REACTIONS OF α-HETEROATOM-SUBSTITUTED ETHERS AND SULFIDES WITH SILYL ENOL ETHERS. CHEMOSELECTIVITY IN THE CLEAVAGE OF HETEROATOM–CARBON BONDS BY. genolysis studies with Ni and other catalysts have hinted at different cleavage mechanisms for the C-O ether bonds in α-keto and α-hydroxy β-O-4 type aryl. Cleavage of Ethers by Organoalkali Metal Compounds. aluminum hydride (see procedure under cis,cis,cis- 1,2,3-tri- methylcyclohexane) to obtain 8 g. THE SEEMS A BETTER PLACE NEW BERN

However, the increased demethylation did not proportionally turn out as more ArOH but led to reduced ArOH, which was probably attributed to the formation of the benzodioxane BD structure, as reported previously. The average molecular weights did not change much, suggesting that the repolymerization was not significantly affected by reaction time, although it was very sensitive to reaction temperature Fig.

Scheme 2 Proposed formation mechanism of the BD structure. Since the concentration of LiBr played an important role in the conversion of creosol, it is plausible that the demethylation of lignin is also greatly impacted by LiBr concentration. The average molecular weight also became higher with increased LiBr concentration, suggesting that the acidity of ACLB was likewise enhanced, which promoted the repolymerization of EPL. As observed above, creosol and 4-methylcatechol underwent severe condensation in concentrated HBr and generated humin-like precipitates.

After the demethylation in ACLB, residual OMe groups were still visible, while the characteristic signals of A, B, and C subunits were no longer visible probably chemically shifted in the aliphatic region due to the cleavage of the ether bonds in these structures Fig. The signal of the characteristic p-hydroxybenzoate unit, however, did not change after the demethylation Fig.

The formation of the BD structure during the acid-catalyzed lignin depolymerization in the ACLB was reported in our previous study, but the formation mechanism of BD was not clearly addressed. Apparently, the formation of BD consumed some of the ArOH generated by the demethylation, which explained why ArOH did not increase proportionally with the demethylation degree, as discussed above. Hibbert's ketone HK was not detected in the ACLB-treated EPL as in the previous studies, 43,45,46 possibly due to the higher acid concentration used in the present study, which promoted the aldol condensation of the ketone.

It was found that acetic acid and acetone did not significantly affect the demethylation performance, suggesting that lignin could be demethylated efficiently in the ACLB even without dissolving the lignin. This is probably due to the etherification between ethanol and ArOH formed from the demethylation, which was supported by the detection of ethoxy groups in the treated lignin by the HSQC spectrum Fig.

It should — these are ideal conditions for an SN2 reaction. Clearly the SN2 is not in play here, as the tertiary carbons are much too hindered for a backside attack. SN1 and SN2 is a continuum. Which way is it going to break? For example, what about t-butyl methyl ether? When you treat it with acid, what happens first? Do you do an SN2 on the methyl group with iodide, or does it ionize to give a tertiary carbocation? This is the type of question that is NOT easy to answer without knowing the results of experiments.

There are, however, a few cases of mixed ethers where there IS a straightforward answer. Take this question for example. What happens? See if you can do it. Next post! Ether cleavage generally requires strong acid and heat, which are forcing conditions. Alternatively, silane reagents can be used, which are reactive at room temperature.

Burwell and Milton E. Below are a variety of papers using silane-based reagents for ether cleavage. The Nobel Laureate late Prof.

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All right. So we're going to react this ether with excess hydrobromic acid. And we're going to heat things up. And when we think about our products, we know that the ether's going to go away. And we know that we're going to get two alkyl halides out of this. So we just need to find our alkyl groups. So if I look over here, here's one of my alkyl groups.

And if I look over here, here's my other alkyl group. So all I have to do is turn those alkyl groups into alkyl halides. And they're going to be alkyl bromides, since we're using hydrobromic acid here. So I'm going to draw one of my alkyl halides like that. So it would be bromine attached to the ring. And then my other alkyl halide will be this methyl group over here. So I take a methyl group and I attach it to bromine. So methyl bromide would be my other product.

And we'd also produce water, as well. But your major organic products would be these two alkyl halides. Let's do another one. This one will make it a little bit tricky. So in three dimensions-- so we're going to have to think about acidic cleavage of ethers in three dimensions. So it makes it much harder. So if I look at an ether, which looks like this, and I add, once again, excess hydrobromic acid, and I heat things up, I'm going to get acidic cleavage of that ether.

And when I'm trying to figure out the products, I know that oxygen's going to go away. And I know that the carbons that are bonded to that oxygen are the ones that are going to form my alkyl halides. So I look at this carbon that's bonded to my oxygen. That's going to be bonded to a halogen.

And if I look at this carbon on the other side of my ether, that carbon is also going to bonded to a halogen. So when I'm trying to draw the products-- so this ring here is going to stay the same. So I'm just going to go ahead and draw my ring like that. And let's first look at this carbon right here, this one in red, which is this one over here. So it was already bonded to a methyl group right here. And the bond between the carbon and the oxygen is going to break. And we're going to form a new bond with our halogen, which happens to be bromine.

So we're going to put bromine there like that. And the opposite side, when we look at this carbon over here-- so let's go ahead and draw out that carbon in. So we're going to get like that. So it's the carbon in blue. The bond between the carbon in blue and the oxygen is going to break. And there's going to be a new bond formed to our bromine, which is our halogen. So we can go ahead and draw our bromine in there like that. And so that will be our product.

So we get two alkyl halides in the same molecule this time. We could draw the product like that, or we could kind of flatten it out a little bit. And let's think about that ring system there. So what kind of a ring system do we have? We have one, two, three, four, five, six carbons present. So if I'm going to draw this molecule in a different way, I would have a six carbon ring-- so cyclohexane.

And if I take a look at-- let's do this carbon right here. So that's this carbon. What is attached to that carbon? Well, this carbon is attached to another carbon, there are two methyl groups, and then a bromine like that. It doesn't really matter the order that you put your methyl groups and your bromine. And this is because this carbon right here is not a chirality center. So you don't have to worry too much about that. Let's look at this carbon down here-- so in green.

So I'll make this one down here in green. There's a methyl group attached to that carbon. So if I make a methyl group attach to that carbon right here. And there's also a bromine attached to that carbon. So these are just two different ways to represent the exact same molecule for your product after the acidic cleavage of the original ether. So kind of a tough one because your thinking in three dimensions here. Let's do one more example for the acidic cleavage ethers. And we'll start with ethyl.

Let's do ethyl phenol ether here. So I'll draw a phenol group. So here is our phenol group. And then we'll have an ethyl group on this side. So what will be the product of ethyl phenol ether reacted with excess hydrobromic acid, and everything is heated up? All right, so let's think about the mechanism. We know that in our mechanism the first step is to protonate the ether.

So the ether's going to function as a base, pick up a proton from hydrobromic acid. So if we go ahead and draw the first step, the acid base reaction there. So we're going to protonate our ether. And that means there's going to be a hydrogen attached to our oxygen now.

Lone pair of electrons still left on our oxygen, plus 1 formal charge on our oxygen like that. Well, we have the bromide anion left over after HBr donates a proton. Br negative is left. The bromide anion is going to function as my nucleophile. And it's going to proceed via an SN2 mechanism. Therefore, it's going to attack the least stericly hindered alkyl group, which is, of course, the alkyl group on the right.

So lone pair of electrons is going to attack this carbon right in here. And these electrons would then kick off onto my oxygen. So if I draw the result of that nucleophilic attack, I now have my ring over here. And I just turned everything into an OH group. Because now I have an oxygen with two lone pairs of electrons on it like that.

However, reactions that would require the formation of unstable carbocations methyl , vinyl , aryl or primary carbon proceed via SN2 mechanism. The hydrohalic acid also plays an important role, as the rate of reaction is greater with hydroiodic acid than with hydrobromic acid. Hydrochloric acid only reacts under more rigorous conditions.

The reason lies in the higher acidity of the heavier hydrohalic acids as well as the higher nucleophilicity of the respective conjugate base. Fluoride is not nucleophilic enough to allow for usage of hydrofluoric acid to cleave ethers in protic media. Regardless of which hydrohalic acid is used, the rate of reaction is comparably low, so that heating of the reaction mixture is required. Cyclic ethers allow for an especially quick concerted cleavage, as seen for THF : Deprotonated acyclic ethers perform beta-hydride elimination , forming an olefinic ether.

Impact[ edit ] Organometallic agents are often handled in etheric solvents, which coordinate to the metallic centers and thereby enhance the reactivity of the organic rests.

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Cleavage of Ethers

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And also water would be produced in this as well-- so H2O, like that. Now, I drew this second part of the mechanism like it's an SN2 mechanism. And it would be an SN2 mechanism if we were starting with a primary alcohol. So if this guy over here is a primary alcohol, and after it gets protonated, a primary alcohol would work the best for an SN2 mechanism because that would be decreased steric hindrance.

However, if we were dealing with something like a tertiary alcohol at this point, things would likely proceed via an SN1 type mechanism. So it's important to look at the structure of the alkyl halide. Let's do an example of the acidic cleavage of ethers. And we'll start with an ether that looks like this.

All right. So we're going to react this ether with excess hydrobromic acid. And we're going to heat things up. And when we think about our products, we know that the ether's going to go away. And we know that we're going to get two alkyl halides out of this. So we just need to find our alkyl groups. So if I look over here, here's one of my alkyl groups.

And if I look over here, here's my other alkyl group. So all I have to do is turn those alkyl groups into alkyl halides. And they're going to be alkyl bromides, since we're using hydrobromic acid here.

So I'm going to draw one of my alkyl halides like that. So it would be bromine attached to the ring. And then my other alkyl halide will be this methyl group over here. So I take a methyl group and I attach it to bromine. So methyl bromide would be my other product. And we'd also produce water, as well. But your major organic products would be these two alkyl halides. Let's do another one. This one will make it a little bit tricky.

So in three dimensions-- so we're going to have to think about acidic cleavage of ethers in three dimensions. So it makes it much harder. So if I look at an ether, which looks like this, and I add, once again, excess hydrobromic acid, and I heat things up, I'm going to get acidic cleavage of that ether. And when I'm trying to figure out the products, I know that oxygen's going to go away. And I know that the carbons that are bonded to that oxygen are the ones that are going to form my alkyl halides.

So I look at this carbon that's bonded to my oxygen. That's going to be bonded to a halogen. And if I look at this carbon on the other side of my ether, that carbon is also going to bonded to a halogen. So when I'm trying to draw the products-- so this ring here is going to stay the same. So I'm just going to go ahead and draw my ring like that.

And let's first look at this carbon right here, this one in red, which is this one over here. So it was already bonded to a methyl group right here. And the bond between the carbon and the oxygen is going to break. And we're going to form a new bond with our halogen, which happens to be bromine. So we're going to put bromine there like that. And the opposite side, when we look at this carbon over here-- so let's go ahead and draw out that carbon in. So we're going to get like that.

So it's the carbon in blue. The bond between the carbon in blue and the oxygen is going to break. And there's going to be a new bond formed to our bromine, which is our halogen. So we can go ahead and draw our bromine in there like that. And so that will be our product. So we get two alkyl halides in the same molecule this time. We could draw the product like that, or we could kind of flatten it out a little bit.

And let's think about that ring system there. So what kind of a ring system do we have? We have one, two, three, four, five, six carbons present. So if I'm going to draw this molecule in a different way, I would have a six carbon ring-- so cyclohexane. And if I take a look at-- let's do this carbon right here.

So that's this carbon. What is attached to that carbon? Well, this carbon is attached to another carbon, there are two methyl groups, and then a bromine like that. It doesn't really matter the order that you put your methyl groups and your bromine. And this is because this carbon right here is not a chirality center. So you don't have to worry too much about that.

Let's look at this carbon down here-- so in green. So I'll make this one down here in green. There's a methyl group attached to that carbon. So if I make a methyl group attach to that carbon right here. And there's also a bromine attached to that carbon. So these are just two different ways to represent the exact same molecule for your product after the acidic cleavage of the original ether. So kind of a tough one because your thinking in three dimensions here. Let's do one more example for the acidic cleavage ethers.

And we'll start with ethyl. Let's do ethyl phenol ether here. So I'll draw a phenol group. So here is our phenol group. And then we'll have an ethyl group on this side. So what will be the product of ethyl phenol ether reacted with excess hydrobromic acid, and everything is heated up? All right, so let's think about the mechanism. We know that in our mechanism the first step is to protonate the ether. So the ether's going to function as a base, pick up a proton from hydrobromic acid.

So if we go ahead and draw the first step, the acid base reaction there. So we're going to protonate our ether. And that means there's going to be a hydrogen attached to our oxygen now. Lone pair of electrons still left on our oxygen, plus 1 formal charge on our oxygen like that. Well, we have the bromide anion left over after HBr donates a proton. Br negative is left.

The hydrohalic acid also plays an important role, as the rate of reaction is greater with hydroiodic acid than with hydrobromic acid. Hydrochloric acid only reacts under more rigorous conditions. The reason lies in the higher acidity of the heavier hydrohalic acids as well as the higher nucleophilicity of the respective conjugate base. Fluoride is not nucleophilic enough to allow for usage of hydrofluoric acid to cleave ethers in protic media.

Regardless of which hydrohalic acid is used, the rate of reaction is comparably low, so that heating of the reaction mixture is required. Cyclic ethers allow for an especially quick concerted cleavage, as seen for THF : Deprotonated acyclic ethers perform beta-hydride elimination , forming an olefinic ether.

Impact[ edit ] Organometallic agents are often handled in etheric solvents, which coordinate to the metallic centers and thereby enhance the reactivity of the organic rests. Here, the ether cleavage poses a problem, as it does not only decompose the solvent, but also uses up the organometallic agent.

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11a: Acidic cleavage of ethers

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