IB Chemistry Quiz: Apply Electron Sharing Reactions
20 questions · exam conditions
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Apply Electron Sharing ReactionsQuestion 1 of 20

Average Bond Enthalpies (kJ mol⁻¹): Cl-Cl: 243 C-H: 414 H-Cl: 431 C-Cl: 346

Using the provided bond enthalpy data, which statement correctly describes a step in the chlorination of methane?

The initiation step (Cl₂ → 2Cl•) is endothermic by 243 kJ mol⁻¹.
The propagation step •CH₃ + Cl₂ → CH₃Cl + Cl• is endothermic.
The propagation step CH₄ + Cl• → •CH₃ + HCl is exothermic.
The termination step •CH₃ + Cl• → CH₃Cl is endothermic.
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IB Chemistry Quiz

IB Chemistry Quiz: Apply Electron Sharing Reactions

Practice Apply 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 Apply Electron Sharing Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

How to use this quiz

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

Average Bond Enthalpies (kJ mol⁻¹): Cl-Cl: 243 C-H: 414 H-Cl: 431 C-Cl: 346

Using the provided bond enthalpy data, which statement correctly describes a step in the chlorination of methane?

  1. The initiation step (Cl₂ → 2Cl•) is endothermic by 243 kJ mol⁻¹.
  2. The propagation step •CH₃ + Cl₂ → CH₃Cl + Cl• is endothermic.
  3. The propagation step CH₄ + Cl• → •CH₃ + HCl is exothermic. (correct answer)
  4. The termination step •CH₃ + Cl• → CH₃Cl is endothermic.
Explanation: Let's analyze the enthalpy change for each step. ΔH = Σ(bonds broken) - Σ(bonds formed). A: Initiation involves breaking one Cl-Cl bond. ΔH = +243 kJ mol⁻¹, so this step is endothermic but by 243 kJ mol⁻¹, not 486. B: Propagation step 2 involves breaking a Cl-Cl bond (+243) and forming a C-Cl bond (-346). ΔH = 243 - 346 = -103 kJ mol⁻¹. This step is exothermic. C: Propagation step 1 involves breaking a C-H bond in methane (+414) and forming an H-Cl bond (-431). ΔH = 414 - 431 = -17 kJ mol⁻¹. This step is exothermic. D: Termination involves forming a C-Cl bond. Bond formation is always exothermic. ΔH = -346 kJ mol⁻¹. Therefore, statement C is the only correct one.

Question 2

A reaction is initiated between a large excess of ethane and a small amount of bromine in the presence of UV light. Which species will have the lowest concentration during the active course of the reaction?

  1. Ethane (C₂H₆)
  2. Bromoethane (C₂H₅Br)
  3. Hydrogen bromide (HBr)
  4. Ethyl radicals (C₂H₅•) (correct answer)
Explanation: Radicals are highly reactive intermediates. They are formed at a low rate and react very quickly in subsequent propagation or termination steps. As a result, their steady-state concentration during the reaction is extremely low. Ethane is a reactant in large excess. Bromoethane and hydrogen bromide are stable products that accumulate as the reaction proceeds. Therefore, the ethyl radical will have the lowest concentration.

Question 3

When propane reacts with a limited amount of bromine in the presence of UV light, two structural isomers of monobromopropane are formed. Which pair of radicals must be formed during the propagation stage to produce these isomers?

  1. CH₃CH₂CH₂• and CH₃CH(Br)CH₃•
  2. CH₃CH₂CH₂• and (CH₃)₂CH• (correct answer)
  3. CH₃CH₂CH₂Br• and (CH₃)₂CBr•
  4. CH₃CH₂• and CH₃CH₂CH₂•
Explanation: The formation of monobromopropane isomers results from the abstraction of a hydrogen atom from propane by a bromine radical. Propane (CH₃CH₂CH₃) has two types of hydrogen atoms: primary (on C1 and C3) and secondary (on C2). Abstraction of a primary hydrogen leads to the primary propyl radical (CH₃CH₂CH₂•). Abstraction of a secondary hydrogen leads to the secondary propyl radical ((CH₃)₂CH•). These two radicals then react with Br₂ in a subsequent propagation step to form 1-bromopropane and 2-bromopropane, respectively. A includes a product, not a radical. C shows radicals of the products, not the intermediates from propane. D shows an ethyl radical, which would come from ethane, not propane.

Question 4

Consider the reaction between benzene and chlorine gas in the presence of FeCl₃ catalyst. If the reaction proceeds via an electrophilic aromatic substitution mechanism, which statement best explains why the intermediate carbocation is stabilized during the rate-determining step?

  1. The positive charge is delocalized across the entire benzene ring through resonance, maintaining full aromaticity throughout the reaction.
  2. The positive charge is localized on one carbon atom, but the remaining four π electrons maintain partial aromatic character through resonance structures. (correct answer)
  3. The FeCl₃ catalyst forms a coordinate covalent bond with the carbocation, providing additional electron density to stabilize the positive charge.
  4. The chlorine atom donates electron density back to the ring through its lone pairs, creating a partial double bond character.
Explanation: In electrophilic aromatic substitution, the intermediate is a cyclohexadienyl cation (arenium ion) where the positive charge is localized but can be delocalized through resonance structures involving the remaining π electrons. The ring loses its full aromaticity temporarily. Choice A is wrong because full aromaticity is lost during the intermediate formation. Choice C is incorrect because FeCl₃ activates the electrophile, not the carbocation. Choice D is wrong because the C-Cl bond hasn't formed yet in the carbocation intermediate.

Question 5

In the bromination of alkenes using Br₂ in CCl₄, the reaction proceeds through a cyclic bromonium ion intermediate. Which experimental observation provides the strongest evidence for this mechanism rather than a carbocation intermediate?

  1. The reaction proceeds rapidly in nonpolar solvents, indicating that charge separation is minimized throughout the mechanism.
  2. The reaction rate is independent of alkene substitution pattern, suggesting that carbocation stability does not influence the mechanism.
  3. The stereochemistry shows anti-addition across the double bond, which is consistent with backside attack on the three-membered ring intermediate. (correct answer)
  4. The reaction shows no rearrangement products, indicating that carbocation intermediates capable of 1,2-shifts are not formed during the process.
Explanation: When you encounter questions about reaction mechanisms, focus on which experimental evidence directly distinguishes between proposed intermediates. The key here is understanding how different intermediates would lead to different observable outcomes. The bromination of alkenes can theoretically proceed through either a planar carbocation or a cyclic bromonium ion intermediate. The strongest evidence comes from stereochemical analysis. If a planar carbocation formed, the incoming bromide could attack from either face of the molecule, leading to both syn and anti addition products. However, experimental observations show exclusively anti addition - the two bromine atoms always end up on opposite faces of the original double bond. This stereospecificity is perfectly explained by the cyclic bromonium ion mechanism: the three-membered ring blocks one face, forcing the second bromide to attack from the opposite side, guaranteeing anti addition. Answer A is incorrect because reaction rate in nonpolar solvents doesn't distinguish between these specific intermediates - both involve some charge separation. Answer B is wrong because reaction rates actually do depend on alkene substitution, as more substituted alkenes form more stable bromonium ions. Answer D is misleading because while rearrangements are rare, their absence alone doesn't definitively prove the mechanism since some carbocations might not rearrange under these conditions. For IB Chemistry mechanism questions, always look for stereochemical evidence first - it provides the most direct proof of reaction pathways. Remember that anti addition is the signature of cyclic intermediate mechanisms, while syn addition typically indicates different pathways.

Question 6

In the reaction between 2-chloro-2-methylbutane and aqueous sodium hydroxide, both substitution and elimination products are observed. If the concentration of hydroxide ion is doubled while keeping temperature constant, which kinetic effect is most likely to occur?

  1. Both substitution and elimination rates increase proportionally, maintaining the same product ratio since both reactions follow second-order kinetics with respect to hydroxide concentration.
  2. The elimination rate increases more than the substitution rate because E2 elimination is more sensitive to base concentration changes than SN2 substitution under these conditions.
  3. The substitution rate increases while elimination rate remains constant because SN1 substitution depends on hydroxide concentration but E1 elimination depends only on substrate concentration.
  4. The elimination rate increases more significantly because the tertiary substrate undergoes E2 elimination with second-order kinetics while substitution proceeds via SN1 with first-order kinetics. (correct answer)
Explanation: 2-chloro-2-methylbutane is a tertiary halide, which typically undergoes SN1 substitution (first-order kinetics, independent of [OH⁻]) and E2 elimination (second-order kinetics, dependent on [OH⁻]). Doubling [OH⁻] will double the E2 rate but not affect the SN1 rate, shifting the product ratio toward elimination. Choice A incorrectly assumes both are second-order. Choice B incorrectly suggests SN2 occurs with tertiary halides. Choice C incorrectly invokes E1 mechanism.

Question 7

In a nucleophilic substitution reaction, 2-bromobutane reacts with sodium methoxide (CH₃O⁻Na⁺) in methanol. If the reaction temperature is increased from 25°C to 65°C, which mechanistic change is most likely to occur and why?

  1. The mechanism shifts from SN1 to SN2 because higher temperature increases the nucleophile's kinetic energy and collision frequency with the substrate.
  2. The mechanism shifts from SN2 to SN1 because higher temperature promotes carbocation formation by providing sufficient activation energy for C-Br bond cleavage.
  3. The mechanism remains SN2 but the stereochemistry changes from inversion to retention because thermal energy disrupts the backside attack geometry.
  4. The mechanism shifts from SN2 to E2 because increased temperature favors elimination over substitution, with methoxide acting as a base rather than nucleophile. (correct answer)
Explanation: Higher temperature generally favors elimination reactions over substitution. Methoxide is both a strong nucleophile and a strong base. At elevated temperature, E2 elimination becomes more favorable, where methoxide abstracts a β-hydrogen while bromide leaves. Choice A is incorrect because SN1 is not favored for secondary halides with strong nucleophiles. Choice B is wrong because SN1 is still unlikely for secondary substrates even at higher temperature. Choice C is incorrect because SN2 stereochemistry doesn't change with temperature.

Question 8

During the acid-catalyzed hydration of 2-methylpropene, the intermediate carbocation can undergo a 1,2-hydride shift. Which statement best explains the thermodynamic driving force for this rearrangement?

  1. The hydride shift converts a less stable secondary carbocation to a more stable tertiary carbocation through hyperconjugation effects from additional alkyl groups. (correct answer)
  2. The hydride shift relieves steric strain in the secondary carbocation by redistributing electron density to a less crowded carbon center.
  3. The hydride shift creates a more electrophilic carbon center that can more readily accept electron density from the incoming water molecule.
  4. The hydride shift aligns the carbocation with the most substituted carbon, allowing for better orbital overlap during the subsequent nucleophilic attack.
Explanation: The initial protonation of 2-methylpropene at the terminal carbon creates a secondary carbocation, which can rearrange via 1,2-hydride shift to form a more stable tertiary carbocation. Tertiary carbocations are more stable due to hyperconjugation from three alkyl groups versus two in the secondary carbocation. Choice B is incorrect because the tertiary position is actually more sterically crowded. Choice C is wrong because greater stability means lower electrophilicity. Choice D incorrectly describes orbital considerations that don't drive the rearrangement.

Question 9

When 1-butanol undergoes acid-catalyzed dehydration at 170°C, both elimination products are observed. Which factor most significantly influences the ratio of 1-butene to 2-butene in the product mixture?

  1. The stability difference between primary and secondary carbocation intermediates formed during the E1 mechanism predominates over kinetic factors.
  2. The relative thermodynamic stability of the alkene products, with more substituted alkenes being favored according to Zaitsev's rule under these conditions. (correct answer)
  3. The steric accessibility of β-hydrogens, where less hindered hydrogens are preferentially abstracted during the elimination process.
  4. The strength of the C-H bonds being broken, with tertiary hydrogens being more easily abstracted than secondary hydrogens during elimination.
Explanation: At 170°C, the reaction conditions favor thermodynamic control, where the more stable (more substituted) alkene predominates according to Zaitsev's rule. 2-butene is more substituted than 1-butene and thus more thermodynamically stable. Choice A is incorrect because primary alcohols don't readily form carbocations via E1. Choice C describes kinetic factors that would favor the less substituted product. Choice D is wrong because the difference in C-H bond strength between secondary and primary positions is not the determining factor under these conditions.

Question 10

In the base-catalyzed aldol condensation between acetaldehyde and benzaldehyde, the reaction produces cinnamaldehyde as the major product. Which mechanistic step is most crucial for achieving the observed cross-selectivity?

  1. The preferential deprotonation of acetaldehyde occurs because it has α-hydrogens that can form enolate ions, while benzaldehyde lacks α-hydrogens entirely.
  2. The enolate ion from acetaldehyde selectively attacks benzaldehyde because aromatic aldehydes are more electrophilic than aliphatic aldehydes due to conjugation effects.
  3. The initial aldol adduct undergoes preferential dehydration because the resulting α,β-unsaturated system can conjugate with the aromatic ring, providing thermodynamic stability. (correct answer)
  4. The base selectively abstracts protons from acetaldehyde because the resulting enolate can be stabilized through resonance with the carbonyl group more effectively than benzaldehyde enolates.
Explanation: Cross-selectivity in mixed aldol condensations often depends on the thermodynamic stability of the final product. Cinnamaldehyde formation is favored because the dehydration step creates an extended conjugated system (alkene conjugated with both carbonyl and aromatic ring), providing significant stabilization. Choice A describes a necessary condition but not the selectivity-determining step. Choice B incorrectly explains electrophilicity differences. Choice D is incorrect because benzaldehyde cannot form enolates due to lack of α-hydrogens.

Question 11

In the reaction between acetone and hydrogen cyanide (HCN) in the presence of a base catalyst, a cyanohydrin is formed. Which mechanistic step determines the regioselectivity of this nucleophilic addition reaction?

  1. The initial deprotonation of HCN by the base creates the nucleophilic cyanide ion, which determines where nucleophilic attack will occur on the carbonyl carbon.
  2. The nucleophilic attack of CN⁻ occurs exclusively at the carbonyl carbon because this is the most electrophilic center, making regioselectivity predetermined by electronic factors. (correct answer)
  3. The protonation of the alkoxide intermediate occurs preferentially at the oxygen atom rather than the carbon, establishing the final regiochemistry of the product.
  4. The reformation of the C=O double bond during the reaction determines which carbon atom retains the cyanide group in the final product structure.
Explanation: In nucleophilic addition to carbonyls, regioselectivity is determined by the electrophilicity of the carbonyl carbon, which is the most electron-deficient center due to the electronegativity difference between carbon and oxygen. CN⁻ attacks only at this position. Choice A describes catalyst activation but not regioselectivity. Choice C is incorrect because the alkoxide intermediate doesn't determine regioselectivity - it's already established by the initial attack. Choice D is wrong because the C=O double bond doesn't reform in cyanohydrin formation.

Question 12

During the acid-catalyzed esterification of benzoic acid with methanol, isotopic labeling experiments show that the oxygen atom in the water byproduct comes exclusively from the carboxylic acid. Which mechanistic detail does this evidence support?

  1. The reaction proceeds through acyl-oxygen cleavage, where the C-O bond of the carboxyl group breaks during the tetrahedral intermediate collapse. (correct answer)
  2. The reaction proceeds through alkyl-oxygen cleavage, where the methanol C-O bond breaks as methanol acts as a leaving group from the tetrahedral intermediate.
  3. The reaction involves protonation of the carboxyl oxygen followed by direct displacement by methanol in a concerted SN2-like mechanism.
  4. The reaction proceeds through alkyl-oxygen cleavage, where the carboxylic acid C-O bond breaks and the oxygen from methanol is incorporated into the water product.
Explanation: The isotopic labeling shows that the oxygen in water comes from the carboxylic acid, indicating that the C-O bond of the carboxyl group breaks (acyl-oxygen cleavage). In the tetrahedral intermediate, the -OH from the original carboxylic acid becomes the leaving group. Choice B incorrectly describes alkyl-oxygen cleavage. Choice C describes a mechanism that doesn't involve tetrahedral intermediates. Choice D contradicts the experimental evidence about oxygen source.

Question 13

Which of the following species would NOT be expected to be present in the reaction vessel during the free-radical chlorination of ethane?

  1. C₂H₅•
  2. HCl
  3. C₄H₁₀
  4. Cl⁻ (correct answer)
Explanation: The free-radical substitution mechanism involves neutral species and radicals (which are also neutral). The species present include reactants (C₂H₆, Cl₂), intermediates (Cl•, C₂H₅•), and products (C₂H₅Cl, HCl). Termination steps can also produce larger alkanes like butane (C₄H₁₀) from the combination of two ethyl radicals. The chloride ion (Cl⁻) is the result of heterolytic fission, where one atom takes both electrons from the bond. This type of fission does not occur in this mechanism, so Cl⁻ ions would not be expected to be present.

Question 14

The free-radical mechanism for the chlorination of methane is supported by experimental observations. Which observation provides the strongest evidence for the occurrence of termination steps?

  1. The reaction quantum yield is very high, with many product molecules formed per photon absorbed.
  2. The reaction only proceeds at a significant rate when initiated by UV light or high temperature.
  3. Trace amounts of ethane are detected in the product mixture alongside chloromethane. (correct answer)
  4. A mixture of chloromethane, dichloromethane, and trichloromethane is typically formed.
Explanation: Termination steps involve the combination of two radicals. The only way to form ethane (C₂H₆) in this reaction is through the combination of two methyl radicals (•CH₃ + •CH₃ → C₂H₆). Since methyl radicals are only present as intermediates in the mechanism, the detection of ethane is direct evidence that they exist and can combine, which is a termination process. A provides evidence for a chain reaction (propagation). B provides evidence for the initiation step's high activation energy. D provides evidence for the continuation of the propagation cycle (further substitution).

Question 15

In the free-radical substitution of methane with chlorine, which statement correctly describes the initiation step?

  1. UV light causes heterolytic fission of a C-H bond to form CH₃⁻ and H⁺ ions.
  2. UV light provides the energy for the homolytic fission of a Cl-Cl bond to form two chlorine radicals. (correct answer)
  3. A chlorine radical attacks a methane molecule, causing homolytic fission of a C-H bond.
  4. UV light provides the activation energy for the homolytic fission of a C-H bond in methane.
Explanation: The initiation step is the first step of the mechanism where radicals are generated from a stable molecule. This requires an input of energy, provided by UV light, to break the weakest bond, which is the Cl-Cl bond. The bond breaks homolytically, meaning one electron from the covalent bond goes to each chlorine atom, forming two chlorine radicals (Cl•). A is incorrect because the fission is homolytic, not heterolytic, and it involves the Cl-Cl bond, not a C-H bond. C describes the first propagation step, not initiation. D is incorrect because the initial bond broken is the Cl-Cl bond, not the stronger C-H bond in methane.

Question 16

In the mechanism for the reaction between methane and chlorine under UV light, what is the role of the chlorine radical (Cl•)?

  1. A catalyst that is regenerated in the termination step.
  2. A final product formed when the chain reaction is stopped.
  3. A reactive intermediate that is consumed in one propagation step and regenerated in another. (correct answer)
  4. An electrophile that is attracted to the electron-rich C-H bond in methane.
Explanation: The chlorine radical (Cl•) is formed in the initiation step. It is then consumed in the first propagation step (CH₄ + Cl• → •CH₃ + HCl) and regenerated in the second propagation step (•CH₃ + Cl₂ → CH₃Cl + Cl•). A species that is formed and then consumed during the reaction mechanism is known as a reactive intermediate. Because it is part of a repeating cycle, it allows the chain reaction to continue. A is incorrect; it is not a catalyst and is consumed, not regenerated, in termination. B is incorrect; it is a reactant in termination steps. D is incorrect; while it seeks an electron, its reactivity is defined by its unpaired electron, characteristic of a radical, not an electrophile in the typical sense.

Question 17

What is the primary role of ultraviolet (UV) light in the free-radical substitution reaction between an alkane and a halogen?

  1. To provide the specific energy required for the homolytic fission of the halogen-halogen bond. (correct answer)
  2. To increase the average kinetic energy of the molecules, thus increasing collision frequency.
  3. To act as a homogeneous catalyst by providing an alternative reaction pathway.
  4. To cause the heterolytic fission of C-H bonds, thereby creating carbocation intermediates.
Explanation: The initiation of a free-radical substitution reaction requires breaking the halogen-halogen bond (e.g., Cl-Cl) to form radicals. This bond fission is homolytic and requires a significant amount of energy, known as the bond dissociation energy. UV light provides photons with energy sufficient to break this bond (E = hν). This is the specific role of UV light in this mechanism. A describes the effect of increasing temperature, not UV light. C is incorrect; UV light is an energy source, not a catalyst. D is incorrect as the fission is homolytic (not heterolytic) and it is the halogen-halogen bond (not the C-H bond) that breaks in the initiation step.

Question 18

The reaction of excess methane with a limited amount of chlorine under UV light produces primarily chloromethane. Why is using excess methane crucial for this outcome?

  1. It increases the probability of a chlorine radical colliding with a methane molecule rather than a chloromethane molecule. (correct answer)
  2. It ensures that all chlorine radicals are immediately consumed in termination steps, stopping the reaction.
  3. It acts as an inert solvent, separating the reactive chlorine molecules from each other.
  4. It shifts the position of equilibrium towards the formation of monosubstituted product according to Le Châtelier's principle.
Explanation: This is a matter of reaction kinetics and probability. Both methane (CH₄) and the product chloromethane (CH₃Cl) can react with chlorine radicals. To favor the formation of the monosubstituted product (CH₃Cl), we want to maximize the chance that a Cl• radical reacts with CH₄ and minimize its chance of reacting with CH₃Cl. By using a large excess of methane, the concentration of CH₄ is much higher than the concentration of CH₃Cl, making a collision between Cl• and CH₄ far more probable. This suppresses the second substitution step and improves the yield of chloromethane. A is incorrect; we want propagation to occur, not just termination. C is incorrect; methane is a reactant, not an inert solvent. D is incorrect as this is an issue of kinetic control, not thermodynamic equilibrium; the reaction is essentially non-reversible.

Question 19

Consider the monochlorination of 2-methylpropane, (CH₃)₃CH, in the presence of UV light. How many different structural isomers of the product C₄H₉Cl can be formed?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: To find the number of possible structural isomers, we must identify the number of chemically distinct hydrogen atoms in the starting alkane. In 2-methylpropane, (CH₃)₃CH, there are two types of hydrogen atoms. Nine hydrogen atoms are primary, bonded to the three methyl carbons. These nine are all equivalent. One hydrogen atom is tertiary, bonded to the central carbon. Substitution of a primary hydrogen gives 1-chloro-2-methylpropane. Substitution of the tertiary hydrogen gives 2-chloro-2-methylpropane. These are two different structural isomers. Therefore, two isomers can be formed.

Question 20

A sealed vessel contains equimolar amounts of methane (CH₄) and bromine (Br₂) vapour. The vessel is kept in complete darkness at room temperature. Which statement best predicts the outcome?

  1. No discernible reaction will occur as the activation energy for the initiation step is too high. (correct answer)
  2. A rapid chain reaction will occur, producing bromomethane and hydrogen bromide.
  3. A slow reaction will occur via heterolytic fission, eventually reaching equilibrium.
  4. The gases will react to form a stable intermediate complex, CH₄Br₂.
Explanation: Free-radical substitution reactions require a significant energy input to start the initiation step, which is the homolytic fission of the halogen molecule (Br₂). This energy is typically supplied by UV light or high temperatures. In the dark at room temperature, there is insufficient energy to overcome this high activation energy barrier. Consequently, no reaction will occur at a noticeable rate. A is incorrect because UV light is absent. C is incorrect because the mechanism is radical-based (homolytic), not ionic (heterolytic). D is incorrect as no such stable complex is formed.