Free radical reactions
Free radical reactions. Created by Sal Khan.
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- Is there any difference between adding heat and putting light on the mix in terms of reactions?(65 votes)
- Yes. In chemistry, when a mixture is heated, all reaction speeds are increased. This is not favored, because the probability of impurities and byproducts formed also increases. So high energy light (UV light) is used to form radicals, and keep the temperature of your mixture low. The radicals broken down by light are called 'photo-initiators'.(148 votes)
- why wont the chlorine just reach with the other free chlorine in the beginning? they are nearer, highly reactive and all.(14 votes)
- That can absolutely happen! The reformed chlorine could be initiated again to make chlorine radicals and this cycle could continue...some will react with other species though. We don't talk about this chlorine reformation as much because it's not as interesting as there is no net reaction, but that doesn't mean it doesn't happen.(22 votes)
- why do we even need radicals?(4 votes)
- Free radical halogenation is one of the main ways to produce chloromethane, dichloromethane, trichloromethane, and tetrachloromathane. They have different boiling points and are therefore separated into different fractions. These chemicals are important in many industrial and commercial uses. This is one of many industrial processes that utilize free radicals.
We don't want them in our bodies, however. They are extremely damaging to our cells as they oxidize anything that they collide with (organelles, cell membrane, etc...). High levels of hydroxyl radicals can be linked to aging. We have enzymes that break them down to deal with them, but people sometimes take foods that are high in anti-oxidants (blueberries, grapes, etc...) to help support these enzymes.(13 votes)
- So I understand how this reacts, but would there be any difference if you were using something other than chlorine, such as fluorine or Bromine?(6 votes)
- For this specific reaction, nope! the X-X (where X is a halogen) bond would be broken the same way, everything would react similarly, though bromination would proceed more slowly (and require some heat). Fluorine would react even faster. I believe the wavelength required to break the bond is different, but I could be wrong on that.
There would be a difference in a more complicated or branched alkane: bromide radicals are more selective for the hydrogens on more highly substituted carbons, while chlorine radicals are far less selective. Bromine is 97 times more likely to abstract a secondary H than a primary H, chlorine is 4.5 times more likely to abstract a secondary H, and fluorine is virtually nonselective.
The reason that bromination prefers the more substituted carbon is because it is endothermic, while chlorination is exothermic. It has to do with the Hammond postulate. I don't know if there's a video on this but there should be.(6 votes)
- Isn't HCl an ionic compound? Isn't it formed when a hydrogen atom gives its electron to a chlorine atom? So, at4:47, why does Sal say that hydrogen isn't giving away the electron, but is contributing its electron to half of a bond? Why is he drawing half an arrow?(5 votes)
- Most bonds have both ionic and covalent character and in many cases it is actually a bit arbitrary to call something one or the other.
HCl is a great example of this, it has a polar covalent bond as a gas – this can be expressed as the bond having a mixture of covalent and ionic character. However, when dissolved in water, HCl almost completely dissociates with the protons binding to water molecules – and is now best described as being ionic.
One way of thinking of this is that the nature of some compounds changes depending on the environment.
You might also find the following link edifying:
- what if you have primary, secondary, tertiary carbons. Which one will the radical, at least the halide radical, preferentially add?(4 votes)
- Radicals prefer to go to the most substituted position (which sometimes is overruled by being able to forma conjugated system with that radical). Addition of X_2 (a general diatomic halide) via radicals from heat/light follow the same pattern. With Br being especially selective for tertiary positions adding almost exclusively there. The selectivity deceases as you increase in electronagtivity of the halide.(4 votes)
- At7:00-8:00, why did Salma say that the CH3 is more likely to bond with another Cl molecule than the radical atom before to from a chloromethane, isn't it the same if the molecule just form a bond with the radical atom?(1 vote)
- You're right, if the methyl radical reacted with either molecular chlorine (Cl2) or one of the chlorine radicals, it would result in chloromethane. However, since the concentration of molecular chlorine (Cl2) is so much higher than the concentration of chlorine radicals, it is much more likely that the mechanism by which the chloromethane is formed involves the methyl radical reacting with molecular chlorine.(7 votes)
- Why was HCl not added to the termination step?(3 votes)
- A termination step occurs when two free radicals collide to form molecules. A reaction with HCl would just form another free radical, e.g., H₃C· + HCl →H₃C-H + ·Cl. This would be a propagation step.(2 votes)
- Can I make this reaction with any halogen??(1 vote)
- The reaction works well with chlorine and bromine.
Fluorine is so reactive that the reaction is difficult to control, and there is the danger of an explosive reaction.
The reaction with iodine is practically non-existent and thermodynamically unfavourable.(4 votes)
- What's the difference between an ion and free radical?(2 votes)
- ions, most importantly, tend to have an even number of valence electrons, as compared to free radicals, which have an odd number of valence electrons. An exception of this case is the ion like Fe^3+, which has an odd number of electrons in its outer shell. Same is the case for many other (I think all of ) the transition elements, as they form coordinate complexes, and seldom bond covalently or ionically.
Ions have an overall (net) positive or negative charge, whereas, a radical doesn't have a net charge on it.(2 votes)
Let's think about what type of reaction we might be able to get going if we had some methane and some molecular chlorine. So if we just let this be and we didn't heat it up or put in any UV light into this reaction, pretty much nothing will happen. Both of these molecules are reasonably happy being the way they are. But if we were to add heat into it, if we were to start making all the atoms and molecules vibrate more and bump into each other more, or we were to add energy in the form of UV light, what we could start doing is breaking some of these chlorine-chlorine bonds. Out of all of the bonds here, those are the weakest. That would be the most susceptible to breakage. So let's say we were to add some heat, what would happen? So let's see. Let me draw the valence electrons of each of these chlorines. This chlorine has one, two, three, four, five, six, seven valence electrons, and this chlorine over here has one, two, three, four, five, six, seven valence electrons. Now, when you add heat to this reaction, enough for these guys to vibrant away from each other, for this bond to break, what's going to happen, and we haven't drawn an arrow like this just yet, but what's going to happen is that each of these chlorines, this bond is going to break. Each of these chlorines are just going to take their part of the bond. So this guy on the left, he's just going to take his electron. And notice, I draw it with this half arrow. It looks like a fish hook. It's just half an arrowhead. This means that this electron is just going to go back to this chlorine, and this other magenta electron is going to go back to the right chlorine, so we can draw it like that. If it was up to me, I would have drawn it more like this. I would have drawn it more like this to show that that electron just goes back to the chlorine, but the convention shows that you can show that half of the bond is going back to the entire atom. Now, after this happens, what will everything look like? Well, we're still going to have our methane here. It hasn't really reacted. So we still have our methane. Let me draw it a little bit. So we still have our methane here. And all that's happened is, because we've put energy into the system, we've been able to break this bond. The molecular chlorine has broken up into two chlorine atoms. So we have the one on the left over here, and then we have the one on the right. And let me draw the left's valence electrons. It has one, two, three, four, five, six, seven. I just flipped it over so that the lone electron is on the left-hand side right here. And then you have the guy on the right. He has one, two, three, four, five, six, seven valence electrons. Now that each of these guys have an unpaired electron, they're actually very, very, very reactive. And we actually call any molecule that has an unpaired electron and is very reactive a free radical. So both of these guys now are free radicals. And actually, the whole topic of this video is free radical reactions. Both of these guys are free radicals. And you've probably heard the word free radical before. In the context of nutrition, that you don't want free radicals running around. And it's the exact same idea. It's not necessarily chlorine that they're talking about, but they're talking about molecules that have unpaired electrons. They'll react with some of your cell's machinery, maybe even with your DNA, maybe cause mutations that might lead to things like cancer. So that's why people think you shouldn't have free radicals in your body. But as soon as we form these free radicals, in this step right here, where we put energy in the system to break this bond, we call this the initiation step. Let me put this. We used energy here. This was endothermic. We use energy. This right here is the initiation step. And what we're going to see in general with free radical reactions is you need some energy to get it started. But once it gets started, it kind of starts this chain reaction. And as one free radical reacts with something else, it creates another free radical, and that keeps propagating until really everything has reacted. And that's why these can be so dangerous or so bad for biological systems. So I've told you that they react a lot. So how will they react now? Well, this guy wants to form a pair with someone else. And maybe if he swipes by this methane in just the right way, with just enough energy, what will happen is he could take the hydrogen off of the carbon, and not just the proton, the entire hydrogen. He will form a bond with the hydrogen using the hydrogen's electrons, so they'll get together and they'll form a bond. The hydrogen will contribute one electron. Notice, I'm drawing the half-arrow again, so the hydrogen isn't giving away the electron to someone else. That would be a full arrow. The hydrogen is just contributing its electron to half of a bond. And then the carbon, the carbon would do the same. I'll do that in blue. So the carbon, this valence electron right here, could be contributed to half of a bond, and then they will bond, and this bond over here will break. And so the carbon over here on the left, this carbon over here will take back its electron. So what does it look like? What does everything look like after that's done? So our methane now, it's no longer methane. It is now, if you think about it-- so we have three hydrogens. It took its electron back. It is now a free radical. It now has an unpaired reactive electron. The hydrogen and this chlorine have bonded. So let me draw the chlorine. It has this electron right over here. It has the other six valence electrons: one, two, three, four, five, six. And we have the hydrogen with its pink electron that it's contributing to the bond. And so we have them bonded now. This chlorine is no longer a free radical, although this one out here is still a free radical. Let me copy and paste it. So it's hanging around. Copy and paste. And now, notice we had one free radical react, but it formed another free radical. That's why we call this a propagation step. So this right here is a propagation step. When one free radical reacts, it created another free radical. Now, what's that free radical likely to do? You might be tempted to say, hey, it's going to just react with that other chlorine, but think about it. These molecules, there's a gazillion of them in this solution, so the odds that this guy's going to react exactly with that other free radical is actually very low, especially early on in the reaction where most of the molecules are still either methane or molecular chlorine. So this guy is much more likely to bump into another molecular chlorine than he is to bump into one of these original free radicals that formed. So if he bumps into another molecular chlorine in just the right way-- so let me draw another molecular chlorine. So that's another molecular chlorine. And each of these one, two, three, four, five six, seven; one, two, three, four, five, six, seven. There is a bond here. If they bump in just the right way, this chlorine electron might get contributed, and this free unpaired electron will be contributed and then this CH3, I guess we could call it, this free radical, this carbon free radical, or this methyl free radical, will then form a bond with this chlorine. What's everything going to look like after that? Well, after that happens this is now bonded to a chlorine. It's now chloromethane. Let me draw it. So it's carbon, hydrogen, hydrogen, hydrogen. Now, it's bonded to a chlorine. Let me draw the electrons so we can keep track of everything. We have that magenta electron right over there. And then we have the chlorine with its one, two, three, four, five, six, seven valence electrons. They are now bonded. This is chloromethane. And now you have another free radical because this guy-- and I should have drawn it there. This guy, that bond was broken, so he gets back his electrons. So he's sitting over here. He is now a free radical. So this is another propagation step. And we still have that original free radical guy sitting out over here. So we keep forming more and more free radicals as this happens. Now, eventually we're going to start running out of methanes and we're going to start running out of the molecular chlorines. So they're going to be less likely to react and you're actually going to have more free radicals around. So once the concentration of free radicals gets high enough, then you might start to see them reacting with each other. So when the concentration of free radicals get high enough, you might see, instead of this step happening-- this will happen a long time until most of the free radicals or most of the non-free radicals disappear. But once we have a soup of mainly free radicals, you'll see things like this. You'll see the methyl free radical. So let me draw it like this. You'll see him maybe reacting with another methyl free radical, where they both contribute an electron to form a bond. And then, once the bond forms, you have ethane. I could just write as CH3, H3C. So you might have something like this. And so this type of a step where two free radicals kind of cancel each other out, this is a termination step because it's starting to lower the concentration of free radicals in the solution, but this is only once the concentration of free radicals becomes really high. You might also see some of the chlorines cancel out with each other again, so a chlorine free radical and another chlorine free radical. I'll only draw the unpaired electron. They can bond with each other and form molecular chlorine again. That again is a termination step. Or you could see something like the methyl free radical. Just for shorthand, I'll write it like this: H3C. The methyl free radical and a chlorine free radical might also just straight-up react and form chloromethane, And form H3C-Cl. So this will all happen once the concentration of free radicals gets really high. Now, another thing that might happen once this reaction proceeds, and we have a lot of the propagation steps, is that you might have a situation where you already have a chloromethane, so it looks like this. You already have a chloromethane. And once you have enough of these, it then becomes more likely that some free radical chlorine might be able to react with this thing, so it might actually add another chlorine to this molecule. And the way it would do it, this chlorine over here-- I'm just drawing the free electron pairs. It would form a bond with this hydrogen right over there. They would both contribute their electrons. And then the carbon would take back its electron. Notice, all of the half-arrows. You'd be left with-- the hydrogen and the chlorine would have bonded. And now, this guy's going to be a free radical, but he's going to be a chlorinated free radical. So it's going to look like this. He has a free electron over there: hydrogen, hydrogen. And then he might be able to react with another chlorine molecule. He contributes an electron. Maybe this guy contributes an electron. This guy-- I don't want to draw a full arrow-- he contributes an electron to a bond, and then this guy takes his electron back and becomes a free radical. And then we're left with what? We're left with a doubly chlorinated methane. So then we have Cl, Cl, and then a hydrogen and a hydrogen. And this could actually keep happening. As the concentration of these get higher, then it becomes more likely that this can react with another chlorine. Of course, this chlorine over here becomes another free radical. But the general idea here that I wanted to show you is that once a free radical reaction starts-- the first step requires some energy to break this chlorine-chlorine bond, but once it happens, these guys are highly reactive, will start reacting with other things, and as they react with other things, it causes more and more free radicals, so it starts this chain reaction. And actually, all in all, this required energy to occur. This step right here, this propagation step, it requires a little bit of energy, but it's almost neutral. It requires energy to break this bond, but it creates energy when this bond is formed. It still requires a little net energy. And then things like this start to become exothermic. And especially once you start getting to the termination steps, you start releasing a lot of energy. So actually, all in all, this reaction is actually going to release energy, but it needed some energy to get started.