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Anti Markovnikov Product of Radical Addition Reactions
| Content Provider | AK Lectures |
|---|---|
| Description | Recall that in a straightforward addition reaction in the presence of hydrogen bromide and an asymmetrical alkene, we form the more substituted Markovnikov Product. This is because the intermediate that leads to the more substituted product undergoes a more substituted (and therefore more stable) carbocation. On the other hand, if we mix the hydrogen bromine with a trace amount of peroxide, we initiation the radical addition reaction that leads to the less substituted anti-Markovnikov product. Why is the less substituted product preferred over the more substituted product? In order to answer this question, we have to examine the propagation step of the chain reaction. In that step, the chain-carrying bromide radical attaches to the less substituted side of the alkene because that will lead to a more stable radical intermediate. This radical intermediate contains hyperconjugation between the electron in the 2p orbital and the two electrons in the adjacent carbon-hydrogen bond. Generally speaking, the bromide radical attaches to the less substituted carbon because it produces a more substituted radical intermediate, which is stabilized by hyperconjugation. |
| Language | English |
| Access Restriction | Open |
| Subject Keyword | Organic Chemistry carbocation peroxide initiation form presence |
| Content Type | Video |
| Educational Role | Teacher Student |
| Educational Use | Self Learning Lecture Reading |
| Resource Type | Video Lecture |
| Education Level | Under Graduate |
| Subject | Organic chemistry |
