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Alcohols, Epoxides and Ethers

By James Ashenhurst

Opening of Epoxides With Acid

Last updated: September 13th, 2019 |

Opening Epoxides With Aqueous Acid

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


Table of Contents

  1. Formation of trans-Diols Through Opening of Epoxides With Aqueous Acid
  2. The Mechanism For Opening of trans-Diols With Acid Is Similar To Opening Halonium Ions
  3. Other Nucleophiles: Alcohols and HX
  4. Acid Only Helps The Reaction If The Nucleophile Is Compatible With Strong Acid
  5. What About Opening of Epoxides With Base?
  6. Notes

1. Formation of trans-Diols Through Opening of  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.


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!

2. The Mechanism for Opening Epoxides With  Acid Is Just Like Opening Halonium and Mercurinium Ions

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]

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.

3. What About Other Nucleophiles? Alcohols And HX Also Work

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

4. When Doesn’t Acid Help? When The Nucleophile Is Not Compatible With Strong Acid

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

5. What About Opening Under Basic Conditions?

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]


Related Posts:


Comment section

29 thoughts on “Opening of Epoxides With Acid

  1. 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 )

    1. 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.

      1. 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?

  2. 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.

  3. 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..

    1. The inductive effects of neighboring alkyl groups only stabilize the positive charge by sharing their electrons through sigma bonds by induction. They do not change the polarity of the affected carbon, only stabilize it. I think Dr. James used a good analogy with the poor man surrounded by rich neighbors who can ‘tolerate’ being poor because the rich neighbors are willing to share what they got. This does not make the poor man rich but better than a poor man with poor neighbors.

    2. 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?

  4. I want to know i want to introduce other group in epoxy like nitro group so what would i do first to open ring and introduce or add when epoxy is make?

  5. 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).

    1. This is my sleep-deprived guesswork and probably wrong, but I’ll put forward some possible factors revolving around the intermediate trying to lower it’s free-energy. Remember that a carbon bonded to a highly electronegative atom is going to be partially positive even if it had THREE alkyl groups attached. Note: when I say carbons in this post I’m referring to the annular atoms as opposed to the methyl substituent.
      Realize this is essentially the same dilemma and reasoning involved in another three-membered ring pathway: Opening an epoxide with Br2 & H2O to form a halohydrin. Keep in mind the drawings don’t capture that the ring bends side to side, or you can think of the electroneg. atom able to tilt towards one carbon or the other, or as resonance structures. There are pictures of this in Klein’s First Semester, 3rd ed. p.290 and here , here ,
      and here
      The intermediate is VERY unstable. The Bromine, or Oxygen in this case, and carbons involved got suckered into this situation, and they have to deal with it the best they can. They have no idea they will be relieved by Nuc attack in the future; they just want to be as stable as possible at the moment. Which means dealing with the ring strain, steric hindrance, and key to understanding here: The positive charge; it’s gotta go SOMEWHERE! And the O will be G*d*mned if it’s gonna be the one to bear it.
      *Mnemonically, OJ doesn’t want a positive charge to stick on him, so he tries whatever he can until they find “The real killer” (Nuc attacker), including leaning on Johnnie Cochran (Carbon). Lots of possibilities for those old enough to remember. (Kato Kaelin ~ Carbocation…eh?)

      1) Think of the Br or O as a suddenly trapped frightened animal whose panicking instinct is to do what it knows, which is to completely go to only one carbon or the other to form a carbocation–a structure more stable with the positive charge at the more substituted carbon. This would alter the more substituted carbon from sp3 to sp2, and from tertiary to secondary, making an SN2 attack possible there…Imagine O tries to do this with its burly arms holding on to each carbon’s electrons the whole time. As O leans and stretches to one side, it pulls the electron density with it (another reason it would be bad if O were instead pulling it towards the carbon with the most original/prior electron density). But as strong as O pulls, this ring…can’t…quite…be broken. Still, it’s enough to render SOME sp2 and secondary character to the more substituted carbon. Though it’s not the desired carbocation, this causes the more substituted carbon to gain just enough of this character and just enough partial positive charge to allow for it to be the electrophilic site (especially as its electrons get more and more repelled by the approaching electrons of the Nuc as it gets closer and closer).

      2) If we’re focusing only on the intermediate here, then maybe steric hindrance between the lone pairs of O and the methyl substituent…?

      3) Maybe moving towards one side will decrease the angle of O further, thereby lessening electron-electron repulsion between it’s lone pairs and bonds (with the further decrease from ideal sp3 angle balanced by the increase in the angle of the less substituted carbon)…?

  6. If an amine is reacted with an epoxy in the presence of acid in an aqueous system, the acid can protonate the amine lone pair nucleophile and actually slow the reaction between the amine and epoxy. How do you account for this phenomenon per your illustrations, in which the acid is shown to accelerate the reaction?

    1. Amines are at least 10 orders of magnitude more basic than epoxides, so any acid added to the system will protonate the amine, not the epoxide. Protonating the amine makes it into an ammonium salt, which is non-nucleophilic. That’s why the reaction slows down. Acid slows down the rate of addition of Grignards to epoxides too.

    1. OK, so what’s the strongest base? The epoxide oxygen. So protonate the oxygen, forming a new O-H bond and breaking the H-Cl bond. Now you have a protonated epoxide with a Cl- floating around. Cl- can be a good nucleophile. So what happens next?

  7. Hello, what if you have a cyclohexane ring with a epoxide on it with a methyl groupattached to one of the carbons bonded to the epoxide and is reacting with a Acetylide anion and water. Will there be stereochemistry even though the acetylide anion is linear.

    1. The acetylide ion is a basic nucleophile and will attack the least substituted carbon in SN2 fashion. The water is added after the reaction is over to quench the alkoxide anion.

  8. Dear James,

    Under Note 1, you state that the nucleophile attacks the MORE substituted carbon. In your examples, you show a tertiary carbon being attacked in the protonated epoxide. How is this possible? I have always learned that a tertiary substrate cannot undergo an SN2 reaction due to steric hindrance.

    In fact, my textbook states that if you react an epoxide with Grignard or organolithium reagents, the reagents would attack the LESS hindered epoxide carbon atom.

  9. Suppose if grignard reagent is first treated with epoxycyclohexane and than protonated.
    What will be the mechanism..

  10. Bro grignard reagents and Gilman reagents (organolithium reagents) give R- which is a base not acid. Bases with epoxides go through sn2

    1. Start drawing resonance forms. You will quickly find a species which is a good nucleophile on carbon. That will tell you where it will react with an electrophile, and from there you can figure out the product.

  11. Hi
    Are carbocation rearrangements possible in acidic opening of epoxides? Or is the carbocation never really formed, just a direct attack on the cyclic intermediate?

    1. No, they wouldn’t be carbocation rearrangements, in acidic opening of epoxides, because there likely won’t be a free carbocation.

      However, there are pinacol-related rearrangements that can occur with epoxides when acid is added in the absence of nucleophiles, such as the epoxide-aldehyde rearrangement.

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