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Alkanes: conformations and halogenation mechanism: 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. What is the magnitude of the energy barrier (difference in potential energy) between the staggered and eclipsed conformations of ethane?

  1. 1.25 kJ mol⁻¹
  2. 25.0 kJ mol⁻¹
  3. 12.55 kJ mol⁻¹
  4. 45.5 kJ mol⁻¹
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Answer: (C) 12.55 kJ mol⁻¹

According to NCERT, the difference in potential energy between the staggered and eclipsed conformations of ethane is approximately 12.55 kJ mol⁻¹. This small barrier is easily overcome by thermal collisions at room temperature.

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2. In the chlorination of alkanes, the presence of a small amount of oxygen (O₂) initially retards the reaction because:

  1. O₂ oxidizes the alkane into an alcohol rapidly.
  2. O₂ directly reacts with Cl₂ molecules to form chlorine oxides.
  3. O₂ absorbs ultraviolet light and prevents homolysis of Cl₂.
  4. O₂ reacts with alkyl radicals to form relatively unreactive peroxy radicals.
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Answer: (D) O₂ reacts with alkyl radicals to form relatively unreactive peroxy radicals.

Oxygen acts as an inhibitor in free-radical chlorination because it combines with alkyl radicals (R• + O₂ → R–O–O•) to form alkylperoxy radicals, which are far less reactive and unable to propagate the halogenation chain effectively.

hard

3. Which elementary reaction step in the free-radical halogenation of methane is endothermic when chlorine (Cl₂) is used?

  1. Cl₂ + hν → 2 Cl•
  2. Cl• + CH₄ → •CH₃ + HCl
  3. •CH₃ + Cl₂ → CH₃Cl + Cl•
  4. •CH₃ + Cl• → CH₃Cl
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Answer: (B) Cl• + CH₄ → •CH₃ + HCl

The first propagation step, in which a chlorine radical abstracts a hydrogen atom from methane (Cl• + CH₄ → •CH₃ + HCl), has a positive enthalpy change (ΔH ≈ +4 kJ/mol) and is mildly endothermic. In contrast, the second propagation step and radical combination steps are exothermic.

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4. Consider the fluorination of methane: CH₄ + F₂ → CH₃F + HF Which of the following statements correctly explains why direct fluorination of alkanes cannot be carried out smoothly like chlorination?

  1. The initiation step requires a very high activation energy because the F–F bond is exceptionally strong.
  2. Both propagation steps in fluorination are strongly exothermic, leading to an extremely high reaction rate that causes non-selective cleavage of C–C and C–H bonds.
  3. The reaction is endothermic overall, so it does not proceed without continuous external heating.
  4. Fluorine free radicals are extremely unreactive due to their high electronegativity, making the propagation step too slow.
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Answer: (B) Both propagation steps in fluorination are strongly exothermic, leading to an extremely high reaction rate that causes non-selective cleavage of C–C and C–H bonds.

Unlike other halogens, both propagation steps for fluorination (abstraction of H by F• and reaction of alkyl radical with F₂) are highly exothermic (ΔH° < 0). This releases tremendous heat rapidly, triggering explosive chain reactions and unselective rupture of C–C and C–H bonds.

easy

5. In the Newman projection of the gauche conformation of n-butane along the C2–C3 bond, what is the dihedral angle between the two bulky methyl (–CH₃) groups?

  1. 60°
  2. 0°
  3. 120°
  4. 180°
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Answer: (A) 60°

The gauche conformation of n-butane is a staggered arrangement where the two terminal methyl groups are positioned at a dihedral angle of 60° relative to each other.

hard

6. Consider the two statements: Statement I: Bromination of alkanes is highly selective compared to chlorination. Statement II: The hydrogen abstraction step by a bromine radical is exothermic, leading to an early transition state. Choose the correct option:

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

Statement I is correct because bromine radicals are less reactive and therefore much more selective. Statement II is incorrect because hydrogen abstraction by a bromine radical is endothermic, which according to the Hammond postulate leads to a late, radical-like transition state, resulting in high selectivity.

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7. Which of the following statements regarding the conformations of ethane is INCORRECT?

  1. In the eclipsed conformation, the C–C bond length is significantly shorter than in the staggered conformation.
  2. Any intermediate conformation between staggered and eclipsed is known as a skew conformation.
  3. The energy difference between eclipsed and staggered conformations is about 12.5 kJ mol⁻¹.
  4. The staggered conformation has the minimum torsional strain.
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Answer: (A) In the eclipsed conformation, the C–C bond length is significantly shorter than in the staggered conformation.

Conformations arise from rotation about the C–C single bond without changing bond lengths or bond angles significantly. The C–C bond length remains virtually unchanged between the staggered and eclipsed forms.

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8. What is the total number of constitutionally isomeric monochloro derivatives that can be formed from 2-methylbutane upon photochemical chlorination?

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

2-Methylbutane is CH₃-CH(CH₃)-CH₂-CH₃. Replacement of hydrogen at different non-equivalent positions gives: 1-chloro-2-methylbutane (C1), 2-chloro-2-methylbutane (C2), 2-chloro-3-methylbutane / 3-chloro-2-methylbutane (C3), and 1-chloro-3-methylbutane (C4). This yields exactly 4 constitutional isomers.

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9. In the free-radical chlorination of methane, which of the following represents a propagation step?

  1. •Cl + CH₄ → •CH₃ + HCl
  2. •CH₃ + •Cl → CH₃Cl
  3. Cl₂ + hν → 2 •Cl
  4. •CH₃ + •CH₃ → CH₃CH₃
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Answer: (A) •Cl + CH₄ → •CH₃ + HCl

Propagation steps consume one radical and generate another radical. The abstraction of a hydrogen atom from CH₄ by a chlorine radical (•Cl + CH₄ → •CH₃ + HCl) is a key chain propagation step.

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10. Which of the following statements is correct regarding the dipole moment and physical properties of the conformations of n-butane?

  1. All staggered conformations of n-butane have identical dipole moments and identical thermodynamic stabilities.
  2. The fully eclipsed conformation has the lowest potential energy and zero net dipole moment.
  3. The gauche conformation possesses lower potential energy than the anti conformation because of attractive dispersion forces.
  4. The anti conformation has zero dipole moment due to centrosymmetry, while the gauche conformation has a non-zero dipole moment.
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Answer: (D) The anti conformation has zero dipole moment due to centrosymmetry, while the gauche conformation has a non-zero dipole moment.

In the anti conformation of n-butane, the two terminal methyl groups and corresponding C–H bonds lie diametrically opposite (dihedral angle 180°), resulting in an inversion center and zero net dipole moment. In the gauche conformation (dihedral angle 60°), bond dipoles do not cancel, giving a non-zero dipole moment.

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11. Consider the following two statements: Assertion (A): In the free-radical halogenation of alkanes, iodination is a reversible reaction and requires the presence of an oxidizing agent such as HIO₃ or HNO₃. Reason (R): The hydrogen abstraction step by iodine radical (I• + R–H → R• + HI) is highly endothermic, and the HI formed is a powerful reducing agent that reduces alkyl iodide back to alkane.

  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)

Assertion is true: iodination is reversible and needs an oxidizing agent. Reason is also true and explains the assertion: the C–H bond cleavage by iodine radical is distinctly endothermic with a high activation energy, and the HI produced readily reduces the alkyl iodide back to the alkane; oxidizing agents remove HI by oxidizing it to I₂, driving the equilibrium forward.

easy

12. In a Sawhorse projection of the anti conformation of n-butane viewing along the C2–C3 bond, the two methyl groups are oriented:

  1. Pointing in opposite directions at an angle of 180° to each other.
  2. Pointing in the same direction along parallel vertical lines.
  3. At an angle of 60° with respect to each other.
  4. Eclipsing adjacent hydrogen atoms.
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Answer: (A) Pointing in opposite directions at an angle of 180° to each other.

In the anti conformation, the two bulky methyl groups are located as far apart as possible, pointing in opposite directions with a dihedral angle of 180°.

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