in this video I'm going to cover the topic of electrophilic addition uh one of the reaction mechanisms in the as the AQ specification um for chemistry so with any of these sort of topics where you've got new words like electrophile and maybe addition within this concept of of this um this reaction mechanism you got to think well what do the terms mean and actually once you sort of get your head around what the terms mean it opens up and and actually makes the mechanisms that much more sort of logical um so we'll start with the term electrophile so an electrophile what's that going to be well if you've already looked at nucleophilic substitution then you should know what a nucleophile is and you should know that a nucleophile uh the way that it can be defined is as an electron paired donor but also it's something that is attracted to areas of positivity and so really electrophile is the complete opposite it's attracted to negativity so I just put attracted to negative and I'll put that in speech M so now in terms of a definition it's an electron pair acceptor so it is the complete opposite of a nucleophile on the other hand um addition with any addition reaction and that could be uh addition polymerization as well any addition reaction we find that things are added together surprise surprise but what that means is when that added together we get nothing else produced so two things are added together so it's almost be like so of A and B go to see there's only one product and that's really really important that's the kind of reasoning for the addition now so we've got one thing being added to one thing one of these is going to be our electrophile creating one product now in the case of electrophilic addition um the organic side of it the organic reactant is it's always an Aline note Aline not alkan and so we're looking for something uh for example ethane so we're looking for a molecule which contains that classic double bond and it's this double bond here that actually links through to this attracted to negative so our electrophile is invariably attracted to this area here the carbon carbon double bond where we have what we can class as um an electron dense area or an area of high electron density and that's the key thing about our carbon carbon double bond as with nucleophilic substitution again there are three particular electrophiles that you must know um that could be asked in your exam the first two we look at are more common the second the third one I should say uh less common but it has come up in the past um and these electrophiles are as follows so we have first of all bromine molecule br2 never ever write bromine as a standalone thing the only time you'll see that bromine on its own would be if it was an ion or if it was a free radical if we're talking about General bromine we're going to be looking at bromine as in br2 the bromine molecule this diatomic molecule so looking like that with a little calent bond in the middle the other one of electrophile hydrogen bromide final one sulfuric acid now what I'm going to do is I'm going to work through these in this order starting with bromine showing you the mechanism which actually is exactly the same for all three um but there's a slight different reasoning between this one and these two as to as to how the mechanism starts but once you've got the mechanism down it'll be no there'll be no issue actually applying it to any sort of scenario whether that be different molecules from these as well You' just got to realize that you're dealing with electrophilic addition and that's where the kind of the alkan see an Alan in the reaction as a reactant it's going to be electroic addition so let's have a look at this mechanism then okay so this is our starting position um for our mechanism we've got in this case a molecule of ethane and we've got the bromine molecule there as well so the br2 c2h4 now at this point you might be thinking or might not but you hopefully would be thinking well how can this occur how can this as a mechanism actually happen if you think back to the nucle fit substitution or if you've not done it look at it in that case we have a difference in polarity which leads us to um for the nucleophilic attack to PL take place in this case there's no difference in polarity within my bromine molecule I've got certainly no permanent dipoles here at all and what actually we find is that if you imagine these two things sort of floating around the screen as they come close to one another these are electrons actually cause repulsion within the bromium molecule and that repulsion means we end up with an induced dipole now what I'm saying in this sense I actually I'll rub this out and make it a little bit clearer so what happens is it pushes the electrons from one toward the other so in this case what I'm saying is the electrons are being pushed from the top toward the bottom and this is in reducing a dipole and we now have this region here of Delta positiveness Delta positivity and this region of Delta negativity at the bottom here and this now means that actually a reaction can take place and this is the starting point now you don't need to include these in the mechanism but I always think it's a good idea to do so to just get it a bit clear bit more clear in your head so first step doing a different color First Step always from from the double bond nice curly Arrow toward the bromine here so what we're going to find is we're forming a bond here and in doing so these electrons are ejected and they move onto this guy here onto this bromine so what we end up with is is an intermediate stage which we must include and it looks something like this single Bond now we've got rid of that double bond this pair of electrons here it's got a different color this pair of electrons has moved and is allowed for a bond to be formed between the carbon and the bromine so we now have a Broman attached as well as obviously our two hydrogen from before one and two there and over here we have a carbon with our two hydrogens now we're missing any sort of bond here at all what that means is at this point we've created what's called a caroa so this here is now a a well it's actually a positive carbon the reason for that is that if you can imagine within this carbon carbon double bond here within this single calent Bond of the two one of those electrons from the pair was from this carbon for example and what we've done is we've taken both of those electrons and we've switched them round so the electrons were here we've now moved them and they're now here well this carbon is still using it still has that same electron from before involved in the bond but this carbon now has lost that electron over to here and it's if you like been donated to bromine to allow for the bond to be taken to be created which means this carbon is deficient now you don't if you didn't understand that I firstly I apologize but secondly you don't need to you just need to be able to include this plus here that's very important because that carries a mark on its own you must have that it must be on the right carbon which in this simple simple example here could be on either but you just switch it completely around it would be the same same molecule what we find then is we now have a bromine negative ion now and it has a cheeky little lone pair and what can happen now is that this comes round and now we are doing the opposite the bromine is now in in essence acting as a nucleophile it's donating those that lone pair to here to create a bond between so and you don't need to draw the product but I will in this case so we go in between there and then here we end up with carbon carbon hydrogen up here another hydrogen up here hydrogen another hydrogen and then finally two bromines now so from what I said at the start you can now see where that addition is coming in we started with two molecules we now end with one molecule there is no waste product there is nothing else produced just this and in this case we have Di bromo ethane so di bromo ethane and I should of course include the numbering here one two di bromo eone oh missed the O there the other time you've actually seen this and this is a reaction that people don't invariably they don't remember this you've seen this back in sort of year 10 time and this actually here is the reaction or the test I should say which is reaction but the test for an alkan so if you remember bromine water goes from Orange to colorless that standard thing test for an alkan well this is it this is actually why it's happening the bromine is is being added onto the double bond across the double bond forming a new molecule thereby changing obviously the alken into the DI bromo alkane and the bromine in doing so the color changes from Orange to colorless so this is that test that you've seen time and time again be aware at this point that although I've used broing here there could be any molecule it could be chlorine it could be any of the well that's about it really just chlorine would be the other one I can think of possibly Florine I guess if they wanted to do that but you're more likely to see bromine or um chlorine in these examples and again all you do there you just changed the bromines here for chlorines everything else would remain the same negative the arrows in the same place lone pair of electrons and all the rest so what you could get for your find for this question here potentially could get four marks of this one mark for your initial Arrow one mark for the second uh one mark for this what's the structure Mark and then finally a mark there as well so four relatively straightforward marks okay what I'm do now is we'll look at the hydrogen bromide example um and then we'll finish off with the sulfuric acid molecule okay so stick with ethane again this time we have hydrogen bromide reacting with ethane and what we find is that this is actually a little bit more straightforward to understand because here we have uh a difference in electro negativity between my hydrogen and my bromine therefore we have a more naturally occurring if you like um polarity there uh a permanent dipole between the hydrogen and the bromine and so exactly the same though as before with the obviously the Delta positive and Delta negative we have the exact same reaction find my red that comes down to here that goes across to there we form the intermediate and the intermediate this time is again slightly differently than before but similar concept and it doesn't matter whether you do this on the other side or not I just tend to prefer doing this positive there carbo cation as before get my bromine involved negative ion lone pair of electrons Arrow across and there we go one mark two Mark structure Mark fourth Mark for the arrow four fairly straightfor marks the mechanism hasn't changed at all we have used exactly the same mechanism there is absolutely no different besides the starting um molecule here but the arrows are exactly the same you could actually have as a template and just put your various um various atoms or ions in the right place and you would be you would be you would be away there is one extra point you can that can be drawn into the hydrogen bromide and that is that whereas with the bromine we've got two of the same atom being added across and so there's no difference and in this case with the simple Ethan starting product it doesn't matter where the hydrogen the bromine goes we're going to get the exact same thing occurring if we started increasing the size of the molecule and say we took propane for an example then actually be find that it changes slightly um and what we would get is that let's take propane so we've got our molecule drawn exactly the same concept as before my hydrogen bromide comes in and it's going to follow exactly the same pattern as before now what we can find is that if we think about this now the hydrogen has two potential options of where it can Bond and that would create two different molecules we could have I'll draw the two options out here the two intermediates so we could have that or we could have that now quite difficult to see just in this case what I'm got here is that this is the hydrogen that came from the hydrogen bromo in this case and this is him in that case so you can see he can either join to this carbon in in this first example or he can join to this one and in doing so we form a different carbocation and that's a very very important difference and if the bromine were to join here we would form one bromopropane if it were to join here we would form two bromopropane two different molecules entirely and so we've got a um positional isomerism occuring there now in reality it's it's not quite as straightforward as it could just join in either place the carbo cat that is produced as the intermediate will be the one that is more stable and the way the stability works is more stable equals more alcohol groups attached I should say more stable carbo CA equals more alkal groups attached to that carbo cation so if we would look at this example here choose a SL different color this carbon here has one alal group this whole thing here whole thing around that is an alkal group yes it's it's an ethal group it doesn't matter it's an alkal group it's just treated as one alkal group and what we could say is we would say that this because it has one alkal group attached this is a primary terrible P there this is a primary carbo cation on the other hand this carbon here has has one two alkal groups both methal groups and you might be thinking well this is an ethy that's bigger that's better no it's the number of alkal groups that makes the difference so two methy groups here means that we have a secondary carbo cation and there's a shorthand way of writing these primary you can write as one degree and secondary you can WR as 2 degrees and it kind of it links exactly into the one is the number of alkal groups that two is the number of alcohol groups and so what we find is this is the preferential caroa that will be produced because it is more stable and the reason it's more stable is because these guys here I'll choose a another color these alkal groups um we get something called the inductive effect occurring and to put it more simply this is more more advanced really than you need to know but these push electrons if you like onto the carbo cat therefore spreading electrons around the molecule more evenly and that creates a more stable overall molecule um so this is the one choose and therefore we will ultimately end up with see if I can find another color yep I will ultimately end up with not the clearest one in the world but the bromine attacking onto this one and so my major product if you like my major product is to bromo propane and my minor product would be this one up here the one that's less likely to be formed minor product one bromo propane so that's the hydrogen bromo so you can see it's a little bit more little bit more advanced but actually it doesn't there's not a huge amount extra it's it's just this part here with the stability and once you get your head round the idea of the primary versus secondary and we could go one further and say oh we've got tertiary which would be um 3 Dee then that would be three alkohol groups joining that's the most we can get to on this point with the carbocation so tertiary if you ever see quaternary that just means four um as opposed to uh the tertiary being three secondary being two primary being one so that's really the key thing here um and that doesn't come up very often on exams um to be perfectly honest with you in a more simple way you would just have to be able to add the hydrogen bromide onto the molecule they be aware they may give you a product which they want you to form so they might give you a starting um point they may give you that you're producing two bromopropane and so you need to draw your mechanism as is appropriate you can't draw it the wrong way around and somehow miraculously form this the product they want you to the mechanism must actually create the product that they're asking for and that's how they can get around needing to ask you or needing to delve into the idea of stability if they were to wish okay so finally um gone with that one long enough we've got sulfuric acid Okay so we've got an Ethan molecule as before um and this time we're going to have a sulfuric acid molecule now what I would advise in an exam and it's I would advise this but if you really can't remember how to draw it then don't I would advise that certainly you learn the structure of the sulfuric acid molecule and it looks like this so we got h24 h204 you could draw it as this um but it's a little bit sort of sloppy I'm not entirely sort of happy with that I would stick with this molecule here in all honesty also because it allows you just to see a little bit more clearer kind of what you've got and similarly to the hydrogen bromide we've got a region where we've got polarity here Electro negative not very Electro negative therefore we've got a Delta negative region we've got a Delta positive region and exactly as with the previous two we get that Bond coming in there curly Arrow we get that moving onto there we have an intermediate exactly as before um and we'll I'll stick with what I've done previously carbon hydrogen carbon hydrogen carbon hydrogen carbon hydrogen oh uh hydrogen there positive region there this time we have obviously a negative oxygen there um going to our sulfur oxygen hydrogen like that and finally we would get this is a lone pair by the way terribly drawn but that's a lone pair we would get that moving across there what we produce here and I'm going to draw this um out for you um sort of anyway so what we get is we have our carbon hydrogen hydrogen hydrogen carbon hydrogen hydrogen lost my hydrogen up there uh and then onto here we have this o and we would have S SO2 o I'm actually going to change this thing here if you want to draw it as a short hand don't draw it like that in fact I apologize for this do it as h o s SO2 oh H H2 S4 draw it like that I would stick to keeping the whole molec drawn out unless you can you know you're going to make mistakes but to be honest there's learning that that there's not a huge amount to learn here just just get by and just do it really U and this is what you would produce here so and this is named um as per like an alkohol group here so it' be called ethy hydrogen sulfate um a couple of things really about this um so we've got our ethy hydrogen sulfate what we can do and this actually comes up sort of in it's kind of in the alcohol's topic as well in the idea of hydrating an Aline into an Al alol if we were to add water now so if we add H2O we would produce in this case ch3 ch2 o we produce ethanol and actually we reform this would reform the sulfuric acid and if you think if you've done the alcohol topic already then then great if you haven't you will come on to this but in the production of ethanol from ethane we use concentrated sulfuric acid or phosphoric acid in this case we could use the sulfuric acid as a catalyst and you can see this is actually the mechanism for how it would work there would be a further mechanism occurring here which you would not need to know in terms of an exam situation if they're asking you to use the sulfuric acid as the electrophile they're expecting you to end here not to draw a mechanism to take it to the alcohol they're expecting to understand that this can go to an alcohol but not to draw the mechanism for that the mechanism stops here doesn't go any further okay quite a lot there but hopefully that's been of some help um and there you have it that's electrophilic addition