Question

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.]

Answer 1

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.

Answer 2

I think you form a carbocation as the intermediate on C3

Answer 3

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.

Answer 4

I think it would proceed in an Sn2-ish manner.

Answer 5

what would the isomers be though? it would be due to the resonance right? i'm just not sure about the order of things.

Answer 6

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

Answer 7

well i think the one with the terminal hydroxyl group would be more stable because the double bond would be disubstituted.

Answer 8

Like aaronq, pointed, there would probably be a hydride shift to form a more stable carbocation. To where the water molecule will attach.