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What will be the major product when conjugated alkene (like 1,3-butadiene) is reacted with $\ce{HBr}$ in the presence of $\ce{R2O2}$?

3‐chloro‐4‐ethenyl‐1‐methylcyclohexa‐1,4‐diene + HBr + R2O2

As we know, two products are formed when no $\ce{R2O2}$ is used because of conjugation in double bond and carbocation. But when $\ce{R2O2}$ is used, free radical is formed.

So, will there be the same case in the following question as well? Also, if answer is yes, then what will be the major product?

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    $\begingroup$ Please convert the image into a better form.. You can use this to do so. $\endgroup$ Jul 16 '20 at 14:47
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    $\begingroup$ The 6-member ring is highly suggestive... $\endgroup$
    – Zhe
    Jul 16 '20 at 15:14

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