Alcohols, Epoxides and Ethers
Cleavage Of Ethers With Acid
Last updated: November 7th, 2022 |
Acidic Cleavage of Ethers Can Proceed Through an SN2 or SN1 Mechanism, Depending On The Structure
- Ethers do not undergo very many reactions.
- One key reaction of ethers is that they can undergo cleavage to alcohols in the presence of strong acids, such as HI, or strong Lewis acids such as boron tribromide (BBr3) .
- These reactions involve protonation of the ether oxygen, followed by either an SN1 or SN2 reaction pathway, depending on structure (ethers of primary alcohols tend to undergo cleavage via SN2, ethers of tertiary alcohols tend to undergo cleavage via SN1).
- If excess HI is used for the cleavage of alkyl ethers, the products tend to be alkyl iodides.
- Because ethers are so unreactive to all but the most acidic conditions, they can find use as protecting groups for alcohols.
Table of Contents
- All The Reactions Of Ethers In Once Place (har)
- The First Step In Ether Cleavage Is Protonation
- For Methyl and Primary Ethers, The Second Step of Ether Cleavage is SN2
- For Tertiary Ethers, The Second Step of Ether Cleavage is SN1
- For Secondary Ethers, The Second Step Could Occur Through a Mixture of Either Pathway
- The Mechanism for Cleavage of Unsymmetrical Ethers Is Hard To Generalize (With One Exception!)
- Summary: Acidic Cleavage of Ethers
- (Advanced) References and Further Reading
1. All The Reactions of Ethers In One Place (har!)
I’ve been looking forward to today’s post for a long time! We’ve gone through so many different ways of synthesizing ethers, and finally we get to talk about all the exciting things we get to do with them.
Here it is, the moment you’ve been waiting for. All the reactions of ethers in one place:
We have now covered the reactions of ethers.
Thank you for your attention.
Is that it? Yes, really: the only significant reaction of ethers you need to know…. is how to break them.
[I was just pulling your leg about the “exciting things we get to do with ethers” line.]
Does this make ethers the most boring functional group there is? Yes!!! (as long as you don’t count alkanes as a “functional group”).
So, you might ask – what’s the point?
All I’ll say for now is that there are some times when “boring is good”. Ethers, as we’ll learn later, can be useful as “protective groups” for masking (reactive) alcohols. But that’s a later discussion. [See: Protecting Groups For Alcohols]
Right now, let’s dig in to how this ether cleavage reaction works, because it actually does have its subtleties. This discussion should be pretty straightforward if you’ve been following along, however, because it’s just going to involve the familiar mechanisms of protonation, SN1 and SN2.
2. The First Step In Acidic Cleavage Of Ethers Is Protonation Of Oxygen
Neutral ethers are generally resistant to nucleophiles in substitution reactions – that’s because the leaving group would have to be RO- , which is a very strong base.
For that reason, the first step in any ether cleavage is protonation by a strong acid. Why does protonation help us? Remember that the “conjugate acid is always a better leaving group” . Protonation of the ether allows for loss of ROH as a leaving group, which is a vastly weaker base than RO- . This is going to set up our next step – cleavage of one of the C–O bonds.
The usual strong acid of choice is usually hydroiodic acid (HI). Not only is it powerful (pKa of –10), as we’ll see the iodide counter-ion plays a role as well.
3. For Methyl And Primary Ethers, The Second Step Of Ether Cleavage Is SN2
After protonation, what happens next? If we start with a primary ether like diethyl ether, we will have a good leaving group (ROH) on a primary carbon in the presence of a decent nucleophile (iodide ion). Sound familiar? It should – these are ideal conditions for an SN2 reaction. And that’s what happens.
The product will be ROH and R-I .
If an excess (2 equiv or more) of HI is present, that alcohol can be converted into an alkyl iodide through two subsequent steps (protonation / SN2).
This “SN2″ pathway will be dominant for primary and methyl ethers.
4. For Tertiary Ethers, The Second Step of Ether Cleavage Is SN1
What about a symmetrical tertiary ether like di-t-butyl ether?
Clearly the SN2 is not in play here, as the tertiary carbons are much too hindered for a backside attack. However, tertiary carbocations are relatively stable – and “ionization” (i.e. loss of a leaving group) leaves us with an alcohol (R-OH) and a tertiary carbocation, which can then be attacked by iodide ion to give R-I
Again, if excess HI is present then that alcohol will be converted into an alkyl halide. We’ll have more about that to say in a few posts actually.
5. For Secondary Ethers, The Second Step Could Proceed Through A Mixture Of Either Pathway
What about secondary ethers? I don’t have a good answer. SN1 and SN2 is a continuum. You’ll likely have a mixture of SN2 and SN1 pathways operating. If someone tells you they can look at an ether like di-isopropyl ether and the SN2 or SN1 pathway will be 100% dominant, that’s just not true.
6. The Mechanism For The Cleavage Of Unsymmetrical Ethers Is Hard To Generalize (With One Exception!)
Just as tricky as the case of secondary ethers is the case of “mixed” ethers. What if you have two different groups attached to the oxygen (“unsymmetrical ethers”). 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? The answer is very clear and it goes 100% one way. See if you can do it.
Click to Flip
If you’re curious about the answer, you’ll have to leave a comment.
7. Summary: Acidic Cleavage of Ethers
OK. So ethers, as we’ve talked about them so far, ARE pretty boring. But (and there’s always a but) – there IS a special class of ethers which is, in fact, very interesting and very reactive. If you’ve covered alkenes, you’ve seen them before – but under a different name. Can you guess what functional group I’m talking about ? Next post!
Next Post – Epoxides, The Outlier Of The Ether Family
Notes
(Advanced) References and Further Reading
Ethers are widely inert to a lot of conditions, and thus find common use as solvents (e.g. diethyl ether, THF (tetrahydrofuran), dioxane, glyme, and others). Ether cleavage generally requires strong acid and heat, which are forcing conditions. Alternatively, silane reagents can be used, which are reactive at room temperature.
- A NEW METHOD FOR THE PREPARATION OF ORGANIC IODIDES
HERMAN STONE and HAROLD SHECHTER
The Journal of Organic Chemistry 1950, 15 (3), 491-495
DOI: 1021/jo01149a008
Instead of using HI, which is expensive, one can use the combination of phosphoric acid + KI for ether cleavage, which generates HI in situ. - 1,4-DIIODOBUTANE
Herman Stone and Harold Shechter
Org. Synth. 1950, 30, 33
DOI: 10.15227/orgsyn.030.0033
This procedure from Organic Syntheses, a reliable source of independently tested synthetic organic laboratory procedures, demonstrates the cleavage of THF with refluxing strong acid. - The Cleavage of Ethers by Hydrogen Bromide
Robert L. Burwell and Milton E. Fuller
Journal of the American Chemical Society 1957, 79 (9), 2332-2336
DOI: 10.1021/ja01566a085
Classic paper on the cleavage of ethers with HBr.Below are a variety of papers using silane-based reagents for ether cleavage. The Nobel Laureate late Prof. George Olah did a lot of work in this area in the middle of his career. - Cleavage of Esters and Ethers with Iodotrimethylsilane
Tse‐Lok Ho Prof. Dr. George A. Olah
Angew. Chem. Int. Ed. 1976, 15 (12), 774-775
DOI: 10.1002/anie.197607741 - Synthetic methods and reactions. 62. Transformations with chlorotrimethylsilane/sodium iodide, a convenient in situ iodotrimethylsilane reagent
George A. Olah, Subhash C. Narang, B. G. Balaram Gupta, and Ripudaman Malhotra
The Journal of Organic Chemistry 1979, 44 (8), 1247-1251
DOI: 10.1021/jo01322a012 - Trichloro(methyl)silane/Sodium Iodide, A New Regioselective Reagent for the Cleavage of Ethers
George A. Olah, Altaf Husain, B. G. Balaram Gupta, Subhash C. Narang
Angew Chem. Int. Ed. 1981, 20 (8), 690-691
DOI: 10.1002/anie.198106901 - Silane/iodine-based cleavage of esters and ethers under neutral conditions
Ho, T-L., Olah, G. A.
Proc. Natl. Acad. Sci. USA. 1978 Jan; 75 (1):4-6.
DOI: 10.1073/pnas.75.1.4 - Synthetic Methods and Reactions; 32: Mild and Effective Cleavage of Esters and Ethers with Phenyltrimethylsilane/Iodine Reagent
G. A., Ho, T-L.
Synthesis 1977, 417
DOI: 10.1055/s-1977-24423 - Mild cleavage of methoxymethyl (MOM) ethers with trimethylsilyl bromide
Stephen Hanessian, Daniel Delorme, Yves Dufresne
Tetrahedron Lett. 1984, 25 (24), 2515-2518
DOI: 10.1016/S0040-4039(01)81219-3
MOM ethers are commonly used as protecting groups for -OH in organic synthesis, and so strategies for selective deprotection of MOM ethers under mild cleavage are invaluable. - Synthetic Methods and Reactions; 951. Ceric Ammonium Nitrate-Catalyzed Oxidative Cleavage of Alkyl and Silyl Ethers with Sodium Bromate
George A. Olah, B. G. Balaram Gupta, Alexander P. Fung
Synthesis 1980; 1980 (11): 897-898
DOI: 10.1055/s-1980-29258
This does not use silane reagents, but is still an interesting reagent for ether cleavage.
If there is a conjugated system ie, (CH3)2-C=CH-O-CH=CH2 how will we decide where should be carbocation, and how will double bond get affected?
Sir can we consider the solvent in the case of bond breaking in unsymmetrical ethers???
To identify sn1 or sn2
It is difficult to predict with unsymmetrical di-alkyl ethers. What often happens is that excess H-I is used such that both alkyl fragments are converted into alkyl iodides.
It is easier with aryl alkyl ethers, which will never result in cleavage of the aryl-O bond
If we take isopropyl methyl ether with HI,then what would be the major product? Please tell🙏
Hard to predict from first principles without doing the experiment.
However, one way to know for sure is to use a large excess of HI, which will give you both isopropyl iodide and methyl iodide.
Phenol + Ch3I
We shall obtain phenol and methyliodide if I’m not wrong.
Correct.
what does the reaction of 1,4-dioxane and 1,3-dioxane in dilute acid?
1,3-dioxane is an acetal, and should cleave in mild acid to give formaldehyde and 1,3-propanediol. 1,4-dioxane is not an acetal (it is an ether) and requires harsh acid to cleave (e.g. HI)
what does the reaction of 1,4 dioxane with excess HI give?
Replace each C-O bond with a C-I bond and you get the idea.
In part 4, “Again, if excess HI is present then that alcohol will be converted into an alcohol.”
it is alkyl halide
Fixed. Thank you!
So why are MOM protecting groups typically removed with HCl?
MOM is not an ether, it’s an acetal. Much easier to cleave. https://www.organic-chemistry.org/protectivegroups/hydroxyl/mom-ethers.htm
can ethers react with hcl?
Generally, HCl will be poorer at cleaving ethers than HI.
What if we proceed hydrolysis of methyl vinyl ether in acidic medium. What I wish to know is should we protonate ether first or hydrolyze alkene??
Thanks for your awesome work
Vinyl ethers are a special case.
The first step is protonation of the alkene, which is unusually reactive due to the presence of pi-donating oxygen. The second step is addition of water, which forms a hemiacetal and is then easily hydrolyzed.
Sir if conc HI is used what would be the change if any?
Any alcohols present would likely be converted to alkyl halides.
Phenol and methyliodide
Is there a difference when aqueous HI and anhydrous HI + ether is used ?
In aqueous HI the strongest acid possible is H3O+, (pka of about -3) whereas in anhydrous HI the strongest acid possible is HI itself (pKa -10). So anhydrous HI is much more powerful.
Answer to the quiz problem is phenol +CH3I
Yes.
Still not too sure how this works
Wouldn’t the methyl/ethyl carbocation react back with the newly formed compound to give back the original compound as they both are excellent bases?
Isn’t iodide a weaker base than water which would make it a better leaving group?
Also, how does iodide being both a good nucleophile (polarizable) and a good leaving group (polarizable) add up? Which take precedence?
In the paragraph under “Case #2 – Tertiary Ethers”, I believe “Again, if excess HI is present then that alcohol will be converted into an alcohol” should read “alcohol will be converted into an alkyl iodide” instead.
Don’t say that this is the one and only possible reaction to an organometallic chemist!
Good article, thanks.
Ha. Well, it’s the one key reaction undergrads learn in introductory courses.
Yeah, sure. To advance that on slightly, you can discuss the mechanism of deprotection of methoxy groups by BBr3, being not far removed from that which you have written about.
True!
The article is great. Thank you for sharing. Hope to hear more from you.
First step is the protonation of the oxygen, leaving a positive charge on the latter.
For the second step there are certainly two possible ways:
– A Sn1 reaction with Phenol or Methanol as leaving group, forming either a rather unstable methyl cation or a very unstable benzene cation, respectivley. Both ways seem to be not as reasonable as the second way…
– A Sn2 reaction. After Protonation of the oxygen, there is a nucleophilic attack of the iodide to the methylcarbon, generating Phenol as leaving group and Methyl-Iodide.
Another possible consideration would be the nucleophilic attack of iodide at the rather electron-rich aromatic carbon connected to the oxygen. That wouldn´t be a reasonable mechanism.
Right on the money!
Why is the benzylic cation unstable? I thought it would be stable because of resonance.
I thought the phenylmethylether would react through an SN1 mechanism so that the phenyl gets the I and the methyl gets the OH
Ah – what you call the “benzylic” carbon is actually the phenyl cation (C6H5+). The p orbital of the carbocation is in the same plane as the hydrogens in the benzene ring – perpendicular to the pi bonds which would be able to stabilize it through resonance.
What we term the “benzylic” carbon is actually C6H5CH2+ . This has a carbocation adjacent to a phenyl ring, which CAN be stabilized by resonance.
I should write a post on this common source of confusion.
Actually, you must since it is a major confusion between people about benzyl and phenyl groups
Yes, it’s just one of those things you have to get used to and move on.
Yeah who decided that “benzene” is C6H6 yet C6H5 is a “phenyl” group and that a “benzyl” group refers to benzene with a methyl attached to the ring? I think it is a source of confusion for many who are new to O-chem.
Yes, absolutely!
I still am unsure why it is assumed in the Sn2 reaction that the methylcarbon (and not the ring) will be attacked by the iodide? Is this something to do simply with the characteristic of the Sn2 reaction that I’m missing? I would think that a carbocation could be formed in either ‘direction’, and therefore the iodide could attack either side.
It’s important to recall the factors affecting carbocation stability that you likely learned during SN1/SN2. Carbocations increase in stability with substitution of carbon, so tertiary carbocations are far more stable than methyl carbocation. The SN2 is sensitive to steric hindrance since the nucleophile must do a backside attack, so methyl (and primary) is much faster than tertiary.
Bottom line is that the methyl carbocation will not form, and an SN2 will dominate and the fastest SN2 will occur at the methyl carbon. An sn2 is not possible on the aromatic ring because the empty sigma-star orbital (where the nucleophile would need to attack) is buried in the middle of the aromatic ring and is therefore inaccessible).
Another informative article. About the question at the end, will HI add to anisole to give Iodobenzene?
Think about the SN2 reaction. It has to go through a backside attack. Is one SN2 more favoured than the other?
One of the products will surely be phenol…cause once phenol is produced, it will not be converted to the iodo compound :)
Phenol is indeed one of the products.
I’d say phenol is formed because a phenyl carbocation or a phenyl carbocation character would be much more unstable than methyl carbo cation. Thus phenol and methyl iodide are formed.
It would not go through a methyl cation, it would go via SN2.