Alcohols, Epoxides and Ethers

By James Ashenhurst

Cleavage Of Ethers With Acid

Last updated: March 26th, 2024 |

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.

summary of methods for ether formation from alkenes with strong acid and alcohols or via oxymercuration

Table of Contents

  1. All The Reactions Of Ethers In Once Place (har)
  2. The First Step In Ether Cleavage Is Protonation
  3. For Methyl and Primary Ethers, The Second Step of Ether Cleavage is SN2
  4. For Tertiary Ethers, The Second Step of Ether Cleavage is SN1
  5. For Secondary  Ethers, The Second Step Could Occur Through a Mixture of Either Pathway
  6. The Mechanism for Cleavage of Unsymmetrical  Ethers Is Hard To Generalize (With One Exception!)
  7. Intramolecular Cleavage of Ethers
  8. Summary: Acidic Cleavage of Ethers
  9. (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:

overview reactions of ethers with acid ether cleavage using strong acid hi leads to roh and ri

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 .

ether cleavage with hi diethyl ether gives ethanol and ethyl iodide ethanol forms ethyl iodide with excess hi

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).

mechanism for ether cleavage of diethyl ether uses hi protonate ether oxygen sn2 by iodide then second iteration with hi

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

cleavage of tertiary ethers occurs through sn1 pathway protonation curved arrow mechanism formation of carbocation giving alkyl iodide

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. 

question is what happens with tert butyl methyl ether with one equivalent of hi is it sn2 or sn1 hard to say

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

7. Intramolecular Cleavage of Ethers

All right. We’ve seen a few examples of ether cleavage with acid.

Now let’s take tetrahydrofuran (THF), a cyclic ether. And now let’s add strong acid to it. What happens?

what is product when tetrahydrofuran is broken up using h i to give new product intramolecular cleavage of ether

Don’t panic! It’s the same sequence of bonds that form and break as before.

First, we protonate the oxygen to give the conjugate acid, which is now a better leaving group. Second, iodide ion then attacks the carbon, forming C–I and breaking O–C  [SN2 in this case].

mechanism for cleavage of tetrahydrofuran thf with strong acid hi to give alcohol iodide further reaction with hi will give diiodide

It’s important to note that the bonds that form and break in intramolecular processes (rings forming or breaking) are fundamentally no different than those that form and break in intermolecular processes.

8. 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


(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.

    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.
    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.
  3. 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.
  4. 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
  5. 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
  6. 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
  7. 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
  8. 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
  9. 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.
  10. 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.


Comment section

53 thoughts on “Cleavage Of Ethers With Acid

  1. 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?

    1. 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

    1. 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.

    1. 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)

  2. In part 4, “Again, if excess HI is present then that alcohol will be converted into an alcohol.”
    it is alkyl halide

  3. 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

    1. 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.

    1. 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.

  4. 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?

  5. 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?

  6. 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.

  7. Don’t say that this is the one and only possible reaction to an organometallic chemist!
    Good article, thanks.

      1. 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.

  8. 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.

      1. 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

        1. 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.

          1. Actually, you must since it is a major confusion between people about benzyl and phenyl groups

          2. 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.

      2. 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.

        1. 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).

  9. One of the products will surely be phenol…cause once phenol is produced, it will not be converted to the iodo compound :)

    1. 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.

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