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Aromatic hydrocarbons, benzene structure and aromaticity: NEET MCQs with answers

50 questions on this topic of Hydrocarbons (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. When benzene is treated with excess hydrogen in the presence of nickel catalyst at elevated temperature and pressure, compound 'A' is formed. What is the total number of hydrogen atoms present in one molecule of 'A'?

  1. 6
  2. 8
  3. 12
  4. 10
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Answer: (C) 12

Catalytic hydrogenation of benzene (C₆H₆) using Ni catalyst under high temperature (473-573 K) and pressure leads to complete reduction of all three double bonds, yielding cyclohexane (C₆H₁₂). Therefore, one molecule of compound 'A' contains 12 hydrogen atoms.

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2. Given below are two statements: Statement I: Cyclopentadienyl anion is aromatic and contains 6 delocalised π-electrons. Statement II: Cyclopentadienyl cation is antiaromatic and contains 4 delocalised π-electrons. In light of the above statements, choose the correct answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

Cyclopentadienyl anion has 6 π-electrons (two C=C double bonds + one lone pair) in a planar cyclic conjugated system, satisfying Hückel's (4n + 2) rule with n = 1 (aromatic). Cyclopentadienyl cation has 4 π-electrons in a planar cyclic conjugated ring, following the 4n rule (antiaromatic).

hard

3. Sodium salt of benzoic acid on heating strongly with sodalime gives a gas 'Y'. What is the molar mass of the product formed when 'Y' is completely hydrogenated in the presence of nickel catalyst at elevated temperature and pressure?

  1. 78 g/mol
  2. 84 g/mol
  3. 92 g/mol
  4. 82 g/mol
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Answer: (B) 84 g/mol

Decarboxylation of sodium benzoate with sodalime (NaOH + CaO) yields benzene (C₆H₆) as compound 'Y'. When benzene is completely hydrogenated using H₂/Ni under heat and pressure, it forms cyclohexane (C₆H₁₂). The molar mass of cyclohexane is (6 × 12) + (12 × 1) = 84 g/mol.

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4. How many delocalised π electrons are present in the aromatic polycyclic hydrocarbon phenanthrene (C₁₄H₁₀)?

  1. 10
  2. 12
  3. 16
  4. 14
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Answer: (D) 14

Phenanthrene consists of three fused benzene rings. It contains seven conjugated double bonds, contributing a total of 14 delocalised π electrons. This fits Hückel's rule: 4n + 2 = 14 (with n = 3), making it an aromatic hydrocarbon.

easy

5. Which of the following cyclic species is antiaromatic in nature?

  1. Cyclopropenyl cation
  2. Cyclopentadienyl anion
  3. Cyclopentadienyl cation
  4. Tropylium cation
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Answer: (C) Cyclopentadienyl cation

Cyclopentadienyl cation is a cyclic, planar, completely conjugated ring with 4 π electrons (4n where n = 1). According to Hückel's rule, systems having 4n π electrons are antiaromatic. Cyclopropenyl cation (2 π e⁻), cyclopentadienyl anion (6 π e⁻), and tropylium cation (6 π e⁻) are all aromatic.

hard

6. Given below are two statements: Assertion (A): Cyclopropenyl cation is aromatic while cyclopropenyl anion is antiaromatic. Reason (R): Cyclopropenyl cation has 2 π electrons obeying (4n + 2) rule (n = 0), whereas cyclopropenyl anion has 4 π electrons following the 4n rule (n = 1). In the light of the above statements, choose the correct answer:

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
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Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)

Cyclopropenyl cation is planar, completely conjugated, and contains 2 π electrons, fitting Hückel's (4n + 2) rule with n = 0. In contrast, cyclopropenyl anion contains 4 π electrons in a conjugated planar cyclic system, which satisfies 4n (n = 1), making it antiaromatic.

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7. Given below are two statements: Assertion (A): Halogens are ortho- and para-directing groups in electrophilic aromatic substitution, yet they deactivate the benzene ring. Reason (R): For halogens, the electron-withdrawing inductive effect (-I) is stronger than the electron-donating resonance effect (+R), but resonance stabilizes the arenium intermediate specifically at ortho and para positions. In light of the above statements, choose the most appropriate answer:

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
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Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A)

Halogens exhibit a strong -I effect that exceeds their weak +R effect, overall decreasing electron density in the ring and deactivating it. However, the resonance effect selectively stabilizes carbocation intermediates formed via ortho and para attack, directing electrophiles to those positions. Thus, both statements are true and (R) explains (A).

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8. Given below are two statements: Assertion (A): Chlorobenzene undergoes electrophilic substitution reactions predominantly at ortho and para positions. Reason (R): The inductive effect (-I) of chlorine increases electron density at ortho and para positions relative to the meta position. In light of the above statements, choose the most appropriate answer from the options given below:

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
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Answer: (C) (A) is true but (R) is false

Chlorobenzene is ortho/para-directing because the resonance effect (+R) releases lone pair electron density into the ring specifically at ortho and para positions. The inductive effect (-I) withdraws electron density from all positions (most strongly from ortho and meta), so Reason (R) is false.

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9. What is the total number of monochloro derivatives theoretically possible for anthracene (C₁₄H₁₀)?

  1. 2
  2. 3
  3. 4
  4. 5
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Answer: (B) 3

Anthracene has three distinct positions for monosubstitution: position 1 (or 4, 5, 8), position 2 (or 3, 6, 7), and position 9 (or 10). Therefore, it yields 3 monochloro positional isomers.

easy

10. Which of the following compounds reacts with benzene in the presence of anhydrous AlCl₃ followed by acidic workup to form acetophenone?

  1. Ethyl chloride
  2. Benzoyl chloride
  3. Chloromethane
  4. Acetyl chloride
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Answer: (D) Acetyl chloride

Benzene reacts with acetyl chloride (CH₃COCl) in the presence of anhydrous AlCl₃ via Friedel-Crafts acylation to yield acetophenone (C₆H₅COCH₃) and HCl. Benzoyl chloride yields benzophenone, while chloromethane gives toluene.

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11. Consider the following statements regarding the stability and heat of hydrogenation of benzene: Statement I: The actual heat of hydrogenation of benzene is less exothermic than the calculated value for a hypothetical 1,3,5-cyclohexatriene. Statement II: The difference in energy between the hypothetical cyclohexatriene and benzene is called resonance energy. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

The experimental heat of hydrogenation of benzene is -208 kJ/mol, whereas the calculated value for 1,3,5-cyclohexatriene is -358.5 kJ/mol. Benzene is more stable by about 150.5 kJ/mol due to delocalisation of π electrons, and this difference represents its resonance energy.

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12. In the electrophilic nitration of benzene using a mixture of concentrated HNO₃ and concentrated H₂SO₄, nitric acid acts as a:

  1. Base
  2. Acid
  3. Catalyst
  4. Reducing agent
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Answer: (A) Base

In nitrating mixture, H₂SO₄ is a stronger acid than HNO₃ and protonates HNO₃. The protonated species eliminates water to generate NO₂⁺; thus, HNO₃ accepts a proton and acts as a Brönsted-Lowry base.

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