Question about ether cleavage by HBr

[CuriousBanker]

TL;DR

Sorry, just having one of those days were things don’t make sense to me

So in this video, this cyclic ether steals a Hydrogen off HBr. That gives it a +1 formal charge, so it breaks the ring open and steals the electrons from the Carbon to alleviate the formal charge. Carbon then has a formal charge which bonds to the Bromide anion that was formed from when it donated the Hydrogen to oxygen.
So my question is...if oxygen does not want to have a +1 formal charge and therefore needs to open the ring to alleviate it, why does it steal the hydrogen in the first place? It was a neutral atom before. Also, how is the oxygen in the ether more negative than a bromide ion which has a negative formal charge, enough to steal a hydrogen from anion?

 

[TeethWhitener]

I assume you’re talking about the part about 4:50 into the video (for the future, if you have a specific question about a 7 minute video, it’s a good idea to point people to the part of the video in question, so they don’t have to sit through the whole thing).

The C-O bond breaks because the nucleophilic bromide anion attacks the partial positive charge on the carbon.

The protonated ether is in equilibrium with the HBr. This step is (usually) fast, so at any given time, a certain fraction of ethers are protonated at the oxygen. The S_N2 bromide attack on the alpha carbon is slower, so it’s not the case that every time an ether is protonated it undergoes ring cleavage. Eventually a bromide will be in the right place at the right time when the ether is protonated to attack the alpha carbon and this will open the ring.

(Also, if I’m being kind of pedantic, the formal reaction mechanism shouldn’t have a bunch of anionic bromide flying around in acidic conditions. Instead, the HBr should attack the alpha carbon directly to generate the S_N2 product and a proton.)