Remember the "fusion" keyword. The correct sequence is:
1. Neutralization (room temp):

\text{C}_6\text{H}_5\text{SO}_3\text{H} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{SO}_3\text{Na} + \text{H}_2\text{O}

2.  **Fusion with excess NaOH** (high temp, ~300°C):

\text{C}_6\text{H}_5\text{SO}_3\text{Na} + 2\text{NaOH} \xrightarrow{\text{fusion}} \text{C}_6\text{H}_5\text{ONa} + \text{Na}_2\text{SO}_3 + \text{H}_2\text{O}

Mistake 4: Forgetting the Final Acidification Step

Students stop at sodium phenoxide ( $\text{C}_6\text{H}_5\text{ONa}$ ) and call it phenol. …

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

←Previous question7.11 — Write the mechanism of hydration of ethene to yield ethanol.→Next question7.13 — Show how will you synthesise: (i) 1-phenylethanol from a suitable alke…

Q.You are given benzene, conc. $H_2SO_4$ and NaOH. Write the equations for the preparation of phenol using these reagents.

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

The Concept: Electrophilic Aromatic Substitution (EAS) as a Route to Phenol

Step-by-Step Preparation

Method: Alkali Fusion of Sodium Benzenesulfonate

Step 1 — Sulfonation of Benzene

Step 2 — Neutralization to Sodium Salt

Step 3 — Alkali Fusion

Step 4 — Acidification to Phenol

Mistake 1: Forgetting the First Step (Sulfonation)

Mistake 2: Writing the Wrong Reagent for Sulfonation

Mistake 3: Forgetting the Fusion Step (High Temperature)

Mistake 4: Forgetting the Final Acidification Step

Q.You are given benzene, conc. H2SO4H_2SO_4H2​SO4​ and NaOH. Write the equations for the preparation of phenol using these reagents.

The Concept: Electrophilic Aromatic Substitution (EAS) as a Route to Phenol

Step-by-Step Preparation

🎓Unlock everything free for 14 days

Q.You are given benzene, conc. $H_2SO_4$ and NaOH. Write the equations for the preparation of phenol using these reagents.

Unlock everything free for 14 days