Opening of epoxides with base
In the last post we discussed the reactions of epoxides under acidic conditions and saw how they resembled the “3-membered ring” family of alkene mechanisms. We left off by noting that the reaction of the epoxide (shown above) with NaOH in H2O gave a different product than that obtained through H3O+. Therefore it must go through a different mechanism.
Our question is, how does this reaction work?
How It Works
Noting the fact that the nucleophile (HO- ) attacks the epoxide at the least substituted position (C2) and results in inversion of stereochemistry at this position, our best evidence is consistent with this reaction proceeding through the familiar SN2 mechanism followed by a transfer of proton from the weakly acidic solvent (H2O) to the alkoxide (RO – ) providing a neutral alcohol.
Along the same lines, if epoxides are treated with alkoxides (RO- ) in alcohol solvent, they lead to essentially the same reaction. Here’s an example with NaOCH3. Note that the proton must come from CH3OH here.
The Role Of Solvent
Now – at the end of the last post, we mentioned that acids are incompatible with some strong nucleophiles because they will be destroyed by irreversible protonation [classic example: Grignard reagents and water]. Should we be worried about NaOH in H2O or NaOCH3 in CH3OH? No! These are examples of bases in equilibrium with their conjugate acids. It’s OK to have NaOH in the same reaction vessel with H2O.
However, if we move to more strongly basic nucleophiles, such as Grignard reagents (RMgX), organolithium reagents (R-Li) or hydrides (H-) we should be concerned, because the conjugate acids of each of these species (pKa of about 50 for alkanes, and about 38 for H2) are much weaker acids than ROH (pKa 16-18). [A good rule of thumb for “reversibility” of an acid base reaction is a pKa difference of 8 or less]. That means that if these strong bases come into contact with ROH, an irreversible acid-base reaction will occur. We thus obtain an alkoxide (RO-) which is insufficient in strength to deprotonate alkanes or H2.
1) Add this 2) THEN, add this!
What this means in practice is that we just move the protonation step to the end, in a process sometimes called a “quench”. Note how each of the following reactions has a Step 1) and a Step 2). Step 1 is addition of nucleophile. After the reaction is done, we add our source of proton (Step 2) to quench. Note that there are many reagents commonly written for this process – H2O, H3O+, H+, NH4Cl, “acid workup” just to name a few. They all mean the same thing!
Here are some examples of reactions of epoxides with strongly basic nucleophiles. Grignard reagents , organolithium reagents, and hydride (e.g. LiAlH4 , a source of H- ).
Is there anything else? [yes – there are other nucleophiles which will react with epoxides, but they are seldom seen at this level. However I have added them in a note at the bottom].
What Doesn’t Work
Just like any SN2 reaction, for this pathway to work, we really must be dealing with a decent nucleophile. Poor nucleophiles, especially neutral species like H2O, ROH, and RCOOH simply won’t cut it. In order for epoxides to react we need either to use strongly acidic conditions (good for weak nucleophiles) or basic conditions (HO- , RO- , RMgBr, RLi, LiAlH4).
This concludes our exploration of epoxides and ethers. However we still have a lot to talk about regarding alcohols – and certainly we aren’t done with examples of SN1 and SN2 reactions. That’s coming up next.
Next Post – Making Alkyl Halides From Alcohols
Note – some other nucleophiles which react with epoxides but are rarely seen in introductory organic chemistry. Note that the second and third examples don’t show the source of proton (H+) that adds to the oxygen, but it can be added afterwards in a “quench” step.