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MCAT Organic Chemistry Tutorial: Radical Reactions

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MgujSMP.png

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\({\bf{Basic~Mechanism:}}\) start with A:B show single headed arrows to indicate the movement of one electron rather than a pair produces two radicals (compound w/ one free electron) |dw:1529687408041:dw|

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process typically requires some form of energy, usu. heat or light. a dialkyl peroxyde can thus be split into two alkoxyl radicals |dw:1529687518504:dw|

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\({\bf{Hydrogen~Abstraction:}}\) a hydrogen is ~abstracted~ or taken from an alkane and onto a halogen X like so: |dw:1529687719117:dw|

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\({\bf{Homolytic~Bond~Dissociation:}}\) When a bond between two atoms is broken to produce two radicals, energy is released (homolytic bond dissociation energy, or DH). More energy released = stronger bond. Recall that when a bond is broken the enthalpy change is positive and when a bond is formed the enthalpy change is negative. (stability is usu. higher for a bond than for the separate atoms) Using the data for bond dissociation also tells us the stability of radicals (tertiary > secondary > primary > methyl cation) following the same principles as the stability of carbocations.

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\({\bf{Chlorination:}}\) Three steps: - initiation (generation of the chlorine radicals - propagation: extension of the alkane chain via attack by the chlorine radical - termination: reaction of radicals to pair up free electrons

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|dw:1529688536225:dw|

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|dw:1529688561532:dw|

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|dw:1529688671029:dw|

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|dw:1529688860287:dw|

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\({\bf{Reactivity~with~Other~Halogens}}\) order of reactivity: fluorine > chlorine > bromine > iodine fluorine is generally too fast and out of control to be practical iodine is so unreactive that it is generally not used for this purpose bromine is less reactive than chlorine, still exothermic overall though, can be used when a slower, more controlled reaction is rqeuired \({\bf{Larger~Alkanes}}\) similar procedure applies, be careful when accounting for the size and orientation of the alkane fragment radicals. Generally tertiary hydrogens are more reactive than secondary hydrogens, etc. and more susceptible for removal by the halogen radical.

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\({\bf{Anti~Markovnikov~Addition~using~Radical~Rxns}}\) start w/ the formation of alkoxyl radicals (see discussion above) attack of radical on HBr |dw:1529689519558:dw|

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|dw:1529690192168:dw|

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|dw:1529690443566:dw|

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this remaining compound will react w/ another HBr to extract the hydrogen and produce another bromine radical which can either terminate or lead to another chain rxn

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Anyway, that's the end of my tutorial, I hope it was a helpful resource. Source material is the 2nd Edition Barron's Prep book for the new MCAT

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