IB Chemistry Quiz: Understand Electron Sharing Reactions
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Understand Electron Sharing ReactionsQuestion 1 of 20

The free-radical substitution of methane occurs readily with chlorine (Cl₂) but is extremely slow with iodine (I₂). What is the best explanation for this difference in reactivity?

The first propagation step, involving the reaction of an iodine radical with methane, is highly endothermic.
The I-I bond is much stronger than the Cl-Cl bond, making the initiation step more difficult for iodine.
Iodine is a solid at room temperature, which limits its ability to mix and react with gaseous methane.
The iodine radical (I•) is significantly less electronegative than the chlorine radical (Cl•), making it a weaker oxidizing agent.
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IB Chemistry Quiz

IB Chemistry Quiz: Understand Electron Sharing Reactions

Practice Understand Electron Sharing Reactions in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Understand Electron Sharing Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The free-radical substitution of methane occurs readily with chlorine (Cl₂) but is extremely slow with iodine (I₂). What is the best explanation for this difference in reactivity?

  1. The first propagation step, involving the reaction of an iodine radical with methane, is highly endothermic. (correct answer)
  2. The I-I bond is much stronger than the Cl-Cl bond, making the initiation step more difficult for iodine.
  3. Iodine is a solid at room temperature, which limits its ability to mix and react with gaseous methane.
  4. The iodine radical (I•) is significantly less electronegative than the chlorine radical (Cl•), making it a weaker oxidizing agent.
Explanation: The correct answer is B. The key difference lies in the energetics of the first propagation step: CH₄ + X• → CH₃• + HX. For iodine, this step is endothermic. The energy required to break the strong C-H bond in methane is not sufficiently compensated by the energy released upon forming the relatively weak H-I bond. This high activation energy makes the step very slow. A is incorrect; the I-I bond is actually much weaker than the Cl-Cl bond, so initiation is easier for iodine. C is a physical property difference but not the fundamental chemical reason for the low reactivity. D is true but is a less direct explanation than the thermodynamic analysis of the propagation step.

Question 2

What is the defining characteristic of any propagation step within a free-radical chain mechanism?

  1. A stable molecule is converted into two or more free radicals.
  2. Two free radicals combine to form one or more stable molecules.
  3. A free radical and a stable molecule react to form a different free radical and a different stable molecule. (correct answer)
  4. A free radical reacts with another free radical to form a different free radical and a stable molecule.
Explanation: The correct answer is C. This is the definition of a propagation step. A radical is consumed, but a new radical is generated, thus keeping the chain reaction going. The net result is the conversion of a stable reactant into a stable product, with the radical acting as a carrier of the chain. A describes an initiation step. B describes a termination step. D describes a very rare process; reactions between two radicals are almost exclusively termination steps.

Question 3

The overall equation for the monochlorination of methane is CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g). Which species is a necessary intermediate in the mechanism but does NOT appear in the overall balanced equation?

  1. Cl₂
  2. HCl
  3. CH₃Cl
  4. CH₃• (correct answer)
Explanation: The correct answer is D. A reaction intermediate is a species that is formed in one step of a reaction mechanism and consumed in a subsequent step. In the free-radical chlorination of methane, the methyl radical (CH₃•) is formed in the first propagation step (CH₄ + Cl• → CH₃• + HCl) and consumed in the second propagation step (CH₃• + Cl₂ → CH₃Cl + Cl•). Because it is both produced and consumed, it does not appear in the overall stoichiometry. A is a reactant, while B and C are products; all appear in the overall equation.

Question 4

The free-radical chlorination of propane (CH₃CH₂CH₃) produces two isomeric radicals: a primary radical (CH₃CH₂CH₂•) and a secondary radical (CH₃ĊHCH₃). Which compound is a possible, but minor, product formed exclusively via a termination step?

  1. 2-chloropropane
  2. 1,2-dichloropropane
  3. 2,3-dimethylbutane (correct answer)
  4. Hydrogen chloride
Explanation: The correct answer is C. A termination step involves the combination of two radicals. The secondary radical is CH₃ĊHCH₃ (an isopropyl radical). If two of these radicals combine, they form 2,3-dimethylbutane. This is a product formed exclusively via termination. A (2-chloropropane) is a major product formed in a propagation step. B (1,2-dichloropropane) is a product of further substitution, which involves more propagation steps. D (Hydrogen chloride) is a product of the first propagation step.

Question 5

If the free-radical bromination of ethane (C₂H₆) is carried out, ethyl radicals (C₂H₅•) and bromine radicals (Br•) are present. Which molecule could be formed in a termination step but is NOT the desired major product, bromoethane?

  1. Dibromomethane
  2. 1,2-dibromoethane
  3. Butane (correct answer)
  4. Hydrogen bromide
Explanation: The correct answer is C. Termination steps involve the combination of any two radicals present in the mixture. While C₂H₅• + Br• → C₂H₅Br is a termination step that forms the desired product, other combinations are possible. The combination of two ethyl radicals (C₂H₅• + C₂H₅•) forms butane (C₄H₁₀), a common side-product. A is incorrect as it involves the wrong carbon skeleton. B would be a product of further substitution, not a primary termination product. D, hydrogen bromide (HBr), is a major product of the reaction, but it is formed during a propagation step (C₂H₆ + Br• → C₂H₅• + HBr), not a termination step.

Question 6

Consider the following proposed mechanism for a free-radical reaction: Step 1: Br₂ → 2Br• Step 2: CH₃CH₃ + Br• → CH₃CH₂• + HBr Step 3: CH₃CH₂• + Br₂ → CH₃CH₂Br + Br• Step 4: CH₃CH₂• + Br• → CH₃CH₂Br Which of the following lists contains all the species that act as free radicals in this mechanism?

  1. Br₂, HBr, CH₃CH₂Br
  2. Br• only
  3. CH₃CH₂• only
  4. Br• and CH₃CH₂• (correct answer)
Explanation: The correct answer is D. A free radical is a species with an unpaired electron, conventionally denoted by a dot (•). In the mechanism provided, the bromine radical (Br•) and the ethyl radical (CH₃CH₂•) are the two species shown with unpaired electrons. They act as the reactive intermediates in the chain reaction. A lists stable molecules (a reactant and products). B and C are both incomplete as they each list only one of the two radical species involved.

Question 7

Free-radical substitution reactions are often considered to have low synthetic utility because they produce a mixture of products. Which feature of the mechanism is the primary reason for this lack of specificity?

  1. The initiation step requires a high-energy input like UV light, which can break many different bonds indiscriminately.
  2. The termination steps are random combinations of any radicals present, leading to various alkane side-products.
  3. The reaction is a chain reaction where propagation steps can occur on already substituted products, leading to polysubstitution. (correct answer)
  4. The activation energy for the initial propagation step is very low, making the overall reaction difficult to stop or control.
Explanation: The correct answer is C. The main reason for the product mixture is that the desired product (e.g., CH₃Cl) can itself undergo further substitution. A chlorine radical can abstract a hydrogen from CH₃Cl just as it can from CH₄, leading to the formation of •CH₂Cl radicals and subsequently CH₂Cl₂, CHC₃, and CCl₄. For longer alkanes, substitution can also occur at different carbon atoms, creating constitutional isomers. While B is also true and contributes to side-products (e.g., ethane), the formation of polysubstituted isomers from ongoing propagation steps (C) is the more significant problem for synthetic control.

Question 8

To favour the formation of chloromethane (CH₃Cl) and minimize the formation of polysubstituted products like dichloromethane (CH₂Cl₂) during the free-radical chlorination of methane, what reaction conditions should be used?

  1. A large excess of chlorine relative to methane and low temperature.
  2. A large excess of methane relative to chlorine. (correct answer)
  3. High pressure to increase collision frequency and a suitable catalyst.
  4. The absence of UV light, using a high temperature instead to initiate the reaction.
Explanation: The correct answer is B. To favor monosubstitution, the concentration of the alkane (methane) should be much higher than that of the halogen (chlorine). This increases the probability that a chlorine radical will collide with a methane molecule rather than a newly formed chloromethane molecule. This minimizes the second substitution step (CH₃Cl + Cl• → •CH₂Cl + HCl) that leads to dichloromethane. Using excess chlorine (A) would strongly favor polysubstitution. High pressure (C) would increase all reaction rates non-selectively. Changing the initiation method from UV to heat (D) does not solve the selectivity problem, which is controlled by reactant ratios.

Question 9

Which statement best describes a free radical in the context of chemical reactions?

  1. A molecule with a net positive or negative charge due to the loss or gain of electrons from its neutral state.
  2. A highly reactive chemical species that possesses at least one unpaired valence electron. (correct answer)
  3. A stable intermediate that is formed during the initiation step and fully consumed during the termination step.
  4. An atom, ion, or molecule that has formed a coordinate bond by accepting a lone pair of electrons.
Explanation: The correct answer is B. This is the precise definition of a free radical. The unpaired electron makes the species highly reactive as it seeks to pair up. A describes an ion, not a radical. C is incorrect because free radicals are highly unstable and reactive, not stable. Furthermore, they are produced in initiation and propagation steps, and consumed in propagation and termination steps. D describes a Lewis acid or a species forming a dative covalent bond, which is unrelated to the concept of a free radical.

Question 10

Which equation represents the initiation step in the free-radical substitution reaction of ethane (C₂H₆) with bromine (Br₂)?

  1. C₂H₆ + Br• → C₂H₅• + HBr
  2. Br₂ → 2Br• (correct answer)
  3. C₂H₅• + Br₂ → C₂H₅Br + Br•
  4. C₂H₅• + Br• → C₂H₅Br
Explanation: The correct answer is B. The initiation step is the step where free radicals are generated from a stable, non-radical molecule. This typically requires an input of energy, such as UV light or heat, to cause homolytic fission. In this case, the bromine molecule breaks into two bromine radicals. A and C are both propagation steps, as a radical is consumed and another is generated. D is a termination step, where two radicals combine to form a stable molecule, thus removing radicals from the reaction.

Question 11

In the free-radical substitution of methane with chlorine, which step leads to the termination of the chain reaction?

  1. CH₃• + Cl₂ → CH₃Cl + Cl•
  2. CH₄ + Cl• → CH₃• + HCl
  3. CH₃Cl + Cl• → •CH₂Cl + HCl
  4. CH₃• + Cl• → CH₃Cl (correct answer)
Explanation: The correct answer is D. Termination steps are those in which free radicals are consumed without generating new ones, thereby ending the chain reaction. This occurs when two radicals collide and combine to form a stable molecule. In choice D, a methyl radical and a chlorine radical combine to form chloromethane. A, B, and C are all propagation steps because in each case, a radical is consumed but another radical is produced, allowing the chain to continue. C specifically shows how polysubstitution can occur.

Question 12

Which statement best describes the relative activation energies (Ea) of the steps in the free-radical chlorination of methane?

  1. The initiation step has the highest Ea and determines the overall rate of the reaction.
  2. The termination steps have the highest Ea because they require the collision of two low-concentration species.
  3. The termination steps have an Ea of approximately zero as they involve the combination of two highly reactive radicals. (correct answer)
  4. The second propagation step (CH₃• + Cl₂) has a higher Ea than the first propagation step (CH₄ + Cl•).
Explanation: The correct answer is C. The combination of two free radicals to form a stable bond is a highly exothermic process with a very small or negligible activation energy barrier. The radicals are extremely reactive and readily form a bond upon collision. A is incorrect because initiation is driven by absorbing a photon of specific energy, not by thermal activation energy in the same sense as propagation steps. B is incorrect because termination steps have very low activation energies; their rate is low only because the concentration of radicals is low. D is incorrect; the first propagation step (CH₄ + Cl• → CH₃• + HCl) involves breaking a strong C-H bond and has a higher activation energy than the second propagation step (CH₃• + Cl₂ → CH₃Cl + Cl•), which involves breaking the weaker Cl-Cl bond.

Question 13

Which statement best explains why a free-radical chain reaction eventually slows down and stops, even if reactants are still present?

  1. The concentration of radical intermediates increases, making termination steps more frequent until they dominate propagation. (correct answer)
  2. An equilibrium is established between the forward propagation steps and the reverse reactions.
  3. The UV light source is depleted over time, preventing further initiation from occurring.
  4. The products formed, such as HCl, act as inhibitors by reacting with the radical intermediates.
Explanation: The correct answer is C. The rate of propagation steps depends on the concentration of one radical (e.g., rate ∝ [Cl•]), while the rate of termination steps depends on the concentration of two radicals (e.g., rate ∝ [Cl•]²). As the reaction proceeds, the concentration of radicals builds up. Because the termination rate has a second-order dependence on radical concentration, it increases much faster than the propagation rate. Eventually, termination steps consume radicals faster than initiation steps create them, and the chain reaction dies out.

Question 14

During the chlorination of methane, several steps occur. Which of the following equations correctly represents a propagation step that continues the chain reaction?

  1. Cl• + Cl• → Cl₂
  2. CH₄ + Cl• → CH₃• + HCl (correct answer)
  3. CH₃• + CH₃• → C₂H₆
  4. CH₄ → CH₃• + H•
Explanation: The correct answer is B. A propagation step is characterized by the reaction of a radical with a stable molecule to produce a new radical and a new stable molecule. This conserves the number of radicals, allowing the chain reaction to continue. In choice B, a chlorine radical (Cl•) reacts with methane (CH₄) to produce a methyl radical (CH₃•) and hydrogen chloride (HCl). A and C are termination steps where two radicals combine. D represents the homolytic fission of methane, which is energetically very unfavorable and is not a step in this mechanism; initiation involves the halogen.

Question 15

A small amount of ethane (C₂H₆) is often detected among the products of the free-radical chlorination of methane. What is the most likely origin of this ethane?

  1. The reaction of a methyl radical with a methane molecule to form ethane and a hydrogen radical.
  2. The combination of two methyl radicals during a termination step. (correct answer)
  3. The decomposition of the main product, chloromethane, under the influence of UV light.
  4. An impurity that was present in the original methane gas sample before the reaction began.
Explanation: The correct answer is B. During the reaction, methyl radicals (CH₃•) are present as intermediates. A termination step can occur when two of these radicals collide and combine, forming a stable ethane molecule (CH₃• + CH₃• → C₂H₆). This is a well-established side-reaction in this mechanism. A is incorrect as the reaction of a methyl radical with methane would typically abstract a hydrogen atom, regenerating the reactants, not form a C-C bond. C is less likely than the direct combination of abundant intermediates. D is a possibility in a practical setting, but the question asks for a mechanistic origin of the product.

Question 16

In the accepted mechanism for the reaction CH₄ + Cl₂ → CH₃Cl + HCl, what are the direct sources of the hydrogen and chlorine atoms in the hydrogen chloride (HCl) product?

  1. The hydrogen atom comes from a methane molecule and the chlorine atom comes from a chlorine radical. (correct answer)
  2. The hydrogen atom comes from a methane molecule and the chlorine atom comes from a chlorine molecule.
  3. The hydrogen atom comes from a methyl radical and the chlorine atom comes from a chlorine molecule.
  4. The hydrogen atom comes from a hydrogen radical and the chlorine atom comes from a chlorine radical.
Explanation: The correct answer is B. The step in which HCl is formed is the first propagation step: CH₄ + Cl• → CH₃• + HCl. In this step, a chlorine radical (Cl•) abstracts a hydrogen atom from a stable methane molecule (CH₄). Therefore, the hydrogen atom originates from methane, and the chlorine atom originates from a chlorine radical. The other options describe incorrect combinations of species for this specific product formation step.

Question 17

In the free-radical substitution of methane by chlorine, what is the primary role of ultraviolet (UV) light?

  1. To provide the activation energy for the reaction between a chlorine radical and a methane molecule.
  2. To cause the homolytic fission of the Cl-Cl bond, generating two chlorine radicals. (correct answer)
  3. To cause the heterolytic fission of the C-H bond in methane, forming a methyl cation and a hydride ion.
  4. To act as a catalyst by providing an alternative reaction pathway with a lower overall enthalpy change.
Explanation: The correct answer is B. The initiation step of a free-radical substitution reaction requires energy to break a covalent bond and form radicals. In this case, UV light provides the energy for the homolytic fission of the relatively weak Cl-Cl bond (Cl₂ → 2Cl•). A is incorrect because activation energy for the propagation steps is provided by the kinetic energy of the colliding particles (thermal energy), not directly by the UV light. C is incorrect because the process is homolytic (even splitting of electrons), not heterolytic (uneven splitting), and it is the Cl-Cl bond, not the C-H bond, that breaks during initiation. D is incorrect because UV light is a condition or energy source, not a catalyst; it is consumed in the process and does not provide an alternative pathway in the catalytic sense.

Question 18

A free-radical chlorination is in progress. What would be the most likely effect of introducing a small amount of a stable radical, such as nitrogen monoxide (NO•), into the reaction mixture?

  1. It would accelerate the reaction by providing an alternative, lower-energy initiation pathway.
  2. It would have no significant effect, as it cannot participate in the specific propagation steps of this reaction.
  3. It would act as an inhibitor by reacting with chain-carrying radicals, causing premature termination of the chains. (correct answer)
  4. It would isomerize the alkane reactant, leading to the formation of different substituted products.
Explanation: The correct answer is C. Species like NO• or O₂ are known as radical scavengers or inhibitors. Because they are themselves radicals, they can readily combine with the highly reactive radicals (like CH₃• or Cl•) that propagate the chain. For example, CH₃• + NO• → CH₃NO. This is a termination step that removes the chain-carrying radical, thus inhibiting or stopping the chain reaction. A is incorrect; it does the opposite. B is incorrect because it will readily react with the unstable radicals. D is not a known function of radical inhibitors.

Question 19

In the mechanism for the reaction between methane and chlorine under UV light, what are the roles of the two reactant molecules in the first two distinct steps of the chain reaction mechanism?

  1. Chlorine undergoes homolytic fission in the initiation step, and methane is attacked by the resulting radical in the first propagation step. (correct answer)
  2. Methane undergoes homolytic fission in the initiation step, and chlorine acts as the radical in the first propagation step.
  3. Both methane and chlorine form radicals in the initiation step, which then combine in the first propagation step.
  4. Chlorine acts as a catalyst for the homolytic fission of methane in the initiation step, then reacts in the propagation step.
Explanation: The correct answer is B. The mechanism begins with the initiation step where the Cl-Cl bond, being weaker than the C-H bond, undergoes homolytic fission under UV light: Cl₂ → 2Cl•. This is the first step. The second step is the first propagation step, where one of these highly reactive chlorine radicals attacks a stable methane molecule: CH₄ + Cl• → CH₃• + HCl. This sequence correctly identifies the roles of the reactants.

Question 20

Consider the cleavage of a covalent bond between atoms X and Y. Which statement correctly distinguishes between homolytic and heterolytic fission?

  1. Homolytic fission occurs in polar solvents and produces a cation and an anion, while heterolytic fission occurs under UV light and produces two radicals.
  2. Homolytic fission involves an uneven distribution of bonding electrons resulting in ions, while heterolytic fission involves an even distribution resulting in neutral species.
  3. Homolytic fission results in two uncharged species, each with an unpaired electron, whereas heterolytic fission results in a cation and an anion. (correct answer)
  4. Homolytic fission is characteristic of nucleophilic substitution reactions, while heterolytic fission is characteristic of free-radical reactions.
Explanation: The correct answer is C. This statement correctly defines both processes. Homolytic fission involves the symmetrical breaking of a covalent bond, where each atom retains one of the bonding electrons, resulting in two free radicals (X• and Y•). Heterolytic fission involves the asymmetrical breaking of a bond, where one atom takes both bonding electrons, resulting in a cation (X⁺) and an anion (Y⁻). A, B, and D all incorrectly swap the definitions, conditions, or associated reaction types for the two kinds of fission.