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# When 1,3 butadiene is treated with HBr in the presence of peroxides and light at very low temperatures two major products are produced how do you draw the 2 resonance structures for the intermediate radical and the final products?

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2
Feb 7, 2018

First I'll describe what happens when HBr is added to a peroxide in the presence of light at low temp

$\left(1\right) R - O - O - R \rightarrow R O + O R$
$\left(2\right) R O + H B r \rightarrow R - O H + B {r}^{-}$

Withut the catalyst the reaction would have the same products as dissociation of HBr also yields Br- but that would be slower

Sorry i cannot show the valence electrons because i cannot edit the letters so much

Because aromatic bonds donate electrons, ${H}^{+}$ originally from the dissociation of the acid joins with either of the 2 aromatic bonds and draws all the electron density towards it the making bond joint's positive .The negative $B {r}^{-}$ joins the positive site to form 3 brom o-1-butene and 1 brom o-2-butene

$\text{Resonance structures}$

In intermediate (A) and (B) ,the aromatic bond will travel between the positive as the positive charge attracts electron density,but this would cause the original aromatic bond which is now no more there also positive . So the double bond would travel between the two carbon bonds

The final product is electron precise so no resonance will take place.Both of them do not have electrons sufficient for resonance

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