Chapter: Radical Reactions

1. Introduction to Radicals

• Definition: Radicals (or free radicals) are atoms, molecules, or ions that have at least one unpaired valence electron. This unpaired electron makes them highly reactive and often unstable.
• Paramagnetism: Due to the presence of unpaired electrons, radicals are paramagnetic, meaning they are attracted to an external magnetic field.
• Reactivity: Radicals are highly reactive intermediates because they seek to achieve a stable octet (or duet for hydrogen) by pairing their unpaired electron.
• Bond Breaking/Forming:
  • Homolytic Cleavage: Bonds break symmetrically, with each atom retaining one electron from the shared pair, forming two radicals. This typically requires energy (heat or light).
    • Example: Cl-Clhν or heat​Cl⋅+⋅Cl
  • Heterolytic Cleavage: Bonds break unsymmetrically, with one atom taking both electrons, forming ions (not radicals).
  • Fishhook Arrows (Half-headed arrows): Used in reaction mechanisms to depict the movement of a single electron.

2. Formation and Stability of Radicals

2.1. Formation (Initiation)

Radicals are typically formed by homolytic cleavage of a weak bond, requiring energy input.

• Heat (Δ): Thermal energy can break bonds.
• Light (hν): UV light provides the energy to break certain bonds (e.g., halogen-halogen, peroxide O-O bonds).
• Peroxides (R-O-O-R’): The O-O bond in peroxides is relatively weak and easily undergoes homolytic cleavage to form two alkoxy radicals. Peroxides are common radical initiators.

2.2. Stability of Radicals

Like carbocations, radical stability increases with increasing substitution. This is primarily due to hyperconjugation (stabilization by electron donation from adjacent C-H σ bonds) and inductive effects.

• Order of Stability (Alkyl Radicals): Methyl (⋅CH3​) < Primary (⋅RCH2​) < Secondary (⋅R2​CH) < Tertiary (⋅R3​C)
• Resonance Stabilization: Radicals can be significantly stabilized if the unpaired electron can be delocalized over multiple atoms through resonance.
  • Allylic Radicals: ⋅CH2​−CH=CH2​ are more stable than tertiary alkyl radicals due to resonance.
  • Benzylic Radicals: ⋅CH2​−C6​H5​ are even more stable due to extensive resonance delocalization into the aromatic ring.
• Order of Overall Stability: Benzylic ≈ Allylic > Tertiary > Secondary > Primary > Methyl
• Factors influencing stability:
  • Hyperconjugation: Alkyl groups can donate electron density to the electron-deficient radical center, stabilizing it.
  • Resonance: Delocalization of the unpaired electron into π systems.

3. General Mechanism of Radical Chain Reactions

Radical reactions often proceed via a three-step chain mechanism:

1. Initiation: Formation of initial radicals from non-radical precursors. This step usually requires energy (heat or light) or a radical initiator (like a peroxide).
   • Example: Cl-Clhν​2 Cl⋅
2. Propagation: Radicals react with stable, non-radical molecules to form new radicals, perpetuating the chain. This is a crucial step as it continues the reaction without the need for further initiation.
   • Example 1: Cl⋅+CH4​→HCl+⋅CH3​ (H abstraction)
   • Example 2: ⋅CH3​+Cl2​→CH3​Cl+Cl⋅ (radical attack on a molecule)
3. Termination: Two radicals combine to form a stable, non-radical molecule, effectively removing radicals from the system and ending the chain. This step consumes radicals and does not produce new ones.
   • Example 1: Cl⋅+⋅Cl→Cl2​
   • Example 2: ⋅CH3​+⋅CH3​→CH3​CH3​
   • Example 3: Cl⋅+⋅CH3​→CH3​Cl (also forms product, but ends chain)

4. Important Radical Reactions

4.1. Halogenation of Alkanes (e.g., Chlorination of Methane)

• Reaction: Alkanes react with halogens (Cl2​, Br2​) in the presence of heat or light to form alkyl halides.
  • Example: CH4​+Cl2​hν​CH3​Cl+HCl
• Mechanism (Chlorination of Methane):
  • Initiation: Cl2​hν​2 Cl⋅
  • Propagation 1: Cl⋅+CH4​→HCl+⋅CH3​ (Rate-determining step, H-abstraction)
  • Propagation 2: ⋅CH3​+Cl2​→CH3​Cl+Cl⋅
  • Termination: (Various combinations of radicals, e.g., Cl⋅+⋅Cl, ⋅CH3​+⋅CH3​, Cl⋅+⋅CH3​)
• Regioselectivity (for higher alkanes):
  • The halogenation reaction proceeds by abstraction of a hydrogen atom by the halogen radical, forming an alkyl radical. The more stable the alkyl radical formed, the faster that pathway will be.
  • Order of H-abstraction (reactivity): Tertiary H > Secondary H > Primary H
  • Chlorination vs. Bromination Selectivity:
    • Chlorination: Less selective. Chlorine radicals are very reactive, and the reaction is less sensitive to the stability of the intermediate radical. Therefore, chlorination often leads to a mixture of products when multiple types of hydrogens are available. (Relative rates: 3∘:2∘:1∘≈5:4:1).
    • Bromination: More selective. Bromine radicals are less reactive and thus more selective. They abstract the most sterically accessible and electronically favorable hydrogen (i.e., the one leading to the most stable radical). Therefore, bromination often gives a single major product. (Relative rates: 3∘:2∘:1∘≈1600:82:1).
• Side Reactions: Polysubstitution (e.g., formation of CH2​Cl2​, CHCl3​, CCl4​) is common, especially with chlorination due to its lower selectivity.

4.2. Allylic Bromination (NBS – N-Bromosuccinimide)

• Reaction: Alkenes react with NBS (a source of Br⋅ at low concentration) in the presence of light or peroxides to selectively brominate at the allylic position (the carbon atom adjacent to a C=C double bond).
• Mechanism:
  • Initiation: Light or peroxide generates Br⋅ radicals from trace Br2​ (NBS keeps Br2​ concentration low).
  • Propagation:
    1. Br⋅ abstracts an allylic H, forming a resonance-stabilized allylic radical.
    2. The allylic radical reacts with Br2​ (generated from NBS and HBr) to form the allylic bromide product and regenerate Br⋅.
• Selectivity: The reaction is highly selective for the allylic position because the resulting allylic radical is resonance-stabilized.

4.3. Addition of HBr to Alkenes (Peroxide Effect)

• Reaction: In the absence of peroxides, HBr adds to alkenes via an electrophilic addition mechanism (Markovnikov’s rule). However, in the presence of peroxides, HBr adds in an anti-Markovnikov fashion via a radical mechanism.
• Mechanism (Peroxide effect):
  • Initiation: Peroxide (ROOR) heat or hν​ 2 RO⋅ ; RO⋅+HBr→ROH+Br⋅
  • Propagation:
    1. Br⋅ adds to the alkene to form the more stable carbon radical (this leads to anti-Markovnikov regioselectivity).
    2. The carbon radical abstracts an H from HBr to form the alkyl bromide product and regenerate Br⋅.
• Regioselectivity: Anti-Markovnikov addition means the bromine adds to the less substituted carbon, and the hydrogen adds to the more substituted carbon. This is because the Br⋅ radical adds to the alkene to form the most stable carbon radical intermediate.

4.4. Radical Polymerization

• Reaction: Formation of long-chain polymers by the successive addition of monomer units, initiated by radicals.
• Mechanism: Involves initiation, propagation (where the growing polymer chain is itself a radical), and termination steps.
  • Example: Polymerization of styrene or vinyl chloride.

5. Stereochemistry of Radical Reactions

• Planar Radical Intermediates: Carbon radicals are often sp2 hybridized and planar (or nearly planar) at the radical center.
• Loss of Stereochemical Information: If a chiral center becomes a radical, and then a new bond forms at that radical center, the product will often be a racemic mixture. This is because the attacking species can approach from either face of the planar radical with equal probability.
• Exceptions: If the radical is not planar or if a specific steric environment dictates attack from one side, some stereoselectivity might be observed.

6. Selectivity in Radical Reactions (Recap)

• Chlorination vs. Bromination: Bromination is much more selective than chlorination due to the higher reactivity of the chlorine radical. The more reactive a species, the less discriminating it is in its choice of reaction site.
  • Chlorine: Highly reactive, abstracts H from almost any position, leading to mixtures (kinetic control dominates).
  • Bromine: Less reactive, more selective, abstracts H that leads to the most stable radical, favoring one major product (thermodynamic control is more influential for radical formation).

Multiple Choice Questions (MCQ) on Radical Reactions

Instructions: Choose the best answer for each question.

1. What is the defining characteristic of a radical species? a) It has a positive charge. b) It has an unpaired valence electron. c) It is always a carbon-based intermediate. d) It has a negative charge.

2. Which term describes the symmetrical breaking of a covalent bond, where each atom retains one electron from the shared pair? a) Heterolytic cleavage b) Homolytic cleavage c) Ionic bond formation d) Dative bond formation

3. What type of arrow is used in reaction mechanisms to show the movement of a single electron? a) Double-headed arrow (↠) b) Curved arrow with a full arrowhead (↷) c) Fishhook arrow (half-headed arrow, ↷) d) Straight arrow (→)

4. Which of the following is a common method for initiating radical reactions? a) Adding a strong acid. b) Heating or exposing to UV light. c) Using a polar protic solvent. d) Adding a strong base.

5. Which of the following radical types is the most stable? a) Methyl radical b) Primary alkyl radical c) Secondary alkyl radical d) Tertiary alkyl radical

6. What is the primary reason for the increased stability of tertiary alkyl radicals compared to primary alkyl radicals? a) Inductive electron withdrawal. b) Resonance stabilization. c) Hyperconjugation from surrounding alkyl groups. d) Steric hindrance.

7. Which of the following radicals is resonance-stabilized? a) Methyl radical b) Ethyl radical c) Allylic radical d) Isopropyl radical

8. What are the three main steps in a radical chain mechanism? a) Addition, Elimination, Substitution b) Initiation, Propagation, Termination c) Oxidation, Reduction, Neutralization d) Protonation, Deprotonation, Rearrangement

9. In a radical chain reaction, what happens during the propagation step? a) Radicals are formed from non-radical species. b) Two radicals combine to form a stable molecule. c) Radicals react with stable molecules to form new radicals. d) The reaction stops.

10. What is the purpose of the termination step in a radical chain mechanism? a) To produce more radicals. b) To generate products and continue the chain. c) To consume radicals and end the chain. d) To form an intermediate.

11. Which of the following is an example of a common radical initiator? a) Sodium hydroxide (NaOH) b) Hydrochloric acid (HCl) c) Benzoyl peroxide d) Water (H2​O)

12. When methane reacts with chlorine in the presence of UV light, what is the major organic product typically formed under controlled conditions? a) Methylene chloride (CH2​Cl2​) b) Carbon tetrachloride (CCl4​) c) Chloromethane (CH3​Cl) d) Chloroform (CHCl3​)

13. In the free-radical halogenation of a higher alkane (e.g., propane), which type of hydrogen atom is most readily abstracted by a halogen radical? a) Primary hydrogen b) Secondary hydrogen c) Tertiary hydrogen d) All types are abstracted equally

14. Why is bromination generally more selective than chlorination in alkane halogenation? a) Bromine radicals are more reactive. b) Chlorine radicals are less reactive. c) The bromine abstraction step is more exothermic. d) Bromine radicals are less reactive and more discriminating.

15. What is NBS (N-Bromosuccinimide) typically used for in radical reactions? a) Electrophilic addition of bromine to alkenes. b) Selective bromination at the allylic position of alkenes. c) Bromination of aromatic rings. d) As a solvent for radical reactions.

16. In the addition of HBr to an alkene, what effect do peroxides have on the regioselectivity? a) It remains Markovnikov. b) It becomes anti-Markovnikov. c) The reaction stops. d) It leads to an elimination product.

17. What is the key reason for the anti-Markovnikov addition of HBr to alkenes in the presence of peroxides? a) The formation of the more stable carbocation. b) The formation of the more stable carbon radical intermediate. c) The nucleophilic attack of the bromide ion. d) The high concentration of peroxides.

18. What is the common stereochemical outcome if a chiral center becomes a radical and then forms a new bond in a subsequent step? a) Inversion of configuration b) Retention of configuration c) Racemization d) Epimerization

19. Why do carbon radicals often lead to racemic products if a new chiral center is formed? a) Carbon radicals are typically tetrahedral. b) Carbon radicals are typically planar (sp2 hybridized). c) Carbon radicals are very stable. d) Carbon radicals are very small.

20. Which radical is formed in the rate-determining step of the free-radical chlorination of methane? a) Chlorine radical (Cl⋅) b) Methyl radical (⋅CH3​) c) Hydrogen radical (H⋅) d) Methylene radical (⋅CH2​⋅)

21. In radical polymerization, the growing polymer chain is what type of species? a) An ion b) A radical c) A molecule with no unpaired electrons d) A catalyst

22. Which of the following is a termination step in the chlorination of methane? a) Cl⋅+CH4​→HCl+⋅CH3​ b) ⋅CH3​+Cl2​→CH3​Cl+Cl⋅ c) Cl⋅+⋅CH3​→CH3​Cl d) Cl2​hν​2 Cl⋅

23. What does it mean for a radical to be “paramagnetic”? a) It is highly colored. b) It is repelled by a magnetic field. c) It is attracted to a magnetic field. d) It has a very long lifetime.

24. What is the relative rate of H abstraction for secondary hydrogens compared to primary hydrogens by a chlorine radical? a) Secondary H reacts faster (e.g., ratio of ≈4:1). b) Primary H reacts faster. c) They react at the same rate. d) Secondary H does not react.

25. Which bond is homolytically cleaved during the initiation step of free-radical chlorination of methane? a) C-H bond in methane b) C-Cl bond in product c) Cl-Cl bond in chlorine molecule d) H-Cl bond in HCl

26. Why is free-radical chlorination often problematic for synthesizing a single desired product from larger alkanes? a) It is too slow. b) It primarily produces elimination products. c) Chlorine’s low selectivity leads to mixtures of mono- and polysubstituted products. d) It requires very expensive catalysts.

27. What functional group is introduced by allylic bromination using NBS? a) An alcohol b) A ketone c) A bromine atom d) An alkene

28. In the propagation step of alkane halogenation, which is typically the rate-determining step? a) Halogen radical abstracts a hydrogen from the alkane. b) Alkyl radical reacts with halogen molecule. c) Two halogen radicals combine. d) Two alkyl radicals combine.

29. Which of the following is the most reactive type of hydrogen for abstraction by a radical? a) Vinylic hydrogen b) Alkane hydrogen c) Allylic hydrogen d) Benzylic hydrogen

30. Which of the following could undergo anti-Markovnikov addition of HBr in the presence of peroxides? a) Benzene b) Cyclohexane c) 1-butene d) Butane

31. What is the role of peroxides in the anti-Markovnikov addition of HBr to alkenes? a) To act as a nucleophile. b) To initiate the radical chain reaction. c) To catalyze the electrophilic addition. d) To terminate the reaction.

32. The carbon radical intermediate formed during the propagation step of anti-Markovnikov addition of HBr to propene (CH3​CH=CH2​) would be: a) A primary radical at C1​. b) A secondary radical at C2​. c) A tertiary radical at C2​. d) A primary radical at C2​.

33. If a mixture of CH4​ and CD4​ (deuterated methane) is subjected to monochlorination, what observation would indicate that the C-H bond breaking is the rate-determining step? a) CH3​Cl and CD3​Cl are formed in equal amounts. b) CH3​Cl is formed faster than CD3​Cl. c) CD3​Cl is formed faster than CH3​Cl. d) No reaction occurs.

34. Which of the following statements about radicals is generally true? a) They are very stable. b) They are highly reactive. c) They always have an octet of electrons. d) They are charged species.

35. What is the typical effect of branching on the stability of an alkyl radical? a) Decreases stability. b) Increases stability. c) Has no effect. d) Makes it paramagnetic.

36. In the propagation step of allylic bromination, the allylic radical abstracts a bromine atom from: a) HBr b) NBS c) Br2​ d) Succinimide

37. Which of the following is NOT a type of propagation step? a) Radical abstracts an atom from a molecule. b) Radical adds to a π bond. c) Molecule reacts with a molecule to form a radical. d) Radical reacts with a molecule to form a new radical and a new molecule.

38. What is the most common reason for the formation of undesired by-products (e.g., alkanes or disproportionation products) in radical reactions? a) Errors in initiation. b) Termination steps. c) Peroxide effect. d) Electrophilic attack.

39. What is the typical bond dissociation energy for a Cl-Cl bond (approximately)? a) Very high (e.g., 400kJ/mol) b) Moderate (e.g., 243kJ/mol) c) Very low (e.g., 50kJ/mol) d) Zero, it breaks spontaneously.

40. If a chiral molecule undergoes a radical reaction where the chiral center becomes the radical, and then a new bond is formed at that radical center, what happens to the optical activity of the product? a) It remains optically active with the original configuration. b) It becomes optically active with inverted configuration. c) It becomes optically inactive (racemic mixture). d) It becomes optically inactive, but only if the radical is tertiary.

Answer Key with Explanations

1. b) It has an unpaired valence electron.
   • Explanation: The defining characteristic of a radical is the presence of at least one unpaired electron in its valence shell.
2. b) Homolytic cleavage.
   • Explanation: Homolytic cleavage is the symmetrical breaking of a covalent bond, where each atom receives one electron from the shared pair, resulting in two radicals.
3. c) Fishhook arrow (half-headed arrow, ↷).
   • Explanation: Fishhook arrows are specifically used in reaction mechanisms to depict the movement of a single electron, distinguishing radical mechanisms from ionic ones.
4. b) Heating or exposing to UV light.
   • Explanation: Radical reactions are initiated by providing energy (heat or light) to homolytically cleave a weak bond, generating the initial radicals.
5. d) Tertiary alkyl radical.
   • Explanation: Radical stability follows the order: Tertiary > Secondary > Primary > Methyl, primarily due to hyperconjugation.
6. c) Hyperconjugation from surrounding alkyl groups.
   • Explanation: Alkyl groups can donate electron density to the electron-deficient radical center through hyperconjugation (overlap of adjacent C-H σ bonds with the half-filled p-orbital), stabilizing the radical.
7. c) Allylic radical.
   • Explanation: Allylic radicals (⋅CH2​−CH=CH2​) are highly stabilized by resonance delocalization of the unpaired electron over the π system.
8. b) Initiation, Propagation, Termination.
   • Explanation: These are the three fundamental steps that characterize a radical chain mechanism.
9. c) Radicals react with stable molecules to form new radicals.
   • Explanation: The propagation step is the “chain-carrying” part of the reaction where a radical reacts with a non-radical molecule, producing a new non-radical product and a new radical, thus continuing the chain.
10. c) To consume radicals and end the chain.
    • Explanation: Termination steps involve the combination of two radicals to form a stable, non-radical molecule, effectively removing reactive radical species from the reaction mixture and stopping the chain.
11. c) Benzoyl peroxide.
    • Explanation: Peroxides, like benzoyl peroxide, have weak O-O bonds that readily undergo homolytic cleavage upon heating or irradiation, forming alkoxy radicals that initiate the chain.
12. c) Chloromethane (CH3​Cl).
    • Explanation: Under controlled conditions (limiting halogen), the monochlorination product is favored. However, due to chlorine’s low selectivity, polysubstitution is a common problem.
13. c) Tertiary hydrogen.
    • Explanation: H-abstraction by a halogen radical proceeds via the formation of an alkyl radical. Since tertiary radicals are the most stable, the abstraction of a tertiary hydrogen is kinetically favored.
14. d) Bromine radicals are less reactive and more discriminating.
    • Explanation: Bromine radicals are less reactive (less exothermic H-abstraction step) than chlorine radicals. This means they are more selective in choosing which type of hydrogen to abstract, showing a strong preference for forming the most stable radical.
15. b) Selective bromination at the allylic position of alkenes.
    • Explanation: NBS is a reagent that provides a low, steady concentration of Br⋅ radicals, which selectively abstract allylic hydrogens due to the resonance stabilization of the resulting allylic radical.
16. b) It becomes anti-Markovnikov.
    • Explanation: In the presence of peroxides, the addition of HBr proceeds via a radical mechanism where the bromine adds to the less substituted carbon, leading to anti-Markovnikov regioselectivity.
17. b) The formation of the more stable carbon radical intermediate.
    • Explanation: The regioselectivity of radical addition is dictated by the step where the Br⋅ radical adds to the alkene to form the most stable carbon radical intermediate (e.g., secondary radical over primary).
18. c) Racemization.
    • Explanation: If a chiral center becomes a radical, it typically adopts a planar, sp2 hybridized geometry. Subsequent attack from either face leads to a racemic mixture (equal amounts of R and S enantiomers), resulting in loss of optical activity.
19. b) Carbon radicals are typically planar (sp2 hybridized).
    • Explanation: The planar geometry of sp2 hybridized carbon radicals allows attacking species to approach from either side, leading to racemization if a new chiral center is formed.
20. b) Methyl radical (⋅CH3​).
    • Explanation: The rate-determining step in the chlorination of methane is the abstraction of a hydrogen from methane by a chlorine radical, producing a methyl radical and HCl.
21. b) A radical.
    • Explanation: In radical polymerization, the growing polymer chain always ends with an unpaired electron, making it a radical that continues to add monomers.
22. c) Cl⋅+⋅CH3​→CH3​Cl.
    • Explanation: A termination step involves the combination of two radicals to form a stable, non-radical product, ending the chain.
23. c) It is attracted to a magnetic field.
    • Explanation: Paramagnetism is the property of a substance containing unpaired electrons that causes it to be weakly attracted to an external magnetic field.
24. a) Secondary H reacts faster (e.g., ratio of ≈4:1).
    • Explanation: For chlorination, the relative rates of H-abstraction are typically 3∘:2∘:1∘≈5:4:1. So, secondary hydrogens are abstracted about 4 times faster than primary hydrogens.
25. c) Cl-Cl bond in chlorine molecule.
    • Explanation: The initiation step involves the homolytic cleavage of the weakest bond present, which is the Cl-Cl bond, by light or heat.
26. c) Chlorine’s low selectivity leads to mixtures of mono- and polysubstituted products.
    • Explanation: Because chlorine radicals are very reactive and not very selective, they can abstract hydrogens from various positions and then react further with the product, leading to complex mixtures of mono-, di-, tri-, and polysubstituted products.
27. c) A bromine atom.
    • Explanation: Allylic bromination using NBS introduces a bromine atom specifically at the allylic position.
28. a) Halogen radical abstracts a hydrogen from the alkane.
    • Explanation: This H-abstraction step is typically the rate-determining step in the propagation phase of alkane halogenation, as it involves breaking a strong C-H bond.
29. c) Allylic hydrogen.
    • Explanation: Allylic hydrogens are the most reactive for abstraction by a radical because their removal leads to the formation of a resonance-stabilized allylic radical. Benzylic hydrogens are also highly reactive for the same reason.
30. c) 1-butene.
    • Explanation: Anti-Markovnikov addition applies to alkenes (C=C double bonds) in the presence of HBr and peroxides. 1-butene is an alkene. Benzene is aromatic, and cyclohexane/butane are alkanes.
31. b) To initiate the radical chain reaction.
    • Explanation: Peroxides decompose under heat or light to form alkoxy radicals, which then react with HBr to generate Br⋅ radicals, thereby initiating the anti-Markovnikov radical chain addition.
32. b) A secondary radical at C2​.
    • Explanation: In the anti-Markovnikov addition to propene (CH3​CH=CH2​), the Br⋅ radical adds to the less substituted C1​ carbon to form the more stable secondary radical at C2​ (CH3​⋅CHCH2​Br), rather than a less stable primary radical at C1​.
33. b) CH3​Cl is formed faster than CD3​Cl.
    • Explanation: This is an example of a kinetic isotope effect. If the C-H bond breaking is rate-determining, the lighter isotope (H) will react faster than the heavier isotope (D), leading to faster formation of CH3​Cl.
34. b) They are highly reactive.
    • Explanation: Due to their unpaired electron and incomplete octet, radicals are typically very reactive species, constantly seeking to pair their electron.
35. b) Increases stability.
    • Explanation: Similar to carbocations, increasing branching (more alkyl groups) around the radical center increases its stability due to hyperconjugation.
36. c) Br2​.
    • Explanation: In allylic bromination, the allylic radical formed abstracts a bromine atom from Br2​ (which is regenerated in small amounts from NBS and HBr) to form the product and regenerate the Br⋅ radical.
37. c) Molecule reacts with a molecule to form a radical.
    • Explanation: Propagation steps involve a radical reacting with a stable molecule to produce a new radical (and a new product molecule). A molecule reacting with a molecule to form a radical is generally part of an initiation step (e.g., homolytic cleavage), not a propagation step.
38. b) Termination steps.
    • Explanation: While propagation steps are productive, termination steps involve the non-selective combination of any two radicals present in the mixture. This can lead to the formation of various undesired by-products (e.g., alkanes from alkyl radical recombination).
39. b) Moderate (e.g., 243kJ/mol).
    • Explanation: The Cl-Cl bond is relatively weak (compared to C-H or C-C bonds), making it susceptible to homolytic cleavage by heat or light to initiate radical reactions. 243kJ/mol is a commonly cited value.
40. c) It becomes optically inactive (racemic mixture).
    • Explanation: When a chiral center becomes a radical, it loses its original stereochemical information by adopting a planar geometry. Subsequent reaction with another species can occur from either face, leading to a racemic mixture and the loss of optical activity.