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what is the DEAl with ROOR im in destryoer and on organic problem 90 it says it is markovnikav and somewhere else it was like anti??? does anyone know what is happening?
wtf roor
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Hi. I understand your confusion, and it is one that I had too. *For those of you who read my original post, there was an error, I found the answer on wikipedia. http://en.wikipedia.org/wiki/Hydrohalogenation
In the presence of peroxides (chemicals containing the general structure ROOR'), HBr adds to a given alkene in an anti-Markovnikov fashion. This is because the reaction proceeds through the most stable carbon radical intermediate (relative stability: 3o>2o>1o>methyl) instead of a carbocation. The mechanism for this reaction is similar to a chain reaction such as free radical halogenation in which the peroxide promote the formation of the Bromide radical (radicals are indicated in bold):
Initiation: RO:OR → 2RO. (proceeds in the presence of light or heat) RO. + HBr → ROH + Br. Propagation: CH3-CH=CH2 + Br. → CH3-ĊH-CH2Br CH3-ĊH-CH2Br + HBr → CH3-CH2-CH2Br + Br. Termination Occurs when two radicals form a covalent bond. [edit] Why only HBr reacts in this way
No other hydrogen halide behaves in the manner described above, this can be explained by a survey of the different halogen acids: HF (hydrogen fluoride), HCl (hydrogen chloride -- more commonly known by the aqueous species hydrochloric acid), and HI (hydrogen iodide).
HF
The hydrogen-fluorine bond is simply too strong and therefore no fluorine radicals can be generated in the propagation step.
HCl
Hydrogen chloride will react in a manner that is so slow that it is essentially synthetically useless. This is because the hydrogen-chlorine bond is strong and thus the second step of the reaction would be extremely slow due to the heat required (it is an endothermic reaction).
In the presence of peroxides (chemicals containing the general structure ROOR'), HBr adds to a given alkene in an anti-Markovnikov fashion. This is because the reaction proceeds through the most stable carbon radical intermediate (relative stability: 3o>2o>1o>methyl) instead of a carbocation. The mechanism for this reaction is similar to a chain reaction such as free radical halogenation in which the peroxide promote the formation of the Bromide radical (radicals are indicated in bold):
Initiation: RO:OR → 2RO. (proceeds in the presence of light or heat) RO. + HBr → ROH + Br. Propagation: CH3-CH=CH2 + Br. → CH3-ĊH-CH2Br CH3-ĊH-CH2Br + HBr → CH3-CH2-CH2Br + Br. Termination Occurs when two radicals form a covalent bond. [edit] Why only HBr reacts in this way
No other hydrogen halide behaves in the manner described above, this can be explained by a survey of the different halogen acids: HF (hydrogen fluoride), HCl (hydrogen chloride -- more commonly known by the aqueous species hydrochloric acid), and HI (hydrogen iodide).
HF
The hydrogen-fluorine bond is simply too strong and therefore no fluorine radicals can be generated in the propagation step.
HCl
Hydrogen chloride will react in a manner that is so slow that it is essentially synthetically useless. This is because the hydrogen-chlorine bond is strong and thus the second step of the reaction would be extremely slow due to the heat required (it is an endothermic reaction).
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yea that question is an odd ball since usually with HBR and light you end up with anti-Mark. But according to the destoyer with HCl it adds normally.
This question is little sketch since in most textbooks they don't mention this specific fact about HCl adding regularly.
This question is little sketch since in most textbooks they don't mention this specific fact about HCl adding regularly.
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yea that question is an odd ball since usually with HBR and light you end up with anti-Mark. But according to the destoyer with HCl it adds normally.
This question is little sketch since in most textbooks they don't mention this specific fact about HCl adding regularly.
My textbook mentioned it. Just another rule with another exception. Weird wild stuff!!!
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anti merchoncov with ROOR only works with HBR, with HCL and HF it adds merchoncov?
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Peroxide with HBr will add anti-markovinov because the radical Br acts like the electrophile and adds first.
Peroxide will have no effect on the addition of HCl or HI. they both will add markovnikov just as they normally would if there were no peroxide present.
This difference is based on the changes in enthalpy in the propagation steps. In the addition of radical Br, both propagation steps are exothermic. However, in the addition of radical Cl or radical I, one of the two propagation steps is endothermic. Thus, these propagation steps cannot compete with the exothermic termination steps, and these radicals will almost always terminate before they are able to get a radical chain reaction going.
Peroxide will have no effect on the addition of HCl or HI. they both will add markovnikov just as they normally would if there were no peroxide present.
This difference is based on the changes in enthalpy in the propagation steps. In the addition of radical Br, both propagation steps are exothermic. However, in the addition of radical Cl or radical I, one of the two propagation steps is endothermic. Thus, these propagation steps cannot compete with the exothermic termination steps, and these radicals will almost always terminate before they are able to get a radical chain reaction going.
