Organic Chemistry Quiz: Radical Addition To Alkenes Hbr Peroxides

What this quiz covers

This quiz focuses on Radical Addition To Alkenes Hbr Peroxides, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic 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.

Question 1

1. A phenyl radical (Ph•)
2. A bromine atom (Br•) (correct answer)
3. A benzoyloxy radical (PhCOO•)
4. An oxygen-centered radical cation (ROOR⁺•)

Explanation: Initiation involves homolysis of the peroxide to form benzoyloxy radicals, which often decarboxylate to form phenyl radicals. A phenyl radical (or an alkoxy radical) then abstracts a hydrogen atom from HBr to form a bromine radical (Br•). This bromine radical is the species that adds across the alkene's π-bond in the first key propagation step, determining the reaction's regiochemistry.

Question 2

1. 1-Hexene
2. Styrene (vinylbenzene)
3. 2-Methyl-2-pentene
4. 2,3-Dimethyl-2-butene (correct answer)

Explanation: The difference between ionic (no peroxide) and radical (with peroxide) addition of HBr lies in regioselectivity. For an alkene to give the same product under both conditions, the distinction between Markovnikov and anti-Markovnikov addition must be irrelevant. 2,3-Dimethyl-2-butene is a symmetrically substituted alkene; the two carbons of the double bond are chemically equivalent. Therefore, addition of H and Br across this bond will always produce 2-bromo-2,3-dimethylbutane, regardless of the mechanism.

Question 3

1. Higher selectivity for anti-Markovnikov addition due to increased kinetic control at low temperature
2. A switch to Markovnikov addition because the peroxide cannot generate radicals effectively at low temperature (correct answer)
3. Slower reaction rate but identical product distribution since regioselectivity depends only on radical stability
4. Formation of different constitutional isomers due to competing ionic and radical pathways at low temperature

Explanation: At 0°C, the peroxide initiator cannot effectively decompose to generate the initial radicals needed to start the chain reaction. Without sufficient radical initiation, the reaction will proceed via the normal ionic mechanism (Markovnikov addition) rather than the radical pathway. This demonstrates the critical role of temperature in radical initiation. Choice A incorrectly suggests temperature affects selectivity rather than mechanism. Choice C assumes the radical mechanism still operates. Choice D incorrectly suggests competing pathways rather than recognizing the switch from radical to ionic mechanism.

Question 4

1. The bromine radical adds to C2, forming a secondary radical that is not resonance-stabilized.
2. The bromine radical adds to C1, forming a secondary, resonance-stabilized allylic radical. (correct answer)
3. The bromine radical adds to C1, forming a secondary radical that immediately rearranges to a more stable tertiary radical.
4. A hydrogen atom from HBr adds first to C1, forming a resonance-stabilized allylic carbocation.

Explanation: The first propagation step involves the addition of a bromine radical to the diene. Addition to an end carbon (C1 or C4) is favored because it generates a resonance-stabilized allylic radical intermediate. This intermediate has radical character at both C2 and C4, which leads to the formation of both 1,2- and 1,4-addition products. Addition to an interior carbon (C2 or C3) would form a non-allylic radical, which is less stable and thus not the major pathway.

Question 5

1. The planar geometry ensures the incoming H-atom adds anti to the bromine atom, resulting in a specific diastereomer.
2. The reaction yields a single enantiomer because the initiator is chiral and directs the H-atom abstraction to one face.
3. The reaction is stereospecific, meaning the E/Z geometry of the starting alkene dictates the product's stereochemistry.
4. If the radical carbon becomes a stereocenter, H-atom abstraction can occur from either face, typically leading to a racemic mixture. (correct answer)

Explanation: When analyzing radical addition reactions like HBr addition to alkenes, you need to understand how the geometry of radical intermediates affects stereochemistry. The key insight is that alkyl radicals have trigonal planar geometry at the radical center, making them achiral even if they contain what will become a stereocenter. The correct answer is D because when the planar radical intermediate abstracts a hydrogen atom from HBr, the hydrogen can approach from either face of the trigonal planar radical with equal probability. Since there's no preference for one face over the other, you get equal amounts of both enantiomers—a racemic mixture. This is a fundamental consequence of the planar, achiral nature of the radical intermediate. Option A is wrong because it confuses radical addition with ionic addition mechanisms. In radical reactions, there's no requirement for anti-addition, and the stereochemistry isn't determined by relative positions of added atoms. Option B incorrectly suggests the reaction produces a single enantiomer due to chiral initiators. While initiators start the chain reaction, they don't remain bound to influence stereochemistry of individual addition steps. Option C confuses this with stereospecific ionic additions (like syn-dihydroxylation). In radical additions, the alkene's E/Z geometry doesn't directly control product stereochemistry because the radical intermediate can freely rotate. Remember: radical intermediates are planar and achiral, so when they become stereocenters through subsequent reactions, expect racemic mixtures unless other chiral influences are present in the molecule.

Question 6

1. Step 1 (Initiation): RO-OR → 2 RO•
2. Step 2 (Initiation): RO• + H-Br → ROH + Br•
3. Step 3 (Propagation): Br• + H₂C=CH₂ → Br-CH₂-CH₂⁺ + Br⁻ (correct answer)
4. Step 4 (Propagation): Br-CH₂-CH₂• + H-Br → Br-CH₂-CH₃ + Br•

Explanation: Step 3 incorrectly depicts the addition of the bromine radical. This step should result in a carbon-centered radical (Br-CH₂-CH₂•), not a carbocation and a bromide ion. The formation of ions is characteristic of a heterolytic (ionic) mechanism, not a homolytic (radical) one. Radical reactions involve the movement of single electrons, not pairs.

Question 7

1. The reaction flask was accidentally exposed to UV light during the reaction.
2. The peroxide used was old and had decomposed, rendering it an ineffective initiator. (correct answer)
3. The reaction was run at a very high temperature, favoring the thermodynamic product.
4. An aprotic solvent was used, which favors the radical mechanism over the ionic one.

Explanation: The desired product, 1-bromo-2-methylcyclohexane, is the anti-Markovnikov product from a radical chain reaction. The observed product, 1-bromo-1-methylcyclohexane, is the Markovnikov product from an ionic electrophilic addition. The most plausible reason for the ionic pathway's dominance is the failure of the radical pathway to initiate. If the peroxide initiator is inactive, no radicals are formed, and the standard electrophilic addition of HBr proceeds via the more stable tertiary carbocation, leading to the Markovnikov product.

Question 8

1. 2-Bromohexane
2. Dodecane
3. Bromine (Br₂) (correct answer)
4. 1,2-Dibromohexane

Explanation: Termination steps halt the chain reaction by consuming radicals. One possible termination event is the combination of two bromine radicals (Br•) to form a stable bromine molecule (Br₂). While other termination products like alkyl radical dimers are possible, the concentration of Br• is often significant. Dodecane (B) is not expected, as the intermediate is a bromohexyl radical, not a hexyl radical. 2-Bromohexane (A) is the Markovnikov product. 1,2-Dibromohexane (D) would result from the addition of Br₂.

Question 9

1. The addition of the bromine radical to the alkene is the slow, rate-determining step. (correct answer)
2. The abstraction of H from HBr by the carbon radical is the slow, rate-determining step.
3. The initiation step (peroxide decomposition) is the overall rate-determining step of the reaction.
4. Both propagation steps have similar rates, and neither is uniquely rate-determining.

Explanation: When you encounter kinetics questions about radical reactions, focus on how the observed rate law reveals which step controls the overall reaction speed. The HBr-alkene reaction with peroxides follows a radical chain mechanism with two key propagation steps: bromine radical addition to the alkene, and hydrogen abstraction from HBr by the resulting carbon radical. The crucial clue here is that the rate depends on alkene concentration but is independent of HBr concentration. In radical chain reactions, the rate-determining step consumes the species whose concentration affects the overall rate. Since the rate correlates with [alkene] but not [HBr], the slow step must be the one that consumes the alkene—bromine radical addition to the alkene double bond. Answer A correctly identifies this: bromine radical addition to the alkene is rate-determining because it's the step that consumes the concentration-dependent species (alkene). Answer B is wrong because if hydrogen abstraction from HBr were slow, the rate would depend on [HBr], which contradicts the observation. Answer C misses the point—we're asked specifically about the propagation cycle's rate-determining step, not the overall reaction initiation. The kinetic data clearly shows concentration dependence, indicating one step is definitively slower. Answer D incorrectly suggests both steps have similar rates, but this would create a more complex rate law than the simple alkene-dependent pattern observed. Remember: In radical mechanisms, the rate-determining propagation step consumes whichever reactant appears in the experimental rate law. Match concentration dependence to the step that uses that reactant.

Question 10

1. Radical rearrangements require higher activation energy than hydrogen abstraction from HBr, making rearrangement kinetically unfavorable (correct answer)
2. The secondary radical is actually more stable than the tertiary radical in this system due to hyperconjugation effects
3. Peroxides specifically inhibit radical rearrangements through coordination with the radical center
4. Radical rearrangements only occur in ionic mechanisms, not in radical chain processes

Explanation: Radical rearrangements (1,2-hydride or alkyl shifts) do occur but require significant activation energy. In the presence of HBr, the hydrogen abstraction step is very fast and has low activation energy, so it occurs before the radical has time to rearrange. This is a kinetic effect - the fast H-abstraction outcompetes the slower rearrangement process. Choice B incorrectly states secondary is more stable than tertiary. Choice C wrongly suggests peroxides prevent rearrangements. Choice D is false - radical rearrangements can occur in radical mechanisms but are usually outcompeted by faster processes.

Question 11

1. The reaction rate would increase because BHT helps generate additional bromine radicals from the peroxide initiator
2. The reaction would shift from anti-Markovnikov to Markovnikov selectivity due to changes in the radical intermediates
3. The reaction rate would decrease significantly because BHT terminates radical chains by scavenging reactive radical species (correct answer)
4. The stereoselectivity would change from random to highly selective due to BHT coordination with the radical intermediates

Explanation: Radical scavengers like BHT are designed to react with and neutralize radical species, effectively terminating radical chain reactions. Since the HBr/peroxide addition relies on a chain mechanism (initiation, propagation, termination), adding a radical scavenger would prematurely terminate chains and dramatically slow the reaction rate. Choice A incorrectly suggests BHT generates more radicals. Choice B is wrong because regioselectivity depends on radical stability, not the presence of scavengers. Choice D incorrectly suggests BHT affects stereochemistry through coordination rather than radical termination.

Question 12

1. The tertiary radical formed from 3,3-dimethyl-1-butene is too stable and reacts too slowly with HBr in the propagation step (correct answer)
2. Steric hindrance around the double bond prevents effective approach of the bromine radical to initiate addition
3. The highly branched structure favors elimination reactions over addition, reducing the yield of the desired product
4. Multiple constitutional isomers are formed due to competing addition at different positions, diluting the yield of any single product

Explanation: While tertiary radicals are more stable than secondary or primary radicals, this stability can actually slow down their reaction rates in subsequent steps. The very stable tertiary radical formed from 3,3-dimethyl-1-butene may have a slower rate of hydrogen abstraction from HBr, leading to longer-lived radicals that can undergo termination reactions, reducing chain length and overall yield. Choice B is unlikely since Br• is a small radical. Choice C incorrectly suggests competing elimination. Choice D is wrong because this alkene would give a single regioisomer under radical conditions.

Question 13

1. They are constitutional isomers formed by competing Markovnikov and anti-Markovnikov addition pathways
2. They are enantiomers formed by equal attack of the radical intermediate from both faces of the molecule (correct answer)
3. They are diastereomers resulting from syn and anti addition of HBr across the double bond geometry
4. They are identical molecules since the radical addition proceeds through a single regiochemical pathway

Explanation: The radical addition creates a new stereocenter at the carbon where the hydrogen ultimately attaches. Since the radical intermediate is planar (sp2), the hydrogen abstraction step can occur from either face with equal probability, generating equal amounts of both enantiomers (racemic mixture). Choice A is wrong because only anti-Markovnikov addition occurs under radical conditions. Choice C incorrectly applies syn/anti concepts from ionic additions; radical additions are stepwise and don't have concerted stereochemistry. Choice D ignores the stereochemical consequences of creating a new chiral center.

Question 14

1. 1-Butene reacts faster because it forms a more stable primary radical intermediate compared to the secondary radical from isobutylene
2. 2-Methyl-1-propene reacts faster because it forms a more stable tertiary radical intermediate compared to the secondary radical from 1-butene (correct answer)
3. Both react at similar rates because they both undergo anti-Markovnikov addition through comparable radical intermediates
4. 2-Methyl-1-propene reacts faster due to increased electron density at the double bond from the methyl substituents

Explanation: In radical additions, the rate-determining step is often the first addition of Br• to form the radical intermediate. 2-Methyl-1-propene forms a tertiary radical when Br• adds to the terminal carbon, while 1-butene forms a secondary radical. The greater stability of the tertiary radical intermediate leads to a lower activation energy and faster reaction rate. Choice A incorrectly states that 1-butene forms a primary radical (it forms secondary) and that primary would be more stable than secondary. Choice C ignores the significant stability difference between secondary and tertiary radicals. Choice D focuses on electronics rather than radical stability.

Question 15

1. A bromine atom attached to the more substituted carbon of the original double bond with the methyl group in an axial position
2. A bromine atom attached to the less substituted carbon of the original double bond with anti-Markovnikov regioselectivity (correct answer)
3. A bromine atom attached to the more substituted carbon following Markovnikov addition with syn stereochemistry
4. A rearranged carbocation intermediate leading to bromine attachment at the tertiary carbon position

Explanation: Under radical conditions (HBr/peroxides), the addition follows anti-Markovnikov regioselectivity. The bromine radical adds to the less substituted carbon of the alkene, generating the more stable secondary radical intermediate at the more substituted position. This secondary radical then abstracts hydrogen from HBr to give the anti-Markovnikov product. Choice A describes Markovnikov addition. Choice C incorrectly invokes both Markovnikov selectivity and syn stereochemistry (radical additions don't have concerted stereochemistry). Choice D incorrectly suggests carbocation formation, which occurs in ionic (not radical) mechanisms.

Question 16

1. A single achiral product, 1-bromo-4-methylhexane, is formed.
2. A racemic mixture of (2R,4R)- and (2S,4S)-1-bromo-4-methylhexane is formed.
3. A mixture of diastereomers, (2R,4R)- and (2S,4R)-1-bromo-4-methylhexane, is formed. (correct answer)
4. The original stereocenter at C4 is inverted, leading to a mixture of (2R,4S)- and (2S,4S)-products.

Explanation: The reaction is a radical addition, so Br• adds to C1 to form the more stable secondary radical at C2. The original stereocenter at C4 is not involved in the reaction and retains its (R) configuration. The intermediate radical at C2 is planar, and subsequent H-atom abstraction can occur from either face, creating a new stereocenter with both R and S configurations. The resulting products, (2R,4R)- and (2S,4R)-1-bromo-4-methylhexane, have the same configuration at C4 but differ at C2, making them diastereomers.

Question 17

1. 1-Bromobutane, because the scavenger catalyzes the anti-Markovnikov pathway.
2. 2-Bromobutane, because the scavenger inhibits the radical pathway, allowing the ionic pathway to prevail. (correct answer)
3. A mixture of cis- and trans-1,2-dibromobutane from a competing halogenation reaction.
4. No reaction will occur, as hydroquinone forms a stable complex with the alkene.

Explanation: Radical scavengers (inhibitors) like hydroquinone terminate radical chains by reacting with radical intermediates. This quenches the anti-Markovnikov radical addition pathway. In the absence of an effective radical reaction, HBr will add to the alkene via the slower, but still viable, electrophilic addition (ionic) mechanism. This pathway proceeds through the more stable secondary carbocation, leading to the Markovnikov product, 2-bromobutane.

Question 18

1. Both conditions yield 1-bromo-3,3-dimethylbutane as the exclusive product.
2. Ionic conditions yield 2-bromo-2,3-dimethylbutane, while radical conditions yield 1-bromo-3,3-dimethylbutane. (correct answer)
3. Ionic conditions yield 1-bromo-3,3-dimethylbutane, while radical conditions yield 2-bromo-2,3-dimethylbutane.
4. Both conditions yield the rearranged product 2-bromo-2,3-dimethylbutane.

Explanation: This question tests a key difference between carbocation and radical intermediates. Under ionic conditions, H⁺ adds to form a secondary carbocation which undergoes a rapid 1,2-methyl shift to a more stable tertiary carbocation, leading to the rearranged product 2-bromo-2,3-dimethylbutane. Under radical conditions, Br• adds to form a secondary radical. Radicals do not undergo these 1,2-shifts, so the radical abstracts a hydrogen atom to give the non-rearranged, anti-Markovnikov product, 1-bromo-3,3-dimethylbutane.

Question 19

1. CH₃CH(•)CH₂Br + HBr → CH₃CH₂CH₂Br + Br•
2. Br• + CH₂=CHCH₃ → CH₃CH(•)CH₂Br
3. 2 CH₃CH(•)CH₂Br → BrCH₂(CH₃)CH-CH(CH₃)CH₂Br (correct answer)
4. RO• + HBr → ROH + Br•

Explanation: Termination steps are reactions that consume radicals without generating new ones, thus ending the chain. The coupling of two carbon-centered radical intermediates to form a stable, non-radical dimer is a classic termination step. Choice C shows this process.

Question 20

1. 1. Dehydration with conc. H₂SO₄; 2. HBr with ROOR
   (correct answer)
2. 1. Dehydration with conc. H₂SO₄; 2. HBr without initiator
3. 1. Conversion to a tosylate (TsCl, pyr); 2. Reaction with NaBr in acetone
4. 1. Direct reaction with PBr₃; 2. Addition of ROOR to the product

Explanation: The question requires a radical addition step. First, 1-propanol must be converted to an alkene. Dehydration with concentrated acid accomplishes this, forming propene. To convert propene to 1-bromopropane (the anti-Markovnikov product), a radical addition of HBr is required, which is achieved by using a peroxide initiator (ROOR).