Read the case carefully and answer the questions that follow. Alcohols undergo a number of reactions involving

Question

Q. Read the case carefully and answer the questions that follow. Alcohols undergo a number of reactions involving the cleavage of C - OH bond. However, phenols do not undergo reactions involving the cleavage of C - OH bond. Alcohols are weaker acids than water. Alcohols react with halogen acids to form the corresponding haloalkanes. Phenols are stronger acids than alcohols. A characteristic feature of phenols is that they undergo electrophilic substitution reactions such as halogenation, nitration, etc. Since - OH group is a strong activating group, phenol gives trisubstituted products during halogenation, nitration, etc.

(a) What happens when phenol is treated with the following ?

(i)

Br_2

water

(ii) Conc.

HNO_3

(b)

(i) Write the mechanism of alcohol reacting as nucleophile in a reaction with

CH_3^{\oplus}

(methyl cation).

(OR)

(b)

(ii) Why do phenols not undergo reactions involving cleavage of C – OH bond ?

ZnCl_2

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Part (a)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. …

Part (b)Concept understanding — Lucas Test

The Lucas Test: Why Some Alcohols React in Seconds and Others Take Days

Imagine you have three alcohols in front of you — one primary, one secondary, one tertiary — and you need to tell them apart without a spectrometer. The Lucas test gives you a simple, visual answer: how fast the mixture turns cloudy.

The cloudiness is not magic. It is a tiny, solid organic compound called an alkyl chloride, which does not dissolve in water. When it forms, it scatters light and the clear solution becomes milky. The question is: why does this happen at very different speeds for different alcohols?

The Chemistry Behind the Cloud

The Lucas reagent is concentrated hydrochloric acid (HCl) with zinc chloride (ZnCl₂) dissolved in it. ZnCl₂ is a Lewis acid — it grabs the oxygen of the alcohol, making the C–O bond much easier to break. Once that bond breaks, a carbocation forms. Then chloride ion (Cl⁻) from HCl attacks the carbocation, giving you the alkyl chloride — the insoluble stuff that makes the solution turbid.

So the entire test hinges on how easily the carbocation forms.

Carbocation Stability: The Deciding Factor

Tertiary carbocations are the most stable (three alkyl groups push electron density toward the positive carbon). Secondary are less stable. Primary carbocations are so unstable they barely exist at room temperature.

Tertiary alcohol: Forms a stable carbocation almost instantly. You see turbidity within seconds.
Secondary alcohol: Forms a less stable carbocation. Turbidity appears in 5–10 minutes (often with gentle warming).
Primary alcohol: A primary carbocation is too unstable to form under these conditions. The reaction is extremely slow — you may see no turbidity even after hours at room temperature. Heating is required, and even then it is sluggish.

Watch out

The Lucas test works only for alcohols that are soluble in the reagent. Methanol, ethanol, and most low-molecular-weight primary alcohols are soluble — but their alkyl chlorides are also somewhat soluble, so turbidity may be faint or absent. The test is most reliable for alcohols with 3–6 carbons.

The Precise Statement

\text{ROH} + \text{HCl} \xrightarrow{\text{ZnCl}_2} \text{RCl} + \text{H}_2\text{O}

How to Perform and Interpret

Take about 1 mL of the alcohol in a test tube.
Add 1 mL of Lucas reagent (conc. HCl + anhydrous ZnCl₂).
Shake and observe at room temperature.

| Observation | Inference |

|-------------|-----------| …

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Part (a): Phenol + Br $_2$ water $\rightarrow$ 2,4,6-tribromophenol; phenol + conc. HNO $_3$ $\rightarrow$ picric acid; an alcohol's O lone pair attacks $CH_3^+$ to give a protonated ether that loses $H^+$ .

Part (b): Phenol's C–O has partial double-bond character (so no C–OH cleavage); Lucas test — butan-1-ol (1°) no immediate turbidity, 2-methylpropan-2-ol (3°) instant turbidity.

Part (a)

(a)(i) Phenol is highly activated toward electrophilic substitution. With bromine water it gives a white precipitate of 2,4,6-tribromophenol (all three activated positions substituted):

C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH\ (\text{2,4,6-tribromophenol}) + 3HBr

(a)(ii) With concentrated nitric acid, phenol is trinitrated to 2,4,6-trinitrophenol (picric acid):

C_6H_5OH + 3HNO_3 \xrightarrow{\text{conc.}} C_6H_2(NO_2)_3OH + 3H_2O

(b)(i) Alcohol acting as a nucleophile toward the methyl cation. The oxygen of the alcohol carries lone pairs and is nucleophilic. It attacks the electrophilic carbon of $CH_3^{\oplus}$ , forming a protonated ether (oxonium ion); loss of a proton then gives the neutral ether:

$R\text{-}\underset{\displaystyle ..}{\ddot{O}}\text{-}H + CH_3^{\oplus} \longrightarrow R\text{-}\overset{\displaystyle +}{O}(H)\text{-}CH_3 \xrightarrow{-H^+} R\text{-}O\text{-}CH_3$ …

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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 38 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Primary, secondary and tertiary alcohols can be distinguished by$
(a) Hinsberg's test
(b) Tollen's reagent
(c) Fehling's solution
(d) Lucas test
›Reveal solutionSolution
The Lucas test converts an alcohol to its (water-insoluble, turbid) alkyl chloride at a rate that tracks carbocation stability, so the time taken to turn turbid tells 1°, 2° and 3° alcohols apart.

Mechanism: $R\text{-}OH + HCl \xrightarrow{ZnCl_2} R\text{-}Cl + H_2O$ . $ZnCl_2$ coordinates to the $-OH$ oxygen, making it a good leaving group as water; the alcohol then ionises to a carbocation (best for 3°, poor for 1°), which is captured by chloride to give the insoluble alkyl chloride, seen as turbidity/cloudiness in the otherwise clear reagent.

Tertiary alcohol: forms a stable 3° carbocation instantly — turbidity appears immediately at room temperature.

Secondary alcohol: forms a less stable 2° carbocation — turbidity appears only after a few minutes (often needs gentle warming). …
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CBSE 2026Set ANNUAL1 mark
$-NH_2$
›Reveal solutionSolution

The $-NH_2$ group is a powerful activating, ortho/para-directing group, so aniline reacts instantly with bromine water at all three activated ring positions to give 2,4,6-tribromoaniline as a white precipitate.

The lone pair on the amino nitrogen delocalises into the aromatic ring by resonance, strongly raising electron density especially at the ortho ( $2,6$ ) and para ( $4$ ) positions. This makes those three positions so reactive toward electrophiles that no Lewis-acid catalyst is required (unlike ordinary benzene bromination, which needs $FeBr_3$ ), and substitution does not stop after one bromination — it proceeds at all three activated sites simultaneously: …

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CBSE 2026Set SEM31 markMCQ
$Q. The reagent which can be used for the following transformation is: phenol (C6H5OH) -> salicylaldehyde (2-hydroxybenzaldehyde, OH and CHO on adjacent ring carbons)$
(a) i) CHCl3, NaOH, 60-80 C ii) dil. HCl
(b) i) CO2, NaOH, 120-140 C ii) dil. HCl
(c) i) CCl4, NaOH, 60-80 C ii) dil. HCl
(d) i) HCHO, NaOH ii) dil. HCl
›Reveal solutionSolution
Phenol + CHCl3 + NaOH (60-80 C) then acidification gives 2-hydroxybenzaldehyde (salicylaldehyde) by the Reimer-Tiemann reaction. Correct option (a).

In the Reimer-Tiemann reaction, chloroform (CHCl3) with aqueous NaOH generates dichlorocarbene (:CCl2), the electrophile. It attacks the phenoxide ring, chiefly at the ortho position; subsequent hydrolysis on acidification (dil. HCl) converts the -CHCl2 group into -CHO, introducing an aldehyde group ortho to -OH.

Product: salicylaldehyde (2-hydroxybenzaldehyde).

CO2/NaOH (option b) is the Kolbe reaction, giving salicylic acid (-COOH), not the aldehyde. …
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CBSE 2025Set ANNUAL1 mark
$Q. Aniline does not undergo Friedel-Crafts reaction. Give reason.$
›Reveal solutionSolution

The catalyst itself reacts with aniline's basic amino group, deactivating the ring before any substitution can occur.

Friedel–Crafts reactions (alkylation/acylation) require the Lewis acid catalyst $AlCl_3$ . Aniline's $-NH_2$ group is strongly basic (it has a lone pair on nitrogen), so it readily reacts with $AlCl_3$ to form a salt/complex ( $C_6H_5\overset{+}{N}H_2-AlCl_3^-$ ).

…

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CBSE 2025Set A1 mark
$Q. Match the following — Column A:$
(iii) Lucas reagent. Match with Column B:
(a) Reducing sugars
(b) Semiconductor
(c) Red ants
(d) Conc. HCl and ZnCl2
(e) Counter ions.
›Reveal solutionSolution

Lucas reagent is the mixture of concentrated HCl and anhydrous ZnCl2, used to distinguish primary, secondary, and tertiary alcohols by the speed of turbidity formation.

Lucas reagent = conc. HCl + anhydrous ZnCl2 (a Lewis acid catalyst that helps generate the carbocation). When an alcohol is shaken with Lucas reagent, it is converted to the corresponding alkyl chloride: 3° alcohols react instantly (turbidity/oily layer appears immediately, via SN1 since a stable 3° carbocation forms easily …

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CBSE 2025Set ANNUAL1 mark
$Q. Fill in the blank: When phenol is reacted with concentrated nitric acid, the product formed is ________.$
›Reveal solutionSolution

Phenol reacts with concentrated nitric acid to give 2,4,6-trinitrophenol (picric acid) via nitration at all three activated ortho/para positions.

The -OH group of phenol is a strong activating, ortho/para-directing group. With concentrated HNO3 (a strong nitrating agent), phenol undergoes exhaustive electrophilic nitration at both ortho pos …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The test used to distinguish between primary, secondary and tertiary alcohol is$
(a) Tollen's test
(b) Lucas test
(c) Fehling's test
(d) Hinsberg's test
›Reveal solutionSolution
The Lucas test uses the differing SN1 reactivity of 1°, 2° and 3° alcohols with Lucas reagent (conc. HCl/ZnCl2) - shown by how quickly a cloudy alkyl chloride layer forms.

Lucas reagent (a mixture of conc. HCl and anhydrous ZnCl2) reacts with alcohols to form alkyl chlorides:

$\text{R-OH} + \text{HCl} \xrightarrow{\text{ZnCl}_2} \text{R-Cl} + \text{H}_2\text{O}$

Tertiary alcohols react immediately (turbidity appears at once) - fastest, via a stable carbocation (SN1). …
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CBSE 2025Set ANNUAL1 mark
$Q. What happens when aniline is treated with bromine water?$
›Reveal solutionSolution

The -NH2 group strongly activates the benzene ring, so aniline reacts with bromine water even without a catalyst, substituting at all three positions ortho/para to -NH2 at once.

Aniline's -NH2 group is a powerful electron-donating, ring-activating group (o,p-director). It makes the ring so reactive that bromine water reacts directly, without needing a Lewis-acid catalyst, substituting simultaneously at both ortho positions and the para position:

$\text{C}_6\text{H}_5\text{NH}_2 + 3\text{Br}_2 (\text{aq}) \rightarrow 2,4,6\text{-tribromoaniline}\downarrow (\text{white ppt}) + 3\text{HBr}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Reaction of bromine water with phenol gives:$
(a) 2, 4, 6-Tribromophenol
(b) o-Bromophenol and p-Bromophenol
(c) o-Bromophenol
(d) p-Bromophenol
›Reveal solutionSolution

Phenol's -OH group strongly activates the ring at all three of the ortho/ortho/para positions, so with excess aqueous bromine (bromine water) all three positions get substituted at once, giving 2,4,6-tribromophenol as a white precipitate — no catalyst needed.

The -OH group donates electron density into the ring by resonance, making the ortho and para positions highly electron-rich. Bromine water (dilute aqueous Br2) is reactive enough on its own (unlike with benzene, which needs a Lewis-acid catalyst like FeBr3) to brominate all three activated positions (2, …

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CBSE 2025Set ANNUAL1 mark
$Q. Identify the structure of the missing component in the given reaction sequence : Toluene --(conc. HNO3 + conc. H2SO4)--> ? --(Fe/HCl)--> 4-aminotoluene$
›Reveal solutionSolution

Nitration of toluene (methyl = o,p-director) followed by reduction of the nitro group gives the target amine — the missing intermediate is the nitro compound before reduction.

Toluene, treated with a nitrating mixture (conc. HNO3 + conc. H2SO4), undergoes electrophilic aromatic substitution. The methyl group is an ortho/para-directing, ring-activating substituent, so nitration occurs mainly at the para (and ortho) position, giving predominantly 4-nitrotoluene (p-nitrotoluene) as the major product. This nitro compound, on reduction with F …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. In the chlorination of benzene, the reactive species is$
(a) Cl+
(b) Cl-
(c) Cl2
(d) Cl2-
›Reveal solutionSolution

Chlorination of benzene proceeds via electrophilic attack by Cl+.

In the presence of a Lewis acid catalyst such as anhydrous FeCl3 or AlCl3, Cl2 is polarised and heterolysed to generate an electrophilic chlorine species, Cl+ (as part of a complex with the catalyst, e.g. [FeCl4]- Cl+). This Cl+ then attacks the electron-rich benzene ring …

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CBSE 2024Set 56/3/11 markMCQ
$Q. For the following question, two statements are given - one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Aniline does not undergo Friedel-Crafts reaction. Reason (R) : Friedel-Crafts reaction is an electrophilic substitution reaction.$
›Reveal solutionSolution
Aniline fails in Friedel-Crafts alkylation/acylation because the amino group forms a complex with the Lewis acid catalyst (AlCl₃), making the ring strongly deactivated. The Reason is true but does not explain this specific failure — it only states a general fact about the reaction type.

Concept first: Electrophilic Aromatic Substitution (EAS) and why aniline is special

Friedel-Crafts reactions are classic EAS reactions. In EAS, an electrophile attacks the electron-rich benzene ring. The more electron-rich the ring, the faster the reaction. Activating groups (like –NH₂, –OH, –OCH₃) donate electrons to the ring, making it more reactive toward electrophiles. So at first glance, aniline (C₆H₅NH₂) should be highly reactive in Friedel-Crafts reactions — the –NH₂ group is a strong activator.

But real chemistry is not that simple. The catalyst in Friedel-Crafts reactions is a Lewis acid, typically anhydrous AlCl₃. AlCl₃ is a strong electron-pair acceptor. The lone pair on the nitrogen of aniline is basic — it readily coordinates to AlCl₃, forming a salt-like complex. This complex changes everything.

Let’s walk through the reasoning step by step.

What the Assertion says: Aniline does not undergo Friedel-Crafts reaction. This is a well-known experimental fact. If you try to alkylate or acylate aniline using AlCl₃ and an alkyl halide or acyl halide, you get either no reaction or a messy tar. The desired product is not formed.

Why the Assertion is true: When aniline is mixed with AlCl₃, the nitrogen’s lone pair donates to the aluminium, forming C₆H₅NH₂·AlCl₃. This complex has a positive charge on nitrogen (or at least a strongly polarised N–Al bond). The –NH₂ group is no longer an electron-donating group — it becomes a strong electron-withdrawing group (–NH₂⁺AlCl₃⁻). This deactivates the ring so severely that even a powerful electrophile like the acylium ion cannot attack it. The ring becomes less reactive than nitrobenzene. So the reaction simply does not proceed.

What the Reason says: Friedel-Crafts reaction is an electrophilic substitution reaction. This is a true statement — it is the textbook definition. Both alkylation and acylation proceed via an electrophilic attack on the aromatic ring. …
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Part (a)

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