Alkene Addition Pattern #2: The “Three-Membered Ring” Pathway
Last updated: July 7th, 2019 |
The “Three-Membered Ring” Pathway In Alkene Mechanisms: Halogenation, Oxymercuration, Halohydrin Formation, and Acidic Epoxide Opening
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”.
In this post we try to show that bromination is but one example of a whole family of reactions in introductory organic chemistry that pass through a positively charged three membered ring, including not just halogenation, but oxymercuration, halohydrin and haloether formation, and even addition to protonated epoxides.
Since all of these reactions follow the same pattern, if you learn any one of these mechanisms, you’ve essentially learned them all!
Table of Contents
- Bromination of Alkenes: The Mechanism
- Chlorination of Alkenes: Mechanism
- Iodination of Alkenes: Mechanism
- Chlorohydrin Formation: Mechanism
- Haloether Formation: Mechanism
- Chlorohydrin Formation With NBS: Mechanism
- Oxymercuration Of Alkenes: Mechanism
- Oxymercuration: The Reduction Step
- A Non-Obvious Cousin Of Halonium Ions: Protonated Epoxides
- Summary: The Key Pattern Of The “Three-Membered Ring Pathway”
This is just a review of what we saw in the previous post. Treating an alkene with Br2 results in a vicinal dibromide with the two bromines oriented anti to each other. The key intermediate is a “bromonium ion”, which contains a positively charged 3-membered ring.
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!
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”]
This is also the case for iodination reactions, which proceed through the “iodonium ion”:
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.
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.
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.
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.
Do some of the images in this post look repetitive? They should!
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]