Ch. 11 - Radical ReactionsWorksheetSee all chapters
All Chapters
Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins

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

Concept #1: The general mechanism of Allylic Halogenation.  

Transcript

So far we've been learning about radical halogenation mechanisms and how the radical is always going to be attracted to the spot that has the most R groups, so the tertiary is going to be more stable than primary. But what we also know is that according to the radical stability trend, allylic radicals are the most stable. So now I want to throw that extra complexity into the mix. What happens if there are allylic positions available to be halogenated. That brings us to allylic halogenation.
Once again, I just want to show you guys, just in case you forgot, that the allylic and benzylic radicals are more stable than primary, secondary or tertiary. What that means is if I have one of these two possibilities, this is going to be my preferred spot, not those three.
Now, what's interesting about these is that these radicals can resonate. What that means is that resonance is going to have to play a central role in this mechanism. We can't just ignore resonance. We're actually going to have to embrace it for this mechanism.
I just want to show you guys what this could look like. Basically, I've got a diatomic halogen. Let's say this is Br2, so it's selective. And we've got a double bond. Well, usually we would just say secondary or primary, we're going to go for the secondary spot. But in this case, I actually have an allylic position. That allylic position is right here. That's the one next to a double bond. That's actually going to be the position that's the most stable and that's going to be the one that I want to react with.
Now let's go ahead and draw the whole mechanism out. I'm going to show you guys how things are a little bit different now.
First of all, my initiation step is really easy. We're just going to use Br2. And where do those first arrows go? We would just make radicals, there and there. So now what I'm going to get is Br radical plus Br radical. That's it for our initiation. We have our target radical.
Now what we have is the propagation step. In the propagation step, I'm going to want to pull off the hydrogen that's going to give me the most stable radical, so obviously, it should be this one. So let's go ahead and react that. I would use three arrows. I would do one into the middle of nowhere. One into the middle of nowhere and then one right there. What I wind up getting is I wind up getting a radical that looks like this. And also we would get HBr.
Now usually what would happen is that my propagation phase would continue and I would have to regenerate the Br radical by reacting with Br2. That's normal. That's what we're expecting. But, wait. There's a little difference. This radical can now resonate. What that means is that I can't actually just continue from here. I need to draw the resonance structures of this radical before I can continue. What that means is that the resonance structure actually looks like this.
What that means is that now, my Br, that's going to continue in the propagation stage, can attach both there and there. So it's going to be able to attack both of those carbons because the radical is hybridized between those two. Remember that resonance structures, it's not in equilibrium, it's a mathematical statistical average of where that radical is.
What that means is that I can go ahead and draw my mechanism with the Br here. And I can draw it from the first one just to keep things easy I'll just to this there, there, and there. What I wind up getting is I'm going to wind up getting a Br right here plus Br radical. But I'm also going to get something else. I'm going to get a mixture of products because the radical on the end could have also attacked. What that means is that I'm also going to get a product that looks like this.
This is one of the complications of allylic halogenation if your double bond isn't perfectly symmetrical, you're going to wind up getting a combination of products. And really we don't deal with major and minor in this section. We're just going to say we're going to get both of them. It kind of sucks.
Then finally, what would the termination step look like? Well, there are a lot of terminations. We could get a Br and a Br. We could get just any of these radicals, I mean any of these. That plus my Br. Sorry, I'm just going to draw this and this. That would give us our product that looked like this. We would also get the other one. The other radical that would like this. The radical here is Br and that would give us the other looking product.
Now I know that there was another one that was the two R groups. That is almost never seen in here. But if you wanted to include it, you could just add R radical, R. And just say that that's going to give you R, R. But honestly, for these reactions, usually professors just neglect this because it's assumed that you're not really going to get much of this at all.
What is important is that now you can detect when you're going to get just one product or when you're going to get a mixture. When you're going to get a mixture is when you're doing an allylic halogenation. If it was just tertiary or secondary, you would only get one, but since it's allylic that means it can resonate.
In this case, you would get two different structures. The way to know how many structures you're going to get is just to draw the resonance structures yourself.
Now what I want to do is go more into depth on allylic halogenation with the specific reagents that we're going to use. 

Remember our friendly addition reaction halogenation? Notice that you achieve a vicinal dihalide in this reaction.  

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

This one added condition will lead to the formation of a mixture of allylic alkyl halides. Here’s the full mechanism:

Concept #2: The products of Allylic Chlorination.

Transcript

So now that we know the mechanism predicting these products is going to be easy the number one thing I'm looking for here that I want you guys to know is these reagents, so it happens to be that when you're doing a little chlorination 400 degrees Celsius is just the most common heat that's used so you should recognize that when you see CO2 400 hundred degrees and a double bond and you have Allylic position available we're going to wind up getting our you know we're going to wind up getting an allylic halogenation with multiple products because I could get the radical here or I could also resonate the radical to here, OK? So I'm going to one of getting both products one looks like this and one looks like this, OK? Easy stuff but you guys should be aware of that.

Concept #3: Mechanism of Allylic Bromination.

Transcript

For Allylic Bromination usually you want low trace amounts of bromine we don't want too much bromine or we might wind up getting like an actual halogenation on the double bond which would be bad so we use NBS and we usually use NBS with a combination of light or heat doesn't matter but in this case, it would be the same exact thing, I would wind up getting a bromine here and I would get...Opps how do I change all this? There we go and I would get a bromine right here, OK? So really these are the same exact reaction just the reagent are a little tiny bit different that's it. Cool so what I want you guys to do is take this problem try to solve it yourself be sure to think about the whole mechanism before you predict your products I don't need the full mechanism as long as you can give me all the products that's what I'm looking for so go for it and then I'll solve it.

Predict the product(s) of the following reaction. 

Example #1: Predict the product(s) of the following reaction.