Bromination of a highly electron-rich alkene such as 2-methoxy...

Question

Bromination of a highly electron-rich alkene such as 2-methoxybut-2-ene has been shown to produce approximately equal mixtures of the trans- and cis-dibromide. Suggest an explanation for this observation.

Bromination reaction of 2-methoxybut-2-ene producing trans- and cis-dibromide products.

Accepted Answer

All right. Hi, everyone. So this question says that when a highly electron rich alkene such as the given four methoxylated is brominated, trans and cysty bromides are produced in approximately equal amounts. Explain this observation. So what I want to point out here is that even though one product is sis and the other is trans, both are known as visional dye bromide, right? Because there are two atoms of bro means on adjacent carbons. Hence the term vicinal and the formation of a vicinal di bromide means that this is going to be an example of the stereos specific brominated of an alkene. But right, this reaction is interesting, right? Because if we take a generic Ain here, which I'm drawing on the screen, if I take this generic Aine and I go ahead and I react it with one equivalent of elemental bromine, then the pie bond and one of the bromine atoms are going to interact with each other at the same time. So the Python interacts with the bromine and that bromine is interacting with, with one of the sp two hybridized carbons at the same time. Thus breaking the bond between the two bromine and BR two. Now, the reason why this reaction is so interesting because the intermediate is known as this bro moi ion. And the bromium ion looks a little bit like an oxide, except that it's a three member ring composed of two carbons and one bromine, which gives bromine a positive charge. Now, in the second step, the bromide ion that was formed in the first step is going to go ahead an attack, one of the two carbons of the bramon ion on the opposite side of the brei ion itself. So as to prevent STAIC hindrance, so that at the very end is going to produce a die line, but that that hide is always going to be trance. It's always going to be tr the these bromine are always going to be on opposite sides of each other. So why then is one of our products this, well, what I also want to bring to your attention is that these two products, right, the sis and the trans product are produced in approximately equal amounts. So what that means right is that instead of having this bro monium intermediate, bro ion intermediate that has its own stereo chemistry. The intermediate in this reaction here must have been plainer, right? Because by having a plainer intermediate, then the bromide ion could attack it from essentially either side of it, which results in a mixture of CIS and trans products. Now recall that one possible plan or intermediate is a carbo car. So in this case, we can go ahead and see whether or not a carbo cion can form in this case. So to do that, I'm going to go ahead and redraw are starting material on the bottom. Here, here's my myox group. This is a pense ring and double one, right? So let's go ahead and proceed with the mechanism. So what I'm going to do here is I'm going to add elemental bromine with my four meth oxy styrene. So the first step is going to be the same because we have to establish a connection between either carbon of the alkene itself with one of our bromine in B two, right? So the pi bond instead is going to go ahead and interact with bromine and it's going to form a bond with bromine, which is therefore going to break the bond in between BR two. Now, this is what's going to result in a carbo car on intermediate because now only one carbon has a new bond to bromine, whereas the other is electron deficient. But of the two carbons here that could form a carbo carion, we have to consider which one would be more stable, right. So if we look here and we examine where this double bond used to be more closely. One of these carbons of the starting double pond is a benzo position. Its a Benzi position because it is adjacent to a benzene ring but not part of it. Now recall that Benzyl carbo CS are actually extremely stable because the positive charge here can participate in resonance with the entire benzene rink. So therefore, right, our bromine is going to be on the next carbon over which is going to lead to the formation of a Ben Zilli Carbo Kion. So now the Carbocaine itself is plainer. So when we go ahead and we introduce bromide as a nucleo file, bromide can attach itself to the carbo cion in either direction, right on either side. And that is what's going to produce the cis and trans mixture that we saw in our final product, right? So by having bromine attack, this planar carbo cion on either side, we end up with this mixture of this and trans products in approximately equal amounts. So here, right, the bottom line is that the reaction involves a highly stable carbo cion via a step wise transition state forming both cysts and trans di bromide. And there you have it. So with that being said, if you stuck around to the end, thank you so very much for watching. And I hope we found this helpful.

Bromination of a highly electron-rich alkene such as 2-methoxybut-2-ene has been shown to produce approximately equal mixtures of the trans- and cis-dibromide. Suggest an explanation for this observation.

Key Concepts

Electrophilic Addition Reactions

Stereochemistry of Alkene Reactions

Regioselectivity and Stereoselectivity

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