all right guys in this video I'm going to do a summary of every single curly Arrow mechanism you need to know for organic chemistry under the AQA a-level chemistry specification all right I'm not going to go into the very fine details of forming reagents catalysts all that sort of stuff but I'm going to go through everything you need to do regarding drawing the actual mechanism to get the marks on your paper so let's start with nucleophilic substitution okay and this is where you're going to start with a Halo alkane or a halogenoalkane however you want to call it completely fine so I'm going to use the example of bromine you could also use chlorine that's completely fine these are the most common ones you're going to see okay bromine or chlorine so nucleophilic substitution it's pretty self-explanatory we're going to have a nucleophile attacking this Central carbon here whatever the relatively positive carbon is attached to the halogen and it's going to substitute that halogen for whatever the nucleophile is okay and there are three main nucleophiles we need to be aware of so as I mentioned there's going to be a Delta positive Delta negative dipole existing here due to the electronegativity of the bromine all right so before I actually jump into the mechanism itself I'm going to give you some really useful knowledge that I just wish I knew when it came on to answering these mechanisms right what is it what is a curly Arrow show us a two-headed Arrow it's showing us the movement of a pair of electrons okay and this is either going to be from a lone pair like this onto something else donating its lone pair so let's say we're using our example of oh minus right you're going to be donating your lone pair onto something else in order to form a dative covalent bond okay now you don't want to say this in your responses because it's not in the spec but this is what is happening okay you're forming a covalent bond specifically a date of bond you're donating both of the electrons in this lone pair to form a covalent bond right so you're either doing that or you're breaking a bond and showing the movement of the electrons in a different way so you have a bond here you're going to be showing this bond breaking and going on to something else I'm just going to show it as the symbol X okay so that's all a curly Arrow mechanism it is it's either showing the forming or breaking of covalent bonds okay so get your head wrapped around that and it should make the rest of these videos pretty simple you just have to remember the nucleophiles and electrophiles involved and we should be fine let's start with the hydroxide we're going to have our o h minus making sure to draw our lone pair so make sure you're drawing the arrows starting from the lone pair don't start over here or over here or something like that or or from the negative always started from the lone pair okay that's where the electrons are traveling from and we're going on to this Delta positive carbon right here now this carbon is now going to have five bonds essentially and it doesn't want that it's group four so it's going to become energetically unstable so this is going to become our leaving group and the electron pair within this covalent bond is going to donated onto this bromide ion okay so we're going to be forming our product here h3cc H H the bromide owner has left in solution and we have formed an alcohol and then we're going to have our bromide ion going off in solution now these Delta positive and Delta negative charges you don't need to show this in your exam but you do need to understand the theory regarding dipoles electronegativity and how it determines the leaving group Etc but I'm not going to cover that in this video okay so that's our hydroxide nucleophile out the way let's go into Cyanide and it's literally as simple as this guys so if I rub this out I can just replace these two with the cyanide ion exactly the same thing right so we'd have our CN here and CN here literally exactly the same thing just a different nucleophile so our third nucleophile is slightly different we're dealing with ammonia here and there's an intermediate stage all right so let's draw this out we'd have our nucleophile ammonia coming in attacking the carbon and then we're going to have it bonding onto here now I'm going to show the hydrogen bonds here just to signify what's going on and then again we're going to have our bromide ion off-in solution now this nitrogen is not going to be happy with having four bonds around it it only wants three okay so it's going to become positive because it formed a dative covalent bond it donated both of the electrons in this lone pair okay so this is positive now one of these hydrogens is going to be like you know what take my electron it's all good and it's going to donate it onto the nitrogen okay it can be any of these free hydrogens it honestly doesn't matter just pick one draw the curly arrow and you'll be fine and this is going to form us our product here do you remember what this functional group is called this is a primary amine okay you're going to have your ammonium bromide Okay the reason that you get ammonium bromide is that this hydrogen that leaves is actually reacting with the NH3 so you've got more ammonia in solution okay this is going to attack this break this off and it's going to form the ammonium ion which is going to react with the bromide ion and it's going to form ammonium bromide okay but you don't need to show that all you need to show is this right here okay now you need to be aware that we can get subsequent nucleophilic substitution reactions occurring here and why is that the reason being is that this product that we form this primary amine because nitrogen is group five we're going to have a lone pair on this nitrogen here because it's just been donated back by this hydrogen okay so you've got to keep that in mind so what can happen here is this can act as a subsequent nucleophile attack the initial reagent here and Carry Out additional substitution reactions okay so let's draw that out quickly and we can look at how we form a secondary amine in this way before we move on to the secondary amine formation in the subsequent nucleophilic substitution reactions I realized that I left this negative charge here when I was dealing with the hydroxide ion and the cyanide ion this is incorrect okay there is no negative charge on the ammonia really keep that in mind my mistake there so with that done then let's draw our product that was formed our primary amine in the previous nucleophilic substitution coming in and attacking this haloalkane so we'd have our nh2 ch2 ch3 that was formed previously and this is going to come in and attack this relatively positive carbon the bromine is going to be our leaving group again nothing else has changed here so very similar to our ammonia example we're going to be having a positive charge on the nitrogen here in the amine so the ammonia that was present initially is going to come in with its lone pair attack this hydrogen this is going to break off onto here donate its electron pair onto the nitrogen to make it neutral again okay and what we're going to form we're going to form a secondary amine okay so it's going to be ch3 ch2 and then we can have our H and then the rest of the molecule here ch2 ch3 again same thing this nh4 that formed is going to react with this forming ammonium bromide okay and this is a secondary amine because there's two carbon chains on nitrogen but it doesn't stop there guys we can actually form a tertiary amine and a quaternary amine as well now quaternary amine is not actually called a quaternary amine it's called a quaternary ammonium salt okay and that's due to the charge of the nitrogen so what I'm going to do is I'm actually going to show how simple this is and keep the mechanism the same but just changing our nucleophile so in this instance from the previous mechanism it was a primary amine I'm going to change this to be our product our secondary amine okay so it's going to be our NH then we're going to have our one carbon chain ch2 ch3 and then the other carbon chain ch2 ch3 all right so this is our secondary amine acting as a nucleophile coming in attacking the carbon bromine leaves exactly the same thing here and all I'm going to switch up in our intermediate is the fact that this hydrogen is now an e-file chains ch2 ch3 and then in our final product this is no longer a hydrogen okay it's going to be another e-file chain ch2 ch3 as we can see in our molecule here so if I rub this out quickly so we can have space to add it in ch2 ch3 okay this is a tertiary amine that is our product in this case as I mentioned it doesn't stop there you can react subsequently with this tertiary amine to form a quaternary ammonium salt okay so let's do that right now we rub this H out and change it for a third e-file chain ch2 ch3 we're not actually going to get our intermediate our intermediate is going to be our product so let's rub this out to make that very clear rub this out as well we're not going to get subsequent reactions occurring this off this is our final product so our e-file chain is going to slot onto here instead of being a hydrogen okay so it's a bit messy but that is essentially how we would form a quaternary ammonium salt with nucleophilic substitution and these are all subsequent reactions okay that occur in the instance of ammonia okay so still on the topic of Halo or halogenoalkanes let's look at elimination reactions okay and this is going to be with hydroxide ions again so let's draw out our starting reactant so we've got a Halo Arcane here right we're going to have our hydroxide ion coming in and attacking one of these hydrogens all right and we're going to be forming a double bond because our product is an alkene so I'm going to do it coming from over here right so I'm going to go this lone pair is going to be attacking this hydrogen donating its lone pair to form a bond this is going to be leaving the molecule and then this covalent bond right here is like what the hell do I do now if this hydrogen is leaving it donates the pair onto this single Bond here this saturated carbon carbon bond to form a double bond okay now very similar to nucleophilic substitution if we add in an additional Bond here an additional pair of electrons this carbon bromine bond is Delta positive and Delta negative there is a dipole here region of electronegativity on this bromine because it's a halogen so this covalent bond is going to break off and the electrons are going to go onto the bromine is going to be our leaving group okay so that's a real easy one there now you just have to get comfortable with the fact that if this hydroxide ion is attacking one of the hydrogens the covalent bond between the carbon and the respective hydrogen is going to break off and form a double bond just remember that because this can happen from either side for example let's say we switch this up and we wanted to attack this hydrogen right here which can definitely happen Okay so let's say we had our hydroxide ion again comes in attacks this hydrogen now at this stage you might be like oh man like what do I do here just remember that the bond between the carbon and the hydrogen that's being attacked is going to break off and form a double bond where does the double bond form the double bond is going to form to the adjacent side of the carbon so in this instance we're attacking this methyl group right here right so the double bond is going to form here now if we were attacking one of the hydrogens on this me file the double bond is going to form here okay just keep that in mind but that is the gist of elimination of a halogenoalkane or a haloalkane with hydroxide ions right so we formed an alkene right here so let's draw that out real quick because I forgot to draw the product here so it's going to be ch3 C in this instance the double bond was formed in the center so it's going to be like this there's going to be a hydrogen here and then a hydrogen here hydrogen here okay and then on bromine again is going to go off in solution and this is going to form water so it would have plus H2O okay simple as that so on the topic of alkenes then what mechanism do we need to know for alkenes we need to know electrophilic addition okay you also need to know acid catalyzed addition for the hydration of ethene but we'll look at that after electrophilic addition is what we're going to focus on here so what is an electrophile the opposite of a nucleophile is an electron pair acceptor and with electrophiles you're always going to be dealing with regions of electron density okay they love that negativity and a alkene is perfect for that because this CC double bond here there are two electrons per Bond okay so this is a region of four electrons essentially which is very electron dense and very negative okay so if we draw out the remainder of this molecule I'm going to be using ethene as our example here to make it simple for us what electrophiles do you need to know okay you need to know diatomic halogens so for example cl2 br2 Etc is going to look like this you need to know a hydrogen halide so for example hydrogen bromide hydrogen chloride so let's draw it like this and you also need to know sulfuric acid h2so4 okay so let's break down those one by one and see what's going on here all right I'm going to show you the way that I remember it and I think it makes it as simple as possible so you can simply just replace one electrophile with another so let's start with our diatomic halogen so I'm going to use the example of bromine but you can use chlorine as well that's completely fine so we've got our diatomic bromine here bonded by a covalent bond there is actually a dipole present you may not think it because they're identical atoms but there is a Delta positive here and a Delta negative again you don't need to show this in your exam I'm just explaining it so you can understand why the arrows are moving where they're moving so the electrons in this double bond are actually going to repel the electrons in this covalent bond right here and induces this dipole okay that's how it works so the closest atom is going to be relatively positive and the atom further away is going to be relatively negative okay so it's going to induce that dipole here now what's going to happen is an electron pair in this double covalent bond is going to be attracted onto this relatively positive bromine and then this other relatively negative bromine is going to break off like this okay so that's our first two curly arrows the electron dense double bond is attracted onto the bromine here this first bromine atom which is relatively positive and then that is going to form a covalent bond here and then this bond is going to break the electrons are going to go onto this bromine and it's going to go off in solution as a bromide ion okay really important to keep that in mind so we're going to form an intermediate here okay this intermediate for these electrophilic additions is always going to be a carbocation okay you're going to get varying types you're going to get primary secondary of tertiary carbocations and they all have varying stabilities which is something you need to know but I'm not going to cover it in this video because it goes into quite a bit of detail so this double bond breaks okay and we're going to be left with a single single carbon Bond but this bromine has now bonded on to one of these carbons so I'm going to draw it right here we have a bromine down there hydrogen here hydrogen here now this poor carbon right it's lost its double bond here so it's only going to have three bonds still got the two hydrogens but it's lost it's double one so it's not a happy guy it wants four bonds two fulfill that octet those eight outer shell electrons but it's going to have a positive charge here because it's essentially lost one of its electrons here so it's been oxidized right you can essentially think of it like that so what happens now is because this is hella positive all right this is not Delta positive it's not relatively positive to the surrounding atoms it is full on a cation right so what's going to happen is this bromide ion that broke off is going to come and be like you know what I'm getting some Revenge here you took my best mate and it is going to attack this carbon to be reunited and form our product what is our product going to be I'm going to draw out real quickly you're gonna have to do it a bit smaller because I ran out of space so it's going to be our c c h and then the bromide ions are reunited all right so dibromo ethane this is our product okay so that's the reaction with a diatomic halogen out of the way let's look at the reaction with a hydrogen halide very similar process and I'm going to try and make it as simple as possible for you guys and you're going to see right now how identical it really is so let's rub out our products real quick so you can see what's going on here so with a hydrogen bromide or a hydrogen chloride all we're going to do is we're going to change one of these starting atoms okay so we're going to switch out one of the diatomic bromines is going to become a hydrogen okay now these are no longer identical so we're actually going to get a dipole existing already because a halogen is far more electronegative than a hydrogen so there's going to be a dipole existing here regardless of this all right but we're going to get the same thing occurring okay this region of electron density is attracted to this relatively positive hydrogen and this bromine is kicked off and accepts the electrons in this covalent bond so it has a lone pair on it right so in our previous mechanism we had a bromine here easy peasy that is just replaced with a hydrogen okay exact same thing occurs here this bromide ion that was kicked off is going to come in it's Mega attracted to this positive carbocation and we're forming our product okay so I'll draw out our product real quick we're going to have ch3c h h b r okay bromo ethane right here we're going from a alkene to a halogenol alkane or a Halo alkane however you want to call it okay so it's not a DI Halo Arcane it's just a single Halo arcade right so next one sulfuric acid now again I'm going to make this as easy as I possibly can for you guys let's rub this out and just replace it with our h2so4 now I like to draw it like this okay so let's rub this out you can draw the full molecule so if you have your h2so4 you can have your s o o h o h okay that's our h2so4 I don't advise doing this just to make it easier for you guys what you want to do is you want to start with the hydrogen here just like with your hydrogen halide and then you want the remainder of the sulfuric acid molecule drawn out like this oso2 oh okay makes it real easy for you because you can think of the oso20h as the halide so for example bromide ion okay just replace that with the oso20h I'm gonna get exactly the same thing occurring here we're going to get a dipole between the relatively positive hydrogen and the relatively negative oxygen within the sulfuric acid molecule pretty simple here hopefully and then we're going to get exactly the same thing occurring whereby the hydrogen is bonded onto one of these carbons the double bond is broken and then instead of a bromide ion coming in we're going to get the remainder of the sulfuric acid coming in that broke off of this molecule here so it's going to be oso2oh minus okay so H so4 minus that is that right there but just written out in a slightly weird format but that's how you can do it you can do that in your exam completely fine AQA will give you the marks and it saves you having to draw out the full structure and then our product is going to just be h3c C H2 and then we're going to have our oso2oh molecule there okay A bit squished up but hopefully you get the idea and what is this product referred to as it's going to be referred to as an alkyl hydrogen sulfate okay right next up we've got another one involving alkenes specifically ethene okay so I'm going to use this as our starting molecule Rubble this out so we've got the hydration of ethene here okay so what's going to happen is we have a h plus here all right and this comes from our strong acid phosphoric acid okay concentrated for suric acid which is h3po4 if you weren't aware and this region of electron density is going to be attracted to this positive hydrogen ion okay so this double bond here one of these bonds is going to break and we're going to form our intermediate here so I'm going to try and stick to the same color so you can see what's going on then we're going to have our hydrogen here that came from this starting hydrogen ion and what's missing here is a positive on the carbon because it's only got three bonds now instead of four so it's a carbocation the next thing that happens is water comes in right so we've got a water molecule here of a lone pair on it and two hydrogens attached this comes in donates its lone pair onto this carbocation forming a dative covalent bond and we're going to get another intermediate here let's draw this out so we've got H through c c h and then we've got the oh that just bonded on let's switch up the color so you can see what's going on now because it formed a dative covalent bond it's lost one electron essentially it's been oxidized so it has a positive charge it doesn't like it guys it wants to be stable it wants to be neutral so one of these hydrogens one of his best Mates is like you know what take my electron it's all good you can get back to being neutral and we're going to form our final product here what is that going to be I know this is a bit messy but hopefully you can follow me is going to be an alcohol okay so we're going to have our ch3 ch2 o h okay and this hydrogen goes off plus h plus and reforms the Catalyst h3po4 okay okay so that's Halo alkanes and alkenes out the way let's take a look at alcohols there's only one main mechanism you need to know here and that is the acid catalyzed elimination reaction for the formation of an alkene from an alcohol okay so let's draw our starting reactant out so we're going to have our oh group on the end here for our alcohol and there's a lone pair on this oxygen okay this is going to be attracted to a hydrogen ion okay this hydrogen ion comes from a concentrated strong acid so it can be sulfuric acid or phosphoric acid similar to our previous mechanism so the hydrogen ion is going to come from that this is attracted to this positive hydrogen ion and it's going to form a dative covalent bond here so let's draw out what our intermediate is going to be I'm going to show all the hydrogens bonded onto the carbon this time instead of just shorten it to a ch3 methyl group so we're going to have our c h and now the oxygen has got an additional hydrogen on it so we're going to have like this okay so the carbons are happy they've got four bonds each but the oxygen is not happy because it just formed a dative covalent bond and essentially lost one of its electrons okay so what we have to do is we have to signify that it now has a positive charge so this carbon's like you know what mate you're really unhappy You're Bringing Down the Vibes I'm just going to donate my electron onto you so you're happy and then this goes off as water in the solution and one of these hydrogens is going to donate its lone pair and form a double bond okay what is our final product going to be it's going to be an alkene okay so in this example it's just going to be ethene but it will change depending on what your starting alcohol is how many carbons there are in the chain Etc and then this hydrogen that left the molecule is going to be a h plus ion and it's going to reform the concentrated strong acid so either our h2so4 or our h3po4 so one of these two is going to act as a catalyst and be reformed okay so next up we're going to be having mechanisms involving ketones and aldehydes right and the only mechanism for this is nucleophilic addition we're going to have two separate nucleophiles you're either going to have a negative hydride ion or you're going to have a CN cyanide ion now let's draw out our starting reactant here I'm going to use the example of propanone right this is the Ketone if you want to switch this up to be an aldehyde we just swap the end carbon chain for a hydrogen to denote this functional group which is an aldehyde but I'm going to show them with a ketone in this example so as I mentioned the first nucleophile we're dealing with here is a hydride ion which is a negative hydrogen ion okay so this is going to come in and very similar to our previous examples oxygen very electronegative atom when you see an oxygen you should straight away think okay this is electronegative and whatever is bonded to it is most likely going to be relatively positive you're going to get a dipole for me so this hydride ion this negative ion is attracted to this region of electron positivity well relative positivity in this dipole and then this double bond here is going to break and donate its pair of electrons onto the oxygen okay so let's look at our intermediate that's formed we're going to be having the ch3 again C ch3 remember this is a ketone in our starting example but if you want to swap it for an alde high it would just be a h remember that okay so this is now going to be a single bonded oxygen it has gained the electron pair here which you need to signify and it's gained one negative charge because one of the electrons originally belonged to him but it's gained the additional one from the carbon so you have to show the lone pair as well as the negative charge and then this hydrogen has bonded onto the carbon right here okay I know I'm keeping my colors pretty inconsistent but hopefully you get the idea now what's going to happen here is this oxygen is going to react with a another hydrogen ion okay and this is going to come from solution either a weak acid or water so this is going to form a dative covalent bond with hydrogen and ultimately form an alcohol okay so let's draw that out down here what is our final product going to be it's going to be h3cch3 h o h okay that is our final product here an alcohol so next up remember I said we're dealing with a cyanide ion so how does this differ very similar okay let's rub out this H and replace it with a c n okay a little tip here that they mentioned in the mark scheme is let's say you were doing the cyanide from this side okay CN you always want to show the lone pair and the negative charge on the respective ion so for example the negative charge and the lone pair is not going to be on the nitrogen ever you always want to show it on the C okay so just keep that in mind that's the know that right here so we're going to have our CN okay exact same thing's gonna happen this is going to come in attack our relatively positive carbon this is going to get kicked off onto the oxygen you're going to need to show the lone pair as well as the negative charge and you need to show the curly Arrow onto the proton here to form our alcohol now the other thing here is that this hydrogen obviously was not involved in the reaction so we have to switch this out for our cyanide ion so we now have a nitrile formed we also have to do that in our final product so let's pop that in CN and what would we refer to the functional group of our product here this is a hydroxy nitrile okay try and remember that now in our previous example I told you that the h plus ion came from water or a weak acid okay that is not the case for the nucleophilic addition of the cyanide ion this h plus comes from a strong acid more specifically sulfuric acid okay so keep that in mind okay next one is nucleophilic addition elimination and we're going to be dealing with acyl chlorides here okay so acyl chloride so let's start the mechanism with our reactant I'm going to use the example of ethanol chloride so that is our starting reactant now there's four nucleophiles we need to be aware of here you're going to need to react with water and alcohol ammonia and a primary amine those are the four nucleophiles so let's begin with water so we've got our water molecule here the lone pair on the oxygen now if I ask you explain what's going to happen next would you be able to be honest would you be able to explain what's Happening Here okay hopefully you remember there's going to be a dipole existing here because the oxygen is more electronegative than this bonded carbon so we're going to get Delta positive Delta negative again you don't need to show this in your exam but you should understand it and there's going to be a curly Arrow going from this oxygen onto the carbon okay and then this oxygen is going to accept the electron pair in this double covalent bond onto itself okay real simple stuff hopefully you're okay with that now intermediate time let's draw this out so so you have to show the lone pair and the negative charge and then we have to show the bonded water molecule that just came on let's do that in green just to signify that h h no can you guess what's going to happen next what happened to this oxygen what type of bond did it form it formed a dative covalent bond so it essentially lost one electron and sacrificed it in order to form that covalent bond so it's going to have a positive charge right now this oxygen is going to reform the double bond and this chlorine is going to be our leaving group and it's going to get the hell out of the way okay it's getting out of there and the reason for that is because this carbon if this oxygen reforms the double bond this carbon is now going to have five bonds which is not energetically favorable there's another dipole here whereby this is Delta positive this is Delta negative this is going to be electron withdrawing and suck the electrons onto itself and leave the molecule okay the carbon has four bonds again because this double bond form so it's a happy guy this oxygen is going to accept the electrons from one of these hydrogens either one is fine I just picked this one and the hydrogen is going to go off in solution it's going to unite with this guy and form HCL okay hydrogen chloride gas that's pretty much what happens there what product do we have here we're going to be having a carboxylic acid okay now I'm going to run through the next three as quickly as possible because they're pretty much identical so for the reaction with an alcohol you just want to change one of the hydrogens on this water molecule to be some sort of carbon chain so I'm going to use the example of an e-file chain so it's going to be ethanol right here and you're going to want to follow the exact same principle whereby we have a ch2 ch3 here okay and then our end product is not going to be a carboxylic acid it's going to be an ester okay just keep that in mind everything else should be completely the same right and when we're dealing with ammonia that's our next the nucleophile let's change this up slightly and put NH3 here that's going to come in attack our carbon the exact same things are going to happen let's roll about this make it a nitrogen here very similar to with the water and the alcohol one of these hydrogens is going to donate its electron back onto the nitrogen to make it neutral again but these two curly arrows exactly the same no need to worry what is our product going to be it's going to be a primary aim layered okay so let's draw that out instead of it being our s the linkage it's going to be an amide linkage whereby we have an nh2 molecule okay this is our a mind right here our primary amide now when we're reacting it with our fourth and final nucleophile this is going to be a primary amine okay so we're reacting it with a primary amine let's see what's going on here so it would be our NH H and then we're going to have some sort of alcohol chains so I'm going to use the example of ethal again so this is our primary aiming coming in here and attacking our relatively positive carbon exactly the same thing this oxygen is going to accept the electron pair in one of these bonds and we're going to form this intermediate now the only difference here is instead of having an NH3 being bonded we have our primary amine bonded so let's rub out one of these hydrogens and replace it to be our e-filled chains ch2 ch3 but exactly the same principles apply positive nitrogen one of these hydrogens is going to donate its electron onto the nitrogen to make it neutral again and we're going to get our product but instead of it being a primary amide this time it's going to replace one of these hydrogens with the e-file chain it's going to be NH ch2 ch3 this is a secondary a mind also referred to as an N substituted amide and that is our mechanism that is nucleophilic addition elimination with our four different nucleophiles hopefully that was helpful let's move on to some aromatics now and that should be our final mechanism that you need to know electrophilic substitution so good old Benzene right there's two different electrophilic mechanisms you need to know we've gone for electrophilic addition for alkenes you also need to be aware of electrophilic substitution okay so this is our good old Benzene molecule right here get comfortable with drawing these all right so you're going to either be having nitration which involves the nitrate electrophile or you're going to be reacting it in a friedel cross acylation reaction where you're going to be reacting it with an R Co plus okay this R can be any kind of alkyl chain and it originates from an acyl chloride so just keep that in mind this Co plus is due to it used to have a CL but that's been kicked off and you need to know the equations for the formation of the electrophile but I'm not going to cover that in this video just be aware that in both nitration and friedel cross isolation we have a catalyst involved in the formation of the electrophile that's really important to know and you need to know the equations for the formation okay just keep that in mind so going back to our mechanism let's rub this out and focus on what we need to know for our curly arrows so we're going to have our nitrate ion right NO2 plus I keep calling it a nitrate ion it's actually referred to as a nitroenium ion okay so keep that in mind but the actual reaction is referred to as the nitration of benzene so what happens here is this ring right here is incredibly electron dense okay so what happens is it is attracted to this nitroenium ion right here and we're going to be drawing a curly Arrow from the edge of this ring onto the electrophile okay that is our first step what is our second step going to be going to be forming our intermediate right here so let's draw this out so first thing that's going to happen is this ring in the center is broken you need to know how to draw that secondly each of these carbons on the edge has a hydrogen bonded so all the way around there's one hydrogen bonded to each of these carbons in the skeletal Benzene structure so what I'm going to do is I'm actually going to denote that right here so we're going to have our nitronium ion that just bonded on and we're going to have a hydrogen ion okay now the way that we draw the broken ring is what we refer to as a horseshoe structure okay so and something you need to keep in mind for AQA is you cannot pass the boundaries of the adjacent carbon so for example if we draw an imaginary line here to the center and then out to this carbon here you cannot cross that boundary so for example this ring would be incorrect because the Horseshoe has crossed the boundaries of the adjacent carbons super weird rule I'm not gonna lie but that is what it is so what I would do is I would draw a horseshoe keeping in mind the boundaries like this okay it doesn't really matter it doesn't have to be perfect as long as you draw a horseshoe-ish shape and do not cross the boundaries you're completely fine the additional thing is because we formed a dative covalent bond with this nitrate on night I keep calling it nitrate our nitronium ion there is positive charge here because it's lost negativity it's lost electrons so you'd have to denote that with a positive charge in the Horseshoe okay and what's going to happen here is this hydrogen which is very common for all these mechanisms is like you know what mate you can fulfill your neutrality again gain an electron from this covalent bond onto yourself and I'll just get the hell out of here okay so that's exactly what happens and we're going to have our nitrated Benzene as our product and then the h plus ion is going to come off okay I'm gonna go over the final mechanism three little cross isolation hella quickly because it's pretty much the exact same thing so let's rub out this nitronium ion right about this NO2 or about this NO2 and replace it with our carbonyl so it's no longer an acyl chloride we're going to have a c o ch3 plus remember this doesn't have to be a methyl it can be any sort of carbon chain but for this example I'm just going to use this so this again this ring this region of electron negativity electron density is going to be attracted to this positive electrophile we're going to be forming our intermediate here with our broken ring our horseshoe structure and then we're bonding on this molecule here and we're actually going to be forming a phenyl ketone okay a ketone because we've got this carbonyl Bond here that's C Double O and we bonded onto a phenol group because this hydrogen has donated its electron into the ring to reform its neutrality reform its electron density so our final product is going to be this guy right here a phenyl ketone okay and then again this h plus that left the molecule is going to reform our Catalyst which in this instance is alcl3 aluminum chloride and fulfill its function as a catalyst okay that is every single mechanism you need to know hopefully it was helpful if you have any questions let me know down below I'm really hoping I didn't miss anything and I really hope that helps you out if you did find it helpful smash the hell out of that like button it really helps the channel grow hopefully through this entire video you realize how similar these mechanisms are and how you just need to do some practice questions doing practice questions is so much better than memorizing the mechanisms okay you're just going to get a bit confused and get modeled up so practice questions memorize the reagents conditions observations and equations that you need to know specifically to do with the Catalyst but other than that just keep practicing keep revising and you'll be completely fine but best of luck with your revision and upcoming exams guys until next time peace