Radical Hydrohalogenation

The presence of radicals in some familiar looking addition reactions can completely change the product. 

Concept: Overview of Hydrohalogention. 

2m
Video Transcript

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.
In case you don't know a lot about this reaction, this is considered 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 that the carbocation would form in the most stable location or the one with the most R groups. That means the H went right there.
Then in the next step, my Br- attached 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. We said that this would be a Markovnikov addition of bromine to the double bond. 

Remember our friendly addition reaction hydrohalogenation? Notice that you achieve Markovnikov alkyl halide in this reaction.  

Concept: How Radical Hydrohalogention is different from typical Hydrohalogention.

8m
Video Transcript

Well now what I want to show you is that just having a radical initiator present can completely change this reaction and the example I want to use is the same reaction same regions 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 there's actually going to be a radical mediated reaction which means that we're going to have to use the three 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, OK?

Now for this initiation step there's 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, OK? But then one of those OR radicals is going to react with HBR, OK? 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, OK? So that was a little bit longer than you're used to for the initiation step but you can consider that the initiation step isn't over until you get your target radical, OK? In this case that now that I have might my BR radical which is my target radical, now what I want to do is I want to react that with my double bond...Opps I forgot to say propagation, let's do it propagation, OK? 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, OK? Now typically in a regular radical reaction I would expect this to react with one of 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 with the double bond directly, OK? 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 that the BR to make a new bond, OK? But now I have to figure out OK 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 those two 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 radicals moving to the tertiary location like this then what that means is that the BR must be attaching to the less substitute position, OK? So I'm going to tell you guys what the significance of that is in a second so then what happens is that we have to generate the original radical, right? This radical here winds up reacting with HBR, OK? 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, OK? I know that my head was a little bit in the way for that but you guys can hopefully see it, OK? So that's our propagation phase notice that I did get now Alkyl halide but it's attached in a weird spot, OK? 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, OK? The reason itÕs Anti markovnikov is because notice that my bromine attached to the least substituted spot, OK? So this direction is very important because it's going to be one of only two reactions we learn in organic chemistry one that are Anti markovnikov just so you guys know it's two reactions one is called this is a radical addition of HBR and another one is hydroboration oxidation, OK? If you go just remember those two are going to be set because later on I'm going to need to know that, OK? 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, OK?

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.... Uhh I said terminal, termination, OK? Professors don't want to see like all the termination products for this they just want to see that you know it doing because there is a lot of radicals at the beginning so all I would do is I would terminate BR with BR, OK? That's definitely a possibility and I mean really honestly the two R groups coming together is going to happen even less than before so I wouldn't even put the two R groups together that would be one termination and that would really be the main termination, OK? So basically what we're going to be doing and I mean another termination would just be.... Yea another very important termination I'm sorry would be this radical just like reacting with an H radical or whatever that would be another one, OK? These are the ones that are favored but other than that the other ones really aren't favored very often so you wouldnÕt have to draw all the different possibilities, OK? So I hope that this makes sense guys you should be able to have drawn 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 in addition of HBR using radicals, OK? Cool so I hope that made sense let's move on.

Now we see this reaction. Note that the only difference is the presence of a radical initiator. 

However, this one added reagent will lead to the formation of an anti-Markovnikov alkyl halide. Here’s the full mechanism:

Example: Provide the complete mechanism for the following radical hydrohalogenation. 

3m

Radical Hydrohalogenation Additional Practice Problems

Fill in the box with the product(s) that are missing from the chemical reaction equation. Draw only the predominant regioisomer product or products (i.e. Markovnikov or non-Markovnikov products) and please remember that you must draw the structures of all the product stereoisomers using wedges and dashes to indicate stereochemistry. When a racemic mixture is formed, you must write "racemic" under both structures EVEN THOUGH YOU DREW BOTH STRUCTURES.

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Provide the products for the following reaction.

a) clearly label each drawing with the correct regio- and stereochemistry.

b) clearly label pairs of enantiomers & diastereomers, if any.

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Predict the product and show the mechanism for the following synthesis. (You are not required to show any additional resonance structures in the problem)

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Propose a mechanism for the following reaction.

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Provide the products for the following reaction.

a) clearly label each drawing with the correct regio- and stereochemistry.

b) clearly label pairs of enantiomers & diastereomers, if any.

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Draw the starting material that, under the given reaction conditions, results in the following products.

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Complete the following reaction and show the complete arrow-pushing mechanism required to produce the product.

(7 steps: 1 Initiation, 3 Propagation, 3 Termination steps)

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Predict the reagents required to complete the following transformation. 

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Using curved arrows, draw the mechanism for radical addition of HBr to 1-butene demonstrating initiation, propagation and termination steps. 

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Predict the major product

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