Organic Chemistry Quiz: Radical Halogenation Selectivity And Mechanism

What this quiz covers

This quiz focuses on Radical Halogenation Selectivity And Mechanism, 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. They are a pair of enantiomers.
2. They are a pair of diastereomers. (correct answer)
3. They are identical meso compounds.
4. They are the same achiral compound.

Explanation: The starting material, (R)-3-methylhexane, is chiral and has a stereocenter at C-3. Radical bromination at C-2 creates a new stereocenter. The radical intermediate at C-2 is planar, and attack by bromine can occur from either face, creating both (R) and (S) configurations at C-2. Since the original stereocenter at C-3 is unaffected, the two products formed are (2R,3R)-2-bromo-3-methylhexane and (2S,3R)-2-bromo-3-methylhexane. These molecules are stereoisomers but not mirror images, so they are diastereomers.

Question 2

1. The reaction rate and product selectivity will be largely unaffected. (correct answer)
2. The reaction rate will increase dramatically due to stabilization of the radical intermediates.
3. The reaction will switch to an SN1/E1 pathway, producing methoxycyclohexane.
4. The selectivity will decrease because the polar solvent will solvate the bromine radical.

Explanation: Radical intermediates and transition states are neutral and have very little charge separation. Consequently, their stability is not significantly affected by solvent polarity. Unlike ionic reactions (such as SN1, SN2, E1, E2) which show strong solvent dependence, free-radical reactions are generally insensitive to the polarity of the solvent. Therefore, changing from a nonpolar solvent like CCl₄ to a polar protic solvent like methanol will have little to no effect on the rate or selectivity of the reaction.

Question 3

1. Chloroethane (CH₃CH₂Cl)
2. Hydrogen chloride (HCl)
3. Butane (CH₃CH₂CH₂CH₃) (correct answer)
4. Molecular chlorine (Cl₂)

Explanation: In radical chlorination of ethane, the propagation steps are: Cl• + CH₃CH₃ → HCl + •CH₂CH₃ and •CH₂CH₃ + Cl₂ → CH₃CH₂Cl + Cl•. These produce the main products HCl and chloroethane. Termination steps involve radical-radical combinations: Cl• + Cl• → Cl₂; Cl• + •CH₂CH₃ → CH₃CH₂Cl; and •CH₂CH₃ + •CH₂CH₃ → CH₃CH₂CH₂CH₃. While chloroethane and Cl₂ can form in termination steps, they are also involved in other parts of the mechanism. Butane can ONLY be formed by the coupling of two ethyl radicals, which exclusively occurs during termination.

Question 4

1. The primary carbon already bearing chlorine has a weakened C-H bond due to the electron-withdrawing effect (correct answer)
2. Chlorine radicals preferentially attack carbons that are already substituted to minimize steric interactions
3. The secondary carbon is deactivated toward radical substitution by the adjacent primary chloride
4. Statistical factors favor substitution at the primary carbon due to the presence of two equivalent hydrogens

Explanation: The electron-withdrawing chlorine atom weakens the adjacent C-H bonds, making them more susceptible to radical abstraction. This electronic effect overrides the normal preference for secondary over primary positions. Choice B incorrectly suggests steric effects favor substituted carbons. Choice C incorrectly suggests deactivation of the secondary position. Choice D mentions statistical factors but there are still more primary hydrogens at other positions that don't react as readily.

Question 5

1. Cl• + Cl• → Cl₂ (regeneration of starting material)
2. CH₃CH₂CHCH₃ + CH₃CH₂CHCH₃ → C₈H₁₈ (radical coupling to form octane)
3. CH₃CH₂CH₂CH₂• + CH₃CH₂CH₂CH₂• → C₈H₁₈ (coupling of primary radicals)
4. CH₃CH₂CHCH₃ + CH₃CH₂CH₂CH₂ → C₈H₁₈ (cross-coupling of different radicals) (correct answer)

Explanation: Cross-coupling between different carbon radicals (primary and secondary butyl radicals) would produce C₈ dimers that are significant side products and could interfere with analysis of the desired chlorobutane products. Choice A just regenerates Cl₂. Choices B and C would each produce only one type of dimer (from secondary or primary radicals respectively). Choice D represents the most problematic termination because it creates multiple different C₈ isomers from the various radical combinations possible.

Question 6

1. Cyclohexane has more C-H bonds available for abstraction, increasing the statistical probability of reaction
2. The chair conformation of cyclohexane provides better orbital overlap for radical formation compared to cyclopentane
3. Ring strain in cyclopentane destabilizes the radical intermediate, making hydrogen abstraction less favorable (correct answer)
4. Cyclohexane C-H bonds are weaker due to the larger ring size and reduced s-character

Explanation: Cyclopentane has significant ring strain (about 6.5 kcal/mol) compared to cyclohexane which is essentially strain-free. When a hydrogen is abstracted from cyclopentane, the resulting radical retains much of this ring strain, making it less stable than a cyclohexyl radical. This destabilization makes the hydrogen abstraction step less favorable. Choice A is incorrect because we're comparing per-molecule reactivity. Choice B incorrectly invokes orbital overlap differences. Choice D incorrectly suggests ring size affects C-H bond strength through hybridization.

Question 7

1. Equal amounts of chlorinated products from both substrates due to similar C-H bond strengths
2. Preferential chlorination of cyclohexane because all positions are equivalent, reducing steric hindrance
3. Preferential chlorination of methylcyclohexane at the tertiary carbon bearing the methyl group (correct answer)
4. Preferential chlorination of methylcyclohexane at secondary positions adjacent to the methyl-bearing carbon

Explanation: The carbon bearing the methyl group in methylcyclohexane is tertiary, and tertiary C-H bonds are weaker and form more stable radicals than secondary C-H bonds (as in cyclohexane). This leads to preferential abstraction at the tertiary position. Choice A ignores the difference between secondary and tertiary positions. Choice B incorrectly emphasizes steric effects over thermodynamic stability. Choice D incorrectly identifies the substitution pattern - the adjacent positions are still secondary, not more reactive than the tertiary position.

Question 8

1. Primary radicals rearrange to more stable secondary radicals through 1,2-hydride shifts during the propagation step (correct answer)
2. The high temperature conditions promote carbocation rearrangements that compete with the radical pathway
3. Bromine radicals catalyze intramolecular hydrogen atom transfer reactions leading to skeletal rearrangement
4. The NBS reagent generates ionic brominating species that proceed through carbocation intermediates prone to rearrangement

Explanation: Radical rearrangements can occur when a less stable radical (like primary) rearranges to a more stable radical (like secondary or tertiary) through 1,2-hydride or 1,2-alkyl shifts. This happens during the radical chain mechanism. Choice B incorrectly invokes carbocations in a radical mechanism. Choice C incorrectly suggests bromine radicals catalyze rearrangement. Choice D confuses radical bromination with ionic bromination mechanisms.

Question 9

1. Cl• + CH₃CH₃ → HCl + CH₃CH₂•
2. CH₃CH₂• + Cl₂ → CH₃CH₂Cl + Cl•
3. Cl• + CH₃CH₂• → CH₃CH₂Cl (correct answer)
4. CH₃CH₂• + HCl → CH₃CH₃ + Cl•

Explanation: A propagation step in a chain reaction must consume one radical and produce one radical, thus propagating the chain. Step A consumes a Cl• radical and produces an ethyl radical. Step B consumes an ethyl radical and produces a Cl• radical. Step D consumes an ethyl radical and produces a Cl• radical (it is the reverse of step A). Step C consumes two radicals (Cl• and an ethyl radical) to form a stable molecule. This is a termination step, as it results in a net decrease in the number of radicals and stops a chain.

Question 10

1. A racemic mixture of (R)- and (S)-2-bromobutane (correct answer)
2. Exclusively (R)-2-bromobutane
3. Exclusively (S)-2-bromobutane
4. A meso compound, as the starting material is achiral

Explanation: The mechanism involves the formation of a sec-butyl radical intermediate at C2. This radical is sp²-hybridized and trigonal planar (or rapidly inverting). The incoming bromine radical can attack from either face of the plane with equal probability. This leads to the formation of both (R)- and (S)-2-bromobutane in equal amounts, resulting in a racemic mixture.

Question 11

1. Initiation
2. Propagation (correct answer)
3. Termination
4. Homolytic cleavage

Explanation: This step is a propagation step because a radical (Br•) is consumed and another radical (•CH₃) is generated, allowing the chain reaction to continue. Initiation is the initial formation of radicals from non-radicals (e.g., Br₂ → 2 Br•). Termination is the net consumption of radicals (e.g., Br• + •CH₃ → CH₃Br). Homolytic cleavage describes bond breaking, but 'propagation' is the specific name for this type of step within the chain mechanism.

Question 12

1. Secondary carbocations are more stable than primary carbocations during the propagation step
2. The secondary radical intermediate is stabilized by hyperconjugation more than the primary radical (correct answer)
3. Bromine radicals are less reactive than chlorine radicals, leading to greater selectivity
4. NBS selectively generates bromine radicals that preferentially attack secondary carbons over primary carbons

Explanation: Radical halogenation proceeds through radical intermediates, not carbocation intermediates. The selectivity arises because secondary radicals are more stable than primary radicals due to hyperconjugation - the overlap of adjacent C-H bonds with the singly-occupied p-orbital. Choice A incorrectly invokes carbocations. Choice C is true but doesn't explain the mechanism of selectivity. Choice D incorrectly suggests the bromine radical itself is selective for secondary carbons, when actually it's the stability of the resulting carbon radical that drives selectivity.

Question 13

1. n-Pentane
2. 2-Methylbutane
3. 2,2-Dimethylpropane (correct answer)
4. Cyclopentane

Explanation: For an alkane to yield only one monobrominated product, all of its hydrogen atoms must be chemically equivalent. Let's examine the isomers of C₅H₁₂:

• n-Pentane has three distinct sets of hydrogens, leading to three possible products.
• 2-Methylbutane has four distinct sets of hydrogens, leading to four possible products.
• 2,2-Dimethylpropane (neopentane) has only one type of hydrogen—all 12 hydrogens are equivalent primary hydrogens. Substitution at any position gives the same product, 1-bromo-2,2-dimethylpropane.
• Cyclopentane has the formula C₅H₁₀, not C₅H₁₂, and all its hydrogens are equivalent, but it's not the correct starting material based on the formula.

Question 14

1. To provide the activation energy for the rate-determining hydrogen abstraction step.
2. To cause the homolytic cleavage of the halogen-halogen bond, generating the initial radical species. (correct answer)
3. To ionize the alkane, forming a carbocation that is susceptible to nucleophilic attack by the halogen.
4. To break the C-H bonds of the alkane, generating alkyl radicals directly in the first step of the reaction.

Explanation: The initiation step of radical halogenation is the cleavage of the diatomic halogen (e.g., Cl₂ or Br₂) into two halogen radicals (2 X•). This bond breaking is a homolytic cleavage, which has a high activation energy and is highly endothermic. The UV light or heat provides the necessary energy to overcome this barrier and generate the radicals that start the chain reaction. The energy required to break C-H bonds is even higher, so this does not happen directly.

Question 15

1. Chlorination would yield a higher proportion of 1-chloropropane than bromination would yield of 1-bromopropane. (correct answer)
2. Bromination would yield almost exclusively 1-bromopropane, while chlorination gives a mixture.
3. Chlorination would yield almost exclusively 2-chloropropane due to its higher speed.
4. Both reactions would yield approximately the same ratio of 1-halopropane to 2-halopropane.

Explanation: Propane has six primary (1°) hydrogens and two secondary (2°) hydrogens. Bromination is highly selective for the more stable secondary radical, so it will yield almost exclusively 2-bromopropane. Chlorination is much less selective. While the secondary C-H bond is more reactive, the statistical advantage of having six primary hydrogens means a significant amount of 1-chloropropane will be formed. Therefore, the ratio of primary to secondary halide will be much higher for chlorination than for bromination.