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Alcohols, Epoxides and Ethers

By James Ashenhurst

Cleavage Of Ethers With Acid

Last updated: June 17th, 2019 |

Acidic Cleavage of Ethers Can Proceed Through an SN2 or SN1 Mechanism, Depending On The Structure

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.

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. Summary: Acidic Cleavage of Ethers

1. All The Reactions of Ethers In One Place (har!)

Here it is, the moment you’ve been waiting for. All the reactions of ethers in one place:

1-overview

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.

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 .

2-primary

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).

3-primary_mech

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

5-tertiary

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. 

6-questionable

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.

4-question

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

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Comments

Comment section

40 thoughts on “Cleavage Of Ethers With Acid

  1. 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.

  2. 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. 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).

      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.

  3. 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.

  4. 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?

  5. 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?

    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.

  6. 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.

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

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