Alcohols, Epoxides and Ethers
By James Ashenhurst
Opening of Epoxides With Acid
Last updated: March 26th, 2019
Here’s what we’re covering in this post.
Opening Epoxides With Aqueous Acid
In the last post, we saw some examples of how epoxides are considerably more reactive towards breakage than are ordinary ethers. For example, aqueous acid [often abbreviated “H3O+”] will open an epoxide under MUCH milder conditions than an “ordinary” ether such as diethyl ether, because epoxides have considerable ring strain [about 13 kcal/mol].
Looking more closely at the reaction, we also noted two interesting patterns:
- the nucleophile attacks at the “more substituted” position of the epoxide (C-1, below)
- inversion of stereochemistry occurs at this position, but not at the other position (note that the C-O bond at C-2 below is a “wedge” in both starting material and product ).
[By the way, how do we “know” that the OH on C-1 is from the nucleophile and is not the epoxide? Using isotopic labels is one way. Another is to use nucleophiles other than water – see below]
It should be noted that in the absence of acid, no reaction occurs. So clearly the H+ plays a key role.
What could be going on?
By analogy to the reaction of ethers with acid, the first step must be reaction of the most basic site on the molecule – the epoxide oxygen – with acid, giving us a protonated epoxide. This will function as a much better leaving group than does the unprotonated epoxide. [the conjugate acid is always a better leaving group]
The next step must then be reaction of the best nucleophile present in solution – H2O, in this case – with our protonated epoxide. And this occurs at the most substituted position, always with inversion of stereochemistry. So it must be performing a “backside attack” at this carbon, as we observe in SN2 reactions. A final deprotonation gives us the neutral product.
Hold on for a second. If you remember the key lesson of the SN2 – that it is disfavoured by steric hindrance – this might seem weird. If this was a “pure” SN2, reaction, wouldn’t we expect the attack to occur as the “least substituted” position?
Clearly something else must be going on here!
We’ve Seen This Before!
Thankfully, you’ve likely encountered reactions like this before! If you think back to the chapter on alkenes, you might see that the protonated epoxide bears an uncanny resemblance to two other reactive intermediates you met in that chapter: “halonium” ions, and “mercurinium” ions, both 3-membered rings bearing a positive charge:
If you think back to how these species reacted with nucleophiles, it was always at the more substituted position with inversion of stereochemistry. In fact, there is a whole family of alkene addition reactions that proceed this mechanism that we called the “3-membered ring pathway“. Halohydrin formation is a perfect example:
So in essence, the addition of nucleophiles to protonated epoxides is just another example of the “3 membered ring pathway” of alkenes!
[Need a review on why the nucleophile attacks the most substituted carbon? See this note below and then come back]
What About Other Nucleophiles?
Changing the solvent from water to an alcohol will result in the alcohol adding instead. For example if we were to use CH3OH as solvent instead of water, then our product would contain OCH3 joined to the most substituted position.
Hydrohalic acids [HCl, HBr, and HI] can also work well, forming halohydrins.
When Doesn’t Acid Help Us?
Now, you might think – if epoxides are made more reactive by treating with acid, then can’t we extend this to other nucleophiles too? For example, what about NaOH, or NaNH2, or even Grignard reagents?
Herein lies the dilemma. Acidic conditions are only compatible with nucleophiles that are protonated reversibly. [in other words, nucleophiles whose conjugate acids are strong acids – think pKa < 0 ]. [Note #2]
Can you see a little problem with adding NaOH to a solution of aqueous acid? What do you think might happen?
Kaboom. Well, that’s an exaggeration. But the acid will protonate NaOH irreversibly, giving us H2O [recall that acid-base reactions are fast]. Similarly, you can imagine what happens on adding NaNH2 to acid or Grignard reagents to acid: the nucleophile is protonated, giving us the conjugate acid.
One Last Mystery
There’s still one mystery to solve. From the last post you might recall that if we just add NaOH – no acid – to the epoxide we met above, we get a different product altogether.
Note how the stereochemistry at C-2 is completely different than with acid.
What might be happening here? Any thoughts? Hint – it’s a reaction we’ve talked about before, and even mentioned in this post.
We’ll talk about this in the next post.
Next Post – Opening Of Epoxides With Base
- In our protonated epoxide, although oxygen bears a positive formal charge, in reality positive charge density mostly resides on carbon [recall that oxygen is more electronegative than carbon].
- Recall that positive charge is best stabilized by carbon in the order tertiary > secondary > primary. So in our case, the tertiary carbon atom will bear more positive charge. The tertiary carbon will be more electron-poor (electrophilic)
- The length of the C-O bonds will NOT be equal – the C-O bond to the tertiary carbon is longer and weaker than that of the secondary carbon.
Bottom line: the tertiary carbon is more electrophilic (electron poor) and the C-O bond on the tertiary carbon is weaker, longer, and easier to break.