[Cooper]

Hello,

I'm doing my homework and I'm stuck on this problem, I've tried it several times but am having trouble coming up with the right answer.

This is the problem,

3-methylbut-1-ene + HBr (in methanol) --> ?

My first thought (I've attached a paint file) was that the double bond acts as a nucleophile and attacks the proton from HBr, giving 2-methylbutane with a carbocation on the third carbon. Then a methyl shift would occur, but the resulting product would be the same.

Then I thought the oxygen on the methanol attacked as a nucleophile to the carbocation and produced 3-methanol-2-methylbutane. Then the bromide from the HBr in the beginning would act as a nucleophile and remove the proton from the methanol, giving the product. EDIT: Actually, technically another methanol would pluck off the proton, because it's a better nucleophile and more abundant, my bad.

That didn't work so I thought that maybe the HBr and methanol reacted together first?

Thanks for any help in advance!

~Cooper

[discodermolide]

Try a proton shift from the end of the molecule to the cation that you drew to give a tertiary cation, MeOH will then attack there to give 2-methoxy-2-methylbutane.

The drawing starts with the cation you drew.

[Cooper]

Thank you! I should've seen that! And thanks for the drawing What did you use to make it? 

PS What do the snacks mean? If I give you one does that mean like a +1 comment? lol

~Cooper

[discodermolide]

I used ChemDraw.
The snacks are there to register your like or dislike of the answer, if you think it helps give one . If you don't like the answer, or it is wrong take one away.

[Cooper]

Oh okay I gave you one Thanks again

~Cooper

[discodermolide]

My pleasure