anonymous
  • anonymous
Upon treatment with Br2/H2O, 1,3-butadiene reacts to form a halohydrin. Over time, in the presence of the HBr generated by the reaction, this halohydrin converts to a more stable isomer. Please provide the structures of these two compounds and provide a mechanism for their formation. [Hint: there is an intermediate common to the formation of both isomers.]
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
what i did: bromium ion over 2nd alkene bond, h2o attacks more substituted carbon in that bond, h2o leaves, allyl carboncation resonance created, one on 3rd carbon where water left, one on terminal carbon due to resonance. water adds to both carboncations, another water deprotonates water, leads to OH group in both cases. therefore the isomers. i guess the intermediate common to formation of both isomers would be the resonance step with the allyl carboncations.
abb0t
  • abb0t
I think you form a carbocation as the intermediate on C3
abb0t
  • abb0t
HBr, being a strong acid (pKa ~ -9), would be the proton donor to donate it's proton on C1. You'd form the carbocation there. Where the Br can attack.

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abb0t
  • abb0t
I think it would proceed in an Sn2-ish manner.
anonymous
  • anonymous
what would the isomers be though? it would be due to the resonance right? i'm just not sure about the order of things.
aaronq
  • aaronq
i think if you were goin to have a hydroxide on the 1C there would be a hydride shift so the hydroxide would end up on the 2C which would be a more substituted carbon, but I'm not sure i haven't actually drawn this out
anonymous
  • anonymous
well i think the one with the terminal hydroxyl group would be more stable because the double bond would be disubstituted.
abb0t
  • abb0t
Like aaronq, pointed, there would probably be a hydride shift to form a more stable carbocation. To where the water molecule will attach.

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