Assertion (A): Toluene on Friedal Crafts methylation gives o- and p-xylene. Reason (R): CH3-group bonded to be

NCERT Exemplar · Q45

Q. Assertion (A): Toluene on Friedal Crafts methylation gives o- and p-xylene. Reason (R): CH3-group bonded to benzene ring increases electron density at o- and p- position.

(i) Both A and R are correct and R is the correct explanation of A.

(ii) Both A and R are correct but R is not the correct explanation of A.

(iii) Both A and R are not correct.

(iv) A is not correct but R is correct.

Bihar BsebShort· 1mImportance★★★★★

93% · 84/90 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 →

The methyl group in toluene is an ortho/para-directing activator that increases electron density at those positions through hyperconjugation and induction. Friedel–Crafts methylation of toluene does yield o- and p-xylene, so both the assertion and reason are correct, and the reason correctly explains the assertion.

Electrophilic aromatic substitution on benzene derivatives depends entirely on what substituent is already attached. Some groups donate electron density into the ring, making it more reactive and directing incoming electrophiles to specific positions; others withdraw density and deactivate the ring. The methyl group is a textbook example of an activating, ortho/para-directing substituent.

When toluene undergoes Friedel–Crafts methylation, a methyl carbocation (or a complex like $\ce{CH3^+–AlCl3}$ ) attacks the aromatic ring. The question is: where does it attack, and why?

Why the methyl group activates and directs ortho/para

Electron donation by hyperconjugation and induction

The $\ce{CH3}$ group has three $\ce{C–H}$ σ-bonds adjacent to the aromatic ring. These σ-electrons can overlap with the π-system of the benzene ring, a stabilising interaction called hyperconjugation. Additionally, alkyl groups are weakly electron-releasing by induction (carbon is less electronegative than hydrogen in this context). Both effects push electron density into the ring.
Ortho and para positions receive the most density

Resonance structures show that the extra electron density from the methyl group is delocalised preferentially to the ortho and para carbons. If you draw the resonance contributors of toluene, the negative charge (representing excess electron density) appears at the 2-, 4-, and 6-positions (ortho and para), not at the 3- and 5-positions (meta).
Electrophile attacks where density is highest

The incoming $\ce{CH3^+}$ is electron-deficient and seeks the most electron-rich sites. Because the ortho and para positions have higher electron density, the electrophile preferentially attacks there. The intermediate σ-complex (arenium ion) formed at these positions is also more stable, because the positive charge can be delocalised onto the carbon bearing the methyl group, which further stabilises it by hyperconjugation.

Tip

A quick mnemonic: electron-donating groups (EDG) are ortho/para directors and activators; electron-withdrawing groups (EWG, except halogens) are meta directors and deactivators. Halogens are the oddball—they deactivate but still direct ortho/para due to lone-pair resonance donation.

The reaction: Friedel–Crafts methylation of toluene

Reagents and mechanism Friedel–Crafts alkylation uses an alkyl halide (e.g., $\ce{CH3Cl}$ ) and a Lewis acid catalyst (e.g., $\ce{AlCl3}$ ). The catalyst generates the electrophile: $\ce{CH3Cl + AlCl3 -> CH3^+ + AlCl4^-}$ …

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

←Previous questionQ44 — Assertion (A): The compound cyclooctatetraene has a cyclic structural …→Next questionQ46 — Assertion (A): Nitration of benzene with nitric acid requires the use …

Q.Assertion (A): Toluene on Friedal Crafts methylation gives o- and p-xylene. Reason (R): CH3-group bonded to benzene ring increases electron density at o- and p- position.

Why the methyl group activates and directs ortho/para

The reaction: Friedel–Crafts methylation of toluene

🎓Unlock everything free for 14 days

Q.Assertion (A): Toluene on Friedal Crafts methylation gives o- and p-xylene.
Reason (R): CH3-group bonded to benzene ring increases electron density at o- and p- position.

Unlock everything free for 14 days