Cleavage of Ethers & Ring opening of epoxides

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

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.

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

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.

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 .

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.

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.

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?