Anti Markovnikov Addition of Br - Online Tutor, Practice Problems & Exam Prep

Hydrohalogenation is an addition reaction where a double bond reacts with HBr, typically following Markovnikov's Rule, leading to a carbocation intermediate. However, the presence of a radical initiator, like peroxide, shifts the mechanism to a radical-mediated process, resulting in an anti-Markovnikov addition. This involves initiation, propagation, and termination steps, where the bromine attaches to the less substituted carbon, creating a unique alkyl halide. Understanding this mechanism is crucial as it highlights the significance of radical reactions in organic synthesis.

Overview of Hydrohalogention.

So now I want to show you guys how just having a radical initiator present in a reaction can completely change the expected product. And the reaction I want to talk about is called hydrohalogenation. So in case you don't know a lot about this reaction, this is an addition reaction. The mechanism of this reaction was that the double bond would grab the H, which is electrophilic, and then it would kick out the Br. So what I would wind up getting is an H on one side and a carbocation on the other. Now there was a rule to figure out where the carbocation went and that was called Markovnikov's Rule. Markovnikov's Rule said the carbocation will form in the most stable location, or the one with the most R groups. That means it went right there. Then in the next step, my Br- attacked the positive charge and what I wound up getting is what we would call a Markovnikov alkyl halide. So this whole reaction was a carbocation intermediated reaction and we said that this would be a Markovnikov addition of bromine to the double bond. Alright.

How Radical Hydrohalogenation is different from typical Hydrohalogenation.

Well, now what I want to show you is that just having a radical initiator present can completely change this reaction. The example I want to use is the same reaction, same reagents, a double bond with HBr. But now notice that there's peroxide present. Now remember that peroxide was a form of radical initiator. So what we want to do in this first step is instead of doing a carbocation mediated reaction, this is actually going to be a radical mediated reaction, which means that we're going to have to use the 3 steps of initiation, propagation, and termination to figure out what this is going to do. So let's go ahead and draw the first step which is initiation.

Now for this initiation step, this is going to be a little bit more complicated than usual just because I'm starting off with peroxides and this is actually not the radical that I want to use for my reaction. So my first step is going to be to generate my peroxide, so OR•2 equivalents of OR• radical. Okay? But then one of those OR radicals is going to react with HBr. So what that's going to do is that's going to make a radical that I can actually use in my reaction. That would be basically I would get, ROH, which is alcohol because I just got the OR attaching to the H and I would get Br• radical.

So I know that was a little bit longer than you used to for the initiation step, but you can consider that the initiation step isn't over until you get your target radical. Okay? In this case, now I have my Br• radical, which is my target radical. Now what I want to do is I want to react that with my double bond. Oops, I forgot to say propagation. Let's do it. Propagation. Okay? So for the propagation step, what we're going to see is we're going to have a double bond and we're going to have that radical. Now typically, in a regular radical reaction, I would expect this to react with one of the hydrogens on the alkane. But it turns out that double bonds are also very good sources of electrons. So instead of pulling off an H, it could just react to the double bond directly. So what I'm going to expect is that I'm going to get this electron moving into the space in between, one electron from my double bond also giving up, you know, also moving towards the Br to make a new bond. But now I have to figure out, okay, which atom does the Br attach to? Does it attach to the red carbon or does it add to the blue carbon? Both of these are attached to the double bond, so which one do I pick? And the answer is that we're going to pick the one that allows us to have the most stable intermediate. So basically, notice that this double bond had 2 electrons in it to begin with. Now, one of them just went out to meet the Br. Where is the other one going to go? And that's going to answer our question. It turns out that the last one would want to go to the place that's going to make it the most stable, which would be the tertiary location. If the radical is moving to the tertiary location like this, then what that means is that the Br must be attaching to the less substituted position.

So then what happens is that we have to generate original radical. This radical here winds up reacting with HBr. Okay? So then what I wind up getting is that, that, and that. And I finish off my product. And what my product looks like is now a bromine here plus Br• radical. Okay? I know my head was a little bit in the way for that, but you guys can hopefully see it. Okay? So that's our propagation phase. Notice that I did get an alkyl halide, but it's attached in a weird spot. And this is the part that's interesting.

This reaction, I haven't drawn the termination step yet, but let's just go ahead and fill in these blanks. What kind of intermediate are we dealing with here? We're dealing with a radical intermediate. So this is no longer carbocation. And because we're dealing with a radical intermediate, what that means is that this is going to be an anti Markovnikov addition of bromine. Okay? The reason it's anti Markovnikov is because notice that my bromine attached to the least substituted spot. So this reaction is very important because it's going to be one of only 2 reactions we learn in Organic Chemistry 1 that are anti Markovnikov. Okay? Just so you guys know, it's 2 reactions. 1 is called this. It's the radical addition of HBr and another one is hydroboration oxidation.

If you guys just remember those 2, you're going to be set because later on you're going to need to know that. Okay? But this is a big deal because now I know how to add halogens Markovnikov through a normal carbocation mechanism, but now I also know how to add halogens in an anti Markovnikov fashion and that would just be to add radicals. Okay? Let's go ahead and finish up this termination step. The termination step for this part, I'm not going to be picky. A lot of times professors don't really want us oh, I said terminal. Termination. Okay? Professors don't want to see all of the termination products for this. They just want to see you know what you're doing because there's a lot of radicals at the beginning. So all I would do is I would terminate Br• with Br•. Okay? That's definitely a possibility. And I mean, really honestly, the the 2 r groups coming together is going to happen even less than than before. So I wouldn't even put the 2 r groups together. That would be one termination and that would really be the main termination. Okay?

So basically, what we're going to be doing I mean, another termination would just be Yeah. Another very important termination. I'm sorry. Would be this radical. Okay. Just like reacting with a h radical or whatever. That would be another one. Okay. These are the ones that are favored. But other than that, the other ones really aren't favored very often. So you wouldn't have to draw all the different possibilities. Okay? So I hope that this makes sense, guys. You should be able to know how to draw the mechanism, but even more than that, you should be able to recognize when a reaction is going to be anti Markovnikov because it's using radicals. And this only happens when we're doing an addition of HBr using radicals. Okay? Cool. So I hope that made sense. Let's move on.

Here’s what students ask on this topic:

What is the mechanism of anti-Markovnikov addition of HBr?

The anti-Markovnikov addition of HBr involves a radical mechanism, which includes three main steps: initiation, propagation, and termination. In the initiation step, a radical initiator like peroxide decomposes to form radicals. These radicals react with HBr to generate Br radicals. During propagation, the Br radical adds to the double bond, forming a more stable radical intermediate. This intermediate then reacts with another HBr molecule, resulting in the addition of Br to the less substituted carbon. The termination step involves the combination of radicals to form stable products. This mechanism contrasts with the Markovnikov addition, which involves a carbocation intermediate.

Why does the presence of peroxide lead to anti-Markovnikov addition of HBr?

Peroxide acts as a radical initiator in the reaction, leading to the formation of Br radicals. These radicals follow a different mechanism compared to the carbocation pathway in Markovnikov addition. In the radical mechanism, the Br radical adds to the double bond, forming a more stable radical intermediate. This intermediate then reacts with HBr, resulting in the addition of Br to the less substituted carbon. This process is known as anti-Markovnikov addition because the bromine attaches to the carbon with fewer substituents, opposite to what is observed in Markovnikov addition.

What are the steps involved in the radical mechanism of anti-Markovnikov addition?

The radical mechanism of anti-Markovnikov addition involves three main steps: initiation, propagation, and termination. In the initiation step, a radical initiator like peroxide decomposes to form radicals, which then react with HBr to generate Br radicals. During propagation, the Br radical adds to the double bond, forming a more stable radical intermediate. This intermediate reacts with another HBr molecule, resulting in the addition of Br to the less substituted carbon. The termination step involves the combination of radicals to form stable products, completing the reaction.

How does the stability of the radical intermediate affect the product in anti-Markovnikov addition?

In the anti-Markovnikov addition of HBr, the stability of the radical intermediate plays a crucial role in determining the product. When the Br radical adds to the double bond, it forms a radical intermediate. The position of this intermediate is chosen based on stability, with more stable radicals (such as tertiary radicals) being favored. This means that the Br radical will attach to the carbon that results in the formation of the most stable radical intermediate. Consequently, the bromine ends up on the less substituted carbon, leading to the anti-Markovnikov product.

What are the differences between Markovnikov and anti-Markovnikov addition of HBr?

The primary difference between Markovnikov and anti-Markovnikov addition of HBr lies in the mechanism and the position where the bromine attaches. In Markovnikov addition, the reaction follows a carbocation mechanism, where the H adds to the less substituted carbon, and the Br attaches to the more substituted carbon, following Markovnikov's Rule. In contrast, anti-Markovnikov addition involves a radical mechanism, initiated by a radical initiator like peroxide. Here, the Br radical adds to the less substituted carbon, resulting in the bromine attaching to the less substituted position. This difference is due to the formation of a more stable radical intermediate in the anti-Markovnikov mechanism.