1. What is the major product of the following reaction?

2. Predict the major product of the following reaction: 2-methylbutane + Br2/light energy ?
A) 1-bromo-2-methylbutane
B) 2-bromo-2-methylbutane
C) 2-bromo-3-methylbutane
D) 1-bromo-3-methylbutane

3. Which of the following alkanes will give more than one monochlorination product upon treatment with chlorine and light? A) 2,2-dimethylpropane
B) cyclopropane
C) ethane
D) 2,3-dimethylbutane

4. Which of the following is the major product of the chlorination of methane if a large excess of methane is used?
A) CH3Cl
B) CH2Cl2
C) CH3CH2Cl
D) CCl4

5. Consider the reaction of 2-methylpropane with a halogen
Consider the reaction of 2-methylpropane with a halogen. With which halogen will the product be almost exclusively 2-halo-2-methylpropane?
A) F2
B) Cl2
C) Br2
D) I2

6. The alllyl carbon is the carbon adjacent to a double bond.
The allylic radical is more stable than a 3 radical.

7. Because allylic C—H bonds are weaker than other sp3 hybridized C—H bonds, the allylic carbon can be selectively halogenated using NBS in the presence of light or peroxides.
NBS contains a weak N—Br bond that is homolytically cleaved with light to generate a bromine radical, initiating an allylic halogenation reaction.
Propagation then consists of the usual two steps of radical halogenation.

9. NBS also generates a low concentration of Br2 needed in the second chain propagation step (Step [3] of the mechanism).
The HBr formed in Step [2] reacts with NBS to form Br2, which is then used for halogenation in Step [3] of the mechanism.

10. Thus, an alkene with allylic C—H bonds undergoes two different reactions depending on the reaction conditions.
How is the vicinal dibromide formed?

11. Halogens add to  bonds because halogens are polarizable.
The electron rich double bond induces a dipole in an approaching halogen molecule, making one halogen atom electron deficient and the other electron rich (X+—X–).
The electrophilic halogen atom is then attracted to the nucleophilic double bond, making addition possible.

12. Why does a low concentration of Br2 (from NBS) favor allylic substitution (over ionic addition to form the dibromide)?
The key to getting substitution is to have a low concentration of bromine (Br2).
The Br2 produced from NBS is present in very low concentrations.
A low concentration of Br2 would first react with the double bond to form a low concentration of the bridged bromonium ion.
The bridged bromonium ion must then react with more bromine (in the form of Br¯) in a second step to form the dibromide.
If concentrations of both intermediates—the bromonium ion and Br¯ are low (as is the case here), the overall rate of addition is very slow, and the products of the very fast and facile radical chain reaction predominate.

13. Predict the products.

14. Halogenation at an allylic carbon often results in a mixture of products. Consider the following example:
A mixture results because the reaction proceeds by way of a resonance stabilized radical.

15. Predict the products.

17. Radical Additions to Double Bonds
HBr adds to alkenes to form alkyl bromides in the presence of heat, light, or peroxides.
The regioselectivity of the addition to unsymmetrical alkenes is different from that in addition of HBr in the absence of heat, light or peroxides.
The addition of HBr to alkenes in the presence of heat, light or peroxides proceeds via a radical mechanism.

18. Hydrohalogenation—Electrophilic Addition of HX
The mechanism of electrophilic addition consists of two successive Lewis acid-base reactions. In step 1, the alkene is the Lewis base that donates an electron pair to H—Br, the Lewis acid, while in step 2, Br¯ is the Lewis base that donates an electron pair to the carbocation, the Lewis acid.

19. With an unsymmetrical alkene, HX can add to the double bond to give two constitutional isomers, but only one is actually formed:
This is a specific example of a general trend called Markovnikov’s rule.
Markovnikov’s rule states that in the addition of HX to an unsymmetrical alkene, the H atom adds to the less substituted carbon atom—that is, the carbon that has the greater number of H atoms to begin with.

21. The more stable 2° radical forms faster, so Path [B] is preferred.
Note that in the first propagation step, the addition of Br• to the double bond, there are two possible paths:
Path [A] forms the less stable 1° radical.
Path [B] forms the more stable 2° radical.
The more stable 2° radical forms faster, so Path [B] is preferred.

22. The radical mechanism illustrates why the regio-selectivity of HBr addition is different depending on the reaction conditions.

23. HBr adds to alkenes under radical conditions, but HCl and HI do not
HBr adds to alkenes under radical conditions, but HCl and HI do not. This can be explained by considering the energetics of the reactions using bond dissociation energies.
Both propagation steps for HBr addition are exothermic, so propagation is exothermic (energetically favorable) overall.
For addition of HCl or HI, one of the chain propagating steps is quite endothermic, and thus too difficult to be part of a repeating chain mechanism.