[Strike]

Hello,

I'm having a little trouble with this acetal question. At first I thought the ring would break and an aldehyde would form on the left and an alcohol on the right but the electron movements don't make that possible.



What I'm thinking is:

1. Lone pair from oxygen picks up H from HCl
2. Ring breaks leaving a secondary carbocation
3. Oxygen from Ethanol attacks the carbocation and Cl- is reprotonted


(Of course the Ethanol could attack from the top or bottom, giving both stereochemistries)

But now I'm thinking what if the other oxygen get's protonated first and leaves as methanol leaving the ring intact? Then the ethanol would attack, but the end product would be back at an acetal, except with an ethanol instead of a methanol. I'm not sure of any other way to approach this question.

Any help would be greatly appreciated!

[discodermolide]

Surely you will end up with a mixture of alpha and beta anomers.

[orgopete]

I doubt there will be any un-ionized HCl in this solution. Chloride is a very weak base and should not abstract any protons. The oxygen atoms are more basic.

If you form an oxygen stabilized carbocation, an intramolecular cyclization is favored over intermolecular reaction for entropic reasons. Are there any other carbocations that can form? What might happen if they do?

Re: alpha and beta isomers
While this is correct, I am guessing that just as the alpha isomer is the starting material and if this problem is designed to follow literature chemistry, is probably easier to form. I further presume that it would have formed in a similar reaction to that of the post and therefore one should presume the same isomer to be the major product. I know there are some sugar chemists on this forum so I will leave the question of how to form an alpha or beta isomer to them.

[Strike]

Hmm.. I'm slightly confused here.

@Discodermolide, are you saying that I have the correct answer, but both "anomers" where the Ethanol attacks from either the top or bottom?

Sorry, I haven't learned carbohydrate chemistry yet so I'm not entirely sure what approach you're going at in solving this.

[Dan]

What you have here is a series of transacetalisation equilibria, and you must predict the most stable product.

2,5-dimethoxyoxane a-2-ethoxy-5-methoxyoxane b-2-ethoxy-5-methoxyoxane 5,5-diethoxy-2-methoxypentan-1-ol

You should be able to write a mechanisms that interconvert any two of those structures. You need to identify the most thermodynamically stable product.

To predict which anomer you'll get, have a read about the "anomeric effect".

[Strike]

So for "anomers" to occur, the ring stays intact?

1. The left oxygen gets protonated and leaves as methanol
2. Ethanol attacks the carbocation on the ring to give a racemic mixture of both anomers?

Is that the correct mechanism? If so, why couldn't my first mechanism occur? The ring isn't aromatic so it can still easily break right?

[discodermolide]

So for "anomers" to occur, the ring stays intact?

1. The left oxygen gets protonated and leaves as methanol
2. Ethanol attacks the carbocation on the ring to give a racemic mixture of both anomers?

Is that the correct mechanism? If so, why couldn't my first mechanism occur? The ring isn't aromatic so it can still easily break right?

For anomers to interconvert the ring must open and re-close, it ends in an equilibrium mixture.

[Dan]

For anomers to interconvert the ring must open and re-close, it ends in an equilibrium mixture.

Not true, the reaction can proceed via a cyclic oxonium intermediate.