[Rutherford]

I was reading a textbook and got confused with the reaction of oxacyclopropanes.

For example:

If we treat the above compound in basic condition with Grignard the Grignard reagend would attack the less sterically hindered C atom which is the one right to the oxygen in the ring.
Then, if we treat the same compound with Grignard in acidic conditions, the oxygen gets protonated and the Grignard will attack the sterically hindered C atom because the transition state is more stable.

Now to the thing that confuses me. If we treat a compound like:

With OH^- regardless of acid/base conditions it should attack the sterically hindered C atom because it has a bigger partial positive charge than the other C atom. This seems to be in contradiction with the previous examples. Why wouldn't Grignard attack that C atom because of the positive charge in the previous example? I need some clarification here.

[discodermolide]

Firstly an "oxacyclopropane" is called an epoxide.
You need to work out the mechanism of the three reactions you described, therein lies your answer.
See if you can come up with something.

[Rutherford]

Don't get it  . What causes the different mechanism, the substrate or the reagent, and how?

[AlphaScent]

This is a good question.  I am trying to think of a orbital justification for this perceived mechanism. 

Water is not as bulky as the grignard and would add to the bromonium ion ans the hindered position to allow the bromine atom to be in less hindered environment for further displacement.  This is just me rambling a bit and trying to make sense of it.

My one question to you Raderford is when would you do a grignard under acidic conditions??  It is a base and would react with said acid and do nothing, right?

[discodermolide]

The reagent. The first reaction is ok, the second, I would suggest that the ring opens first and the carbocations is attacked by water

[Rutherford]

The reagent. The first reaction is ok, the second, I would suggest that the ring opens first and the carbocations is attacked by water

It's not that way presented in Vollhard but it would rationalize things.

This is a good question.  I am trying to think of a orbital justification for this perceived mechanism. 

Water is not as bulky as the grignard and would add to the bromonium ion ans the hindered position to allow the bromine atom to be in less hindered environment for further displacement.  This is just me rambling a bit and trying to make sense of it.

My one question to you Raderford is when would you do a grignard under acidic conditions??  It is a base and would react with said acid and do nothing, right?

That reaction I made up, didn't want to change reagent for simplicity. Sorry for the silly mistake.

I like the idea of bromine being in a less hindered environment. But when epoxyde gets protonated it reacts in the same way as bromonium ion, so it can't be that.

[AlphaScent]

Well it is the idea that the carbocation is more stable on a tertiary carbon.  That is what drives the ring opening for both the bromonium and a protonated epoxide.  i dont know what reaction though would proceed in any type of acidic condition for alkylation.  diols can be made this way.

[AlphaScent]

I should have added, as disco said, the ring opens, the cation moves to the tertiary position and then attack by water.  not hydroxide.

[Rutherford]

Okay, I will remember it that way. It's best to take it as s_N1.
Thanks to you two for solving this.

[TwistedConf]

Okay, I will remember it that way. It's best to take it as s_N1.

It's actually not.

The exact nature of a bromonium ion is quite complex, and depends upon the structure of the ion itself (ie. what's attached to the two carbons) and other reaction variables.

In a case like the one you drew, experimental evidence suggests that a carbocation is not involved. For one thing, these ring opening reactions are stereoselective- so there's no rotation around the C-C bond in the intermediate. In related (but not identical) bromonium ions, favorable rearrangements are possible but not observed.

You're better off thinking of that intermediate as having a partial polarization at the more substituted carbon, which is then attacked from the back side, rather than an actual carbocation/SN1.

[Rutherford]

If I take it that way, then my first question from my first post arises again. Why the epoxyde won't be attacked the same way?

[AlphaScent]

Is a bromonium that much more complex than a protonated epoxide ion?  I have made bromohydrins before where the hydroxide always winds up on the more substituted carbon.  For Sn2 type reactions, such as grignards, the attack is on the less hindered carbon in a asymmetrical situation. Adding water to an epoxide needs an acid catalyst, the water then adds to the more hindered carbon in an Sn1 type reaction.  I understand bromonium ions are stereoselective in nature but they are also regioselective based upon the nucelophile.

[Rutherford]

I was comparing non-protonated epoxyde with protonated and bromonium. The s_N1 mechanism for bromonium and s_N2 for epoxyde would explain the difference, but in reality I don't know what happens with bromonium.

As Twisted said, carbocations are not involved in bromonium reactions.

[TwistedConf]

Is a bromonium that much more complex than a protonated epoxide ion?

In the practical final product sense, not really.  But comments above mentioned bromonium ring opening prior to attack and carbocations being involved in the process, and there's literature and experimental evidence out there that indicates otherwise- at least in cases like the one drawn in the original post.

[TwistedConf]

Why the epoxyde won't be attacked the same way?

They are attacked the same way, except when you have a really powerful nucleophile (like your Grignard) which doesn't require activation (protonation) of the oxygen and just attacks directly via the path of least resistance.