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Alkene Reactions

By James Ashenhurst

Alkene Addition Pattern #2: The “Three-Membered Ring” Pathway

Last updated: March 27th, 2019

The “Three-Membered Ring” Pathway In The Mechanisms of Reactions of Alkenes

In the last post we walked through a proposal for how the bromination of alkenes works and showed that it adequately explains many of the experimental observations made for this reaction. Namely, the reaction proceeds with anti addition of substituents across the alkene, and (where relevant) the reaction proceeds with Markovnikov regioselectivity. These observations are best explained through the intermediacy of a “bromonium ion”.


Taking bromination of alkenes as a starting point, we might ask: “do variants of this mechanism operate for other reactions of alkenes as well?”

The answer is yes!

Chlorination of Alkenes

Take, for example, the chlorination  of alkenes. The products of this reaction has identical patterns of stereoselectivity and regioselectivity to those of bromination. Therefore we might surmise that they proceed through the same type of reaction intermediate! [This intermediate is called the “chloronium ion”]


Iodination of Alkenes

This is also the case for iodination reactions, which proceed through the “iodonium ion”:


Chlorohydrin Formation

This also applies to reactions where the intermediate chloronium ion is trapped with solvent (water in this case). After deprotonation of R–OH+ to give R–OH,  the product is referred to as a “chlorohydrin”.


Note that bromohydrin and iodohydrin formation work exactly the same way and if you merely replace “Cl” with either of those halogen atoms you’ll obtain the indicated product.

“Haloether” Formation

As described in the last few posts, what’s notable about bromination is that by using a solvent which can act as a nucleophile, we can obtain products which incorporate that solvent. For example by using an alcohol as solvent, we obtain the following “chloroetherification” product. It likewise proceeds through the exact same mechanism described above.


Furthermore, these reaction pathways are not confined to the dihalogens Cl2, Br2, and I2 [nor F2, the Tiger of Chemistry, which is a very difficult beast to keep on its leash]. And a good thing too, since Cl2 and Br2 are to various extents vile and inconvenient to work with.

Chlorohydrin Formation Using N-Chloro Succinimide

A convenient source of “electrophilic” chlorine is the crystalline salt N-chlorosuccinimide (NCS), an innocuous appearing white crystalline solid. Alkenes react rapidly with NCS to form chloronium ions, which can then be intercepted to form a variety of useful products by analogy to those shown above. With the exception of this more convenient source of halogen, the reaction is otherwise the same. N-bromosuccinimide (NBS) and N-iodosuccinimide (NIS) likewise find use.


Moving beyond the halogens, are there other reagents that form these cyclic intermediates? Why, yes indeed.


When alkenes are treated with mercury (II) salts (such as mercuric acetate) in the presence of water or alcohols, we obtain products with the same pattern of stereochemistry and regiochemistry that we’re accustomed to seeing by now. What’s a likely intermediate here? A three-membered ring called the “mercurinium ion”.


Organomercury compounds find very little application in themselves, but can be used as intermediates in subsequent reactions. To replace mercury with hydrogen, sodium borohydride (NaBH4) is added. In this case,  rather than being “anti” , the stereochemistry of this reaction ends up being a wash: treatment with NaBH4 leads to cleavage of the C-Hg bond and formation of a free radical. The free radical can react from either face with hydrogen, leading to scrambling of the stereocenter. Mechanism link.


How many other reactions go through this type of mechanism? There is actually a sizable list. For instance, there are electrophilic sources of sulfur and selenium that can likewise form three membered cationic rings just like those we’ve seen; we won’t go into those.

A Non-Obvious Cousin: Protonated Epoxides

One last example that is worth going into is one that might not immediately seem obvious: protonated epoxides.

Treatment of an epoxide with acid leads to a positively charged intermediate that resembles a bromonium ion. As you might guess, the nucleophile attacks the backside of the most substituted carbon and the resulting product has anti stereochemistry. Just as we’ve seen numerous times above.

7-epoxide opening

The lesson for this very long post from today is that one can group together a sizable number of different reactions by identifying their common mechanism. Just as there is a family of reactions that pass through the carbocation pathway, there is likewise a “family” of reactions that pass through a three-membered ring.  Instead of learning a dozen different mechanisms, we merely learn one – and merely change the actors to suit the occasion.

NEXT POST – Hydroboration of Alkenes

P.S. Besides the dihalides, there are also such things as mixed dihalides, such as iodine monochloride. We have all the tools at our disposal to answer how this reaction might proceed. What do you think the product is?


[answer in comments]


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Comment section

20 thoughts on “Alkene Addition Pattern #2: The “Three-Membered Ring” Pathway

  1. Would you consider writing a post about the demercuration mechanism? It’s one of the mechanisms that has bugged me most since it was not taught in any of my organic classes.

    Or, if not, do you know of any papers or other sources that show it?

    Thanks a lot!

    1. Included a link – thanks for mentioning this, it was definitely something I wondered about when learning this reaction.

        1. Good point – but the drawing is correct! The drawing shows that we go from Hg in the (II) oxidation state to Hg in the (I) oxidation state. Similarly, at the end of the complete process we have Hg (0) . We *could* draw in those electrons, but generally do not. For example, when we draw FeCl3 we don’t draw in all the electrons on the Fe atom, we can figure that out by knowing where Fe is on the periodic table and that it is in the (III) oxidation state.

          1. Is there evidence of radical and/or Hg(I) intermediates in the demercuration step? I would think C-H bond forming reductive elimination from Hg(II) would be a possibility after making the mercury hydride. Do you have a link to the primary lit. that shows the scrambling of stereochem with the isotopic label?

          2. Yes, there’s lots of evidence that demercuration is a free radical process.
            Whitesides: J Am Chem Soc 1974, vol 96 p 870

            evidence – can be diverted by oxygen; also, cyclic products were observed when hex-5-enylmercury compounds were reduced with sodium borohydride [ this doesn’t occur in the presence of oxygen, indicating that trapping is much faster than cyclization]
            JACS 1976 vol 98 p. 5973

            It’s possible to demercurate w/o scrambling by using sodium amalgam.

            My source for this is Carey & Sundberg B 4th ed.

  2. why is markovnikov relevant in a few reactions above and not in others ,markovnikov is relevant when th ydrigen bonds to the most carbon with most hydrogen’s but here markovnikov is relevant even when hydrogen’s are not being bonded

    1. Also through a 3 membered intermediate, and also with anti stereochemistry. Key is to figure out which of the halogen atoms is the electrophile, which you can do by comparing their relative electronegativities.

  3. I don’t really understand how two enantiomers can be formed during an epoxide opening. Since the epoxide is already the three-membered ring, I would expect that attack of the alcohol is only possible from underneath the epoxide ring. If it would attack from the same angel as the epoxide is than the reaction is not “anti” anymore. I would therefor expect that only 1 structure can be formed from a specific epoxide.

    1. Or does it mean that the alcohol can attack both at the tertiary carbon and the secondary carbon. I wouldn’t expect this because this would be against the Markovnikov rule, is it not?

    1. The double bond attacks iodine, and chlorine is the leaving group. [it attacks iodine because iodine is more electron poor due to the more electronegative chlorine taking electron density away]. Then Cl- attacks the most substituted carbon from behind. Answer here

      1. With ICl, you actually get close to a 50:50 mixture on 1-methylcyclohexene derivatives. With BrCl, you get good selectivity for the bromine at the more substituted carbon, which suggests that the chloronium forms, then bromide acts as the nucleophile. This is explained by the fact that the Br- is a more stable leaving group, even though it is less electronegative.

    1. I disagree. That would give a nitrogen-centered radical, which in turn is not stabilized by resonance. NCS in the presence of acid can provide a source of Cl2, which itself can dissociate and be a source of Cl free radical.

  4. In the last reaction, why would the final product have Cl bonded with the tertiary carbon and not with either the tertiary or the secondary one?

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