Write equations of the following reactions: (i) Fr – Class 12 Chemistry

Exercises · 7.31

Q. Write equations of the following reactions:

(i) Friedel-Crafts reaction - alkylation of anisole.

(ii) Nitration of anisole.

(iii) Bromination of anisole in ethanoic acid medium.

(iv) Friedel-Craft's acetylation of anisole.

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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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Anisole undergoes electrophilic aromatic substitution (EAS) at the ortho/para positions because the methoxy group ( $-\text{OCH}_3$ ) is a strong activating and ortho/para-directing group. The four reactions below show how the electrophile attacks the activated ring, giving major ortho/para products.

Concept and Intuition

Anisole (methoxybenzene) is a classic example of an activated aromatic ring. The oxygen atom in the $-\text{OCH}_3$ group has lone pairs that can donate electron density into the benzene ring via resonance. This makes the ring much more reactive than benzene itself toward electrophiles. Crucially, this resonance effect concentrates electron density at the ortho and para positions relative to the methoxy group. So, in any electrophilic aromatic substitution (EAS) reaction, the electrophile will preferentially attack these positions.

The reactions below are standard EAS transformations. The key is to remember that the methoxy group is so strongly activating that even mild conditions (like bromination without a Lewis acid) work, and that the ortho/para orientation is consistent across all four.

Step-by-Step Solutions

1. Friedel-Crafts alkylation of anisole

Here, we introduce an alkyl group (say, methyl) using an alkyl halide and a Lewis acid catalyst (usually $\text{AlCl}_3$ ). The catalyst generates a carbocation electrophile.

Reaction: Anisole reacts with methyl chloride ( $\text{CH}_3\text{Cl}$ ) in the presence of anhydrous $\text{AlCl}_3$ .
Mechanism: $\text{AlCl}_3$ pulls the chlorine from $\text{CH}_3\text{Cl}$ , forming a methyl carbocation ( $\text{CH}_3^+$ ). This electrophile attacks the ortho or para position of anisole. After deprotonation, the product is a mixture of o- and p-methoxytoluene.
Equation:

\text{C}_6\text{H}_5\text{OCH}_3 + \text{CH}_3\text{Cl} \xrightarrow{\text{AlCl}_3} \text{o-CH}_3\text{C}_6\text{H}_4\text{OCH}_3 + \text{p-CH}_3\text{C}_6\text{H}_4\text{OCH}_3 + \text{HCl}

Watch out

In Friedel-Crafts alkylation, the product (e.g., p-methoxytoluene) is more reactive than anisole itself. This can lead to polyalkylation (multiple alkyl groups adding) unless the reaction is carefully controlled with excess anisole or low temperature.

2. Nitration of anisole

Nitration introduces a nitro group ( $-\text{NO}_2$ ) using a mixture of concentrated nitric and sulfuric acids. The nitronium ion ( $\text{NO}_2^+$ ) is the electrophile.

Reaction: Anisole is treated with a cold mixture of $\text{HNO}_3$ and $\text{H}_2\text{SO}_4$ .
Mechanism: $\text{HNO}_3$ is protonated by $\text{H}_2\text{SO}_4$ , then loses water to form $\text{NO}_2^+$ . This attacks the ortho/para positions. The major product is p-nitroanisole (less steric hindrance than ortho), along with some o-nitroanisole.
Equation:

\text{C}_6\text{H}_5\text{OCH}_3 + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} \text{o-O}_2\text{NC}_6\text{H}_4\text{OCH}_3 + \text{p-O}_2\text{NC}_6\text{H}_4\text{OCH}_3 + \text{H}_2\text{O}

Tip

The ortho/para ratio in nitration of anisole is roughly 30-40% ortho and 60-70% para. The ortho product is slightly less favoured due to steric hindrance from the bulky methoxy group.

3. Bromination of anisole in ethanoic acid medium

Bromination of anisole is so easy that it occurs without a Lewis acid catalyst (unlike benzene, which needs $\text{FeBr}_3$ ). Ethanoic acid (acetic acid) simply provides a polar solvent.

Reaction: Anisole reacts with bromine ( $\text{Br}_2$ ) in glacial acetic acid at room temperature.
Mechanism: The polar solvent polarises the $\text{Br}_2$ molecule, making one bromine slightly positive (the electrophile). This attacks the ortho/para positions. The major product is p-bromoanisole (again, less steric hindrance).
Equation:

\text{C}_6\text{H}_5\text{OCH}_3 + \text{Br}_2 \xrightarrow{\text{CH}_3\text{COOH}} \text{o-BrC}_6\text{H}_4\text{OCH}_3 + \text{p-BrC}_6\text{H}_4\text{OCH}_3 + \text{HBr}

Note

In practice, bromination of anisole in ethanoic acid gives predominantly the para product (over 90%) because the ortho position is somewhat blocked by the methoxy group's bulk. This is a classic example of steric control overriding electronic preference. …

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Method: Electrophilic Aromatic Substitution (EAS) Mechanism

Core Concept

Anisole (methoxybenzene) has a strong activating and ortho/para-directing group ( $-\text{OCH}_3$ ). The lone pair on oxygen donates electron density into the ring via resonance, making the ortho and para positions more nucleophilic.

General Steps for Each Reaction

Generation of the electrophile (varies by reaction)
Attack by the aromatic ring — the $\pi$ electrons of anisole attack the electrophile, forming a arenium ion (sigma complex)
Loss of a proton — a base removes $H^+$ from the arenium ion, restoring aromaticity

(i) Friedel–Crafts Alkylation of Anisole

Electrophile generated: Methyl carbocation ( $\text{CH}_3^+$ ) from $\text{CH}_3\text{Cl}$ and $\text{AlCl}_3$

Equation:

\ce{ \text{Anisole} + CH3Cl ->[AlCl3] \text{ortho- and para-Methoxytoluene} + HCl }

Step-by-step:

$\text{CH}_3\text{Cl} + \text{AlCl}_3 \rightarrow \text{CH}_3^+ + \text{AlCl}_4^-$
$\text{CH}_3^+$ attacks the ring at ortho/para positions
$\text{AlCl}_4^-$ removes $H^+$ to regenerate $\text{AlCl}_3$

(ii) Nitration of Anisole

Electrophile generated: Nitronium ion ( $\text{NO}_2^+$ ) from $\text{HNO}_3$ and $\text{H}_2\text{SO}_4$

Equation:

\ce{ \text{Anisole} + HNO3 ->[H2SO4] \text{ortho- and para-Nitroanisole} + H2O }

Step-by-step:

$\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+$
$\text{NO}_2^+$ attacks ortho/para positions
$\text{HSO}_4^-$ removes $H^+$ to restore aromaticity

(iii) Bromination of Anisole in Ethanoic Acid Medium

Electrophile generated: Bromonium ion ( $\text{Br}^+$ ) from $\text{Br}_2$ — ethanoic acid acts as a polar solvent, polarising the $\text{Br}_2$ molecule

Equation:

\ce{ \text{Anisole} + Br2 ->[CH3COOH] \text{ortho- and para-Bromoanisole} + HBr }

Step-by-step: …

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Here are the common mistakes students make when writing equations for the electrophilic aromatic substitution (EAS) reactions of anisole, along with how to avoid each.

Mistake 1: Forgetting the Role of the Methoxy Group ( $-OCH_3$ )

The Mistake: Students treat anisole like benzene, writing the product as if the substituent goes to a random position. They often forget that the $-OCH_3$ group is a strong activating and ortho/para-directing group.
Why it’s wrong: The lone pair on the oxygen donates electron density into the ring via resonance. This makes the ortho and para positions much more electron-rich (and thus more reactive) than the meta position.
How to Avoid: Always draw the resonance structures of anisole first. You will see the negative charge (and thus the site of attack) appears only at the ortho and para carbons. The major product will always be a mixture of ortho and para isomers.

Mistake 2: Writing the Wrong Reagents for Nitration

The Mistake: Writing the nitrating agent as simply " $\text{HNO}_3$ " or " $\text{HNO}_3 + \text{H}_2\text{O}$ ".
Why it’s wrong: The actual electrophile is the nitronium ion ( $\text{NO}_2^+$ ). This is generated by protonating nitric acid with a stronger acid. Without the correct acid, the reaction won't proceed effectively.
How to Avoid: Always write the reagent as a mixture of concentrated nitric acid and concentrated sulfuric acid ( $\text{HNO}_3 + \text{H}_2\text{SO}_4$ $HNO_{3} + H_{2} SO_{4}$ ). The sulfuric acid acts as a catalyst and a dehydrating agent to form the $\text{NO}_2^+$ $NO_{2}^{+}$ ion.
- Correct Equation:

\text{C}_6\text{H}_5\text{OCH}_3 + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} \text{o-Nitroanisole} + \text{p-Nitroanisole} + \text{H}_2\text{O}

Mistake 3: Using the Wrong Halogenating Agent for Bromination

The Mistake: Writing the reagent as " $\text{Br}_2$ " alone, or using " $\text{Br}_2/\text{FeBr}_3$ " (which is correct for benzene but not for anisole in ethanoic acid).
Why it’s wrong: Anisole is so activated that it reacts instantly with $\text{Br}_2$ even without a Lewis acid catalyst. Using $\text{Br}_2$ in a non-polar solvent leads to polybromination (multiple bromines attaching). The question specifies "in ethanoic acid medium" to control the reaction.
How to Avoid: Recognize that ethanoic acid (acetic acid, $\text{CH}_3\text{COOH}$ $CH_{3} COOH$ ) is a polar protic solvent. It helps to stabilize the polar transition state and slows down the reaction, allowing for mono-bromination at the para position (the major product due to steric hindrance at the ortho position).
- Correct Equation:

\text{C}_6\text{H}_5\text{OCH}_3 + \text{Br}_2 \xrightarrow{\text{CH}_3\text{COOH}} \text{p-Bromoanisole} + \text{HBr}

Mistake 4: Confusing Alkylation with Acylation

The Mistake: Using the same reagent (e.g., $\text{CH}_3\text{Cl}/\text{AlCl}_3$ ) for both alkylation and acetylation, or writing the acylating agent incorrectly.
Why it’s wrong: Friedel-Crafts alkylation uses an alkyl halide ( $\text{R-X}$ ) and a Lewis acid ( $\text{AlCl}_3$ ). Friedel-Crafts acylation uses an acyl halide ( $\text{RCO-X}$ ) or an acid anhydride. The product of acylation is a ketone ( $\text{C}_6\text{H}_5\text{COR}$ ), not an alkylbenzene.
How to Avoid: Memorize the specific reagents.
- Alkylation: $\text{R-Cl} + \text{AlCl}_3$ (generates $\text{R}^+$ carbocation).
- Acetylation: $\text{CH}_3\text{COCl} + \text{AlCl}_3$ (generates $\text{CH}_3\text{CO}^+$ acylium ion).
- Correct Equation for Acetylation: …

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Q.Write equations of the following reactions:

Concept and Intuition

Step-by-Step Solutions

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