Opening Of Epoxide With Base

by James

in Alcohols

Opening of epoxides with base

a summary of what we talk about today:
1-summary

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.

2-first

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.

3-ROH

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.

7-equilibrium

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.

8-equilibrium 2

Bye-bye nucleophile!

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- ).

4-basic nucleophiles

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).

6-what doesnt work

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.

5-note at bottom

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{ 7 comments… read them below or add one }

SS1998

Excellent post!
I’ve a doubt though. What if I tried to add a tertiary alcohol to an asymmetrical epoxide. Since they are the most weakly acidic alcohols, would they follow a basic mechanism (sn2)? would the outcome depend on the solvent or medium used? Would they even react in the first place? If they did, would they prefer to undergo a C-O bond cleavage thereby forming a stable carbocation? I’m really confused on this, it would be a lot of help if I could get this cleared…
Thanks

Reply

James

This post covers opening epoxides with base. Adding a neutral tertiary alcohol does not fall into this category. You’d need to form the conjugate base of the tertiary alcohol – say, for example, sodium t-butoxide. Sodium t-butoxide can act as a nucleophile to attack the least substituted position of an epoxide and open it, just like other alkoxides do.

Carbocations are not relevant to this discussion because they don’t form under these basic conditions.

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SS1998

Would the nucleophile be RO- or OH- ?

Reply

James

In which case? It could be either, depending on conditions.

Reply

Li

This is probably a stupid question. From my understanding, for an epoxide, the more substituted carbon would bear more positive charge. Why wouldn’t the nucleophile attack the more substituted carbon instead? Is it something has to do with steric hindrance?

Reply

James

Hi – not a stupid question at all. You are right that the more substituted carbon would bear more positive charge. However, when we have a relatively strong nucleophile (such as RO- , HO-, Grignard reagents, or LiAlH4) what turns out to be more important is steric hindrance.

Think of it like an SN2 reaction, where the least substituted carbon will react most quickly.

When a weaker nucleophile is involved (such as H2O) we need to add an acid catalyst. In this case, the reaction will primarily occur on the carbon best able to stabilize positive charge. In this respect it is somewhat like an SN1 reaction, although it still occurs with inversion of configuration in this case.

Hope this helps! James

Reply

Stephen

Typo in “The Role of Solvent” first example… The first equation should have NaOH and not NaOR.

Reply

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