Unit I MCQ Practice: Benzene and Its Derivatives
Test your understanding of benzene structure, aromaticity, electrophilic aromatic substitution, reaction mechanisms, reactivity, and orientation through topic-wise multiple-choice questions with answers and brief explanations.
How to Use This Practice Page
Choose a topic, attempt each question before viewing the answer, read the explanation carefully, and note the concepts that require revision. Reattempt the questions after completing the unit.
Choose an MCQ Topic
Select a topic and attempt the questions before checking the correct answers and explanations.
Practise questions on molecular formula, degree of unsaturation, Kekulé structure, experimental evidence, and limitations of the proposed structures.
Test orbital structure, resonance, delocalisation, aromatic character, stability of benzene, and applications of Hückel’s rule.
Practise the general electrophilic aromatic substitution mechanism, electrophile formation, sigma complex, deprotonation, and nitration of benzene.
Test reagents, electrophile formation, reaction conditions, mechanisms, products, reversibility, and applications of sulphonation and halogenation.
Practise Friedel–Crafts alkylation and acylation, catalyst functions, electrophile formation, limitations, rearrangement, and product formation.
Test activating and deactivating groups, ortho–para and meta directors, the halogen exception, steric effects, and sigma-complex stability.
Topic 1: Benzene Structure and Evidence MCQs
Attempt each question before opening the answer. Each explanation highlights the concept commonly tested in university examinations.
1. What is the molecular formula of benzene?
A. C₆H₁₂
B. C₆H₁₀
C. C₆H₆
D. C₆H₁₄
Correct answer: C. C₆H₆
Explanation: Benzene contains six carbon atoms and six hydrogen atoms. Its molecular formula indicates a high degree of unsaturation.
2. What is the degree of unsaturation of benzene?
A. 2
B. 3
C. 4
D. 6
Correct answer: C. 4
Explanation: Degree of unsaturation = (2C + 2 − H) ÷ 2. For C₆H₆, the value is (12 + 2 − 6) ÷ 2 = 4.
3. How many hydrogen atoms are deficient in benzene compared with the corresponding saturated hydrocarbon?
A. 4
B. 6
C. 8
D. 10
Correct answer: C. 8
Explanation: The corresponding saturated hydrocarbon is C₆H₁₄, whereas benzene is C₆H₆. Therefore, benzene contains eight fewer hydrogen atoms.
4. What arrangement did Kekulé propose for benzene?
A. A straight chain with three double bonds
B. A six-membered ring with alternating single and double bonds
C. A five-membered ring with one side chain
D. A six-membered ring containing only single bonds
Correct answer: B.
Explanation: Kekulé proposed a six-membered carbon ring containing alternating single and double bonds.
5. Which observation contradicts a fixed Kekulé structure?
A. Benzene contains six carbon atoms
B. Benzene forms substitution products
C. All carbon–carbon bonds in benzene have equal length
D. Benzene burns with a smoky flame
Correct answer: C.
Explanation: A fixed alternating structure should contain different single- and double-bond lengths. Experimentally, all six carbon–carbon bonds have the same length.
6. What is the approximate carbon–carbon bond length in benzene?
A. 1.20 Å
B. 1.34 Å
C. 1.39 Å
D. 1.54 Å
Correct answer: C. 1.39 Å
Explanation: Each carbon–carbon bond in benzene has an intermediate length between a typical C–C single bond and a C=C double bond.
7. Why does benzene generally undergo substitution rather than addition?
A. Benzene contains no π electrons
B. Addition would destroy its aromatic stability
C. Benzene contains sp³-hybridised carbon atoms
D. Substitution requires no reagent
Correct answer: B.
Explanation: Addition disrupts the delocalised π-electron system and destroys aromaticity. Substitution allows benzene to restore and retain aromatic stability.
8. Which experimental behaviour distinguishes benzene from ordinary alkenes?
A. Benzene contains carbon and hydrogen
B. Benzene undergoes combustion
C. Benzene does not readily decolourise bromine solution without a catalyst
D. Benzene forms monosubstituted derivatives
Correct answer: C.
Explanation: Ordinary alkenes readily undergo addition with bromine. Benzene resists this reaction because its delocalised π system provides exceptional stability.
9. What is the hybridisation of each carbon atom in benzene?
A. sp
B. sp²
C. sp³
D. dsp²
Correct answer: B. sp²
Explanation: Each carbon forms three sigma bonds through sp² orbitals and contributes one unhybridised p orbital to the delocalised π system.
10. How many π electrons are present in benzene?
A. 2
B. 4
C. 6
D. 8
Correct answer: C. 6
Explanation: Six parallel p orbitals overlap around the ring, and each carbon contributes one electron. Therefore, benzene contains six delocalised π electrons.
Topic 2: Orbital Structure, Resonance, Delocalisation, Aromatic Character, Stability of Benzene & Hückel’s Rule MCQs
Attempt each question before opening the answer. Each explanation highlights the concept commonly tested in university examinations.
1. What is the hybridisation state of each carbon atom in benzene?
A. sp
B. sp²
C. sp³
D. dsp²
Correct answer: B. sp²
Explanation: Each carbon atom in benzene is sp² hybridised. The three sp² hybrid orbitals form three σ-bonds, while the remaining unhybridised p orbital participates in the delocalised π-system.
2. The geometry around each carbon atom of benzene is:
A. Linear
B. Tetrahedral
C. Trigonal planar
D. Trigonal pyramidal
Correct answer: C. Trigonal planar
Explanation: sp² hybridisation gives each carbon atom a trigonal planar arrangement. The bond angles around the carbon atoms are approximately 120°.
3. The unhybridised p orbital of each carbon atom in benzene is oriented:
A. In the plane of the ring
B. Perpendicular to the plane of the ring
C. At 45° to the ring plane
D. Toward the centre of the ring
Correct answer: B. Perpendicular to the plane of the ring
Explanation: Each carbon atom retains one unhybridised p orbital. These p orbitals are parallel to one another and perpendicular to the plane of the benzene ring, allowing effective sideways overlap.
4. The π-system of benzene is formed mainly by:
A. Head-on overlap of sp² orbitals
B. Overlap of s orbitals
C. Sideways overlap of unhybridised p orbitals
D. Overlap of sp³ orbitals
Correct answer: C. Sideways overlap of unhybridised p orbitals
Explanation: The unhybridised p orbitals on the six carbon atoms overlap sideways to form a continuous delocalised π-electron system above and below the plane of the ring.
5. How many π-electrons are present in benzene?
A. 2
B. 4
C. 6
D. 8
Correct answer: C. 6
Explanation: Each of the six sp²-hybridised carbon atoms contributes one electron from its unhybridised p orbital. Therefore, benzene contains six delocalised π-electrons.
6. The delocalised π-electron cloud of benzene is located:
A. Only inside the ring
B. Only below the ring
C. Above and below the plane of the ring
D. Only between carbon and hydrogen atoms
Correct answer: C. Above and below the plane of the ring
Explanation: Continuous overlap of the six parallel p orbitals produces two regions of π-electron density, one above and one below the plane of the benzene ring.
7. Which statement best describes the resonance structures of benzene?
A. They are separate compounds present in equilibrium
B. They rapidly convert into one another
C. They are contributing canonical structures of one molecule
D. They represent different molecular formulas
Correct answer: C. They are contributing canonical structures of one molecule
Explanation: The Kekulé structures of benzene are resonance or canonical structures. The actual benzene molecule is not rapidly switching between them; it exists as a resonance hybrid.
8. The actual structure of benzene is best described as:
A. A single Kekulé structure
B. A resonance hybrid
C. A cyclohexane structure
D. A structure containing three isolated double bonds
Correct answer: B. A resonance hybrid
Explanation: Benzene cannot be represented adequately by one Kekulé structure. Its actual structure is a resonance hybrid in which the π-electrons are delocalised over all six carbon atoms.
9. What is the major consequence of π-electron delocalisation in benzene?
A. Formation of unequal C–C bonds
B. Increased instability
C. Equalisation of C–C bond lengths
D. Loss of aromatic character
Correct answer: C. Equalisation of C–C bond lengths
Explanation: Because the π-electrons are delocalised over the entire ring, all six carbon–carbon bonds become equivalent. Their bond length is approximately 1.39 Å, intermediate between typical C–C single and C=C double bonds.
10. The approximate C–C bond length in benzene is:
A. 1.20 Å
B. 1.34 Å
C. 1.39 Å
D. 1.54 Å
Correct answer: C. 1.39 Å
Explanation: The C–C bond length in benzene is approximately 1.39 Å. It lies between the typical bond lengths of a C–C single bond and a C=C double bond because of π-electron delocalisation.
11. Which of the following is primarily responsible for the unusual stability of benzene?
A. sp³ hybridisation
B. Localisation of π-electrons
C. Delocalisation of π-electrons
D. Presence of six C–H bonds
Correct answer: C. Delocalisation of π-electrons
Explanation: Benzene gains considerable stabilisation from the delocalisation of its six π-electrons over the cyclic conjugated system. This stabilisation is a major feature of aromaticity.
12. Benzene generally prefers electrophilic substitution reactions rather than addition reactions because substitution:
A. Eliminates aromaticity
B. Preserves the aromatic ring
C. Converts all carbon atoms into sp³ carbon atoms
D. Removes all π-electrons permanently
Correct answer: B. Preserves the aromatic ring
Explanation: An addition reaction would normally disrupt the aromatic π-system. Electrophilic substitution allows the aromatic system to be restored after the reaction, thereby retaining aromatic stabilisation.
13. Aromaticity is best described as:
A. Instability caused by conjugation
B. Special stabilisation of certain cyclic conjugated systems
C. Presence of only single bonds
D. Presence of a benzene ring in every molecule
Correct answer: B. Special stabilisation of certain cyclic conjugated systems
Explanation: Aromaticity refers to the special stability associated with a cyclic, planar, fully conjugated π-system that satisfies the appropriate electron-count requirement, commonly expressed by Hückel’s rule.
14. Which of the following is NOT a requirement for a classical Hückel aromatic system?
A. Cyclic structure
B. Continuous conjugation
C. Planarity
D. Presence of 4n π-electrons
Correct answer: D. Presence of 4n π-electrons
Explanation: A classical aromatic system must have (4n + 2) π-electrons, not 4n π-electrons. A cyclic, planar, fully conjugated system containing 4n π-electrons is associated with antiaromaticity.
15. According to Hückel’s rule, an aromatic system contains:
A. 4n π-electrons
B. (4n + 2) π-electrons
C. 2n π-electrons
D. (2n + 2) π-electrons
Correct answer: B. (4n + 2) π-electrons
Explanation: Hückel’s rule states that a cyclic, planar, fully conjugated system is aromatic when it contains (4n + 2) π-electrons, where n is 0, 1, 2, 3, and so on.
16. For n = 1, the number of π-electrons required by Hückel’s rule is:
A. 2
B. 4
C. 6
D. 8
Correct answer: C. 6
Explanation: Applying Hückel’s equation: (4n + 2) = (4 × 1 + 2) = 6. Benzene contains six π-electrons and therefore satisfies this electron-count requirement.
17. For n = 2, Hückel’s rule gives:
A. 6 π-electrons
B. 8 π-electrons
C. 10 π-electrons
D. 12 π-electrons
Correct answer: C. 10 π-electrons
Explanation: When n = 2, (4n + 2) = (4 × 2 + 2) = 10. Therefore, a suitable cyclic, planar and fully conjugated system containing 10 π-electrons can be aromatic.
18. Which of the following π-electron counts satisfies Hückel’s rule?
A. 4
B. 6
C. 8
D. 12
Correct answer: B. 6
Explanation: Six π-electrons satisfy the expression (4n + 2) when n = 1. Therefore, a suitable cyclic, planar and fully conjugated system containing six π-electrons can exhibit aromaticity.
19. A cyclic, planar and fully conjugated system containing 8 π-electrons is expected to be:
A. Aromatic
B. Anti-aromatic
C. Saturated
D. Non-cyclic
Correct answer: B. Anti-aromatic
Explanation: Eight π-electrons correspond to 4n when n = 2. A cyclic, planar and fully conjugated system with 4n π-electrons is anti-aromatic and therefore destabilised.
20. Which π-electron count corresponds to the anti-aromatic electron configuration?
A. 2
B. 6
C. 10
D. 8
Correct answer: D. 8
Explanation: Anti-aromatic systems follow the 4n π-electron pattern. For n = 2, 4n = 8. Thus, eight π-electrons can lead to antiaromaticity when the other structural requirements are satisfied.
21. A cyclic, planar, fully conjugated compound contains 10 π-electrons. According to Hückel’s rule, it is:
A. Aromatic
B. Anti-aromatic
C. Non-aromatic
D. Saturated
Correct answer: A. Aromatic
Explanation: Ten π-electrons satisfy (4n + 2), where n = 2. If the system is cyclic, planar and fully conjugated, it meets the classical requirements for aromaticity.
22. Which of the following systems is most likely to be aromatic?
A. Cyclic, planar, fully conjugated system with 4 π-electrons
B. Cyclic, planar, fully conjugated system with 6 π-electrons
C. Cyclic, planar, fully conjugated system with 8 π-electrons
D. Acyclic system with 6 π-electrons
Correct answer: B. Cyclic, planar, fully conjugated system with 6 π-electrons
Explanation: Six π-electrons satisfy Hückel’s (4n + 2) rule with n = 1. The system must also be cyclic, planar and fully conjugated for classical aromaticity.
23. Which condition is essential for continuous π-electron delocalisation in benzene?
A. All carbon atoms must be sp³ hybridised
B. Continuous overlap of adjacent p orbitals
C. Presence of only σ-bonds
D. Absence of conjugation
Correct answer: B. Continuous overlap of adjacent p orbitals
Explanation: Delocalisation requires continuous overlap of neighbouring p orbitals. In benzene, all six carbon atoms possess parallel unhybridised p orbitals, allowing continuous π-electron delocalisation around the ring.
24. Which observation provides strong evidence that benzene does not contain three completely localised C=C bonds?
A. Benzene contains six carbon atoms
B. Benzene contains hydrogen atoms
C. All six C–C bonds have approximately the same length
D. Benzene is a colourless liquid
Correct answer: C. All six C–C bonds have approximately the same length
Explanation: If benzene contained three isolated C=C and three C–C bonds, the bond lengths would be different. Experimentally, all six C–C bonds are equivalent, supporting the delocalised structure.
25. The resonance stabilisation of benzene is mainly associated with:
A. σ-electron localisation
B. π-electron delocalisation
C. Hydrogen bonding
D. Ionic bonding
Correct answer: B. π-electron delocalisation
Explanation: The six π-electrons are delocalised over the entire benzene ring. This delocalisation lowers the overall energy of the molecule and contributes significantly to its stability.
26. Which statement about the π-electrons of benzene is correct?
A. They are localised between three pairs of carbon atoms
B. They are completely localised on individual carbon atoms
C. They are delocalised over all six carbon atoms
D. They are present only between carbon and hydrogen atoms
Correct answer: C. They are delocalised over all six carbon atoms
Explanation: The six p orbitals overlap continuously around the ring. Therefore, the π-electrons are not confined to individual C=C bonds but are delocalised over the complete six-carbon system.
27. Why does benzene have greater stability than an equivalent hypothetical structure containing three isolated double bonds?
A. Benzene has more hydrogen atoms
B. Benzene has fewer carbon atoms
C. Benzene gains stabilisation from π-electron delocalisation
D. Benzene contains sp³ carbon atoms
Correct answer: C. Benzene gains stabilisation from π-electron delocalisation
Explanation: Delocalisation allows the six π-electrons to spread over the entire ring, producing additional stabilisation. This is one of the fundamental reasons for the exceptional stability of benzene.
28. A student says, “The two Kekulé structures of benzene are two different molecules.” Which statement correctly corrects the student?
A. Both structures exist as separate compounds
B. They are resonance contributors of the same molecule
C. One structure is aromatic and the other is non-aromatic
D. They have different molecular formulas
Correct answer: B. They are resonance contributors of the same molecule
Explanation: Kekulé structures are canonical representations used to describe electron delocalisation. The actual benzene molecule is a resonance hybrid rather than either structure individually.
29. Which of the following best explains why all six carbon atoms of benzene are chemically equivalent?
A. Each carbon atom is sp³ hybridised
B. The π-electrons are delocalised around the ring
C. Benzene contains only single bonds
D. Hydrogen atoms prevent conjugation
Correct answer: B. The π-electrons are delocalised around the ring
Explanation: Delocalisation distributes π-electron density throughout the ring. Consequently, the six carbon atoms occupy equivalent positions in the symmetric benzene molecule.
30. Which combination correctly represents the requirements for classical aromaticity?
A. Cyclic + planar + fully conjugated + (4n + 2) π-electrons
B. Acyclic + planar + saturated + 4n π-electrons
C. Cyclic + non-planar + saturated + 6 π-electrons
D. Acyclic + fully conjugated + 4n π-electrons
Correct answer: A. Cyclic + planar + fully conjugated + (4n + 2) π-electrons
Explanation: Classical Hückel aromaticity requires a cyclic, planar and continuously conjugated π-system containing (4n + 2) π-electrons. All these conditions are important when applying Hückel’s rule.
31. If a cyclic conjugated system cannot remain planar, it may avoid antiaromaticity by becoming:
A. More aromatic
B. Non-aromatic
C. More saturated
D. Ionic
Correct answer: B. Non-aromatic
Explanation: Antiaromaticity requires a cyclic, planar, fully conjugated system with 4n π-electrons. If the molecule becomes non-planar and interrupts effective continuous overlap, it can become non-aromatic rather than anti-aromatic.
32. Which of the following is an important consequence of aromatic stabilisation in benzene?
A. Benzene readily undergoes addition reactions
B. Benzene is highly unstable
C. Benzene tends to preserve its aromatic ring during reactions
D. Benzene cannot undergo substitution reactions
Correct answer: C. Benzene tends to preserve its aromatic ring during reactions
Explanation: Aromatic stabilisation is energetically important. Therefore, benzene commonly undergoes electrophilic substitution reactions in which aromaticity is restored rather than permanently lost.
33. A molecule has a cyclic, planar, fully conjugated π-system containing 14 π-electrons. What is the value of n in Hückel’s equation?
A. 2
B. 3
C. 4
D. 5
Correct answer: B. 3
Explanation: Using (4n + 2) = 14, we get 4n = 12 and therefore n = 3. Thus, 14 π-electrons satisfy Hückel’s electron-count requirement.
34. Which of the following π-electron counts does NOT satisfy the Hückel aromaticity rule?
A. 2
B. 6
C. 10
D. 12
Correct answer: D. 12
Explanation: The Hückel series is 2, 6, 10, 14, 18 and so on. Twelve π-electrons do not fit the (4n + 2) pattern.
35. The best overall explanation for the stability of benzene is:
A. Presence of six sp³-hybridised carbon atoms
B. Localisation of three independent double bonds
C. Delocalisation of six π-electrons over a planar cyclic system
D. Absence of π-electrons
Correct answer: C. Delocalisation of six π-electrons over a planar cyclic system
Explanation: Benzene contains six sp²-hybridised carbon atoms whose p orbitals overlap continuously. The resulting delocalised six-π-electron system satisfies Hückel’s rule and provides significant aromatic stabilisation.
36. Which statement about Hückel’s rule is most appropriate?
A. It can be applied to every molecule containing a double bond
B. It is used to assess aromaticity in suitable cyclic conjugated systems
C. It determines the molecular weight of aromatic compounds
D. It applies only to saturated hydrocarbons
Correct answer: B. It is used to assess aromaticity in suitable cyclic conjugated systems
Explanation: Hückel’s rule is specifically useful for evaluating aromaticity in cyclic, planar, fully conjugated systems. The electron count alone is not sufficient; the structural requirements must also be considered.
37. A cyclic system contains 6 π-electrons but is not fully conjugated. It should be classified as:
A. Automatically aromatic
B. Automatically anti-aromatic
C. Not aromatic on the basis of Hückel’s rule
D. Aromatic because it contains six electrons
Correct answer: C. Not aromatic on the basis of Hückel’s rule
Explanation: Having six π-electrons alone does not guarantee aromaticity. Continuous conjugation is essential because the p orbitals must overlap around the complete cyclic system.
38. Which factor allows benzene to maintain a continuous delocalised π-system?
A. sp³ hybridisation of carbon atoms
B. Parallel alignment and overlap of p orbitals
C. Tetrahedral geometry
D. Complete localisation of double bonds
Correct answer: B. Parallel alignment and overlap of p orbitals
Explanation: The six carbon atoms are sp² hybridised and retain parallel unhybridised p orbitals. Their continuous sideways overlap creates the delocalised π-system characteristic of benzene.
39. If the six π-electrons of benzene were completely localised, which important feature would not be adequately explained?
A. Molecular formula C₆H₆
B. Equal C–C bond lengths
C. Presence of carbon atoms
D. Presence of hydrogen atoms
Correct answer: B. Equal C–C bond lengths
Explanation: A localised model would predict alternating short and long C–C bonds. The experimentally observed equal bond lengths are better explained by π-electron delocalisation.
40. Which statement correctly connects orbital structure, delocalisation and aromatic stability in benzene?
A. sp³ carbon atoms prevent π-overlap and destabilise benzene
B. sp² carbon atoms provide p orbitals that overlap continuously, allowing six π-electrons to delocalise and stabilise the ring
C. Localised σ-bonds produce aromaticity
D. Hydrogen atoms form the aromatic π-system
Correct answer: B. sp² carbon atoms provide p orbitals that overlap continuously, allowing six π-electrons to delocalise and stabilise the ring
Explanation: The sp²-hybridised carbon atoms of benzene retain unhybridised p orbitals. Their continuous overlap produces a delocalised six-π-electron system, which satisfies Hückel’s rule and contributes to the exceptional stability of benzene.
Topic 3: Electrophilic Aromatic Substitution (EAS) and Nitration MCQs
Attempt each question before opening the answer. Each explanation highlights the concept commonly tested in university examinations.
1. The characteristic reaction of benzene is:
A. Nucleophilic addition
B. Electrophilic aromatic substitution
C. Free-radical addition
D. Elimination
Answer: B. Electrophilic aromatic substitution
Explanation: The π-electron cloud of benzene attracts electrophiles. Substitution preserves aromaticity in the final product, whereas addition would destroy the aromatic stabilization.
Reaction: Ar–H + E⁺ → Ar–E + H⁺
Exam Point: Benzene generally undergoes substitution rather than addition because aromaticity is restored.
2. The attacking species in an electrophilic aromatic substitution is:
- Nucleophile
- Electrophile
- Free radical only
- Carbanion
Answer: B. Electrophile
Explanation: The electron-rich aromatic π system reacts with an electron-deficient species called an electrophile.
Reaction: C₆H₆ + E⁺ → σ-complex → C₆H₅E + H⁺
Exam Point: EAS always begins with generation or availability of a sufficiently strong electrophile.
3. The intermediate formed during EAS is called:
- Carbanion
- Arenium ion or sigma complex
- Benzyne
- Alkoxide
Answer: B. Arenium ion or sigma complex
Explanation: Attack of the electrophile forms a non-aromatic carbocation intermediate in which the positive charge is resonance-delocalized.
Reaction: C₆H₆ + E⁺ → [C₆H₆E]⁺
Exam Point: The sigma-complex has only two ring double bonds and temporarily loses aromaticity.
4. The slow, rate-determining step in most EAS reactions is:
- Electrophile generation
- Formation of the sigma complex
- Loss of H⁺ from sigma complex
- Product isolation
Answer: B. Formation of the sigma complex
Explanation: Formation of the sigma complex disrupts aromaticity and therefore requires substantial activation energy.
Reaction: Ar–H + E⁺ → [Ar(H)E]⁺
Exam Point: Loss of aromaticity makes sigma-complex formation the high-energy step.
5. The final step of EAS involves:
- Addition of H⁺
- Loss of H⁺ and restoration of aromaticity
- Loss of the electrophile
- Formation of a carbanion
Answer: B. Loss of H⁺ and restoration of aromaticity
Explanation: A base removes the proton from the carbon bearing the electrophile. The π bond reforms and aromaticity returns.
Reaction: [Ar(H)E]⁺ → Ar–E + H⁺
Exam Point: Restoration of aromaticity strongly drives the second step.
6. The electrophile in nitration of benzene is:
- NO₃⁻
- NO₂⁺
- NO⁺
- NH₂⁺
Answer: B. NO₂⁺
Explanation: The nitrating mixture generates the nitronium ion, which is the active electrophile.
Reaction: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O
Exam Point: Nitronium ion, NO₂⁺, is linear and strongly electrophilic.
7. The nitrating mixture for benzene consists mainly of:
- HNO₃ and H₂SO₄
- HCl and H₂SO₄
- HNO₂ and HCl
- NaNO₂ and HCl
Answer: A. HNO₃ and H₂SO₄
Explanation: Concentrated sulfuric acid protonates nitric acid and helps generate the nitronium ion.
Reaction: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O
Exam Point: Concentrated HNO₃/H₂SO₄ is the standard nitrating mixture.
8. Nitration of benzene gives:
- Aniline
- Nitrobenzene
- Benzoic acid
- Phenol
Answer: B. Nitrobenzene
Explanation: Nitronium ion substitutes for a ring hydrogen through the EAS mechanism.
Reaction: C₆H₆ + HNO₃ →(H₂SO₄) C₆H₅NO₂ + H₂O
Exam Point: Product of benzene nitration is nitrobenzene.
9. In the generation of NO₂⁺, sulfuric acid acts mainly as:
- A reducing agent
- A stronger acid/protonating agent
- A nucleophile
- A radical initiator
Answer: B. A stronger acid/protonating agent
Explanation: H₂SO₄ protonates HNO₃, enabling loss of water and formation of NO₂⁺.
Reaction: HNO₃ + H₂SO₄ → H₂NO₃⁺ → NO₂⁺ + H₂O
Exam Point: Know the role of H₂SO₄, not just the reagent combination.
10. During sigma-complex formation in nitration, benzene temporarily:
- Becomes more aromatic
- Loses aromaticity
- Becomes an alkyne
- Forms six double bonds
Answer: B. Loses aromaticity
Explanation: One ring carbon becomes sp³-like after bonding to the electrophile, interrupting cyclic conjugation.
Reaction: C₆H₆ + NO₂⁺ → σ-complex
Exam Point: The energy cost of losing aromaticity explains the activation barrier.
11. The sigma complex formed in EAS is stabilized mainly by:
- Hydrogen bonding
- Resonance
- Ionic crystallization
- van der Waals forces only
Answer: B. Resonance
Explanation: The positive charge is delocalized over several ring carbons through resonance contributors.
Reaction: σ-complex ↔ resonance contributors
Exam Point: Substituents affect EAS rate and orientation by changing sigma-complex stability.
12. Which statement best describes the energy profile of EAS?
- The first transition state is usually associated with loss of aromaticity
- No activation energy is required
- Product formation destroys aromaticity permanently
- The sigma complex is lower in energy than benzene
Answer: A. The first transition state is usually associated with loss of aromaticity
Explanation: The highest barrier is associated with electrophile attack and formation of the non-aromatic sigma complex.
Reaction: Ar–H + E⁺ → TS1 → σ-complex → TS2 → Ar–E
Exam Point: TS1 is generally higher because aromaticity is being lost.
13. Which species removes the proton from the sigma complex during nitration?
- A base such as HSO₄⁻
- NO₂⁺
- Benzene
- HNO₃ only
Answer: A. A base such as HSO₄⁻
Explanation: Deprotonation reforms the aromatic π system and regenerates acid catalyst.
Reaction: σ-complex + HSO₄⁻ → nitrobenzene + H₂SO₄
Exam Point: The catalyst is regenerated in the final step.
14. Nitration is classified as substitution because:
- NO₂ adds without loss of any atom
- A ring hydrogen is replaced by –NO₂
- The ring opens
- Two hydrogens are added
Answer: B. A ring hydrogen is replaced by –NO₂
Explanation: The electrophile ultimately replaces one hydrogen while the aromatic ring remains intact.
Reaction: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O
Exam Point: EAS changes a substituent while preserving the aromatic ring.
15. Compared with benzene, nitrobenzene undergoes further EAS:
- Much faster
- More slowly
- At exactly the same rate
- Only by radical mechanism
Answer: B. More slowly
Explanation: The –NO₂ group strongly withdraws electron density by –I and –R effects, destabilizing the sigma complex.
Reaction: C₆H₅NO₂ + E⁺ → slower EAS
Exam Point: –NO₂ is strongly deactivating and meta directing.
Topic 4: Sulphonation and Halogenation of Benzene MCQs
Attempt each question before opening the answer. Each explanation highlights the concept commonly tested in university examinations.
1. The electrophile in sulphonation of benzene is commonly represented as:
- SO₃ or protonated SO₃
- SO₄²⁻
- HSO₄⁻ only
- S²⁻
Answer: A. SO₃ or protonated SO₃
Explanation: Sulfur trioxide present in fuming sulfuric acid is strongly electrophilic; under strongly acidic conditions protonated forms can also participate.
Reaction: C₆H₆ + SO₃ → C₆H₅SO₃H
Exam Point: SO₃ is the key electrophilic species in aromatic sulphonation.
2. Benzene treated with fuming sulfuric acid gives:
- Benzoic acid
- Benzenesulphonic acid
- Phenol
- Benzaldehyde
Answer: B. Benzenesulphonic acid
Explanation: Sulphonation substitutes –SO₃H for a ring hydrogen.
Reaction: C₆H₆ + SO₃/H₂SO₄ → C₆H₅SO₃H
Exam Point: Sulphonation is an EAS reaction.
3. Sulphonation of benzene is important because it is:
- Completely irreversible
- Reversible
- A radical addition
- A nucleophilic substitution
Answer: B. Reversible
Explanation: Heating benzenesulphonic acid with dilute aqueous acid/steam can remove the sulfonic acid group and regenerate benzene.
Reaction: C₆H₅SO₃H + H₂O ⇌ C₆H₆ + H₂SO₄
Exam Point: Reversibility makes –SO₃H useful as a temporary blocking group.
4. Desulphonation is favoured by:
- Fuming H₂SO₄
- Hot dilute acid/steam
- AlCl₃ only
- NaNO₂/HCl
Answer: B. Hot dilute acid/steam
Explanation: Water and heat shift the reversible sulphonation equilibrium toward removal of –SO₃H.
Reaction: C₆H₅SO₃H + H₂O →(heat) C₆H₆ + H₂SO₄
Exam Point: Sulphonation conditions add –SO₃H; hot aqueous acid removes it.
5. The electrophile in chlorination of benzene is generated using:
- Cl₂/FeCl₃
- Cl₂/NaOH
- HCl only
- NaCl/H₂O
Answer: A. Cl₂/FeCl₃
Explanation: FeCl₃ polarizes chlorine and produces a strongly electrophilic chlorinating species.
Reaction: Cl₂ + FeCl₃ ⇌ Cl⁺-like electrophile + FeCl₄⁻
Exam Point: A Lewis acid catalyst is needed because benzene does not readily react with Cl₂ alone.
6. Chlorination of benzene gives:
- Chlorobenzene
- Benzyl chloride
- Cyclohexyl chloride
- Dichloromethane
Answer: A. Chlorobenzene
Explanation: A ring hydrogen is replaced by chlorine through EAS.
Reaction: C₆H₆ + Cl₂ (FeCl₃) → C₆H₅Cl + HCl
Exam Point: Differentiate ring chlorination from side-chain chlorination.
7. Bromination of benzene commonly uses:
- Br₂/FeBr₃
- Br₂/NaOH
- HBr only
- KBr/H₂O
Answer: A. Br₂/FeBr₃
Explanation: FeBr₃ activates Br₂ to generate a stronger electrophilic brominating species.
Reaction: C₆H₆ + Br₂ (FeBr₃) → C₆H₅Br + HBr
Exam Point: FeBr₃ is the common Lewis acid catalyst for bromination.
8. The main role of FeCl₃ in chlorination is to:
- Act as a base only
- Act as a Lewis acid and activate Cl₂
- Reduce benzene
- Provide chlorine atoms by decomposition
Answer: B. Act as a Lewis acid and activate Cl₂
Explanation: FeCl₃ accepts electron density from Cl₂, polarizing the Cl–Cl bond and increasing electrophilicity.
Reaction: Cl₂ + FeCl₃ → activated chlorine electrophile
Exam Point: Lewis-acid activation is central to aromatic halogenation.
9. Halogenation of benzene proceeds through:
- Arenium ion formation
- Carbanion formation
- Grignard reagent formation
- Peroxide intermediate only
Answer: A. Arenium ion formation
Explanation: Like other EAS reactions, halogenation forms a sigma complex followed by deprotonation.
Reaction: C₆H₆ + X⁺ → σ-complex → C₆H₅X
Exam Point: The same two-step ring mechanism applies to nitration, sulphonation and halogenation.
10. Which halogenation is difficult by direct EAS because fluorine is excessively reactive?
- Fluorination
- Chlorination
- Bromination
- Iodination only
Answer: A. Fluorination
Explanation: Direct fluorination is difficult to control because F₂ is extremely reactive.
Reaction: C₆H₆ + F₂ → difficult to control directly
Exam Point: Aryl fluorides are often prepared by alternative methods such as Balz–Schiemann.
11. Direct iodination of benzene generally requires:
- An oxidizing condition to generate/maintain I⁺-like electrophile
- Only water
- NaOH only
- No activation under any condition
Answer: A. An oxidizing condition to generate/maintain I⁺-like electrophile
Explanation: Iodination is less favorable and commonly requires an oxidant to create an effective electrophile and drive the reaction.
Reaction: C₆H₆ + I₂ →(oxidizing conditions) C₆H₅I
Exam Point: Halogen reactivity differs; do not assume identical conditions for F₂, Cl₂, Br₂ and I₂.
12. Which reaction can be used as a temporary blocking strategy in aromatic synthesis?
- Sulphonation followed by desulphonation
- Hydrogenation only
- Combustion
- Polymerization
Answer: A. Sulphonation followed by desulphonation
Explanation: The reversible –SO₃H group can occupy a ring position during another substitution and later be removed.
Reaction: Ar–H ⇌ Ar–SO₃H
Exam Point: The blocking-group concept is a higher-order application of sulphonation.
13. In chlorination, the proton lost from the sigma complex ultimately forms:
- HCl
- H₂
- Cl₂
- H₂O only
Answer: A. HCl
Explanation: Deprotonation restores aromaticity, and the proton combines with chloride-containing species to give HCl while catalyst is regenerated.
Reaction: σ-complex + FeCl₄⁻ → C₆H₅Cl + HCl + FeCl₃
Exam Point: Show catalyst regeneration in a full mechanism.
14. Which is the correct product of benzene bromination?
- C₆H₅Br
- C₆H₅CH₂Br
- C₆H₁₁Br
- C₆H₄Br₂ necessarily
Answer: A. C₆H₅Br
Explanation: One aromatic hydrogen is replaced by bromine under controlled EAS conditions.
Reaction: C₆H₆ + Br₂ →(FeBr₃) C₆H₅Br + HBr
Exam Point: Monobromobenzene is the standard product under normal conditions.
15. Which statement is correct for both nitration and halogenation of benzene?
- Both proceed through a sigma complex
- Both are nucleophilic additions
- Both permanently destroy aromaticity
- Both require a carbanion intermediate
Answer: A. Both proceed through a sigma complex
Explanation: Although electrophile generation differs, both reactions share the fundamental EAS pathway.
Reaction: Ar–H + E⁺ → σ-complex → Ar–E
Exam Point: Learn the common EAS framework and then the specific electrophile for each reaction.
Check Your Performance
Count one mark for every correct answer.
9–10 correct: Excellent — Your fundamental concepts are strong.
7–8 correct: Very Good — Revise the explanations for the questions you missed.
5–6 correct: Developing — Review benzene structure, degree of unsaturation, and experimental evidence.
Below 5 correct: Revision Required — Revisit the topic notes before attempting this practice set again.
Your first score shows your present understanding. Your second score, after revision, shows your learning progress.