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Why does the following addition reaction use radicals rather than using the normal transfer of hydrogen (seen in last picture)? I would think that because the reaction has no light to indicate homolytic cleavage to begin the free-radical process, it would not proceed this way.

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    $\begingroup$ The $\ce{O-O}$ bond in peroxides is weak $(\Delta H^{\circ} = 150~\pu{kJ} = 36~\pu{kcal})$. So, slight heating can break the peroxy bond. And radicals are very reactive species. $\endgroup$
    – Apoorv Potnis
    Commented Feb 7, 2018 at 4:34
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    $\begingroup$ Thank you. The first image is the question (which is asking us to draw the mechanism) while the second image is the answer. However, I believe the first image does not indicate that there is a peroxide in the reaction. Is there any way to know that there is a peroxide in the reaction just from looking at the first image? $\endgroup$
    – Sarah Smith
    Commented Feb 7, 2018 at 4:37
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    $\begingroup$ See this. $\endgroup$
    – Apoorv Potnis
    Commented Feb 7, 2018 at 4:42
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    $\begingroup$ impure is there $\endgroup$
    – Avyansh Katiyar
    Commented Feb 7, 2018 at 4:44
  • $\begingroup$ @ApoorvPotnis Re: "\ce{O-O} bond in peroxides" as Sarah Smith noted, I don't see why you're bringing in peroxides into the question. Yes, the solution mentions diethyl ether peroxide (to which you later linked) but there doesn't seem to be any immediate connection to the question. Quoting Sarah Smith again, "Is there any way to know that there is a peroxide in the reaction just from looking at the first image?" Please consider elaborating. Thank you! $\endgroup$
    – Gaurang Tandon
    Commented Feb 13, 2018 at 5:12

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