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Benzene Watch

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    I have been given a question and I have no idea how to do it!
    "Predict by showing the probable mechanism what you think would be the reaction on mixing and warming benzene and ICl and with an explanation, how the reaction could be speeded up at room temperature.

    THANK YOU FOR YOUR HELP! MUCH APPRECIATED!!
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    Do you know how to do a general electrophilic substitution mechanism for benzene?
    Bear in mind chlorine is more electronegative than iodine, so the electrophile in this case is going to be I+
    What increases rate of reaction? Temperature is one factor, but there are other factors
    I've tried not to spoon feed you the answer, but hopefully I've given you enough info to solve it
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    Russell, I get the mechanism now, but to speed up the reaction at the same temperature, would you shine UV light at it? so that it homolytically splits into two radicals? But would this work? because its not an alkane, would it work?
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    (Original post by Shinjimonkey)
    Russell, I get the mechanism now, but to speed up the reaction at the same temperature, would you shine UV light at it? so that it homolytically splits into two radicals? But would this work? because its not an alkane, would it work?
    this isn't actually homolytic, it's heterolytic. The bond is splitting to form an I+ and a Cl- two different species, rather than an I radical and a Cl radical (which you'd get from heterolytic). So to increase rate of reaction at room temperature stuff you can do is, increase the concentrations of reactants. This will increase the rate of collisions leading to an increase in the rate of successful collisions i.e. a increase in reaction rate.
    Alternatively, there are certain substituent groups which will make the aromatic system more reactive, for example phenol is more reactive than benzene due to the electron withdrawing -OH increasing the susceptibility of certain carbons to electrophilic attack.
    Can I see your mechanism? I'll check whether it's right if you want
 
 
 
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