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Hydrogen Halide Radical Addition Reactions
| Content Provider | AK Lectures |
|---|---|
| Description | Out of all the hydrogen halides, the only one that will react with an asymmetrical alkene via a radical reaction (in the presence of peroxide) is hydrogen bromide. When we add hydrogen chloride or hydrogen iodide (with or without peroxide) to an asymmetrical alkene, the reaction takes place via a straightforward addition mechanism forming a Markovnikov product. In the first step, the pair of electrons in the pi-bond abstracts hydrogen onto the less substituted carbon, forming the more substituted carbocation intermediate. In the second step, the halide (either chloride or iodide) uses its pair of electrons and attacks the carbon with the positive charge, forming the more substituted halide hydrocarbon. But why does this take place and why don’t they undergo a radical addition. The answer to this question lies in an examination of the two propagation steps. If we were to replace the bromide with iodide, the first step of the propagation would become endothermic. On the other hand, if we replace the bromide with chloride, the second step of the two propagation steps will become endothermic. Recall that endothermic simply means that the products of the reaction have less stable bonds than the reactants. Therefore since hydrogen bromide addition in the presence of peroxide is the only compound that has both exothermic propagation steps, it is the only one that will induce a radical addition reaction forming the less substituted anti-Markovnikov product. |
| Language | English |
| Access Restriction | Open |
| Subject Keyword | Organic Chemistry attacks carbon reactants bonds hydrogen bromide addition electrons compound pair products |
| Content Type | Video |
| Educational Role | Teacher Student |
| Educational Use | Self Learning Lecture Reading |
| Resource Type | Video Lecture |
| Education Level | Under Graduate |
| Subject | Physical chemistry |
