welcome back today we're going to discuss what's something called radicals and so radicals can be involved in a reaction in order to understand radical reactions we need to remind ourselves the difference between how bonds are broken so if we have an imaginary molecule here like this and we add some heat to it and oftentimes heat is represented with just a Delta symbol like that we can break this X Y bond by taking both of these electrons and putting them onto the Y atom to give us two ionic species a positive charge and a negative charge when both electrons represented by this double Barbed Arrow when both the electrons go to one element over another that's called a heterolytic cleavage but in radical chemistry we could have kind of the same thing the same imaginary molecule but instead of adding heat we could add light and then we could break that Bond homololytically let's see that's a c there okay homolytically now when you break a bond homolytically we draw a different Arrow a single Barbed Arrow and then we have to draw another one so the two electrons in this Bond right here one is going to go to the x one is going to go to the Y which is going to give us species that look like this with a unpaired electron you can also notice that but we don't have charges they're still neutral so that's homolytic cleavage and so it's these right here those species are called radicals because they have an unpaired electron and so now we can take radicals and do some really cool chemistry with them but as we before we can jump into the really cool reactions we need to understand the geometry of radicals where is that unpaired electron what orbital and how does that influence the geometry of the species so in order to do that let's let's uh take a look at what we already know about certain molecules so let's take a carbocation okay now let's take a tertiary carbocation a carbocation with three methyl groups now what I want to do is you see that these methyl groups are all in the plane the same plane and they're all in the same plane because this carbon is SP2 hybridized and I want to make demonstrate this that with an SP2 the electron configuration looks like this we have our SP2 hybridized orbitals right here those three and these three orbitals are used to form these three bonds but then we also have this empty p orbital right here so what we have are the three electrons in there and those three electrons are used to form a bond with these carbons but there's no electrons in this p orbital and this p orbital is perpendicular to this plane so that means that the the lobes of the orbital are coming out of the glass and into the glass so the lobes are perpendicular to this plane but maybe if we take this molecule and just take it and tip it 90 degrees all right so like take it and tip it 90 degrees it's going to kind of look like this like that okay and what I want to represent here is the fact that we have this molecule that's represented this one right here is represented like this and then if I tip it 90 degrees looking like this the p orbital is above and below my hand right and so that's what I want to try to represent so I'm going to take this molecule and just rotate it a little bit and hopefully get that concept that I'm trying to draw this flat and so when I have it flat I'm going to have my p orbital and that's the phase right there and so that is a carbocation all right now what's interesting is that if we have an anion like this let's see here let's let's use this right here actually so if this is an anion now we're converting it to an anion so that would be a negative charge and we have now the p orbital right here has two electrons in it what's going to happen to the J geometry now it's going to be trigonal planar the these electrons are going to take these three bonds right there and squish them down put them down so now these three bonds right here are not in the same plane and we're going to get a trigonal planar type of molecule that looks like this foreign and then we are going to have so we have a planer trigonal pyramidal right there so those are speed the ionic species that we've seen before okay we have this in the plane SP2 this species is sp3 so I've misspoke I've got to fix one thing that I did say in incorrect this right here is not a p orbital it is a sp3 orbital all right so let's go back and change this let's change that back to the cation and this guy to the anion okay so let's just rehash this one more time to make it clear so we'll have a carbocation we have a empty p orbital and these bonds right here are all in the same plane and it's flat it's planar when you have a anion the carbon is sp3 hybridized so we have no P orbitals they're just all sp3 hybridized and you have a lone pair in the sp3 orbital which then takes it out of planarity it's going to be trigonal plane or trigonal by trigonal pyramidal words today okay and so that affects the shape now what does this have to do with radicals okay well with radicals what's interesting is a radical is a neutral species with just one electron right so what is the geometry of a radical is a planer or is it more trigonal pyramidal what is the shape or what is the geometry and evidence suggests that it's in between these two extremes so we could say we'll cut put that back to the cation put that back to the anion the radical is going to be right here in between those two extremes so it's not going to be completely trigonal pyramidal and it's not going to be really planar it's in between but for most purposes we make the assumption that radicals are going to be planar that's not entirely true but that's just kind of what we go off of foreign but just know this this idea right here all right now another thing that we need to remember is carbocation stability remember if you had a methyl cation that pen is not working very well is it so there's our methyl cation and then we could have a primary and then a secondary and then a tertiary and these are all cations which one is the most stable cation well it's going to be this one so we increase stability from the methyl to the primary to the secondary to the tertiary now why was this the case because when our R groups are alkyl like a methyl group all right and I'll just draw the other two hydrogens like that why are tertiary carbocations more stable because of hyper conjugation remember that we have a empty p orbital on this carbon so if I could draw that but understanding that that p orbital is perpendicular because I'm drawing this all in the plane so the p orbital is actually perpendicular to this plane but I'm going to draw it like this so we can see it so I'm going to get rid of this I'm going to kind of represent the p orbital like that implying that it's perpendicular to this plane and this is phase and this is an empty p orbital Y is the tertiary carbocat the most stable because these these electrons in that Bond right there can kind of go over into that orbital to help stabilize that carbocation and this phenomenon is called hyper conjugation and so when we take a look at radicals we're going to use the same idea to explain why a methyl radical is less stable than a tertiary radical so we could do the same thing here and just say hey we're looking at radicals now so we have our methyl radical all the way to our tertiary radical and the stability is also in the same order because of hyper conjugation because we have a unpaired electron here carbon wants eight total electrons around it right it wants that octet and so hyper conjugation kind of helps well it does it helps stabilizes that radical same concept same idea I think that's very convenient here now we have what's called bond dissociation Energies Bond Association energies right there and these values are going to help us understand the stability of these radicals okay so I'm going to pause the video for a moment to erase this board so bond dissociation energies is looking at how much energy it takes to go from this methane to this methyl radical and so if we zap it with some energy we want to break this carbon hydrogen bond homolytically and how much energy does it take to break it homolytically 435 kilojoules per mole but when we look at this uh carbon that has three alkyl groups and we zap it with some energy here to form this radical you see that it only takes 381 kilojoules per mole of energy so what that's saying is foreign this radical right here is more stable than this radical because it's so much easier to rip off this Bond or rip off this hydrogen than it is for that hydrogen okay so those numbers are just telling us that hey this is lower in energy so that makes that easier to break that's what we're seeing here all right so we have I've told you the trend and now you have some data to help support that Trend okay with radicals you can also have resonance structures and typically it's between allylic and radicals so if we have something that looked like this let's recall that is an allylic cation and that makes that very stable but what if we have a radical instead so we have one electron on that carbon can we draw a resonance structure here and we can but with these rest instructors now we have to be careful and do the fish hooked arrows or the single Barbed arrows well so this this electron could come in and go there we could take one of the pi electrons here like that and then the other one going there and so what we now have would be that okay relatively straightforward but when you have allylic radicals this is very very stabilizing you could also have a benzylic radical and a benzylic radical is when you have a benzene and you have a a carbon there and that carbon came from a methyl group like so all right but if you convert that into a radical like so we would have ch2 with the one electron there and benzylic radicals are very stable because of resonance because we could take that one electron there one there and then the other goes right there and so you would get a resonance structure looking like this if so so that stabilizes it very very much so let's take a look here at the bond dissociation of these species what if I take a look at this right here so now it's a whole new problem here I have that hydrogen that I want to break homolytically to get us over here we want to form the radical how much energy does that take to rip off this hydrogen hololytically to get a radical and the number is 300 the 64 kilojoules per mole so let's pause for a second here we were looking at a tertiary radical but now we have an allylic radical you see that number right there compared to there so that's telling us that this hydrogen is easier to remove so that means that this allylic radical is going to be more stable than the tertiary radical now granted this isn't the radical but it turns into the radical right get that so just Envision the tertiary radical so the allylic radical is going to be more stable than the tertiary radical now what about this benzylic one how much energy does it take to rip off that benzylic hydrogen so we'll go like this and draw that hydrogen how much energy does it take to rip off that hydrogen right there and the data says 356. and that would be kilojoules per mole so what does that number tell us hey the benzylic carbocation the benzylic radical is going to be more stable than the allylic which is then more stable than the tertiary tertiary radical isn't that pretty cool so just looking at these bond dissociation energies that can help us understand which radical is going to be more stable [Music] so let's see what's next another radical to be aware of is a vinelic radical so if we have something like this let's see here okay and [Music] I want to rip off this hydrogen right here to give us a Vine lick a radical so you can see that this is allylic because the uh this of electron can resonate in to give us this or it can resonate in here we can't do any resonance over here on this carbon right because this carbon already has three bonds so there's no resonance like this that can't happen because that would form a Texas carbon so this is a vine lick radical all right and so what is the energy required to rip this hydrogen off right there that one is 464. kilojoules per mole so that is even what that's going to be less stable than that guy isn't that crazy yeah wait wait the the 435 yeah look at that this vinelik radical does not form this is so unstable look at this our methyl we've learned that methyl cations are so unstable they don't form and it's very similar to methyl radicals very unstable but this Violet guy even worse off okay [Applause] so please note that when you have a Vine lick there's no resonant structures okay it's not stabilized by resonance so it's very unstable so recall if we had a molecule that looks like this and we have a carbocation right there what's going to happen what you always always have to look out for well always double check what type of carbocation you have that is the secondary carbocation then you have to ask can it rearrange in such a way that can give you a more stable carbocation so we could have a one two alkyl shift so that would give us this molecule right here and now what's the driving force or the incentive for that to occur well it goes from a secondary carbocation to a more stable tertiary carbocation that's a phenomenon we see when we have carbocations but when we have a molecule that looks very similar but now it is a radical there's going to be no rearrangements and that's a really cool feature or that's that's just a feature of radical reactions radicals don't rearrange typically so what I want to do now is I want to go through six different uh cat what do you want to call it classes no six patterns of radical reactions that if you understand these six patterns then you can do most if not all of the radical reactions that we're going to discuss so let's take a look at them one at a time this one we've seen already where we have the X and Y right there and if we add some light light is going to be represented with the Planck's constant times new or you can use Heat and that is represented by the delta and so we could break these bonds homolytically using these two conditions which we said looks like this so that's homolytic Bond cleavage which would give us two radicals so that's pattern one and we've seen that already another uh pattern is addition to a pi Bond so addition to Pi Bond that is when you have a pi Bond all right so here's our PI Bond and you have a radical and you're going to add that radical to that Pi Bond so the way that happens is we're going to take one fish hook Arrow there and then you'll well the way let me redraw it this is kind of weird okay we're going to draw this like in space and then one of the electrons in this Pi Bond are going to come and meet it like so and so when these two arrows meet together that's saying we're going to take one electron here and one electron in the pie and we're going to form a bond and and then the other electron in this Pi bond is going to come and go there so the reaction looks like this now we're going to have this our radical there and then we're going to add our X just like that so this Bond right here between the X and the carbon is forming because we're using this electron there and this one electron that's coming from the pi Bond so that's called an addition to a pi bond pattern now we can also abstract hydrogens off of molecules here so this is called the hydrogen abstraction what if we have a radical like so and we have a molecule like this a generic one and we're going to rip off this hydrogen how are we going to do that well the way this is going to happen is this is going to come and then one electron in the sigma bond is going to come like that okay and then the other electron in that Sigma bond is going to go to the R Group and so what we're going to have now is that x h we've abstracted that hydrogen off and then when and then we are left with a radical alkyl group kind of crazy so these arrows are a little bit different than what we're used to seeing right we're used to seeing the the attacking species go to where it's attacking but now it's not so much like that we're drawing the arrows into space and so when these two arrows come together we're using that one electron one electron to form that Sigma Bond right there you can do a very similar reaction as this one at the bottom but with halogens so we could look at a halogen what if we had a alkyl radical like that and we just treated it with a normal halogen like bromine all right so what can happen here is let's use a different color that can go like this that can go like that and then one can go to that X so now what we've done is formed a bond with the RX and then the leftover on this one is going to be a radical so that's a halogen abstraction the next one is called a elimination that is the opposite of the addition to a pi Bond so the so the elimination is the opposite which means we're going to take this one electron stick it there one electron here go like that and then the other one we'll go back on the X and so that is going to give us a x radical plus let's see double bond there are alkene and then the last one or the sixth one is called the coupling pattern and that's when you have two radicals next to each other so here's a radical there's another radical like that and those two can react with one another we could put that there that there so what's that saying that's going to form a covalent bond so we would have something like that so we're going to go through reactions now and in those reactions we could use all six of these patterns or just a few of them but instead of just in general terms now with r's and X's we're going to now start looking at some real examples so we can get a better feel for what's going on okay so here on the board I have the six patterns that we just went over now these six patterns could be classified into three groups and so this right here this piece right there is just that first one right there is going to be called the initiation step so when we do reactions or when we do radical reactions there's going to be three steps and so we have three classes here so right here is another step called propagation and then the last step right here is called the termination steps so we have those the six patterns and now they're organized under these three other terms called initiation propagation and termination and this is basically just showing you the reaction process in order to get a radical reaction to go you have to initiate it which means you have to get it started so in the initiation step you have to get a radical generated because molecules don't just hey ought to be radical you have to force them to do it so we use the homolytic cleavage to generate a radical and that's going to initiate the reaction once you have a radical then you can do these steps and they just these steps just are a reap on a repeating cycle they just keep happening and happening and happening until you stop the reaction by doing a coupling and what a coupling step does if you recall it gets rid of the radicals is going to take two radicals like that and when these two radicals couple they form this and so now there's no more radicals and when you have no more radicals the reaction is terminated reaction stops so when we do the mechanism of our radical reactions we have to realize there's going to be initiation phase propagation and then it's going to stop or be terminated and so now it looks like we want to we're going to start looking at some reactions here okay so let's take a look at that