Opening of Epoxides With Acid

by James

in Alcohols

Here’s what we’re covering in this post.

1-summary

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 ).

2-trans diol labelled

[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.

5-mechanism

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:

3-halonium

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:

4b-bromination

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]

Now – we’ve seen that this works with aqueous acid [H3O+]. Can we extend this to other nucleophiles? Sure! With some reservations that we’ll get to in a second.

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.

6-other nucleophiles

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.

6b caveat

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.

7-next post

We’ll talk about this in the next post.

Next Post – Opening Of Epoxides With Base


Note 1: Why does the nucleophile attack the more substituted carbon?

  • 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.

4-positive charge

These diagrams by Matt McIntosh in the same context are very helpful. [back to discussion]

 

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{ 9 comments… read them below or add one }

Kureshii

Hello James-san
A small question: when opening epoxides with acid, the first step is protonation, then the attack. But if I added excess acid ( like HI or HBr) to an unopened epoxide, would I observe the usual opening of epoxide? Can the final product ( an opened epoxide ) get protonated further by the acid?

( thanks loads for the cool site )

Reply

James

Addition of HBr or HI to alcohols will, eventually, convert them to alkyl halides, so yes, if you opened the epoxide to get a halohydrin, excess HBr or HI would eventually convert the R-OH to R-X where X is Br or I. Not sure why you’d want to do that but it’s doable.

Reply

Sophia

Hi James,
What if the epoxide was treated in excess water after the addition of the Br from HBr? Could the excess water convert the Br to R-OH?

Reply

James

With water, not likely. You’d need a stronger nucleophile. If it was slightly basic, then some hydroxide ions could add as a nucleophile.

Reply

Spencer

Your comparison to opening an epoxide under basic conditions was very helpful. My textbook did a poor job at explaining the difference in the product. Thank you.

Reply

James

Happy to help!

Reply

Jay

hi james,

i really love your blog. its helped me a lot.
just had a question,

u have said that the charge on teriary carbon will be most,
whats wrong in saying that the positive charge on the tertiary carbon will be reduced due to the inductive effect of the alkyl groups..?and so it will have the least positive charge..

Reply

Idan

This has confused me too ever since I learned about it..
How does the tertiary carbon that is stabilized through inductive effects (by electron donating groups) have less electron density than the secondary carbon?
Shouldn’t it be the other way around?

Reply

Idan

Just pointing it out, the article refers to Note #2 which doesn’t exist. Also, the link at the bottom is broken.
Would be great if someone could answer how come the carbon that gets more electron density from its surrounding groups (thus better able to stabilize positive charge) is more positive (even though it gets more electron density).

Reply

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