- Which equation represents a propagation step in the free-radical bromination of ethane?
Answer: C₂H₆ + Br• → C₂H₅• + HBr
A propagation step in a free-radical chain reaction is characterized by the reaction of a radical with a non-radical molecule to produce a new radical and a new non-radical molecule. Option C fits this pattern: the bromine radical (Br•) reacts with an ethane molecule (C₂H₆) to produce an ethyl radical (C₂H₅•) and a hydrogen bromide molecule (HBr). The number of radicals is conserved.
A is the initiation step.
B and D are termination steps, where two radicals combine to form a stable molecule, reducing the number of radicals.
- Which of the following species would NOT be expected to be present in the reaction vessel during the free-radical chlorination of ethane?
Answer: Cl⁻
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.
- 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?
Answer: The mechanism shifts from SN2 to E2 because increased temperature favors elimination over substitution, with methoxide acting as a base rather than nucleophile.
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.
- Using the provided bond enthalpy data, which statement correctly describes a step in the chlorination of methane?
Answer: The propagation step CH₄ + Cl• → •CH₃ + HCl is exothermic.
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.
- 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?
Answer: It increases the probability of a chlorine radical colliding with a methane molecule rather than a chloromethane molecule.
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.
- 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?
Answer: The positive charge is localized on one carbon atom, but the remaining four π electrons maintain partial aromatic character through resonance structures.
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.
- In the mechanism for the reaction between methane and chlorine under UV light, what is the role of the chlorine radical (Cl•)?
Answer: A reactive intermediate that is consumed in one propagation step and regenerated in another.
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.
- Which statement best explains why free-radical substitution is not a preferred method for synthesizing a specific, pure chloroalkane?
Answer: The propagation steps can continue, substituting multiple hydrogen atoms and creating a mixture of polychlorinated products.
The primary problem with free-radical substitution for synthesis is its lack of selectivity. Once the first substitution has occurred (e.g., forming CH₃Cl), this product can compete with the original alkane (CH₄) for reaction with chlorine radicals. This leads to further substitution, producing dichloromethane (CH₂Cl₂), trichloromethane (CHCl₃), and tetrachloromethane (CCl₄). This results in a mixture of products that is difficult to separate, making it a poor synthetic route for a pure, specific chloroalkane.
A is a practical consideration but not the main chemical reason. B is true, but the amount of termination products is typically small compared to the mixture from propagation. D is incorrect as other halogens can be used.
- 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?
Answer: The relative thermodynamic stability of the alkene products, with more substituted alkenes being favored according to Zaitsev's rule under these conditions.
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.
- 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?
Answer: The reaction proceeds through acyl-oxygen cleavage, where the C-O bond of the carboxyl group breaks during the tetrahedral intermediate collapse.
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.
- A student proposes several steps for the free-radical reaction between ethane and chlorine. Which of the following proposed steps is NOT a plausible part of the mechanism?
Answer: C₂H₅• + HCl → C₂H₅Cl + H•
Step D is not plausible. It involves an ethyl radical (C₂H₅•) reacting with HCl to produce chloroethane and a hydrogen radical (H•). This is unlikely for two reasons: 1) The H-Cl bond is very strong (431 kJ mol⁻¹) and difficult to break. 2) The concentration of HCl product is initially low, and the ethyl radical is much more likely to react with a Cl₂ molecule, which is a reactant present in higher concentration and has a weaker bond (Cl-Cl, 243 kJ mol⁻¹). The plausible propagation step is C₂H₅• + Cl₂ → C₂H₅Cl + Cl•.
A is initiation. B is the first propagation step. C is a termination step. All are plausible.
- 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?
Answer: The stereochemistry shows anti-addition across the double bond, which is consistent with backside attack on the three-membered ring intermediate.
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.
- 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?
Answer: CH₃CH₂CH₂• and (CH₃)₂CH•
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.
- What is the primary role of ultraviolet (UV) light in the free-radical substitution reaction between an alkane and a halogen?
Answer: To provide the specific energy required for the homolytic fission of the halogen-halogen bond.
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.
- During the free-radical chlorination of methane, a small amount of a gaseous alkane with a molar mass of 30.07 g mol⁻¹ is produced. Which step is responsible for its formation?
Answer: •CH₃ + •CH₃ → C₂H₆
First, identify the alkane. The molar mass of C₂H₆ (ethane) is (2 × 12.01) + (6 × 1.01) = 30.08 g mol⁻¹, which matches the given value. The question asks for the step that forms ethane. Ethane can only be formed by the combination of two methyl radicals (•CH₃). This is a termination step, as two radicals combine to form a single stable molecule. Therefore, the correct step is •CH₃ + •CH₃ → C₂H₆.
A is a propagation step. B and C are also termination steps, but they produce chloromethane and chlorine, respectively, not ethane.
- Which statement provides the most accurate chemical definition of the methyl radical, •CH₃?
Answer: A species with a central carbon atom that has an incomplete valence shell and one unpaired electron.
A radical is a species with at least one unpaired electron. The methyl radical, •CH₃, is formed by homolytic fission of a C-H bond in methane. The carbon atom has three bonding pairs of electrons and one unpaired electron, totaling 7 valence electrons. This is an incomplete valence shell (octet), making it highly reactive.
A is incorrect; radicals are neutral, not ions. It describes a methyl carbocation, CH₃⁺.
C is incorrect; •CH₃ is a reactive intermediate, which is a species that exists for a finite lifetime in a potential energy minimum between steps, not a transition state, which is a maximum energy point during a single step.
D describes a property of methane due to electronegativity differences but does not define a radical.
- 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?
Answer: 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.
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.
- 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?
Answer: Trace amounts of ethane are detected in the product mixture alongside chloromethane.
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).
- 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?
Answer: The hydride shift converts a less stable secondary carbocation to a more stable tertiary carbocation through hyperconjugation effects from additional alkyl groups.
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.
- The free-radical substitution mechanism is described as a chain reaction. Which feature of the mechanism is the primary reason for this description?
Answer: The propagation cycle consumes a radical and a molecule, but regenerates a radical that can repeat the process.
A chain reaction is characterized by a self-sustaining cycle of steps. In this mechanism, the propagation stage consists of a cycle where one radical reacts to form a product and a new radical. This new radical can then participate in the next cycle. For example, Cl• reacts to form •CH₃, and then •CH₃ reacts to regenerate Cl•. This cycle can repeat thousands of times from a single initiation event, which is the 'chain' aspect of the reaction.
A describes how the chain starts. B describes how it stops. D describes the thermodynamics, not the mechanistic feature of a 'chain'.
