(1) Halogenation (Cl₂/FeCl₃, Br₂/FeBr₃). (2) Nitration (HNO₃/H₂SO₄). (3) Sulphonation (oleum, SO₃). (4) Friedel-Crafts alkylation (RX/AlCl₃) and acylation (RCOX/AlCl₃).
-- NCERT Class 11 Chemistry, Ch. 9, p. 322Electrophilic Aromatic Substitution
8 MCQs9-step worked example
Source: NCERT HydrocarbonsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Electrophilic Aromatic Substitution, explained for NEET
Benzene does not undergo addition reactions the way alkenes do — its aromatic sextet is too stable to break. Instead, benzene reacts by electrophilic aromatic substitution (EAS): an electrophile replaces one hydrogen while the aromatic ring is preserved.
The general mechanism has three steps:
- Electrophile generation. A Lewis acid catalyst (AlCl₃, FeBr₃, anhydrous AlCl₃ for Friedel-Crafts) activates the reagent to produce a strong electrophile (NO₂⁺, Cl⁺, R⁺, RCO⁺).
- Electrophilic attack. The π-electron cloud of benzene attacks the electrophile, forming a non-aromatic carbocation intermediate (sigma complex / arenium ion). This step is rate-determining.
- Proton loss. The sigma complex loses H⁺ to restore aromaticity, yielding the substituted product.
The key named reactions under EAS are:
| Reaction | Electrophile | Catalyst/Reagent | Product |
|---|---|---|---|
| Nitration | NO₂⁺ | Conc. HNO₃ + conc. H₂SO₄ | Nitrobenzene |
| Halogenation | Cl⁺ or Br⁺ | Anhydrous AlCl₃ or FeBr₃ | Chloro-/Bromobenzene |
| Friedel-Crafts alkylation | R⁺ (carbocation) | Anhydrous AlCl₃ | Alkylbenzene |
| Friedel-Crafts acylation | RCO⁺ (acylium ion) | Anhydrous AlCl₃ | Aryl ketone |
| Sulphonation | SO₃ / SO₃H⁺ | Fuming H₂SO₄ (oleum) | Benzenesulphonic acid |
High-frequency trap for NEET: confusing the electrophile identity. Students pick Cl₂ as the electrophile instead of Cl⁺, or write HNO₃ instead of NO₂⁺ (nitronium ion). The catalyst's job is specifically to generate the active electrophile — the neutral reagent itself is not the attacking species.
A second common confusion: Friedel-Crafts alkylation can give polyalkylation (the alkyl group activates the ring for further substitution), while Friedel-Crafts acylation stops at monosubstitution (the acyl group is deactivating). NEET questions exploit this by asking which Friedel-Crafts reaction gives a cleaner monosubstituted product.
NCERT reference: Class 11 Chemistry Chapter 9 (Hydrocarbons), page 322.
Can you answer these Electrophilic Aromatic Substitution MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the nitration of benzene, the electrophile that attacks the benzene ring is:
Show answer and why every option is right or wrong
Answer: D. D is correct. The active electrophile in nitration is the nitronium ion (NO₂⁺), generated from the reaction of conc. HNO₃ with conc. H₂SO₄. NCERT Class 11 Chemistry Chapter 9, Part 2.
Why A is wrong: A is wrong because HNO₃ is the reagent, not the electrophile. It must first react with H₂SO₄ to form NO₂⁺.
Why B is wrong: B is wrong because NO₃⁻ is an anion and cannot act as an electrophile (electrophiles are electron-deficient species).
Why C is wrong: C is wrong because NO₂ is a neutral free radical (nitrogen dioxide), not the cationic electrophile involved in aromatic nitration.
MCQ 2Easy RecallPractice
Which catalyst is used in the Friedel-Crafts alkylation of benzene?
Show answer and why every option is right or wrong
Answer: B. B is correct. Friedel-Crafts alkylation requires anhydrous AlCl₃ as a Lewis acid catalyst to generate the carbocation (R⁺) electrophile from the alkyl halide. NCERT Class 11 Chemistry Chapter 9, Part 2.
Why A is wrong: A is wrong because conc. H₂SO₄ acts as a protonating acid in nitration and sulphonation, not as the Lewis acid catalyst for Friedel-Crafts reactions.
Why C is wrong: C is wrong because FeBr₃ is the Lewis acid catalyst used specifically for bromination of benzene, not for Friedel-Crafts alkylation.
Why D is wrong: D is wrong because Pd/C is a hydrogenation catalyst used in catalytic reduction, not in electrophilic aromatic substitution.
MCQ 3Easy RecallPractice
Sulphonation of benzene is carried out using:
Show answer and why every option is right or wrong
Answer: D. D is correct. Sulphonation requires fuming H₂SO₄ (oleum, which contains free SO₃) to generate the electrophilic species SO₃ or SO₃H⁺. NCERT Class 11 Chemistry Chapter 9, Part 2.
Why A is wrong: A is wrong because dilute H₂SO₄ is a weak proton donor and cannot generate the SO₃ electrophile needed for sulphonation of benzene.
Why B is wrong: B is wrong because concentrated H₂SO₄ alone sulphonates benzene very slowly; fuming H₂SO₄ (which contains excess SO₃) is the standard reagent for effective sulphonation.
Why C is wrong: C is wrong because SO₂ is sulphur dioxide — a reducing agent, not the electrophilic SO₃ species required for sulphonation.
MCQ 4Direct ApplicationPractice
During electrophilic aromatic substitution, the rate-determining step is:
Show answer and why every option is right or wrong
Answer: B. B is correct. The formation of the sigma complex (arenium ion) is the slow, rate-determining step because it requires breaking the aromatic stability of benzene to form a non-aromatic carbocation intermediate.
Why A is wrong: A is wrong because electrophile generation is typically fast in the presence of a strong Lewis acid. The bottleneck is the ring's attack on the electrophile, not the electrophile's formation.
Why C is wrong: C is wrong because proton loss from the sigma complex is fast — it restores the thermodynamically favourable aromatic system, so it has a low activation energy.
Why D is wrong: D is wrong because π-complex formation is a loose, reversible association and is not the rate-determining step. The commitment step is the sigma complex formation.
MCQ 5Direct ApplicationPractice
Friedel-Crafts acylation of benzene gives a monosubstituted product cleanly, while Friedel-Crafts alkylation often gives polysubstituted products. The reason is:
Show answer and why every option is right or wrong
Answer: A. A is correct. The acyl group (–COR) withdraws electron density from the ring through the carbonyl, making the ring less reactive toward further electrophilic substitution. The alkyl group, being electron-donating, activates the ring and allows polyalkylation.
Why B is wrong: B is wrong because the alkyl group is an activating group (electron-donating via hyperconjugation and +I effect), which is why polyalkylation occurs — the opposite of what this option claims.
Why C is wrong: C is wrong because the acyl group is a deactivating group, not an activating group. It is electron-withdrawing (−I and −M through the C=O), which is precisely why monosubstitution predominates.
Why D is wrong: D is wrong because the electrophile strength is not the reason. In fact, acylium ions (RCO⁺) are stabilised by resonance and are effective electrophiles. The key difference is the electronic effect of the product on the ring, not the electrophile's strength.
MCQ 6Direct ApplicationPractice
In the bromination of benzene using Br₂/FeBr₃, the role of FeBr₃ is to:
Show answer and why every option is right or wrong
Answer: C. C is correct. FeBr₃ is a Lewis acid that complexes with Br₂, polarising it to generate Br⁺ (or a Br⁺-like electrophilic species) which then attacks the benzene ring. NCERT Class 11 Chemistry Chapter 9, Part 2.
Why A is wrong: A is wrong because the EAS mechanism for benzene halogenation is ionic, not radical. Free-radical halogenation occurs in alkanes (with UV light/heat), not in catalysed aromatic bromination.
Why B is wrong: B is wrong because FeBr₃ does not stabilise the sigma complex. Its role is complete once the electrophile is generated. The sigma complex is stabilised by delocalisation of the positive charge over the ring carbons.
Why D is wrong: D is wrong because FeBr₃ is a Lewis acid, not a base. The H⁺ removed in the final step is captured by FeBr₄⁻ (the counter-ion formed when FeBr₃ polarised Br₂), regenerating the catalyst.
MCQ 7Concept TrapPractice
Benzene undergoes electrophilic substitution rather than electrophilic addition because:
Show answer and why every option is right or wrong
Answer: C. C is correct. Benzene possesses significant aromatic stabilisation (resonance energy ~150 kJ/mol). An addition reaction would break the cyclic conjugation, losing this stabilisation. Substitution preserves the aromatic sextet, making it the thermodynamically favoured pathway.
Why A is wrong: A is wrong because benzene has six delocalised π electrons — it is the π cloud that attacks the electrophile. The issue is not absence of π electrons but preservation of the aromatic system.
Why B is wrong: B is wrong because σ bond strength is not the deciding factor. Both addition and substitution involve σ bonds. The key factor is the loss of aromatic stabilisation energy upon addition.
Why D is wrong: D is wrong because benzene carbons are sp² hybridised (planar, 120° bond angles), not sp³. This is fundamental to the aromatic π system.
MCQ 8CalculationPractice
Consider the following sequence: benzene is first treated with CH₃Cl/anhydrous AlCl₃, and the product is then treated with Br₂/FeBr₃. The methyl group on the ring is an ortho,para-director. Which statement about the final major product is correct?
Show answer and why every option is right or wrong
Answer: A. A is correct. Step 1 (Friedel-Crafts alkylation) gives toluene. Step 2: the –CH₃ group is an electron-donating ortho,para-director (as stated in the stem), so bromination occurs at the ortho and para positions, giving a mixture of o-bromotoluene and p-bromotoluene as major products.
Why B is wrong: B is wrong because the methyl group is an ortho,para-director (electron-donating via hyperconjugation and +I effect), not a meta-director. Meta-directing groups are electron-withdrawing (e.g., –NO₂, –COOH).
Why C is wrong: C is wrong because electrophilic aromatic substitution replaces a ring hydrogen, not an existing substituent. The methyl group remains intact; a new C–Br bond forms at an ortho or para position.
Why D is wrong: D is wrong because the methyl group activates the ring (makes it more electron-rich), so toluene is actually more reactive than benzene toward electrophilic attack, not less.
Free NEET study resources
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
How do you solve a Electrophilic Aromatic Substitution question? A worked example
- 1
Given
• Substrate: benzene (C₆H₆)• Reagents: conc. HNO₃ + conc. H₂SO₄ (nitrating mixture)
- 2
Required
Identify: (a) the electrophile, (b) the rate-determining step, (c) the major organic product.
- 3
Concept
This is electrophilic aromatic substitution — nitration. Conc. H₂SO₄ protonates HNO₃ to generate the nitronium ion (NO₂⁺), which is the true electrophile.
- 4
Mechanism outline
1. H₂SO₄ + HNO₃ → NO₂⁺ + HSO₄⁻ + H₂O
2. NO₂⁺ attacks benzene π cloud → sigma complex (arenium ion) — rate-determining step
3. Sigma complex loses H⁺ → nitrobenzene + aromaticity restored - 5
Substitution (identification, not numerical)
• Electrophile: NO₂⁺ (nitronium ion)• The slow step is sigma complex formation (Step 2 above)
- 6
Reasoning
The sigma complex is a non-aromatic cyclohexadienyl cation. Its formation requires overcoming the aromatic stabilisation barrier (~150 kJ/mol resonance energy), making it the highest-energy transition state in the pathway. The subsequent proton loss is fast because it restores the thermodynamically stable aromatic sextet.
- 7
Final answer
(a) Electrophile: NO₂⁺ (nitronium ion)
(b) Rate-determining step: formation of the sigma complex (electrophilic attack on the ring)
(c) Major product: nitrobenzene (C₆H₅NO₂) - 8
Common trap
Students often write HNO₃ or NO₂ as the electrophile. HNO₃ is the reagent; NO₂ is a neutral radical (nitrogen dioxide gas). The electrophile must be a cationic species — NO₂⁺.
- 9
Similar NEET-style question
"When benzene is treated with Br₂ in the presence of anhydrous AlCl₃, identify the electrophile and the organic product formed." (Answer: electrophile = Br⁺; product = bromobenzene.)
---
What to remember before solving Electrophilic Aromatic Substitution questions
More in Hydrocarbons: 4 exam traps and mistakes · 1 formula · 1 question pattern from its other lessons.
Electrophilic Aromatic Substitution questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Electrophilic Aromatic Substitution?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Predict major product of electrophilic substitution on substituted benzene. -OH/-NH2/-OR are o,p-directors and activators; -NO2/-COOH are m-directors and deactivators.
InterpretationMedium
Common distractors
classifies halogen as deactivating ortho para incorrectly
Misses halogens are weak deactivators but o,p-directors
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →