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Electrophilic Addition and Free Radical Sub.-Need some clarifying. watch

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    Okay, so during my revision, I came up with two things I couldn't understand.

    The first one is about electrophilic addition of hydrogen halides+unsymmetrical alkenes. From what I understand, the halide of the hydrogen halide has a higher affinity towards the more stable carbocation, and when they bond, they form the major product. So how come there's a minor/less common product? Wouldn't the halide ALWAYS bond with the more stable carbocation?

    The second question I have is about free radical substitution. Okay, so, UV rays break down chlorine to form two free radicals, right? So how come, during the termination stage you can get a reaction between two chlorine free radicals to form chlorine? Wouldn't the chlorine just break down from the UV rays again?
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    (Original post by tehsponge)
    Okay, so during my revision, I came up with two things I couldn't understand.

    The first one is about electrophilic addition of hydrogen halides+unsymmetrical alkenes. From what I understand, the halide of the hydrogen halide has a higher affinity towards the more stable carbocation, and when they bond, they form the major product. So how come there's a minor/less common product? Wouldn't the halide ALWAYS bond with the more stable carbocation?

    The second question I have is about free radical substitution. Okay, so, UV rays break down chlorine to form two free radicals, right? So how come, during the termination stage you can get a reaction between two chlorine free radicals to form chlorine? Wouldn't the chlorine just break down from the UV rays again?
    part 1
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    The most stable carbocation is formed faster as the activation nergy required to reach the more stable intermediate is lower, but there is always some of the less stable carbocation formed as well.

    There is no reason why the halide ion has a greater affinity towards the more stable carbocation. And in any case this phase of the mechanism is not rate determining, so it really does not matter.

    part 2
    ------

    The apposite word here is 'CAN'.

    Any two free radicals 'can' terminate, but on the whole it isn't very likely. The collision of two chlorine FRs is not a probable event, and as you say another dose of UV will start them off again.
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    What my understanding of your first question is:

    A secondary carbocation is favourably formed when a hydrogen halide adds on to unsymmetrical alkenes because the electron pushing effect of the CH3 groups stabilises the central carbon atom. I.e. If HBr added to propene 2 bromo propane would be formed as the hydrogen adds on the carbon atom with the most hydrogen atoms attached - markovnikoffs rule.

    The second question:

    I think it is because during the termination stage they remove the UV light so that the Cl2 isn't instantly broken down again. So once the UV light has been removed the remaining free radicals will combine.
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    (Original post by Avinash123)
    What my understanding of your first question is:

    A secondary carbocation is favourably formed when a hydrogen halide adds on to unsymmetrical alkenes because the electron pushing effect of the CH3 groups stabilises the central carbon atom. I.e. If HBr added to propene 2 bromo propane would be formed as the hydrogen adds on the carbon atom with the most hydrogen atoms attached - markovnikoffs rule.
    That is true,markovnikov rule is based on stability of the intermediate, tertiary>secondary>primary carbocation in terms of stability of the intermediate. The CH3 increases electron density of alkene by inductive effect only.


    The second question:

    I think it is because during the termination stage they remove the UV light so that the Cl2 isn't instantly broken down again. So once the UV light has been removed the remaining free radicals will combine.
    Radical chemistry involves very fast rate that might be difficult to study sometimes, as far as I know. It is chain reactions, so chlorine radicals are continuously being produced by propagation step, I would think this would continue even without the UV light after initiation, and they would most likely recombine after they have exhausted all the reactants, ie no more to perform radical substitution. So, no i don't think termination occurs immediately after uv light is removed.

    Charco, please correct me if i am wrong about this, cheers.
 
 
 

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