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Author: Subject: What is the mechanism for catalytic cleavage of ethers by Lewis Acid's?
Leben
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[*] posted on 7-12-2014 at 01:46
What is the mechanism for catalytic cleavage of ethers by Lewis Acid's?


I understand how the reaction works for mineral/halide acids...

But how does it work for Lewis Acids? When Sulfuric acid is used, Dimethyl Sulfate forms as an intermediate, which can react with water, to form the alcohol and sulfuric acid, repeating the cycle, and I can't quite find any info on how this works with lewis acids.

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deltaH
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[*] posted on 7-12-2014 at 02:07


Quote: Originally posted by Leben  
I understand how the reaction works for mineral/halide acids...


I do not think you do. Strong acids protonate the ether resulting in the formation of a better leaving group (an alcohol). The protonated ether then undergoes nucleophilic attack by water. (1)

The reaction between alcohols and sulfuric acid is another reaction in its own right, don't confuse the two.

(1) http://www.masterorganicchemistry.com/reaction-guide/acidic-...




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Chemosynthesis
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[*] posted on 7-12-2014 at 02:23


What part are you stuck on? I am kind of hindered by being on my phone for this, but the gist of it is that your Lewis acid should accept the ether oxygen's lone pairs, forming a positively charged adduct species. This concludes with a halide transfer.

Seems like there was some contention on the mechanism of some dealkylation last year according to DOI: 10.1002/ejoc.201300337.
Probably a good scheme in one of the citations of Tetrahedron Letters 25(32) (1984) 3497-3500, Tetrahedron Letters 24 (5) 2289–2292, or perhaps US 4695659 A.
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[*] posted on 7-12-2014 at 03:05


Quote: Originally posted by deltaH  
Quote: Originally posted by Leben  
I understand how the reaction works for mineral/halide acids...


I do not think you do. Strong acids protonate the ether resulting in the formation of a better leaving group (an alcohol). The protonated ether then undergoes nucleophilic attack by water. (1)

The link you provided is in agreement with my explanation.

2 Methoxybenzene + Sulfuric Acid -> 2 Phenol + Dimethyl Sulfate
Dimethyl Sulfate + 2 Water -> 2 Methanol + Sulfuric Acid

and the cycle repeats.

Or in the case of the link you provided....

Methoxybenzene + HI -> Phenol + Iodomethane
Iodomethane + Water -> Methanol + HI


In both of these cases, the Dimethyl Sulfate, or Iodomethane, are intermediates in the catalytic cycle. They can both be hydrolyzed by water back to their respective acids giving off an alcohol in the process.


Quote: Originally posted by Chemosynthesis  
What part are you stuck on? I am kind of hindered by being on my phone for this, but the gist of it is that your Lewis acid should accept the ether oxygen's lone pairs, forming a positively charged adduct species. This concludes with a halide transfer.

Seems like there was some contention on the mechanism of some dealkylation last year according to DOI: 10.1002/ejoc.201300337.
Probably a good scheme in one of the citations of Tetrahedron Letters 25(32) (1984) 3497-3500, Tetrahedron Letters 24 (5) 2289–2292, or perhaps US 4695659 A.


I will look up that Tetrahedron letter. I am a little confused by your description because I am not seeing how the lewis acid regenerates, or if it does at all, as a bronsted acid would.


[Edited on 7-12-2014 by Leben]
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[*] posted on 7-12-2014 at 03:20


Quote: Originally posted by Leben  

2 Methoxybenzene + Sulfuric Acid -> 2 Phenol + Dimethyl Sulfate

Where are you getting this from? Please provide a link. It looks very unlikely to me. There is no way that this is happening under vaguely-normal conditions. At extremes of temperature and pressure, then maybe it is possible, but it is not going to happen under normal conditions.

[Edited on 7-12-2014 by forgottenpassword]
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Leben
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[*] posted on 7-12-2014 at 03:41


Quote: Originally posted by forgottenpassword  
Quote: Originally posted by Leben  

2 Methoxybenzene + Sulfuric Acid -> 2 Phenol + Dimethyl Sulfate

Where are you getting this from? Please provide a link. It looks very unlikely to me. There is no way that this is happening under vaguely-normal conditions. At extremes of temperature and pressure, then maybe it is possible, but it is not going to happen under normal conditions.

[Edited on 7-12-2014 by forgottenpassword]


I can't say whether it will happen at room temperature but this reaction easily happens at elevated temperatures. Cleavage of ethers (I've only looked into aromatic ethers) with sulfuric acid is an extremely common reaction. If you take a few seconds to google ether cleavage sulfuric acid, you will see a lot of what comes up says ethers are "easily" cleaved by sulfuric acid.

The above link that deltaH provided details this reaction taking place, but with Hydroiodic acid. The electro and nucleophilicities of the acid and the reagent will obviously play an important role in determining the easy of reactivity.... but that goes without saying for nearly.... all of organic chemistry?
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[*] posted on 7-12-2014 at 03:45


If you know it to be a fact, fair enough. Look at it a different way: If you have hot dimethyl sulfate and phenol mixed together -- they won't react?! I happen to know that they will... Easily... To give the products that you believe to be reactants. :D

[Edited on 7-12-2014 by forgottenpassword]
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[*] posted on 7-12-2014 at 03:45


Quote: Originally posted by Leben  


I will look up that Tetrahedron letter. I am a little confused by your description because I am not seeing how the lewis acid regenerates, or if it does at all, as a bronsted acid would.
Check the first DOI if you want to see the contention on aryl methyl ethers. The halide transfer regenerates the catalyst, but I know I am not explaining well.
Edit-oh, sorry. I see what you mean. Workup regenerates your catalyst kind of like Friedel Crafts acylation gives an adduct before hydrolysis.
Sorry for the confusion.

[Edited on 7-12-2014 by Chemosynthesis]
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[*] posted on 7-12-2014 at 03:47


You missed the note in the link in regards to the sulfuric acid/water system that says it can be written as H3O[+], what they mean by that is H3O[+] protonates the ether and then the H2O attacks the ether, as with their example of HI and an ether (note, no water in that example!).

(CH3CH2)O + H3O[+] <=> (CH3CH2)OH[+] + H2O <=> 2CH3CH2OH

This is what occurs when catalytic amounts of acid is used, usually in an excess of water. If excess acid is used, then yes, indeed you can form dimethyl sulfate, but that is another reaction. You specifically talked about acid catalysed hydrolysis.

[Edited on 7-12-2014 by deltaH]




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Leben
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[*] posted on 7-12-2014 at 04:00


Quote: Originally posted by deltaH  
You missed the note in the link in regards to the sulfuric acid/water system that says it can be written as H3O[+], what they mean by that is H3O[+] protonates the ether and then the H2O attacks the ether, as with their example of HI and an ether (note, no water in that example!).

(CH3CH2)O + H3O[+] <=> (CH3CH2)OH[+] + H2O <=> 2CH3CH2OH

This is what occurs when catalytic amounts of acid is used, usually in an excess of water. If excess acid is used, then yes, indeed you can form dimethyl sulfate, but that is another reaction. You specifically talked about acid catalysed hydrolysis.

[Edited on 7-12-2014 by deltaH]


Looking at the link, they seem to clearly demonstrate that Iodomethane forms when Hydriodic acid is used, even in the absense of water. Why would dimethyl sulfate not form if sulfuric acid is used (assuming reactivity is high enough)? When HI is used, one of the products of the hydrolysis is Iodomethane. That is denoted by CH3-I.

It's the very first example on the page.
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[*] posted on 7-12-2014 at 04:19


The difference is that I- is an excellent nucleophile, whereas HSO4- is not.

[Edited on 7-12-2014 by forgottenpassword]
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[*] posted on 7-12-2014 at 06:20


Exactly forgottenpassword, I wanted to compare the nucleophillicity of the two by some nucleophilicity scale, but couldn't find one with HSO4- and H2O upon a cursory search, so glad you said it :D

Leben, I do not know how else to explain it so that it makes sense to you. If you're still not convinced, then I am afraid we shall have to agree to disagree :)




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[*] posted on 8-12-2014 at 10:56


Quote: Originally posted by forgottenpassword  
If you know it to be a fact, fair enough. Look at it a different way: If you have hot dimethyl sulfate and phenol mixed together -- they won't react?! I happen to know that they will... Easily... To give the products that you believe to be reactants. :D

[Edited on 7-12-2014 by forgottenpassword]



And that is where Le Chatelier's principle comes into play ;)

In the case of OP's phenol, the equilibrium will favor the dimethyl sulfate reacting with the water (moving the reaction forward), than reacting with the phenol (which would regenerate the reactants).

Phenol is more acidic (and electron deficient) than water. Dimethyl Sulfate being an electrophile, I think the reaction with water is much more favored, which would drive the reaction to the right instead of regenerating the reactants. The more acidic/electrophilic the reactant is the more easily to the right the reaction should go...

I think....

[Edited on 8-12-2014 by FireLion3]
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[*] posted on 9-12-2014 at 04:58


What?!

By all means, try the reaction yourself. If it doesn't work try waving a magic wand and telling it that "Le Chatelier's principle's coming into play now!"

That'll kick-start it. ;)

Unfortunately for me I still have this reference that tells me that hot dimethyl sulfate and phenol react to give your "reactants". Not the other way around! Now you're saying that adding water will push it in the right direction? Water?! Adding water?! H2O?! Seriously? And you're dragging Le Chatelier's good name into this?

Feel free to prove me wrong with a simple reference. One is enough. Go ahead!

[Edited on 9-12-2014 by forgottenpassword]
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[*] posted on 9-12-2014 at 10:53


forgottenpassword,

If you had been paying attention to the thread you would see that such a reference has already been posted by deltaH above.

You are right that Dimethyl Sulfate does react with Phenol, but when you introduce water into the equation, you change the entire dynamic. Water is much more reactive towards DMS than phenol, and due to its low molar mass is almost certain to be in a molar excess vs phenol.

http://www.masterorganicchemistry.com/reaction-guide/acidic-...

Quote:
Notes: Common acids for this purpose are HI and other hydrogen halides, as well as H2SO4 in the presence of H2O.


The presence of water and the high degree of reactivity DMS towards it will push the equilibrium to the right. Sulfuric Acid will cleave the methoxy, and the formed DMS could indeed re-react with the phenol, but it is almost certainly much more likely to react with water forming methanol and reforming the sulfuric acid. The sulfuric acid could react with the methanol to reform DMS, but then we are looking at the same reaction probability scenario, of Water vs Phenol. Electron withdrawing groups on the ring will almost certainly push the equation further to the right of the demethylation, as the reactivity of the aryl towards the electrophile would decrease even further.

Almost every step in this reaction is reversible so Le Chatelier's plays an essential role in approximating the equilibrium.


forgottenpassword,

There is nothing wrong with wanting to see references, and I admit the above reference isn't any scholarly paper, but you cannot just start talking about magic wands and pretend that theoretical chemistry and simple chemical electrophilic/nucleophilic dynamics do not exist. This is freshmen level organic chemistry here.

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[*] posted on 9-12-2014 at 11:09


I agree with you: it certainly IS a freshman level reference! No experimental data in sight!

You're saying that people cleave methyl ethers with DILUTE sulfuric acid? Incredible! Why did I never learn that?! I've been using BBr3 while people have been laughing behind my back with a gentle 1M sulfuric acid boil?! :D

Read this paper. Educate yourself.

"Sulfuric acid: In spite of its high acidity, sulfuric acid is not of much use in the cleavage of ethers, probably because of the relatively slight nucleophilic displacing tendency of the bisulfate ion. Thus, SN2 reactions are slow and SN1 reactions are favored. In strong sulfuric acid, conjunct polymerization results."

Attachment: The_Cleavage_of_Ethers.pdf (4.5MB)
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There is a serious problem here: Your understanding of the THEORY is in direct contrast with the FACTS. Try to align the two, because your future studies will be much easier if your understanding has some basis in REALITY. If you are being taught incorrect things, and being taught to rationalize them as the 'truth', then how does that help you at all? What good is your theoretical 'knowledge' if it is bullshit?





[Edited on 9-12-2014 by forgottenpassword]
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[*] posted on 9-12-2014 at 12:02


Quote:
"Sulfuric acid: In spite of its high acidity, sulfuric acid is not of much use in
the cleavage of ethers, probably because of the relatively slight nucleophilic displacing tendency of the bisulfate ion. Thus, SN2 reactions are slow and SN1 reactions are favored. In strong sulfuric acid, conjunct polymerization results."


We don't dispute that, this is cleavage (not hydrolysis), as in cleavage to form methylsulfuric acid or dimethyl sulfate. In fact this supports our argument... "relatively slight nucleophilic displacing tendency of the bisulfate ion." has exactly been said. It does not proceed significantly.

That is not to say that the ether cannot be protonated and then attacked by water as the nucleophile, that proceeds fairly rapidly, but not necessarily to good conversion because of equilibrium.

Also, kinetics and thermodynamics should not be confused here. Many ethers (certainly diethyl ether) are fairly water insoluble (having ionising acids in there lowers solubility even further), this can help the equilibrium in favour of the ether because of phase separation decreasing the equilibrium concentration of the ether in aqeous solution greatly.

I agree, if you wanted to quantitatively convert your ether into alcohols in good yield, it's best to go a circuitous route with other reagents.

That is not to say, that when you hydrolyse in the presence of H2SO4, the mechanism doesn't follow exactly what we have said. The equilibrium position for the hydrolysis might be poor though if the ether is very water insoluble and so yes, it would appear that not much hydrolysed, but that has nothing to do with the mechanism by which the hydrolysis occurs (even if it occurs only slightly).

This is a classic example of confusing kinetics with thermodynamics.

[Edited on 9-12-2014 by deltaH]




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