Why is -NH₂ group of aniline acetylated before carrying out nitration?

NCERT Exemplar · Q29

-NH_2

Telangana TsbieShort· 2mImportance★★★★★

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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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Acetylation of the $-NH_2$ group in aniline converts it into a less activating, bulkier amide group ( $-NHCOCH_3$ ), which reduces the ring's reactivity and provides steric control — this prevents unwanted poly-substitution and oxidation, directing nitration predominantly to the para position.

The key to understanding this lies in Electrophilic Aromatic Substitution (EAS). Aniline has a free amino group ( $-NH_2$ ) that is a very strong activating and ortho/para-directing group. This sounds good for nitration, but it creates two serious problems.

Problem 1: Over-activation and poly-substitution. The $-NH_2$ group donates electrons into the ring so powerfully that the ring becomes too reactive. Under typical nitrating conditions (conc. $HNO_3$ / conc. $H_2SO_4$ ), aniline would undergo multiple nitrations, giving a messy mixture of di- and tri-nitro products. You cannot stop at mono-nitration.

Problem 2: Oxidation by the nitrating agent. The amino group is also a strong reducing agent. Concentrated nitric acid ( $HNO_3$ ) is a powerful oxidising agent. It will oxidise the $-NH_2$ group before it can nitrate the ring, destroying the starting material and forming tarry, black oxidation products. This is the more immediate disaster.

Acetylation solves both problems in one elegant step. Let's see how.

Protection from oxidation. The acetyl group ( $-COCH_3$ ) is attached to the nitrogen, forming acetanilide ( $C_6H_5NHCOCH_3$ ). The lone pair on nitrogen is now delocalised into the carbonyl group ( $C=O$ ) of the acetyl group. This makes the nitrogen much less basic and much less nucleophilic — it is no longer easily oxidised by $HNO_3$ . The ring is now safe to nitrate.
Taming the activating power. The $-NHCOCH_3$ group is still activating and ortho/para-directing, but it is much weaker than $-NH_2$ . The delocalisation of the nitrogen's lone pair into the carbonyl group reduces the electron density it can push into the benzene ring. The ring is still reactive enough for mono-nitration, but not so reactive that it undergoes uncontrolled poly-substitution.
Steric control for para-selectivity. The $-NHCOCH_3$ group is a bulky group. The ortho positions are sterically hindered — the incoming nitronium ion ( $NO_2^+$ ) finds it harder to attack there. The para position is wide open. As a result, the major product is p-nitroacetanilide, with only a small amount of the ortho isomer. This gives you excellent regioselectivity. …

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Here are the common mistakes students make on this exact question, along with how to avoid each — explained with the why first, then the correct answer.

✗ Mistake 1: Saying “Acetylation reduces the reactivity of aniline”

Why it’s wrong:

This is partially true but misses the real reason. Aniline is too reactive — it undergoes unwanted side reactions (like oxidation and polymerisation) under nitrating conditions. Acetylation doesn’t just “reduce reactivity”; it protects the amino group from being oxidised and controls the orientation of the incoming nitro group.

How to avoid:

Always mention both protection and regiocontrol. The correct answer is:

Acetylation converts the strong activating $-NH_2$ group into a moderate activating $-NHCOCH_3$ group, which:

Prevents oxidation of the amino group by the nitrating mixture.

Directs the nitro group to the para position (with some ortho), avoiding unwanted meta substitution.

✗ Mistake 2: Saying “Acetylation deactivates the ring”

Why it’s wrong:

The $-NHCOCH_3$ group is still activating (due to resonance donation from the nitrogen lone pair), just less strongly than $-NH_2$ . It is not deactivating. Deactivating groups are electron-withdrawing (like $-NO_2$ , $-CN$ , $-COOH$ ).

How to avoid:

Remember the activating/deactivating hierarchy:

Group	Effect
$-NH_2$	Strongly activating
$-NHCOCH_3$	Moderately activating
$-NO_2$	Strongly deactivating

So acetylation moderates activation, not deactivates.

✗ Mistake 3: Forgetting to mention ortho/para directing change

Why it’s wrong:

Both $-NH_2$ and $-NHCOCH_3$ are ortho/para directors, but the extent differs. The free $-NH_2$ group gives significant ortho substitution (due to steric hindrance, ortho product is less, but still possible). The acetyl group bulkier and reduces ortho attack, favouring para product — which is easier to separate and more useful industrially.

How to avoid:

State clearly:

Acetylation increases the para selectivity because the $-NHCOCH_3$ group is bulkier than $-NH_2$ , sterically hindering ortho positions.

✗ Mistake 4: Not mentioning reversibility of acetylation

Why it’s wrong: …

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

TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
$\mathrm{p-CH_3C_6H_4COCl}$
›Reveal solutionSolution

A fails on temperature (diazonium salts need 273–278 K, not 285 K); C fails on identity (Hinsberg's reagent is benzenesulphonyl chloride, not an acyl chloride); only B — acetanilide being less reactive than aniline towards nitration — is correct. Option (C), "B only".

The concept first

All three statements orbit one theme: how the nitrogen lone pair of an aromatic amine behaves, and how we control it.

Diazonium chemistry. Aniline $+$ $\mathrm{NaNO_2/HCl}$ gives $\mathrm{C_6H_5N_2^+Cl^-}$ . The ion is stabilised by resonance with the ring — which is precisely why the aromatic diazonium salt can be isolated at all, whereas aliphatic ones decompose instantly. But even so, it is only kinetically stable in ice-cold conditions. Above about $278$ K, the excellent leaving group $\mathrm{N_2}$ departs and water attacks, giving phenol.

Protecting the amino group. Aniline is so strongly activated that direct nitration is a mess: the strongly acidic nitrating mixture protonates the $-\mathrm{NH_2}$ to $-\mathrm{NH_3^+}$ (a meta-directing deactivator), and oxidation side-reactions occur. The classical fix is to acetylate aniline to acetanilide (N-phenylethanamide). Now the nitrogen lone pair is shared with the acetyl carbonyl, so the ring is still activated but much less so, the amine can't be protonated, and nitration gives clean para-nitroacetanilide, which is then hydrolysed back. Understanding why we bother acetylating is exactly what makes statement B obviously true.

Hinsberg's reagent. Benzenesulphonyl chloride, $\mathrm{C_6H_5SO_2Cl}$ , distinguishes the three classes of amine: $1^\circ$ amines give a sulphonamide with an acidic N–H, soluble in alkali; $2^\circ$ amines give a sulphonamide with no N–H, insoluble in alkali; $3^\circ$ amines do not react. Note the $\mathrm{SO_2}$ — that is the fingerprint.

Step-by-step

Statement A — "Aniline forms a stable benzene diazonium chloride at 285 K".

Diazotisation is carried out at $273$ – $278$ K ( $0$ – $5\,^\circ\mathrm{C}$ ). $285$ K is $12\,^\circ\mathrm{C}$ — above the safe window. At that temperature the salt decomposes:
$\mathrm{C_6H_5N_2^+Cl^- + H_2O \longrightarrow C_6H_5OH + N_2\uparrow + HCl}$
So it is not stable at 285 K. A is incorrect. ✗ …
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TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
$\text{C}_6\text{H}_6$
›Reveal solutionSolution
Friedel–Crafts alkylation needs an electron-rich ring. The methyl group of toluene is activating, while $-$ NO $_2$ and $-$ COCH $_3$ are strongly deactivating (rings carrying them do not undergo Friedel–Crafts at all). Toluene — option (D).

The concept first

In Friedel–Crafts alkylation, a Lewis acid generates a carbocation:
$\text{R}\!-\!\text{Cl} + \text{AlCl}_3 \longrightarrow \text{R}^+ + [\text{AlCl}_4]^-$
That carbocation is an electrophile; it is attacked by the aromatic ring's $\pi$ electrons. So the reaction rate is governed by one thing: how much electron density the ring can offer.

Substituents fall into two camps:

Activating (ring-enriching): $-$ CH $_3$ and other alkyls ( $+I$ , hyperconjugation), $-$ OH, $-$ OR, $-$ NH $_2$ ( $+R$ ). They raise the rate.

Deactivating (ring-depleting): $-$ NO $_2$ , $-$ CHO, $-$ COR, $-$ COOH, $-$ CN, $-$ SO $_3$ H ( $-I$ and $-R$ ). They lower the rate — and in the special case of Friedel–Crafts they lower it so much that the reaction fails entirely. (Nitrobenzene is even used as a solvent for Friedel–Crafts reactions, precisely because it will not react.)

Step-by-step

Step 1 — Rank the substituents by their electronic effect.

(A) Benzene — no substituent; the baseline.

(B) Nitrobenzene, $-$ NO $_2$ — one of the strongest deactivators known ( $-I$ and $-R$ ). Ring badly electron-poor.

(C) Acetophenone, $-$ COCH $_3$ — a carbonyl; $-I$ and $-R$ , strongly deactivating. Also, its lone pair can complex with AlCl $_3$ , poisoning the catalyst.

(D) Toluene, $-$ CH $_3$ — electron-donating by $+I$ and hyperconjugation. Ring electron-rich.

Step 2 — Apply the rule. Rate of electrophilic substitution increases with ring electron density: …
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TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ

\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{Cu_2Br_2/HBr}} \mathrm{C_6H_5Br}

›Reveal solution

The Sandmeyer reaction replaces $-\mathrm{N_2^+}$ using a cuprous salt: $\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{Cu_2Br_2/HBr}} \mathrm{C_6H_5Br}$ . That is option (A); the others are the Gattermann, hydrolysis, and Balz–Schiemann reactions respectively.

The concept first: one diazonium salt, four different name reactions

Aryl diazonium salts are a crossroads. From the same $\mathrm{C_6H_5N_2Cl}$ you can go in many directions, and each direction has its own name and its own reagent. Examiners test whether you can tell them apart, so learn them as a set, not one at a time:

Reagent	Product	Name of reaction
$\mathrm{Cu_2Cl_2/HCl}$ , $\mathrm{Cu_2Br_2/HBr}$ , $\mathrm{CuCN/KCN}$	$\mathrm{ArCl}$ , $\mathrm{ArBr}$ , $\mathrm{ArCN}$	Sandmeyer
Cu powder + $\mathrm{HCl/HBr}$	$\mathrm{ArCl}$ , $\mathrm{ArBr}$	Gattermann
$\mathrm{H_2O}$ , warm	$\mathrm{ArOH}$ (phenol)	hydrolysis
$\mathrm{HBF_4}$ , then $\Delta$	$\mathrm{ArF}$	Balz–Schiemann
$\mathrm{KI}$	$\mathrm{ArI}$	(direct, no Cu needed)
$\mathrm{H_3PO_2}/\mathrm{H_2O}$ or $\mathrm{C_2H_5OH}$	$\mathrm{ArH}$	deamination

The distinguishing mark of Sandmeyer is the cuprous (Cu(I)) salt, $\mathrm{Cu_2X_2}$ , in the corresponding halogen acid.

Step-by-step: test each option

$\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{Cu_2Br_2/HBr}} \mathrm{C_6H_5Br}$ — cuprous bromide in HBr, giving bromobenzene. This is exactly the Sandmeyer prescription. ✓
$\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{Cu/HBr}} \mathrm{C_6H_5Br}$ — the product is the same, but the reagent is copper powder, not the cuprous salt. This is the Gattermann reaction — a close cousin, deliberately placed here as the trap. ✗
$\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{H_2O}} \mathrm{C_6H_5OH}$ — this is simple hydrolysis to phenol (which is why diazonium salts must be kept cold). Not Sandmeyer. ✗ …

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TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
$Q. The major product formed in the following reactions is$
i) NaOH, C $_2$ H $_5$ I
ii) CH $_3$ COCl, anhyd. AlCl $_3$
iii) Zn-Hg, conc. HCl (A) [Image of benzene ring with OC $_2$ H $_5$ and C $_2$ H $_5$ substituents at 1,2 positions] (B) [Image of benzene ring with OC $_2$ H $_5$ and C $_2$ H $_5$ substituents at 1,3 positions] (C) [Image of benzene ring with OH and C $_2$ H $_5$ substituents at 1,2 positions] (D) [Image of benzene ring with OC $_2$ H $_5$ and C $_2$ H $_5$ substituents at 1,4 positions]
›Reveal solutionSolution
The sequence involves Williamson ether synthesis, Friedel–Crafts acylation, and Clemmensen reduction; the final product is 1-ethoxy-4-ethylbenzene, which corresponds to the 1,4‑disubstituted pattern — option (D).

Concept & Intuition

This problem tests your ability to predict the outcome of a multi‑step synthesis on an aromatic ring. The three steps are classic transformations:

Williamson ether synthesis – converts a phenol into an aryl alkyl ether.

Friedel–Crafts acylation – introduces an acyl group (here, acetyl) ortho/para to the activating ether group.

Clemmensen reduction – reduces the carbonyl (C=O) to a methylene (CH₂), giving an alkyl side chain.

The key insight: the ether group (–OC₂H₅) is a strong activating and ortho/para‑directing substituent. So the acylation will occur at the para position (less steric hindrance than ortho). The final reduction then turns the –COCH₃ into –CH₂CH₃, leaving the ether untouched. The product is a 1,4‑disubstituted benzene with –OC₂H₅ and –C₂H₅.

Step‑by‑Step Reasoning

Step (i): NaOH, C₂H₅I The starting material is phenol (C₆H₅OH). NaOH deprotonates it to phenoxide (C₆H₅O⁻), which then attacks ethyl iodide (C₂H₅I) in an Sₙ2 reaction. This is a Williamson ether synthesis, giving ethoxybenzene (phenetole):

$\text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5\text{O}^- \xrightarrow{\text{C}_2\text{H}_5\text{I}} \text{C}_6\text{H}_5\text{O}\text{C}_2\text{H}_5$

Step (ii): CH₃COCl, anhyd. AlCl₃ This is a Friedel–Crafts acylation. The –OC₂H₅ group is a strong ortho/para‑director because the oxygen lone pairs donate electron density into the ring.

The ortho positions are sterically hindered by the ethoxy group.

The para position is more accessible. Therefore, the acetyl group (–COCH₃) attaches para to the ethoxy group, forming 4‑ethoxyacetophenone:

$\text{C}_6\text{H}_5\text{O}\text{C}_2\text{H}_5 + \text{CH}_3\text{COCl} \xrightarrow{\text{AlCl}_3} p\text{-}\text{C}_2\text{H}_5\text{O}\text{-}\text{C}_6\text{H}_4\text{-}\text{COCH}_3$

Step (iii): Zn‑Hg, conc. HCl …
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TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
$_3$
›Reveal solutionSolution
Friedel–Crafts alkylation of benzene with n-propyl chloride proceeds via a carbocation rearrangement, so the major product is isopropyl benzene, not n-propyl benzene.

The reaction described is a classic Friedel–Crafts alkylation. In this reaction, an alkyl halide reacts with benzene in the presence of a Lewis acid catalyst (here, anhydrous AlCl₃) to form an alkylbenzene. The key mechanistic step is the generation of a carbocation from the alkyl halide. That carbocation then attacks the benzene ring.

The trap in this specific case is that the n-propyl carbocation (CH₃CH₂CH₂⁺) is primary and therefore relatively unstable. Under the reaction conditions, it rapidly rearranges to the more stable secondary carbocation (CH₃CH⁺CH₃) via a 1,2-hydride shift. This rearranged carbocation is what actually attacks benzene, giving isopropyl benzene as the major product.

Let’s walk through the reasoning step by step.

Generation of the electrophile. Anhydrous AlCl₃ coordinates with the chlorine atom of n-propyl chloride, weakening the C–Cl bond. This leads to the formation of a carbocation and an AlCl₄⁻ complex:

$\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{AlCl}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2^+ + \text{AlCl}_4^-$

Carbocation rearrangement. The primary carbocation (CH₃CH₂CH₂⁺) is not very stable. A hydrogen atom from the second carbon (along with its bonding pair) shifts to the positively charged carbon. This 1,2-hydride shift converts the primary carbocation into a more stable secondary carbocation:

$\text{CH}_3\text{CH}_2\text{CH}_2^+ \xrightarrow{\text{1,2-hydride shift}} \text{CH}_3\text{CH}^+\text{CH}_3$

Electrophilic attack on benzene. The rearranged secondary carbocation acts as the electrophile. It attacks the electron-rich benzene ring, forming a sigma complex (arenium ion): $\text{C}_6\text{H}_6 + \text{CH}_3\text{CH}^+\text{CH}_3 \rightarrow \text{C}_6\text{H}_6^+(\text{CH}(\text{CH}_3)_2)$ …
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TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
$0\,^\circ\mathrm{C}$
›Reveal solutionSolution
The strongly activating $-\mathrm{OH}$ controls the substitution, so two equivalents of bromine enter the positions ortho and para to it — giving 3,5-dibromo-2-hydroxybenzaldehyde, option (B).

The concept first

When a ring carries two substituents, the more powerfully activating one decides where the electrophile goes.

$\mathrm{-OH}$ : strongly activating via its lone pair ( $+R \gg -I$ ), ortho/para directing.

$\mathrm{-CHO}$ : deactivating ( $-R, -I$ ), meta directing. The phenol group is far stronger, so it wins. Conveniently, the positions ortho/para to the OH of salicylaldehyde are also meta to the CHO — the two groups' preferences agree, which is why the reaction is so clean.

Step-by-step

Number the ring. Salicylaldehyde is 2-hydroxybenzaldehyde: $\mathrm{CHO}$ at C-1, $\mathrm{OH}$ at C-2.

Find the positions activated by the OH (at C-2).

ortho to C-2 → C-1 (blocked by the CHO) and C-3 ✓

para to C-2 → C-5 ✓ So the two available activated sites are C-3 and C-5.

Cross-check against the CHO (at C-1). Meta to C-1 → C-3 and C-5 — exactly the same two carbons. Both directing effects therefore point to C-3 and C-5. …
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←Previous questionQ28 — What is the role of HNO₃ in the nitrating mixture used for nitration o…→Next questionQ30 — What is the product when C₆H₅CH₂NH₂ reacts with HNO₂?

Q.Why is $-NH_2$ group of aniline acetylated before carrying out nitration?

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

Method: Protection-Deprotection Strategy in Electrophilic Aromatic Substitution

Why This Method is Needed

The Solution: Acetylation

Step-by-Step Reasoning

✗ Mistake 1: Saying “Acetylation reduces the reactivity of aniline”

✗ Mistake 2: Saying “Acetylation deactivates the ring”

✗ Mistake 3: Forgetting to mention ortho/para directing change

✗ Mistake 4: Not mentioning reversibility of acetylation

The concept first

Step-by-step

The concept first

Step-by-step

The concept first: one diazonium salt, four different name reactions

Step-by-step: test each option

Concept & Intuition

Step‑by‑Step Reasoning

The concept first

Step-by-step

Q.Why is −NH2-NH_2−NH2​ group of aniline acetylated before carrying out nitration?

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Q.Why is $-NH_2$ group of aniline acetylated before carrying out nitration?

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