Ethers are interesting organic compounds unreactive in non-acidic environment (the only one exception are epoxides a particular class of ethers) and towards nucleophilic species; for these reasons and the possibility to use them as aprotic solvents (they are good in solvate cations) they are quite common in organic chemistry.
Even in acidic environment though is not that easy to make them undergo a nucleophilic substitution. Indeed if we added hydrochloric acid (HCl) or hydrobromic acid (HBr) to an ether solution we wouldn't get the expected product. This is because strong acids ( as we saw in the article nucleophilic substitution on alcohols) they usually have a rather unreactive counter-ion and then little nucleophilic. The species Cl- , Br-, HSO4- etc. are very little reactive in acqueous solution. There's just one valid exception that is the hydrogen iodide (HI). In spite of is even stronger than HCl and HBr, the species coming from its dissociation, I- is a good nucleophile.
I- is under chlorine and bromine in the periodic table of elements (same group), is therefore a quite large atom with a wide and polarizable electron cloud. The electrons shielding effect as a consequence that electrons belonging to outer orbitals are quite far from the nucleus and also little attracted; this lead to their possibility to react with electrophilic centers. In terms of energy this means that outer orbitals and consequently their electrons are higher in energy and therefore reactive enough.
Let's have a look to what happens when we expose an ether to hydrogen iodide:
Reaction mechanism
1) In the first step the hydrogen iodide, that's a strong acid completely dissociates in acqueous solution releases a proton accepted from the ether: the cations is called oxonium ion.
2) We can write for the protonated ether different resonace formulas. As we saw in stability of carbocations, the more stable resonance formula is the one where the positive charge is on the heteroatom (oxigen in this case). For this reason usually the protonated ether loses the proton (H+) and gets back to the previous "configuration".
But when the protonated ether is in the secondary carbocation form it can be attacked by the nucleophile I- to give the substitution product (the primary carbocation, too unstable, is the structure less contributing to the description of the resonance hybrid):
3) The product is a iodine-alcohol (yields are not that excellents). This product can be used as substrate for different kinds of reactions. For example the hydroxyl can be oxidated while the iodide is a good leaving group and can be easily substitude by whatever kind of nucleophlic species.
We have to point that the reaction actually follows the route we talked about, in other words its evolution follows the orientation given by the carbocation but isn't really neither an SN1 nor an SN2 reaction (as the addition of bromine Br2 to the double bond), and on the most of substrates the carbocations doesn't actually form.
Other examples
- A class of ethers whose reactivity is particularly well known (compounds quite spread in nature) is that of phenyl - methoxy ethers (on interesting compound belonging to this classi is "vanillin"). In this reaction the hydrogen iodide is obligated to follow an SN2 mechanism, considering that neither an aromatic carbocation nor a methyl carbocation can actually form ( the methyl carbocation doesn't exist for normal conditions and structures because is too much unstable, while an "aromatic carbocation" isn't possible because that would mean to have a positive charge on an sp2 orbital and consequently an unstable and asymmetric carbocation).
- Another border-line case is the one that contemplates the competition between the benzyl carbocation and the tertiary carbocation:
Theoretically the benzyl carbocation, thanks the delocalization on the aromatic ring is more stable than the tertiary; indeed the iodide attacks the carbon adjacent the ring.
- Lastly, particolarly instructive is the addition of hydrogen iodide to an enol-ether;
As we have already saw the reaction can't pass through an sp2 hybridized carbon , considering that the p orbital of the carbon is occupied by an electron pair and again we would have a positive charge on one ot the three sp2 but this is definitely impossible , too high in energy.
