Write following conversions: (i) nitrobenzene → acetanilide (ii) acetanilide → p-nitroaniline

NCERT Exemplar · Q50

Q. Write following conversions:

(i) nitrobenzene → acetanilide

(ii) acetanilide → p-nitroaniline

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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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The key idea is to use Electrophilic Aromatic Substitution (EAS) with careful protection/deprotection of the amino group. For (i), reduce nitrobenzene to aniline, then acetylate to acetanilide. For (ii), nitrate acetanilide (which directs to para), then hydrolyse the amide to get p-nitroaniline.

The Concept: Why This Approach Works

Both conversions involve the aniline functional group ( $-NH_2$ ). Aniline is highly reactive in EAS because the amino group is a strong activating and ortho/para-directing group. However, this very reactivity is a problem: if you try to nitrate aniline directly, you get a messy mixture of poly-substituted products and tars, because the amino group is too activating and also gets oxidised by the nitric acid.

The solution is to protect the amino group by converting it into an amide (like acetanilide). The amide group ( $-NHCOCH_3$ ) is still activating and ortho/para-directing, but much less so than the free amine. This allows for clean, mono-nitration, predominantly at the para position (due to steric hindrance at the ortho position). After the EAS step, you can deprotect (hydrolyse) the amide to get back the free amine.

Step-by-Step Solution

(i) Nitrobenzene → Acetanilide

This is a two-step reduction-acetylation sequence.

1. Reduction of nitrobenzene to aniline

Nitrobenzene ( $C_6H_5NO_2$ ) is reduced to aniline ( $C_6H_5NH_2$ ). The classic reducing agent in the lab is tin (Sn) and concentrated hydrochloric acid (HCl). The reaction proceeds through a series of intermediates (nitrosobenzene, phenylhydroxylamine) before giving the amine.

C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2

After reduction, the aniline is liberated from its salt by adding a base like NaOH.

2. Acetylation of aniline to acetanilide

Aniline is treated with acetic anhydride ( $(CH_3CO)_2O$ ) or acetyl chloride ( $CH_3COCl$ ) in the presence of a base (like pyridine or aqueous sodium acetate). The base neutralises the HCl (or acetic acid) formed and prevents the protonation of the aniline.

C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH

The product is acetanilide.

Watch out

A common mistake is to use only acetic acid for acetylation. Acetic acid is too weak an acylating agent; you need the anhydride or the acid chloride for a good yield.

(ii) Acetanilide → p-Nitroaniline

This is a three-step protection-nitration-deprotection sequence.

1. Nitration of acetanilide …

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Electrophilic Aromatic Substitution (EAS) — Conversion Method

Method: Two-Step EAS with Protecting Group Strategy

This method uses acetylation as a protecting group to control orientation and prevent unwanted side reactions.

(i) Nitrobenzene → Acetanilide

Concept: Nitrobenzene has a strong deactivating meta-directing group ( $-NO_2$ ). To get acetanilide, we must first reduce the nitro group to aniline, then protect the amino group.

Steps:

Reduction of nitro group
- Reagent: Sn / HCl (or Fe / HCl)
- Reaction:

C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2

Why: The $-NO_2$ group is reduced to $-NH_2$ , giving aniline.

Acetylation (protection)
- Reagent: Acetic anhydride $((CH_3CO)_2O)$
- Reaction:

C_6H_5NH_2 \xrightarrow{(CH_3CO)_2O} C_6H_5NHCOCH_3

Why: The $-NH_2$ group is highly activating and ortho/para-directing. Acetylation converts it to $-NHCOCH_3$ , which is still ortho/para-directing but less activating, preventing over-substitution.

Result: Acetanilide ( $C_6H_5NHCOCH_3$ )

(ii) Acetanilide → p-Nitroaniline

Concept: We need to introduce a nitro group para to the amino group, but the free $-NH_2$ would oxidize under nitration conditions. So we keep the acetyl protecting group, nitrate, then deprotect.

Steps:

Nitration of acetanilide
- Reagent: Conc. $HNO_3$ + Conc. $H_2SO_4$ (nitrating mixture)
- Reaction:

C_6H_5NHCOCH_3 \xrightarrow{HNO_3/H_2SO_4} p\text{-}O_2NC_6H_4NHCOCH_3

Why: The $-NHCOCH_3$ group is ortho/para-directing. The para product is major due to steric hindrance at ortho position. …

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Here is a breakdown of the common mistakes students make in these specific Electrophilic Aromatic Substitution (EAS) conversions, along with the correct reasoning to avoid them.

(i) Nitrobenzene → Acetanilide

This is a reduction followed by a protection step. The target is an amide.

Common Mistake 1: Direct Nitration or Halogenation of Nitrobenzene

The Error: Students try to add an acetyl group ( $-\text{COCH}_3$ ) directly to the benzene ring of nitrobenzene via Friedel-Crafts acylation.
Why it's wrong: The nitro group ( $-\text{NO}_2$ ) is a strong deactivating and meta-directing group. It makes the benzene ring so electron-poor that Friedel-Crafts reactions (which require a strong electrophile and a reactive ring) fail completely. You cannot perform acylation on nitrobenzene.
How to Avoid: Recognize that the first step must be to remove the deactivating effect of the nitro group. This is done by reducing it to an amino group ( $-\text{NH}_2$ ), which is strongly activating.

Common Mistake 2: Forgetting to Protect the Amino Group

The Error: After reducing nitrobenzene to aniline ( $\text{C}_6\text{H}_5\text{NH}_2$ ), students stop there, thinking the product is aniline.
Why it's wrong: The question asks for acetanilide ( $\text{C}_6\text{H}_5\text{NHCOCH}_3$ ), not aniline. You must convert the free amine into an amide.
How to Avoid: Always check the final functional group. If the target has an amide ( $-\text{NHCOCH}_3$ ), you must perform an acetylation reaction on the aniline using acetyl chloride ( $\text{CH}_3\text{COCl}$ ) or acetic anhydride ( $(\text{CH}_3\text{CO})_2\text{O}$ ) in the presence of a base (like pyridine or aqueous $\text{NaOH}$ ).

Correct Sequence:

Reduction: Nitrobenzene $\xrightarrow{\text{Sn/HCl or Fe/HCl}}$ Aniline
Acetylation (Protection): Aniline $\xrightarrow{\text{CH}_3\text{COCl/Pyridine}}$ Acetanilide

(ii) Acetanilide → p-Nitroaniline

This is a nitration followed by a deprotection (hydrolysis). The key is controlling the directing effect.

Common Mistake 1: Nitrating Aniline Directly

The Error: Students try to nitrate aniline ( $\text{C}_6\text{H}_5\text{NH}_2$ ) directly to get p-nitroaniline.
Why it's wrong: The $-\text{NH}_2$ group is a very strong activating and ortho/para-directing group. However, in the strongly acidic conditions of nitration (conc. $\text{HNO}_3$ /conc. $\text{H}_2\text{SO}_4$ ), the amino group gets protonated to form anilinium ion ( $\text{C}_6\text{H}_5\text{NH}_3^+$ ). This is a strongly deactivating and meta-directing group. The major product becomes m-nitroaniline, not the desired para isomer.
How to Avoid: Never nitrate a free aromatic amine. You must first protect the amino group by converting it into an amide (acetanilide). The amide group ( $-\text{NHCOCH}_3$ ) is still activating and ortho/para-directing, but it is much weaker than the free amine and is stable in acidic conditions.

Common Mistake 2: Ignoring the Ortho/Para Ratio

The Error: Students assume that nitration of acetanilide gives only the para product.
Why it's wrong: Acetanilide is ortho/para-directing. Nitration will yield a mixture of ortho and para isomers. The para isomer is the major product (due to steric hindrance at the ortho position), but the ortho isomer is also formed.
How to Avoid: In an exam, you must state that the product is a mixture, but the major product is the para isomer. You can then separate them (e.g., by fractional crystallization or column chromatography) to obtain pure p-nitroacetanilide.

Common Mistake 3: Forgetting the Final Hydrolysis Step

The Error: Students stop at p-nitroacetanilide, thinking it is the final product. …

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

KEAM 2026Set eng-2026-04174 marksMCQ
$Br_2$
›Reveal solutionSolution

Using $Br_2$ in the non-polar solvent $CS_2$ at low temperature (273 K) suppresses polybromination and gives p-bromophenol as the major monobrominated product.

Phenol is strongly activated, so bromine water (polar, ionising) gives 2,4,6-tribromophenol.

To stop at monobromination, a low-polarity solvent and low temperature are used, which lowers the electrophilicity/availability of $Br^+$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$Q. When phenol is treated with excess of bromine water, it gives (A) o-bromophenol (B) o- and p-bromophenol (C) 1,3,5-tribromophenol (D) 2,4-dibromophenol (E) 2,4,6-tribromophenol$
›Reveal solutionSolution

Phenol with excess bromine water undergoes electrophilic substitution at the ortho and para positions, giving a white precipitate of 2,4,6-tribromophenol. …

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KEAM 2026Set pha-2026-0418F4 marksMCQ
$Br_2$
›Reveal solutionSolution

Aniline $\rightarrow$ acetanilide (acetic anhydride/pyridine) $\rightarrow$ bromination gives mainly p-bromoacetanilide.

Aniline is acetylated to acetanilide $C_6H_5NHCOCH_3$ . The acetamido group is an activating ortho/para director, but the bulky $-NHCOCH_3$ hinders the ortho positions, so electrophilic bromination with $Br_2$ / $CH_3COOH$ occurs predominantly at the para po …
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KEAM 2026Set pha-2026-0420F4 marksMCQ
$Q. Which of the following reaction yieldstarry oxidation products? (A) Sulphonation of aniline (B) Nitration of aniline (C) Firedel-Crafts alkylation aniline (D) Firedel-Crafts alkylation of aniline (E) Bromination of aniline$
›Reveal solutionSolution

Aniline is readily oxidised; direct nitration with HNO $_3$ /H $_2$ SO $_4$ oxidises it to dark tarry products, so the amino group is protected (acetylated) first.

Aniline is very easily oxidised because the ring is electron-rich. When it is subjected to direct nitration with the strongly oxidising nitrating mixture ( $\mathrm{HNO_3/H_2SO_4}$ ), a large part of it is oxidised to dark, tarry products rather than cleanly nitrated. This is exactly why, in practice, aniline is first acetylated ( …
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KEAM 2025Set eng-2025-04264 marksMCQ
$H_2SO_4$
›Reveal solutionSolution

Sulphonation of phenol gives phenol-2,4-disulphonic acid; subsequent nitration replaces the $–SO_3H$ groups to yield picric acid (2,4,6-trinitrophenol).

Treating phenol with concentrated $H_2SO_4$ introduces sulphonic acid groups, giving phenol-2,4-disulphonic acid (X). On treatment with concentrated $HNO_3$ , the readily displaceable sulphonic groups are replaced by nitro groups and the ring is further nitrated, producing picric acid (2,4,6-trinitrophenol, Y). …
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KEAM 2025Set eng-2025-04284 marksMCQ
$C_6H_5NH_2 \xrightarrow[Pyridine]{(CH_3CO)_2O} A \xrightarrow[CH_3COOH]{Br_2} B$
›Reveal solutionSolution

Acetylation moderates aniline; the acetamido group is an o/p-director and bromination gives mainly the para product.

$\mathrm{C_6H_5NH_2}$ + $(\mathrm{CH_3CO})_2\mathrm{O}$ /pyridine $\to$ acetanilide (A), $\mathrm{C_6H_5NHCOCH_3}$ . The acetamido group is a strong ortho/para director; steric factors make the para product dominant. With $\mathrm{Br_2/CH_3COOH}$ the final product B is **p-bromoacetan …
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KEAM 2025Set eng-2025-04294 marksMCQ
$+ Br_2 \xrightarrow{FeBr_3}$
›Reveal solutionSolution

$\text{-}OH$ activates and directs ortho/para. With the para position occupied by $\text{-}CH_3$ , electrophilic bromination goes ortho to the $\text{-}OH$ , yielding 2-bromo-4-methylphenol.

p-Cresol is 4-methylphenol, with $\text{-}OH$ and $\text{-}CH_3$ para to each other. Both substituents are ortho/para directors, but the $\text{-}OH$ group is a much stronger activator and controls the orientation. Its para position is already occupied by the methyl group, so electrophilic aromatic bromination (with $Br_2/FeBr_3$ ) occurs at the position ortho to the $\text{-}OH$ . The major product is **2-bromo-4-methylphen …
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KEAM 2025Set pha-2025-0424A4 marksMCQ
$AlCl_3$
›Reveal solutionSolution

$AlCl_3$ generates the acylium electrophile $CH_3CO^+$ , which substitutes a ring hydrogen (para to Cl). This is electrophilic aromatic substitution.

$CH_3COCl + AlCl_3 \rightarrow CH_3CO^+ + AlCl_4^-$ . The acylium ion attacks the electron-rich benzene ring of chlorobenzene at the para position (Cl is $o,p$ -directing), for …
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KEAM 2024Set eng-2024-06084 marksMCQ
$C_6H_6O$
›Reveal solutionSolution

X = phenol, Y = benzene, Z = acetophenone: phenol is reduced by Zn dust to benzene, which undergoes Friedel–Crafts acylation to give acetophenone ( $C_8H_8O$ ).

Identify X: $C_6H_6O$ is phenol. Heating phenol with zinc dust reduces it (removes the –OH):
$C_6H_5OH + Zn \longrightarrow C_6H_6 + ZnO$
so Y = benzene.

Benzene then undergoes Friedel–Crafts acylation with acetyl chloride and anhydrous $AlCl_3$ : …
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Q.Write following conversions:

The Concept: Why This Approach Works

Step-by-Step Solution

(i) Nitrobenzene → Acetanilide

(ii) Acetanilide → p-Nitroaniline

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