|Ch. 1 - A Review of General Chemistry||4hrs & 48mins||0% complete|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 19mins||0% complete|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete|
|Ch. 8 - Elimination Reactions||2hrs & 25mins||0% complete|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete|
|Ch. 10 - Addition Reactions||3hrs & 32mins||0% complete|
|Ch. 11 - Radical Reactions||1hr & 55mins||0% complete|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 42mins||0% complete|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 20mins||0% complete|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 56mins||0% complete|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete|
|Ch. 23 - Amines||1hr & 43mins||0% complete|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete|
|Ch. 25 - Phenols||15mins||0% complete|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete|
|Ch. 26 - Transition Metals||5hrs & 33mins||0% complete|
|Radical Reaction||8 mins||0 completed|
|Radical Stability||7 mins||0 completed|
|Free Radical Halogenation||19 mins||0 completed|
|Radical Selectivity||21 mins||0 completed|
|Calculating Radical Yields||19 mins||0 completed|
|Anti Markovnikov Addition of Br||11 mins||0 completed|
|Free Radical Polymerization||9 mins||0 completed|
|Allylic Bromination||12 mins||0 completed|
|Radical Synthesis||9 mins||0 completed|
Alkanes are the backbone of organic molecules, yet they are almost completely unreactive.
The only reaction alkanes undergo is radical halogenation, the gateway to the rest of organic synthesis.
Concept #1: The one reaction that alkanes will actually undergo.
Back when I taught you guys about functional groups, I told you guys that alkanes actually don't count as a functional group even though they're superabundant. They're everywhere. The reason why I said that is because it's true. Functional group, that implies function, that they actually do something. And alkanes really don't react with much at all. Alkanes just come from underground. You dig them up in petroleum. That's what oil is. It's alkanes.
You can't really react with it a whole much. They're super stable. All you can do is blow them up. You can put them in your car and combust them, but you can't really react with them a whole lot.
They seem kind of worthless on first glance, but it turns out that there is one thing that they actually can undergo and that is that they can undergo a radical reaction because radicals are very high energy, so they're going to be able to react with something that's seemingly unreactive, which is alkanes. I want to show you guys the mechanism by which they do that.
As I just said, alkanes are the backbone of organic molecules, but they're almost completely unreactive. That's why they last for millions of years underground because they don't react with shit. But there is one thing that they can do in the presence of radicals and they can add halogens.
Here I have an unreactive hydrocarbon, like I said, that's from the dinosaurs. It didn't do anything that whole time. Now I bring it up to the science lab and I react it with a radical reaction and lo and behold, I get a halogen on that alkane.
Now what's cool about that is now I can do a bunch of other types of reactions to it. This is now called a functionalized hydrocarbon. Why? Because now I have a functional group, an alkyl halide. Once you have an alkyl halide, that's the gateway towards organic synthesis. Because now guess what? I can do a bunch of stuff to that. I can do substitution reactions, elimination reactions, addition reactions, all kinds of stuff because I first added that halogen.
What I'm going to show you right now is really the first step of all organic synthesis.
Concept #2: Radical Chain Reaction Mechanism.
So let's just go ahead and talk about it. It turns out that radicals are so high energy that once they react with something, they're going to keep trying to give away that high-energy intermediate. What winds up happening is that it's like a game of hot potato, where no one wants to have the hot potato, so they keep passing it along and it forms what's called a radical chain reaction.
Now the chain reaction actually does mean that. It means that once you start it, it actually can't end until its fully reacted, until you fully react it with all of your alkane. That is useful for us because, remember, alkanes aren't that great to begin with, so if we can react them completely, that's a useful reaction as an organic chemist.
Let's go ahead and see how this works. Out first step is going to be the initiation step. The initiation step is where I get that first radical because I can't play a game of hot potato without the hot potato itself, so I have to create that first radical. Now what you notice is that this mechanism is broken down into three different steps. We're actually going to need to write all three of these steps. In fact, it's smart that you actually write the words if you do have to draw this mechanism for a test, that you write these three words: Initiation, propagation, and termination.
Let's look at the initiation step and let's say that we're just using the easiest radical initiator which is X2. Let's use X2 over heat. Now what I taught you guys is that in the initiation step, what we're going to wind up getting is electrons from – two electrons, one on each side, jumping on to each X. What I'm going to wind up getting is X radical, plus X radical. That's the end of my initiation step. All I need for the bare minimum of my initiation step to work, all I need is one radical. In this case, I have two, so I'm great.
Now that I have that radical in place, that radical is free to react with other molecules. It turns out that it happens to react really well with alkanes. Now for the sake of a really simple mechanism, let's just use the simplest alkane possible, which is methane. Methane just being a one-carbon hydrocarbon, CH4.
So now I've got CH4 and I'm reacting that with X radical. This X radical hates itself right now. It's super high energy, super unstable. It's saying how can I get rid of this hot potato. How can I get rid of it? Then it sees all these electrons in the methane and it's thinking maybe I can take one of the electrons from one of those carbon-hydrogen bonds. That's exactly what it does.
It turns out that radicals are going to react with hydrogens in alkanes. The way we draw these arrows is just so you know, radical reactions are always going to have three arrows. I'm going to draw one fish hook into the middle of nowhere. Then I'm going to draw another fish hook from the CH bond, meeting that one. What that's implying is that now there's going to be a new bond between the H and the X that's going to form from those two electrons.
That's looking great, but I still have one electron left over. Notice that the bond between the CH had two electrons, so where do you think that last electron goes? It goes onto the C. It goes onto the carbon backbone. What that's going to do is it's going to give me a structure that now looks like this, C, H, H, H, radical. Notice that I'm drawing the geometry different because now this would be trigonal planar, so you should draw it like a triangle. And that would be plus HX. Making sense so far?
Notice that the reason this is called a propagation step is because propagation means I'm reproducing myself, propagating. Notice that the radical just reproduced itself. Now, it kind of moved through my medium and now I've got a radical on a new species.
Well, it turns out that your propagation step isn't done yet because you're not done with a propagation step until you fully reproduce yourself, 100%. What that means is not only do I need to have a radical at the end, I need to have the same exact radical that I started with. So if I started off with an X radical, I need to end off with an X radical.
What that means is that what could I – we're actually just in the middle of our propagation step right now. What could we react my C radical with to generate that original X radical that we had at the beginning? Can you think of anything? It turns out the easiest thing to do is just to react it with another X, X diatomic halogen.
Now you might be wondering, “Well, Johnny, why isn't this already radicalled because we just did that in the first step? We made it radicalled.” Well, it turns out that not all of the diatomic halogen is going to cleave at the same time. So some of it's going to do the initiation step, but some of it isn't going to be hit by enough light or enough heat to split up yet. So what that means is that the one I'm reacting with here hasn't cleaved yet. This is one that's just waiting around for enough energy to finally do that homolytic cleavage.
But wait, before the light can even get to it, another radical just did. Instead, we're going to propagate to the X, X. And the way we draw these arrows is once again three arrows. So I'm going to take – the radical always starts it. I'm going to take the radical. I'm going to put that one out into the middle of nowhere between the C and the X. Then I'm going to take one electron from this bond and make it go there. This represents that now there's going to be a new bond of two electrons between the C and the X. Unfortunately, I've got one electron left over. I'm going to dump that one onto the X.
So what I'm going to get at the end of this step is now notice, C, H, H, H, but now I've got an X. What do we call that functional group when you have a carbon and a halogen attached to each other? Alkyl halide. Notice that now I'm going to have X radical. That's the end of my propagation step because now I have the same exact radical that I started with.
Notice that my propagation step just created a very useful byproduct, which is that it made an alkyl halide. Which, like I said, alkyl halides do a whole lot more stuff than alkanes do. So this is a very useful reaction for me as an organic chemist.
Now we're going to do our termination step. Now, what is termination? Termination means that you're producing all these radicals. These radicals are spreading. Radical, radical, radical. You're generating all of them. What happens in termination is that you finally have so many radicals that instead of propagating and forming new radicals, they're actually extinguishing each other faster than they can make new radicals. Because think about it. If you have two radicals coming together, if they hit each other, what do you think is going to happen?
Well, I've got one electron here, one electron here, they're going to form a new bond. The termination phase is what happens when there's so many radicals that they actually bump into each other. They collide into each other more often than they collide into unreacted species.
My termination phase, what I do is I look at all the different possibilities of radicals that could have collided. The easiest radicals to think of that could have collided are just X plus X because these are being formed in high amounts. What happens if they hit each other? Well, I would just wind up getting the same thing all over again. I would do this. I would do that. And I would get X, X. That's one termination product.
But there's actually some other termination products possible. Can you think of another one? Another one would be if I had an X radical and a CH3 radical. That would be like the radical that I had up here in black, where maybe instead of hitting an X, X, that wasn't reacted yet, maybe it reacts with a radical. Once again I get my arrows. And what I get as a product here would be CH3, X, so I get an alkyl halide.
Can you guys think of any more combinations? There actually is one more and that last one would be well, what if I got CH3, radical plus, CH3 radical. Now this is tricky because now it looks like I'm going to get a completely different product. It's true. What I would get now is actually ethane, CH3CH3. Crazy, right? So now what I just did is I just made a larger hydrocarbon than before. Before I just started off with methane, now I'm getting ethane as a product.
Well, a few notes about these termination products. First of all, the first one is not going to really matter. I'm just going to say, I don't know. I'm trying to think of a symbol that I can use. But just, let's say I have termination product one, two and three. Number one isn't really going to matter that much and the reason is because after I make that X2, it can then just react with heat or light again and make radicals again. So really, this is a termination phase that happens and then it dissociates again.
So you're not going to get a whole lot of X2 forming because the X2 keeps breaking up. That's the first thing. So actually X2 isn't going to form a whole lot. Now your professor probably still wants to see that termination step, but just as your final product, you're not going to get a whole lot of X2 as a final product. So you can cancel out one.
Another one that we can cancel out is actually three. The reason is because the only way that I can form ethane here is if I have a CH3, a methyl radical and a methyl radical colliding. Well, the chances of having methyl radicals is actually way lower than the chances of having a halogen radical. Why? Because you're forming the halogen radical directly through my energy, through my light or my heat, whereas these methyl radicals can only be formed after they've already reacted with a halogen. So the chances of them colliding and making a bigger hydroboration are actually very small compared to the chances of them colliding with another halogen.
So what I would say is that I'm going to get I'm going to get a very tiny amount of this, like scarce. Let me try that one more time. Scarce. Wow, I'm really bad at spelling guys, sorry. Scarce. I will get a tiny amount of this, but it's not going to be a huge amount.
It turns out that there's actually a way that we can even lower this even more. We can lower the amount of my larger alkanes that we get. The way we can do that is just by reacting with excess halogen. If I just react with a lot of halogen and a very low quantity of alkane, then that even reduces the chances more that two alkane radicals are going to collide. So depending on what my concentration is of alkane, I could basically take this down to zero. If I really wanted to, I could take this down to pretty much full zero.
So we're going to cross out one and three. That means that what's my real main product? Well, the real product is going to be my alkyl halide. That's the whole point of these. Let me just move out of the way. The whole point of this reaction is that I'm taking an alkane and I'm making alkyl halides out of it.
Now the reason I taught you guys about this is because your professor is going to want to see – if you're asked to draw this mechanism, they want to see all three terminations. But that doesn't mean that you should draw three products. You should really only draw one major product because that's going to be like 99.99 percent of it and that's the alkyl halide.
Alkanes will react with diatomic halogens in the presence of heat, light or any other radical initiator.
Example #1: Explaining the following example problem.
Now I want you guys to practice the general mechanism for radical halogenation all on your own and I want us to notice that this alkane that I'm reacting with has carbons of different stabilities, OK? I want you guys just to assume that we're going to react with the most stable carbon in this case so you're going to have to think back to what I talked about with radical stability to figure out which of those hydrogens to pull off in the radical halogenation, I think you guys I can get this though so I'm just going to let you guys loose all on your own trying to draw all three steps and then I'll give us the answer so go for it.
Example #2: Show the entire chain reaction mechanism for the following Radical Halogenation reaction, predicting which would be the structure of the major alkyl halide produced.
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