Explain the process (with reactions) by which phenol can be converted to aspirin.

NCERT Exemplar · Q54

Q. Explain the process (with reactions) by which phenol can be converted to aspirin.

Andhra Pradesh BieapLong· 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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Aspirin is synthesised from phenol via a two-step electrophilic aromatic substitution sequence: first, Kolbe-Schmitt carboxylation introduces a carboxyl group ortho to the –OH, giving salicylic acid; then, acetylation of the phenolic –OH with acetic anhydride yields aspirin (acetylsalicylic acid).

The key to understanding this conversion is recognising that phenol’s –OH group is a strongly activating, ortho/para-directing substituent. This makes the aromatic ring highly reactive toward electrophilic attack, especially at the ortho position. The Kolbe-Schmitt reaction exploits this by using carbon dioxide as a weak electrophile under basic conditions — the phenoxide ion is even more activated than phenol itself, and the carboxylation occurs specifically ortho to the –OH.

Once salicylic acid is formed, the second step is not an aromatic substitution but an acylation of the phenolic –OH group. The –OH in salicylic acid is still a phenol (not a carboxylic acid –OH), so it can be esterified. Acetic anhydride is the preferred reagent because it is more reactive than acetic acid and avoids the harsh conditions of direct esterification. The product is aspirin, where the phenolic –OH is now an acetate ester.

Let’s walk through the two steps in detail.

1. Kolbe-Schmitt carboxylation: phenol → salicylic acid

Phenol is first converted to sodium phenoxide by treatment with NaOH. The phenoxide ion is a much stronger nucleophile than phenol because the negative charge is delocalised into the ring, making it even more electron-rich.

\ce{C6H5OH + NaOH -> C6H5ONa + H2O}

The sodium phenoxide is then dried and heated with carbon dioxide under pressure (about 100 °C, 4–7 atm). CO₂ acts as an electrophile; the phenoxide attacks the carbon of CO₂, and the resulting intermediate undergoes a rearrangement that places the carboxyl group at the ortho position.

Watch out

A common mistake is to think the carboxylation occurs at the para position. In the Kolbe-Schmitt reaction, the ortho product (salicylic acid) is the major product because the sodium ion coordinates with both the phenoxide oxygen and the incoming CO₂, directing the attack to the ortho site. Para-hydroxybenzoic acid is a minor byproduct.

The reaction scheme:

\ce{C6H5ONa + CO2 ->[125°C, 4-7 atm] o-HOC6H4COONa ->[H3O+] o-HOC6H4COOH}

The final acidification step (with dilute HCl or H₂SO₄) gives salicylic acid.

Tip

The Kolbe-Schmitt reaction is specific to phenols. Ordinary alcohols do not undergo this reaction because the alkoxide ion is not sufficiently stabilised to allow the necessary rearrangement.

2. Acetylation: salicylic acid → aspirin …

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Here is a breakdown of the common mistakes students make when converting Phenol to Aspirin (a classic Electrophilic Aromatic Substitution + Esterification problem), and how to avoid each.

1. Forgetting the First Step: Kolbe-Schmitt Reaction

The Mistake: Students often try to directly nitrate or brominate phenol, or they jump straight to esterification without introducing the carboxyl ( $-COOH$ ) group.

Why it’s wrong: Aspirin (acetylsalicylic acid) has a carboxylic acid group ortho to the phenolic $-OH$ . You cannot get this from phenol in one step without a specialized reaction.

How to Avoid:

Memorize the sequence: Phenol $\rightarrow$ Sodium phenoxide $\rightarrow$ Salicylic acid (via Kolbe-Schmitt).
The Concept: The $-OH$ group is strongly activating. To force $CO_2$ (a weak electrophile) to attack the ring, you must first convert the $-OH$ to the much stronger activating $-O^-$ group using NaOH.
The Reaction:

\text{Phenol} \xrightarrow{\text{NaOH}} \text{Sodium phenoxide} \xrightarrow{CO_2 (4-7 \text{ atm}), 125^\circ C} \text{Sodium salicylate} \xrightarrow{H^+} \text{Salicylic acid}

2. Getting the Regiochemistry Wrong (Ortho vs. Para)

The Mistake: Assuming the $-OH$ group directs the incoming $CO_2$ to the para position, or thinking the product is meta-substituted.

Why it’s wrong: $-OH$ and $-O^-$ are ortho/para directing (strongly activating). However, in the Kolbe-Schmitt reaction, the sodium ion ( $Na^+$ ) coordinates with the oxygen of the phenoxide ion, forcing the $CO_2$ to attack the ortho position due to steric and chelation effects.

How to Avoid:

Draw the intermediate: The $CO_2$ attacks the ortho carbon because the $Na^+$ holds the $CO_2$ close to that site.
Key fact: The product is Salicylic acid (2-hydroxybenzoic acid), not 4-hydroxybenzoic acid.
Check: Aspirin’s structure has the $-COOH$ and $-OH$ next to each other (ortho).

3. Skipping the Protection Step (or Doing it Wrong)

The Mistake: Trying to directly acetylate salicylic acid without realizing the $-OH$ group will react, but the $-COOH$ group might also react (forming an anhydride or mixed product).

Why it’s wrong: You want to acetylate only the phenolic $-OH$ to form the ester, leaving the $-COOH$ free. If you use acetic anhydride directly on salicylic acid, the reaction works, but students often forget the catalyst (conc. $H_2SO_4$ or $H_3PO_4$ ) or use the wrong reagent (e.g., acetyl chloride without a base).

How to Avoid:

The correct reagent: Acetic anhydride $((CH_3CO)_2O)$ in the presence of a few drops of conc. $H_2SO_4$ .
The Concept: The $-OH$ on the ring is more nucleophilic than the $-COOH$ under acidic conditions. The acid catalyst protonates the carbonyl of acetic anhydride, making it a better electrophile for the phenolic $-OH$ .
The Reaction:

\text{Salicylic acid} + (CH_3CO)_2O \xrightarrow{H_2SO_4} \text{Aspirin} + CH_3COOH

4. Confusing Esterification with Acetylation

The Mistake: Writing the final step as a simple esterification between the $-COOH$ of salicylic acid and methanol/ethanol.

Why it’s wrong: Aspirin is an ester of the phenol group, not an ester of the carboxylic acid group. The $-COOH$ must remain free for the drug to work.

How to Avoid: …

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

Showing the 12 most recent of 23 on this concept.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$CH_3COCl$
›Reveal solutionSolution
B (from nitration) = p-chloronitrobenzene; hydrolysing its (activated) C–Cl with hot NaOH then acidifying gives C = 4-nitrophenol; separately, Friedel–Crafts acylation of chlorobenzene gives the para product D = 4-chloroacetophenone.

Concept and Intuition

Two independent transformations of the same starting material, chlorobenzene: (i) nitration followed by an activated nucleophilic aromatic substitution of Cl by OH (only possible because the ring is now strongly deactivated/activated-for-SNAr by an ortho/para nitro group), and (ii) a direct Friedel–Crafts acylation, both controlled by the $o,p$ -directing, weakly deactivating nature of Cl, with sterics favouring the para product in each case.

Step-by-Step Solution

Chlorobenzene nitration: Cl is $o,p$ -directing (weak deactivator); despite modest steric differences, the major product is the para isomer, B = 1-chloro-4-nitrobenzene.

Ordinary (unactivated) chlorobenzene needs extremely harsh conditions (Dow process: ~623 K, 300 atm) to convert Cl to OH. But once a strongly electron-withdrawing $NO_2$ group sits ortho/para to the Cl, it stabilises the Meisenheimer intermediate of nucleophilic aromatic substitution, so B reacts with NaOH at a much milder 443 K to displace Cl by OH, giving (after acidifying the phenoxide) C = 4-nitrophenol ( $p\text{-}O_2N\text{-}C_6H_4\text{-}OH$ ). …
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AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$\xrightarrow[CH_3CO_2H]{Br_2}$
›Reveal solutionSolution
Anisole undergoes ring (electrophilic) bromination to give mainly para-bromoanisole; toluene under UV light undergoes side-chain (radical) bromination at the benzylic carbon, and hydrolysis of that benzylic bromide gives benzyl alcohol.

Concept and Intuition

The conditions decide whether bromination happens on the ring or on the side chain: Br $_2$ in acetic acid (no light) with an activated ring (anisole) is classic electrophilic aromatic substitution; Br $_2$ with UV light on an alkyl-substituted arene (toluene) instead generates bromine radicals that abstract the weak benzylic C–H, giving side-chain substitution.

Step-by-Step Solution

Anisole: the $-OCH_3$ group is a powerful $o,p$ -activator via resonance donation of a lone pair into the ring. With molecular $Br_2$ in acetic acid (ionic mechanism, electrophilic aromatic substitution), bromination occurs mainly at the position para to $-OCH_3$ (steric hindrance disfavours ortho): X = 4-bromoanisole ( $p\text{-}Br\text{-}C_6H_4\text{-}OCH_3$ ).

Toluene + $Br_2$ /UV light: UV light homolyses $Br_2$ into radicals; the benzylic C–H bond of the methyl group is weak and gives a resonance-stabilised benzylic radical, so substitution happens on the side chain, not the ring, giving benzyl bromide ( $C_6H_5CH_2Br$ ). …
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AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$Me = -CH_3$
›Reveal solutionSolution
Friedel–Crafts methylation of anisole gives predominantly the para product (X = 4-methylanisole); HBr cleaves anisole's ether linkage at the methyl carbon to give phenol (Y) and methyl bromide (Z).

Concept and Intuition

$-OCH_3$ on benzene is a strong activating, ortho/para-directing group. In Friedel–Crafts alkylation, both ortho and para products form, but the para isomer is favoured as the major product because of steric hindrance to ortho attack from the existing $-OCH_3$ group.

Separately, aryl alkyl ethers like anisole are cleaved by strong acids such as HBr/HI on heating. Because the $C_{aryl}-O$ bond cannot easily undergo $S_N2$ (aryl carbons resist backside attack, and aryl cations are high energy), the nucleophile ( $Br^-$ ) instead attacks the methyl carbon ( $S_N2$ on $CH_3$ ), breaking the $O-CH_3$ bond. This releases the phenolic $-OH$ intact on the ring (giving phenol) and generates methyl bromide.

Step-by-Step Solution

Anisole + $CH_3Cl$ , anhydrous $AlCl_3$ → Friedel–Crafts alkylation; $-OMe$ directs ortho/para; steric bulk favours para as the major product → X = 4-methylanisole (OMe, Me para).

Anisole + HBr, $\Delta$ → ether cleavage; $Br^-$ attacks the less hindered methyl carbon ( $S_N2$ ) → products are $CH_3Br$ (Z) and phenol (Y). …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\mathrm{C_6H_5OCH_3}$
›Reveal solutionSolution
Aniline is the classic exception that fails to undergo Friedel–Crafts reactions, because its $-\mathrm{NH_2}$ lone pair complexes with the $\mathrm{AlCl_3}$ catalyst, killing both the catalyst's activity and deactivating the ring.

Concept and Intuition

Friedel–Crafts alkylation/acylation needs a Lewis-acid catalyst (commonly anhydrous $\mathrm{AlCl_3}$ ) to generate the electrophile. This only works smoothly on arenes whose ring is not strongly deactivated and whose substituents don't sequester the catalyst. Aniline's amino group is a strong Lewis base — its nitrogen lone pair readily donates into $\mathrm{AlCl_3}$ 's empty orbital, forming a stable acid–base adduct. This (a) uses up the catalyst so it can't generate the required electrophile, and (b) converts the ring substituent into an electron-withdrawing $-\mathrm{NH_2}\to\mathrm{AlCl_3}$ group (rather than the strongly activating free $-\mathrm{NH_2}$ ), strongly deactivating the ring toward electrophilic attack. As a result, aniline famously fails to give normal Friedel–Crafts products.

Step-by-Step Solution

Anisole ( $\mathrm{C_6H_5OCH_3}$ ): the ether oxygen is a much weaker Lewis base than an amine nitrogen and still activates the ring strongly (o,p-director); anisole undergoes Friedel–Crafts reactions normally.

Aniline ( $\mathrm{C_6H_5NH_2}$ ): the amine nitrogen's lone pair is a strong Lewis base, complexing irreversibly with $\mathrm{AlCl_3}$ , disabling the catalyst and deactivating the ring — Friedel–Crafts fails. …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$Q. Consider the following Statement-I: In the nitration of aniline, more amount of m-nitroaniline is formed than expected. Statement-II: In the presence of a strongly acidic medium, aniline is protonated to form anilinium ion, which is meta directing. The correct answer is (A) Both statement-I and statement-II are correct (B) Both statement-I and statement-II are not correct (C) Statement-I is correct, but statement-II is not correct (D) Statement-I is not correct, but statement-II is correct$
›Reveal solutionSolution
Protonation of aniline to anilinium ion under strongly acidic nitration conditions explains the unexpectedly high m-nitroaniline yield; both statements are true. Answer: (A).

Concept and Intuition

Aniline's $\mathrm{-NH_2}$ group is a lone-pair donor and is normally a strong ortho/para director via resonance. However, nitration is carried out in strongly acidic media (conc. $\mathrm{HNO_3}$ /conc. $\mathrm{H_2SO_4}$ ), and aniline is a reasonably basic amine — a substantial fraction of it is protonated under these conditions to the anilinium ion, $\mathrm{C_6H_5NH_3^+}$ . The $\mathrm{-NH_3^+}$ group has no lone pair available to donate into the ring (it is used in the N–H bonds/protonation) and instead withdraws electron density inductively, behaving like other positively-charged substituents ( $\mathrm{-NR_3^+}$ ) as a deactivating, meta-directing group. So nitration of aniline under these conditions is really a mixture of two competing pathways — direct nitration of the small residual free amine (ortho/para-directing) and nitration of the dominant anilinium ion (meta-directing) — and the meta contribution pushes the overall m-nitroaniline yield well above what pure amine-directed reasoning would predict.

Step-by-Step Solution

Aniline + conc. $\mathrm{HNO_3}$ /conc. $\mathrm{H_2SO_4}$ : strongly acidic medium protonates much of the aniline to anilinium ion.

Anilinium ion ( $\mathrm{-NH_3^+}$ ) is electron-withdrawing and meta-directing, unlike the free amine.

Nitration therefore proceeds partly through the anilinium ion pathway, contributing meta product. …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$\mathrm{Br_2}$
›Reveal solutionSolution
COOH (meta director) sends Br to the meta position in benzoic acid; CH $_3$ (ortho/para director) sends Br predominantly to the para position in toluene — so X is meta-substituted and Y is para-substituted.

Concept and Intuition

Both reactions are electrophilic aromatic bromination catalysed by Fe (which generates the electrophile Br $^+$ from Br $_2$ ); "dark" conditions rule out the alternative photochemical free-radical side-chain bromination pathway, confirming these are ring substitutions. The regiochemistry is governed entirely by the directing effect of the substituent already on the ring:

$-$ COOH is strongly electron-withdrawing (by resonance and induction), so it deactivates the ring but directs the incoming electrophile to the meta position (the position where the resulting arenium-ion intermediate avoids placing positive charge on the carbon bearing the electron-poor COOH group).

$-$ CH $_3$ is electron-donating (hyperconjugation/+I), activating the ring and directing ortho/para; because Br is a moderately bulky electrophile, steric hindrance disfavours the ortho position relative to the CH $_3$ group, so the para isomer is the major product.

Step-by-Step Solution

Benzoic acid + Br $_2$ /Fe: COOH is a meta director $\Rightarrow$ Br enters meta to COOH, giving X = 3-bromobenzoic acid (meta relationship between COOH and Br). …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$\mathrm{CH_3Cl}$
›Reveal solutionSolution
Friedel-Crafts alkylation with CH $_3$ Cl/AlCl $_3$ works on chlorobenzene (weakly deactivated) and anisole (strongly activated) but fails on aniline (N lone pair complexes AlCl $_3$ ) and benzoic acid (too strongly deactivated).

Concept and Intuition

Friedel-Crafts reactions need the aromatic ring to be nucleophilic enough to attack the AlCl $_3$ -generated electrophile ( $^+$ CH $_3$ here). Two situations cause the reaction to fail:

Deactivation of the AlCl $_3$ catalyst itself: groups with a lone pair that is basic enough to coordinate directly with the Lewis acid — most notably $-$ NH $_2$ , $-$ NHR, $-$ NHCOR — tie up AlCl $_3$ as a salt/complex, leaving none to generate the electrophile. Aniline is the textbook example of this failure mode.

Strong deactivation of the ring itself: strongly electron-withdrawing meta-directing groups (COOH, NO $_2$ , CN, SO $_3$ H, etc.) make the ring too electron-poor for the (only moderately electrophilic) Friedel-Crafts species to attack. Benzoic acid falls in this category.

Halogens (as in chlorobenzene) are only weakly deactivating (net effect: deactivating overall by induction, but still ortho/para-directing by resonance donation of lone pairs) — the ring remains nucleophilic enough for alkylation to proceed. Ethers like anisole ( $-$ OCH $_3$ ) are strongly activating (electron-donating by resonance) and undergo Friedel-Crafts alkylation readily.

Step-by-Step Solution

Aniline (a): $-$ NH $_2$ lone pair coordinates with AlCl $_3$ , forming a complex that both consumes the catalyst and makes the ring effectively bear a positively-charged, strongly deactivating $-$ NH $_2^+$ AlCl $_3^-$ -like group $\Rightarrow$ Friedel-Crafts fails. …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$\mathrm{C_6H_5CH_3} \xrightarrow{X} \mathrm{C_6H_5CHO} \xrightarrow{\text{Nitration}} Y$
(i) $\mathrm{CrO_2Cl_2}$ ,
(ii) $\mathrm{H_3O^+}$ ; Y = 3-nitrobenzaldehyde ( $m$ - $\mathrm{O_2N}$ - $\mathrm{C_6H_4}$ -CHO) (C) X =
(i) $\mathrm{CrO_2Cl_2}$ ,
(ii) $\mathrm{H_3O^+}$ ; Y = 4-nitrobenzaldehyde ( $p$ - $\mathrm{O_2N}$ - $\mathrm{C_6H_4}$ -CHO) (D) X = $\mathrm{KMnO_4}\,/\,\mathrm{OH^-}$ ; Y = 4-nitrobenzaldehyde ( $p$ - $\mathrm{O_2N}$ - $\mathrm{C_6H_4}$ -CHO)
›Reveal solutionSolution
This tests the Etard reaction (the selective route from toluene to benzaldehyde) and the meta-directing effect of the –CHO group during nitration.

Concept and Intuition

Strong oxidants like alkaline $\mathrm{KMnO_4}$ oxidise a toluene methyl group all the way to $-COOH$ , overshooting the aldehyde stage. The Etard reaction ( $\mathrm{CrO_2Cl_2}$ followed by aqueous hydrolysis) selectively stops the oxidation at the aldehyde. Once formed, the electron-withdrawing $-CHO$ group deactivates the ring and directs incoming electrophiles to the meta position.

Step-by-Step Solution

X: converting toluene directly to benzaldehyde (not benzoic acid) requires the Etard reaction — (i) treatment with $\mathrm{CrO_2Cl_2}$ to form a chromium complex, (ii) hydrolysis with $\mathrm{H_3O^+}$ to release $C_6H_5CHO$ . $\mathrm{KMnO_4/OH^-}$ would instead push the oxidation all the way to $C_6H_5COOH$ , so it is the wrong reagent here.

Y: nitrating benzaldehyde. The $-CHO$ group is deactivating and meta-directing (like other carbonyl-bearing groups), so the nitro group is introduced predominantly at the meta position, giving 3-nitrobenzaldehyde ( $m\text{-}O_2N\text{-}C_6H_4\text{-}CHO$ ). …
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AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
$\mathrm{C_6H_5OCH_3 \xrightarrow[anhy.\ AlCl_3]{CH_3Cl} X\ (Major) \xrightarrow[(ii)\ H_3O^+]{(i)\ KMnO_4/OH^-,\ \Delta} Y}$
›Reveal solutionSolution
Friedel-Crafts methylation of anisole gives the para isomer as the major product;

oxidizing its methyl side chain to -COOH gives 4-methoxybenzoic acid, keeping the

methoxy and carboxyl groups para to each other.

Concept and Intuition

The methoxy group (-OCH $_3$ ) is a powerful ortho, para-director (through resonance

donation of a lone pair into the ring) in electrophilic aromatic substitution. In

Friedel-Crafts alkylation, the bulky alkyl electrophile combined with steric hindrance

near the -OCH $_3$ group means the para product dominates as the major isomer.

Separately, hot alkaline $\mathrm{KMnO_4}$ is a powerful oxidant that converts any

benzylic alkyl side chain (regardless of its length) all the way down to a -COOH group

directly attached to the ring — it does not affect the ether linkage itself.

Step-by-Step Solution

$\mathrm{C_6H_5OCH_3}$ (anisole) + $\mathrm{CH_3Cl}$ /anhydrous $\mathrm{AlCl_3}$ (Friedel-Crafts alkylation): methoxy directs the incoming methyl group ortho/para; the major product (less steric strain) is X = 1-methoxy-4-methylbenzene (p-cresol methyl ether), with $\mathrm{-OCH_3}$ and $\mathrm{-CH_3}$ para to each other.

X + $\mathrm{KMnO_4/OH^-}$ , $\Delta$ , then $\mathrm{H_3O^+}$ : vigorous oxidation …
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AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$\mathrm{S_2Cl_2}$
›Reveal solutionSolution
Sulphonation of benzene is an electrophilic aromatic substitution in which the attacking electrophile is $\mathrm{SO_3}$ (or its protonated form), generated from fuming sulphuric acid.

Concept and Intuition

Electrophilic aromatic substitution needs a species that is electron-deficient enough to be attacked by the benzene ring's $\pi$ -electron cloud. In sulphonation, benzene is treated with fuming sulphuric acid ( $\mathrm{H_2SO_4}$ containing dissolved $\mathrm{SO_3}$ , i.e. oleum). The sulphur atom in $\mathrm{SO_3}$ is highly electrophilic because it is bonded to three highly electronegative oxygens, leaving it electron-poor and eager to accept electron density from the aromatic ring.

Step-by-Step Solution

Fuming sulphuric acid supplies $\mathrm{SO_3}$ as the reactive species (sometimes considered as the protonated form $\mathrm{HSO_3^+}$ in strongly acidic medium).

The ring's $\pi$ -electrons attack the electrophilic sulphur of $\mathrm{SO_3}$ , forming a sigma complex (arenium ion) with a $\mathrm{-SO_3^-}$ group attached to the ring carbon. …
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AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$\mathrm{NH_2}$
›Reveal solutionSolution
Tests the classic amine-protection strategy before electrophilic aromatic substitution. Answer: X = acetanilide, Y = 4-bromoacetanilide (option D).

Concept and Intuition

Free aniline is such a strong activator (via the lone pair on N) that direct bromination is hard to control — it gives 2,4,6-tribromoaniline even with dilute bromine water, because the ring is so highly activated. To get a clean mono-bromination product, the amine is first "protected" by acetylation (acetic anhydride/pyridine), converting -NH2 to the less-activating, sterically bulkier -NHCOCH3 group (acetanilide). This still directs ortho/para (via resonance donation from N), but the bulky acetamido group sterically disfavours the ortho position, so bromination occurs predominantly at the para position.

Step-by-Step Solution

Aniline + (CH₃CO)₂O / pyridine → acetanilide (X), $\mathrm{C_6H_5NHCOCH_3}$ , via N-acetylation.

Acetanilide + $\mathrm{Br_2}$ / $\mathrm{CH_3COOH}$ : the acetamido group is an ortho/para director but bulky, so bromination occurs mainly at the position para to it.

Product Y = p-bromoacetanilide (4-bromoacetanilide), with Br para to the -NHCOCH3 group. …
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AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
$\mathrm{C_6H_5OCH_3}$
›Reveal solutionSolution
Anisole undergoes para-selective electrophilic bromination (X = 4-bromoanisole) and, with HI, ether cleavage at the methyl carbon (not the aryl carbon), giving phenol + $\mathrm{CH_3I}$ (Y).

Concept and Intuition

Bromination (X): The methoxy group is a powerful activating, ortho/para-directing substituent. Electrophilic aromatic substitution with $\mathrm{Br_2}$ in acetic acid proceeds readily; the para product dominates over ortho because of steric crowding near the bulky $\mathrm{OCH_3}$ group, giving 4-bromoanisole as the major product.

Ether cleavage (Y): Aryl alkyl ethers resist cleavage at the aryl–oxygen bond because that bond has partial double-bond character from resonance delocalisation of the oxygen lone pair into the ring — breaking it would require attacking an $sp^2$ aromatic carbon (very unfavourable for $S_N2$ , and $S_N1$ would give a highly unstable aryl cation). Instead, the iodide ion (from HI) performs an $S_N2$ attack on the far less hindered, non-aromatic methyl carbon, displacing the phenoxide, which is protonated to phenol. This is the basis of the Zeisel method for estimating methoxy groups.

Step-by-Step Solution

Identify the directing effect of $\mathrm{-OCH_3}$ : strongly activating, o/p-director.

Steric bulk favours para substitution as the major product $\Rightarrow$ X = 4-bromoanisole. …
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Q.Explain the process (with reactions) by which phenol can be converted to aspirin.

1. Kolbe-Schmitt carboxylation: phenol → salicylic acid

2. Acetylation: salicylic acid → aspirin …

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