Not all radicals are born equal! Some of them are going to make pretty smart decisions from an energy perspective, while others are a little on the crazy side. Let’s turn to some potential role models for guidance on this topic.
Concept: Radical selectivity: Alcoholics Anonymous Version11m
Hey guys, well now we understand that radical halogenation prefers to occur at the most stable radical intermediate, which in most cases would be tertiary. But it turns out that not all halogens are created equal and some aren't going to play by the rules at all. This is going to introduce a brand new topic that we have to talk about which is called radical selectivity.
Guys, don't get distracted by the pictures yet. We're going to talk about them, but first I just want to talk about what is radical selectivity. Well, radical selectivity is defined by the ability of a radical to only halogenate the carbons with the most stable radical intermediates. Basically, if a radical tends to only halogenate the best part of the molecule, the most stable part of the molecule, that's called selective.
You would think molecules are smart, why wouldn't they all make the right decision. Why wouldn't they all pick the best spot? It turns out, guys, that, like I said, some halogens just don't like to play by the rules. They like to break the rules. Instead of picking the best spot, sometimes they're going to pick some really terrible spots.
What I want to do is I want to bring an analogy to this, so we can figure out what's an analogy of halogens making good decisions versus halogens making really bad decisions. One analogy that comes to mind is a dinner party on a Wednesday night.
Imagine that you're out with your friends and you go to a little dinner get-together at a local restaurant and it's just a normal night. You're sitting with your friends, you're talking about the weather, all that fun stuff. But there's that guy at the party that is the wild card. You don't know what he's like. He could be a total disaster or maybe he's just a straight A student. You're waiting to find out.
It turns out that the way we can predict whether he's going to make some good decisions or bad decisions is really related to how exothermic of a halogen he is. This trend that I'm about to tell you, isn't actually completely true. In a following video, I'm going to explain more rigorously the relationship between how exothermic a reaction is and how selective it is. But this is just going to help you remember the trend.
Basically, if you're talking about radicals, the more exothermic the reaction, the more non-selective the reaction tends to be. So the more energy is released, the more non-selective. And in the terms of your dinner party on a Wednesday night, that means the more hungover this guy is going to get, the more hungover he's going to be in the morning.
Let's go ahead and just go radical by radical. Once again, there's four halogens, four radical halogenations that are possible. Let's talk about each one and see which one makes the best choices and which one just ends up being a big mess at the end of the night.
Let's talk about fluorination. Guys, fluorination is the most exothermic reaction of all. Fluorination releases 432 kilojoules per mole of reactant that's reacted with. This is an insane amount of energy. Now in terms of our friend that we talked about, does that mean it's going to be selective about his alcohol consumption or non-selective? Probably very non-selective.
Perhaps you've heard this phrase before. It says, beer before liquor makes you sicker. Well, how about if you just drink the entire bar. That's what fluorination is. Fluorination just says, I'll take one of everything you've got. Basically, what this means is that fluorination is not used in radical halogenations because it's just too crazy. You can't control it. Call 911.
It turns out that radical fluorination is so reactive, so exothermic, that it just blows up on the spot. You could literally die. We don't use radical fluorination in the lab. And for the same reason that you don't invite that guy to your party. He's just going to drink the whole bar. You're going to have to drag him out. He's probably going to have to go to the hospital.
This guy, fluorine, F2, and light or heat, we don't want use. We're just going to put a big X on that because it's not selective enough.
Let's go on to chlorination. We can see chlorination releases 101 kilojoules per mole. That's a lot between than 432 in terms of it's not quite as exothermic. But guys, it's still a ton of energy.
Imagine that you're at this dinner party and this guy, he's got his act together a little bit more. He's not going to pick all the alcohol, but the alcohol he picks is like Four Loko. If you haven't heard about this alcohol, it's because it was banned by the FDA. Four Loko, just a bad decision. Maybe he'll be able to walk out on his own, but he's not going to be in good shape. This is not the kind of person you want to invite to your dinner party.
Again, very exothermic. Not very selective. Kind of just picking all the flavors of Four Loko. That's not what I mean by selective.
In the same way, if we were to talk about Cl2 and heat or light, let's just put heat, this is a rare reaction. We don't use it very much because it's not very useful. It's not very selective. So the only type of radical chlorination that we'll do is if it has a single type of H. Basically meaning that as long as your molecule only has one type of H on it, you could use chlorination. But, in general, not the best choice. Chlorination, once again, too non-selective. He's just going to pick the worst alcohol of the entire bar.
Let's go on to bromination. Ah. The voice of reason. Bromination is that dude that he's going to be a great dinner guest. He's responsible. He's got his homework done. He probably meal prepped already. And he's having one single glass of wine with dinner because he found out that it's good for his cardiac health.
As you can see, it's still exothermic. It's still going to be most likely a spontaneous reaction, so 26 kilojoules per mole, but it's low enough so that it's going to be selective. It's going to look at its options and he's going to say I'm going to take the Merlot. I'm going to take the best wine that you've got because I just want one glass. I want to have a good time. I'm not going to get crazy blasted tonight and get hammered.
What we say about bromination or simply Br2 and heat or light is that this is your ideal halogenation. This is what we consider the only useful method for selectively halogenating alkanes. What I'm trying to say is that if you have a molecule that has primary, secondaries and tertiaries on it, bromination is the one you want to use because bromination is always going to pick the best spot. It's going to pick the tertiary, not the secondary. If there's no tertiaries around, it's going to pick the secondary. Basically, they're just saying it's always going to make the best decision.
Bromination – awesome. Chlorination, fluorination - huge mess. We want to use chlorination rarely and fluorination you'll never use.
Then finally, iodination. Iodination, very well intentioned. She actually just didn't show up to the dinner party because she was too busy praying for the world. So she didn't show up. Part of this is that iodination as you see actually is endothermic, so it actually is going to take energy to put it into the system to make it happen. So non-spontaneous. This reaction is not going to be spontaneous. As you can see, I say, not a spontaneous reaction, don't even try it.
Basically, iodination has a great heart, but just doesn't show up to the dinner party at all because she has a higher calling. She's not going to do radical halogenation.
What is the summary here? The summary, hopefully maybe this analogy sticks a little bit. But what I'm basically trying to say is that the best halogenation, the most selective, is bromination and then sometimes you can use chlorination. If you've just got one type of hydrogens, you could use chlorination. But you're never, ever going to use fluorination and you're never, ever going to use iodination for totally opposite reasons because one is too crazy and one isn't crazy enough.
Then finally, this last point here, just says keep in mind that chiral products are always racemized. What does that mean? It means that if your tertiary – let's say you're doing a bromination. And let's say that your hydrogen or whatever, this is the one that's going to be reacted with, is chrial. Let's say that you had a methyl group and an ethyl group and a propyl group. Let's say that you were to take away this hydrogen and replace it with a halogen, so Br because Br is so great.
We're going to do this more later. But what I'm trying to say is that if this is a chiral center, you're going to get racemized products because you could get the bromine either attacking from the front or you could get it attacking from the back, so you're going to get enantiomers. You're going to get 50% of one and 50% of another. Awesome, guys.
Anyway, I hope that this video is taken the right way. I'm not promoting underage drinking, by the way. Just letting you guys know, if you're going to make decisions about alcohol, let's make the right decisions. Go with bromination, not these other more wild situations that are going on on the top of the page. Let's go ahead and move on to the next video.
Selectivity is defined as the ability to only halogenate the carbons with most stable radical intermediates.
Example: Predict the product of the following Radical Halogenation. Would the following reaction be synthetically useful? (Yielding only one product).1m
Example: Predict the product of the following Radical Halogenation. Would the following reaction be synthetically useful? (Yielding only one product).2m
Example: Predict the product of the following Radical Halogenation. Would the following reaction be synthetically useful? (Yielding only one product).2m
Early transition states could care less what they look like, whereas late transition states have to be much more careful about the arrangements they take.
Concept: Using the Hammond Postulate to describe radical chlorination.4m
Now that you guys know the funny way to memorize which halogens are going to be more selective and which ones are less selective, now I want to teach you just really quickly, the rigorous actual reason why bromine is so much more sensitive than chlorination. It doesn't really technically have to do with how exothermic it is, in fact, it has to do with the Hammond postulate.
Remember what we learned about Hammond postulate, it is a way that describes what transition states look like. And the Hammond postulate is really the reason why bromination is so much more selective than chlorination. Let me show you guys that now.
Once again, my definition of selectivity hasn't changed, but it turns out that I'm going to use the Hammond postulate to explain why bromination is so much more selective. First of all, I just want to show you the energy diagram for radical chlorination. Let's go ahead and look at the x-axis first and just look at the coordinates.
Notice that all of these things should look familiar. Basically, this is the first step of my propagation phase. This is all propagation here. I'm just going to put prop. I've got my propagation phase. We would expect that that takes energy because you're breaking a bond, so that's going to be basically, this activation energy here. That's going to be the activation energy required to do that first step.
Then what that's going to make is an intermediate that looks like that. That's a radical. That's my intermediate. Notice up here I put rate-determining step. Remember that your slow step, the one that makes the intermediate, is always going to be your rate determining step. Does that make sense so far? Cool.
Then what happens is that in the next part of my reaction diagram, there's like a story, then what happens? Well, remember our propagation phase isn't done yet. I need to react my – well, in this case, this would be a termination step here. Let me just say here, this would be termination. And in my termination step, what I get is now two radicals colliding to make a new single bond.
Now just so you guys know, technically, this might come up conceptually, the activation energy for two radicals to collide and make a new bond is always going to be exactly zero. This is just a definition that you guys should know, that the activation energy of two radicals colliding is always zero. It doesn't take energy for that to happen. That's why there is actually no hump here. Usually, you would expect to see two humps, but there's no hump there because it doesn't take any energy for that to happen.
Now notice here that my overall enthalpy at the end is negative 101, like I told you guys. But what's really important that I want you guys to note is what the transition state looks like. Now I'm not going to go through this just yet, I'm just going to let you guys know, pay attention to what that transition state looks like, Hammond's postulate and now we're going to look at bromination and see how bromination is different.
Concept: Using the Hammond Postulate to describe radical bromination.6m
So for bromination notice that my energy diagram is a little bit different, I get the same first step, OK? Same first step and what I noticed is that I get my same intermediate, OK? Now there's just one major difference between my intermediate here and my intermediate up there, notice that the energy level for my rate determining step here is actually going to be endothermic, OK? What that means is that I actually having to put in energy in order to become this radical, Ok? Now look at the energy difference for my chlorination, for chlorination it actually was exothermic it actually released energy to make that radical, OK? Isn't that crazy so for chlorination it was exothermic to make that radical, for bromination it was endothermic well what's the significance there? What the significance is that remember that what did Hammond postulate say? It said that basically Hammond postulate said that your intermediate is going to look most like the side of your reaction or the side of......Yea the side of the reaction that has the highest energy level, OK? So in this case what that means is that the transition state for chlorination and bromination is going to look radically different, for chlorination the thing with the highest energy.... IÕm just going to use green as the highest energy is going to be the starting molecule, OK? So what that means is that my transition state looks a lot like my starting molecule, notice that my bond between the H and the stage two is really short that means that the intermediate looks a lot like the original alkane, OK? Does that make sense so far? So my intermediate really looks like the original, now we notice is that for radical bromination my highest energy state is actually the radical so what that means is that my intermediate looks a lot more like the radical than like the alkane notice that in this case my H is really far away meaning that this CH2 almost already has a radical on it, OK? Is that making sense so far so really the biggest difference is the way that the intermediate looks, OK? So why is that important? Well because if you think about it if the alkane if the transition state looks a lot an alkane then the energy savings of making a more stable intermediate isn't going to matter that much, notice that I have here that this is the primary this is the energy savings that I get as a primary radical but if you make a secondary radical or tertiary radical you save even more energy, right? So you would think OK it really wants to be tertiary so that it can make it dip all the way down here and be more stable but no chlorine doesn't care, why? Because chlorine doesn't really look at the radical once it's in its transition state really looks like it's alkane and the alkane was always stable to begin with so it doesn't really matter whether it's primary, secondary or tertiary because it looks like the alkane anyway so it doesn't really need to have a stable intermediate, OK?
Now let's look at bromination, for bromination what we notice is that it actually looks a lot like the radical already in the transition state, OK? So if I can make it secondary or if I can make it tertiary even better that's going to make my transition state a lot easier to make, it's going to make my transition state a lot more stable so the rate determining step can happen faster, OK? So basically that's the whole point, bromination cares a whole lot more about making the graph look like this, OK? Because the intermediate according to HammondÕs postulate looks a lot like the radical, OK? So what's this difference? Well the difference is that remember that if you had a transition state that looks like the original starting product you called that an early transition state, OK? And notice that over here what we wind up getting is a late transition state because late means that it looks more like the second step, OK? Once you have a late transition state you're going to care a lot more about how stable that transition state is or how stable that intermediate is because it's going to stabilize your transition state as well, OK? So what I'm trying to say is that my simplification of the exothermic rule works, OK? But technically the way that we explain selectivity is through Hammond postulate, OK? Now for some of you guys this is just going to be theoretically just helps you know more about this class, some professors are going to be really picky about it and actually want you to be able to describe this, OK? So it's really up to you guys if you didn't totally get this it's not the end of the world it's probably not going to be in your tests, OK? But I just want you guys to know it because I like to be thorough and I like you guys to know everything, OK? So let me know if you have questions but if not let's go ahead and move on.
Provide the reagents or starting materials to perform the indicated transformation. More than one reaction may be required.
Predict the major constitutional isomer(s) formed in each of the following reactions. Write "No reaction" where appropriate.
Predict the major product obtained upon radical bromination of t-butylcyclohexane.
Which of the following products could be formed in good yield through radical chlorination in excess alkane?
1. A and D only
2. C and D only
3. A, C and D only
4. A, B, C and D
5. Only C
Draw all possible products of the following reaction. Indicate which is likely to be the major product.
Predict the major product(s) of the following reaction.
Select the answer that best completes the following statement:
Radical halogenation reactions using ______ are the most ______ and often lead to multiple products. While radical halogenation reactions using ______ are the most ______ and produce primarily the major product.
(A) bromine, selective, chlorine, reactive (B) bromine, reactive, chlorine, selective
(C) chlorine, reactive, bromine, selective (D) chlorine, selective, bromine, reactive
Draw all possible monochlorination products of 3,3-dimethylpentane, including stereoisomers.
Predict the major product(s) of the following reaction:
Reaction of bromine radicals with the following polycyclic alkane would lead to the formation of how many different alkyl halides (you may neglect stereoisomers)
Which of the following statements is TRUE regarding the radical chain reaction of diatomic iodine in excess alkane:
a. Would lead to a fully polyiodonized alkane products unless reacted in excess alkane.
b. Would yield mostly monoiodonized alkanes due to the high bond dissociation energy of diatomic iodine and its very high selectivity.
c. Iodine is NOT used in radical halogenation due to its extremely high reactivity. It can lead to lab explosions.
d. More than one of the above is true
e. None of the above are true