10.4 Addition of HBr and Peroxide | Organic Chemistry
TLDRThe video script delves into the hydrobromination of alkenes with peroxide, a reaction that results in an anti-Markovnikov addition where bromine attaches to the less substituted carbon. The lesson revisits the mechanism of the reaction, emphasizing the formation of a more stable and substituted carbon radical intermediate. The process involves two initiation steps and propagation steps that repeat to produce the desired product. The first initiation step generates alkoxide radicals from peroxide, which then react with HBr to form a bromine radical necessary for the propagation steps. These steps are characterized by a radical being both the starting and ending species, with no net decrease in radical concentration. Termination steps, though possible, are less likely and typically result in side products rather than the desired product. The video provides a detailed look at the mechanism, which is often reserved for discussions on radical reactions, offering a comprehensive understanding of the process.
Takeaways
- ๐ The hydrobromination of alkenes with peroxides adds an H and a Br in an anti-Markovnikov manner, with bromine on the less substituted carbon and hydrogen on the more substituted side.
- โ๏ธ The mechanism involves a carbocation intermediate, which is more stable when more substituted, and this trend also applies to carbon radicals.
- ๐ The first step in the reaction is the attack of the alkene by a bromine radical, which ends up on the less substituted side, forming a more stable intermediate.
- ๐ The propagation steps involve a sequence of two steps that repeat to produce the desired product, starting and ending with one radical, without a net decrease in radical concentration.
- ๐ฌ The initiation step in this reaction is unique, involving two steps: the spontaneous breaking of a peroxide bond to form alkoxide radicals, and the reaction of an alkoxide radical with HBr to form a bromine radical.
- ๐ The propagation steps are characterized by the repetition of two steps that produce what is needed to redo the first step, allowing the process to continue.
- ๐ Termination steps involve radicals combining to form stable products, which do not contribute to the formation of the desired product but lead to side products.
- ๐งช The presence of alkoxide radicals is typically low, so their involvement in termination steps is less likely and often omitted in discussions.
- ๐ The reaction and its mechanism are sometimes included in alkene chapters of textbooks, but may also be reserved for discussions on radical reactions.
- ๐ค Understanding the initiation, propagation, and termination steps is crucial for grasping how the reaction proceeds to form the product.
- ๐ The stability of intermediates plays a significant role in determining the outcome of the reaction, favoring more substituted species.
- ๐ The lesson emphasizes the importance of recognizing patterns in reaction mechanisms, especially the distinction between initiation and propagation steps.
Q & A
What is the primary difference between the hydrobromination of alkenes with HBr alone and with HBr in the presence of peroxide?
-The primary difference is that with HBr alone, the addition follows Markovnikov's rule, with hydrogen (H) adding to the more substituted carbon and bromine (Br) to the less substituted one. In contrast, with HBr and peroxide, the addition is anti-Markovnikov, with bromine ending up on the less substituted side and hydrogen on the more substituted side.
What is the role of the peroxide in the hydrobromination reaction with HBr?
-The peroxide initiates the reaction by breaking down spontaneously to form alkoxide radicals. These radicals then react with HBr to abstract a hydrogen, forming an alcohol and generating the bromine radical necessary for the propagation steps of the reaction.
How does the stability of the intermediate in the reaction influence the product formed?
-The stability of the intermediate significantly influences the product formed. The reaction follows a trend where more substituted intermediates are more stable. In the presence of peroxide, a more stable and more substituted carbon radical is formed, leading to the anti-Markovnikov product.
What is the significance of the first step in the propagation phase of the reaction?
-The first step in the propagation phase involves the bromine radical reacting with the alkene to form a new bond on the less substituted side. This step is significant because it sets the stage for the formation of a more substituted and stable intermediate, which is crucial for the overall reaction mechanism.
What happens during the termination steps in the reaction?
-During the termination steps, radicals combine to form stable molecules, effectively ending the chain reaction. Possible termination steps include two bromine radicals combining to form Br2, two carbon radicals forming a new carbon-carbon bond, or a carbon radical reacting with a bromine radical to form a dibromo compound.
Why is the initiation step considered tricky in the context of this reaction?
-The initiation step is considered tricky because it involves two distinct steps rather than one. The first step generates two alkoxide radicals from peroxide, while the second step involves one of these radicals reacting with HBr to form a bromine radical, which is necessary for the propagation steps. The second step does not increase the net concentration of radicals, which can be confusing.
How can you identify the propagation steps in a radical reaction mechanism?
-Propagation steps can be identified by a sequence of two steps that repeat over and over again, starting and ending with one radical. These steps do not result in a net decrease in the concentration of radicals and are responsible for the formation of the desired product.
What is the role of the alkene in the propagation steps of the hydrobromination reaction with HBr and peroxide?
-The alkene acts as a reactant that reacts with the bromine radical to form a new bond on the less substituted carbon. This leads to the formation of a more substituted and stable radical intermediate, which then reacts with HBr to continue the propagation steps and form the final product.
Why are alkoxide radicals often omitted in the discussion of termination steps in this reaction?
-Alkoxide radicals are often omitted in the discussion of termination steps because they are present at fairly low concentrations compared to other radicals like bromine radicals. Their involvement in termination steps is less likely, and thus, they are usually ignored in the overall mechanism.
What is the general outcome of termination steps in radical reactions?
-The general outcome of termination steps is the formation of stable molecules or side products. These steps do not contribute to the formation of the desired product of the reaction but instead signal the end of the radical chain reaction.
Why might some textbooks reserve the discussion of the mechanism for the addition of HBr with peroxide until later in the course?
-Some textbooks might reserve the discussion of this mechanism until later because it involves radical reactions, which may be covered in more depth in subsequent chapters. Introducing the mechanism earlier might be overwhelming if the student has not yet encountered radical reactions.
Outlines
๐ Hydrobromination of Alkenes with Peroxide: Mechanism Overview
This paragraph delves into the specific mechanism of hydrobromination of alkenes using peroxide. It contrasts the addition of HBr with peroxide to the standard hydrobromination, emphasizing the formation of a more substituted and stable carbocation intermediate. The process begins with the bromine radical reacting with the alkene, leading to the formation of a new bond on the less substituted carbon. The mechanism involves two initiation steps, the first producing two alkoxide radicals from the peroxide, and the second generating the necessary bromine radical through the reaction of an alkoxide radical with HBr. The propagation steps are characterized by the repetition of two steps that produce the desired product without a net decrease in the concentration of radicals. Termination steps are also discussed, which involve the combination of radicals to form side products and do not contribute to the formation of the desired product.
๐งฌ Initiation and Termination Steps in Radical Reactions
The second paragraph focuses on the initiation and termination steps involved in the radical addition of HBr to alkenes. It explains the two-step initiation process: the first step involves the spontaneous breaking of the peroxide bond to form two alkoxide radicals, and the second step involves one of these radicals abstracting a hydrogen from HBr to form a bromine radical, which is essential for the propagation steps. The paragraph clarifies the distinction between propagation and initiation steps, noting that the latter does not lead to the formation of the desired product but sets the stage for the former. Termination steps are also outlined, where radicals combine to form side products such as Br2, new carbon-carbon bonds, or additional bromine-substituted products. These steps are less likely due to the lower concentration of certain radicals but are still considered in the overall mechanism.
Mindmap
Keywords
๐กHydrobromination
๐กPeroxide
๐กAnti-Markovnikov
๐กCarbocation
๐กRadical
๐กPropagation Steps
๐กInitiation Steps
๐กTermination Steps
๐กAlkene
๐กBromine Radical
๐กSubstitution
Highlights
Hydrobromination of alkenes with peroxide involves an anti-Markovnikov addition of H and Br, resulting in bromine on the less substituted carbon and hydrogen on the more substituted side.
The mechanism for this reaction involves a carbocation intermediate, which is more stable and substituted.
The first species to react with the alkene ends up on the less substituted carbon, leading to a more stable intermediate.
In the presence of peroxide, the reaction forms a more stable and substituted carbon radical instead of a carbocation.
The bromine radical reacts with the alkene to form a new bond on the less substituted side, creating a radical intermediate.
The radical intermediate then reacts with HBr to form a bond to hydrogen, generating another bromine radical.
The propagation steps involve a sequence of two steps that repeat to produce the desired product without a net decrease in the concentration of radicals.
The initiation steps in this reaction are unique, involving two steps to generate the necessary bromine radical for propagation.
Peroxide is crucial for the first initiation step, where its weak O-O bond breaks to form two alkoxide radicals.
An alkoxide radical reacts with HBr to abstract a hydrogen, forming an alcohol and the essential bromine radical.
The initiation steps increase the net concentration of radicals, setting the stage for the propagation steps.
Termination steps involve radicals combining to form products like Br2, new carbon-carbon bonds, or additional bromine attachments.
Termination steps do not contribute to the formation of the desired product but lead to side products.
The reaction mechanism is complex and is often reserved for discussion when other radical reactions are covered.
The lesson provides a detailed look at the mechanism of hydrobromination with peroxide, which is crucial for understanding radical reactions.
The importance of the more substituted and stable intermediate in determining the product of the reaction is emphasized.
The lesson also covers the role of peroxide in generating the necessary radicals for the reaction to proceed.
Understanding the propagation and initiation steps is key to grasping how the reaction produces the desired product.
The potential side reactions and their impact on the overall reaction mechanism are also discussed.
Transcripts
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