[magician4]

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.

what do you mean by "basic conditions"  / "acid conditions" ?
in the presence of Grignard reagents, it is very difficult to have substances more basic than the Grignard - substance in its own right around ( BuLi , maybe) ...
... and those wouldn't interfer with the Grignard ./. epoxide situation at all, IMHO  (except of competing, maybe)

if, on the other hand, you had an acid condition with an oxonium - ion (which would be most difficult to achieve , to start with), all that would happen was the destruction of the Grignard reagent:
R-MgX + [epoxonium]Y  R-H + MgXY

to me, the example given hence seems to be completely meaningless

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.

same problem again: it is impossible to have something like a bromonium-structure under alkaline conditions
... and under acid conditions, general expectation would be that the attacking particle would be water (and not hydroxy) and would be "added" (with subsequent release of H⁺) to the more stable carbocation  (which is the tertiary , and not the secondary)

regards

Ingo

[Rutherford]

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.

If they are attacked the same way, why in one case the less hindered C atom is attacked and in the other the more hindered C atom is attacked? That's again the original question I posted.

what do you mean by "basic conditions"  / "acid conditions" ?

I just wanted to emphasize that in the first case the epoxyde won't be protonated and in the second that it will, and I wrote in one of my previous posts that I made a mistake when putting together acid and Grignard:
Me:

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.

No carbocation would be formed in the reaction of bromonium with H₂O.

[magician4]

I just wanted to emphasize that in the first case the epoxyde won't be protonated and in the second that it will, and I wrote in one of my previous posts that I made a mistake when putting together acid and Grignard (...)

I just picked up your example, as it really doesn't matter what kind of negatively charged nucleophile  will try to attack: they're all subject to the same fate
replace "Grigneard " with "cyanide" , for example , and the outcome would be the same

No carbocation would be formed in the reaction of bromonium with H2O.

I think you're definitively wrong here insofar, as for the relevant moment of the reaction the "easiness" of the formation of a stabilized cation is of highest importance, and determines the orientation resulting thereof.


that's where all the difference you're wondering about originates from: that this pathway is relevant in case of -onium situation under acid conditions
...and irrelevant in all other cases, including the alkaline situation, where steric factors then become the name of the game.

you might not like my answer, you even might keep on giving me negative mole snacks for it again and again, but that's just what the chemical situation is like

regards

Ingo

[Rutherford]

Why is it relevant in one situation and irrelevant in the other?

[Rutherford]

I see that this is so complex that no accepted explanation exist. I found a conclusion: Reactions of protonated epoxyde and bromonium ion with nucleophilic reagents obey s_N1 regioselectivity but s_N2 stereoselectivity. This means that the true reaction mechanism should be something between s_N1 and s_N2 ! About the non-protonated epoxyde, a simple s_N2 reaction takes place.

[magician4]

Why is it relevant in one situation and irrelevant in the other?

[ let's stay with expoxide / oxonium for this (as I don't like the idea of a bromonium-ion under alkaline conditions):]
because we're talking different mechanisms in different situations leading to different products !

(i) attack of a negatively charged particle on the non-protonated epoxide  (i.e. alkaline conditions)
here, the epoxide is quite stable and won't open up all by itself in any meaningfull degree (i.e. wouldn't form any carbocations, neither secondary nor tertiary, and hence will not react via S_N1 )
hence, we need S_N2 to achieve something, with the sterically less hindered side being the one most likely to be the center of the successfull action

(ii) addition of a neutral "nucleophile" on the protonated epoxide (with subsequent extrusion of H⁺)

oxonium (! the oxygen must be charged for this) being an excellent leaving group, there'll be an equilibrium of three (relevant) species present:
- the epoxide with the H⁺ bearing, hence positively charged, oxygen
- one open structure where the positive charge is located at the -CH2- group
- one open structure where the positive charge is located at the (CH₃)₂-C- group

from those three, only the last two are able to result in something meaningfull when adding the nucleophile, and the last one of those will be much much more likely by far (due to the methylgroups stabilizing this structure)
this statistical advantage will lead to the product described, when adding the nucleophile
there might be some minor neighbouring group effects still involved , though, as the hydroxygroup first will need to move away and make place for the nucleopile to rush in (from one side, not from the other), which could result in some kind of pseudo - stereoselectivity  (if we studied respective species, like one with one of the two methyls being CD₃- group in exchange)

I see that this is so complex that no accepted explanation exist.

this is your opinion

I am deeply convinced to the opposite (as shown above), and am very confident that there must be a bunch of deep and deeper mechanistic studies out there somewhere, supporting my general analysis, with such a simple system (and hence inviting to those studies at that)


regards

Ingo

[Rutherford]

I like your explanation, but one more thing is unclear for me.
(Now talking only about reactions happening in acidic conditions.) What would happen if we had an optically active epoxyde? Would the configuration inverse or remain unchanged when the reaction happens?

[magician4]

factors that might "feign" retention (with adequate epoxides) here should be

- as I said, there might be some "blocking" effects due to the hydroxylgroup that initially has to move away beforehand (at least for some directions from which the nucleophile might wish to reach the carbon) for the subsequent tertiary carbocation to react.
this is a statistical disadvantage to the respective "side"

- second, any very fast reaction at the tertiary cation taking place before the attached groups were able to revolve several (undefined) times around the XYC⁺-R' bond

hence , you might find some "incomplete" retention of stereochemistry here (I called it "pseudo" in my last post, as this is NOT due to some Walden inversion type effects)


regards

Ingo

[Rutherford]

Okay, thank you for the help.