hello everybody and welcome to this a-level chemistry explanations video about organic mechanisms in this video we'll take a look at the major mechanisms in year one chemistry so we're talking about electrophilic addition nucleophilic substitution and elimination we'll look at why mechanisms are used generally what happens in each of the different types of mechanism and how you can learn to spot which mechanism you're actually being asked about in an exam question when we draw mechanisms we use them to show the movement of electrons during a chemical reaction this allows us to see how a reaction proceeds by identifying the bonds that are breaking and those bonds that form as a general rule curly arrows which we use to represent mechanisms must begin at either a lone pair of electrons on an atom which you must show or the curly Arrow must begin at a bond that is about to be broken and then the curly arrows need to point at either a new bond that is forming that might be a single Bond or a double bond or at an atom that is taking both electrons from a bond and it is therefore gaining a lone pair of electrons which again should be shown after that particular mechanism step now when we draw curly arrows for most courses we are typically going to be drawing curly arrows which are double-headed so they show the movement of a pair of electrons some mechanisms that are on very few courses require you to to show curly arrows that have got a single head that looks like this this is showing the movement of a single electron and most courses don't require this type of curly Arrow AQA for instance only use the double-headed curly arrows electrophilic addition reactions are going to be reactions that are happening for the organic molecules called alkenes these alkenes will typically react with one of three electrophiles the first and most common is hydrogen bromide or some other hydrogen halide the next one that is most common is bromine as a molecule or it could technically be any halogen and then the last type of electrophile that you need to be able to recognize that could react with an alkene is concentrated sulfuric acid this means that if you see an alkene as one of the reactants particularly with one of these three other molecules you are going to be being asked about electrophilic addition so what is electrophilic addition well an electrophile is a negative seeking species that will join on to another molecule the etymology of the word Electro well Electro is like electrons electrons are negative and so philic is seeking so electrophiles are negative seeking and alkenes are particularly attractive to electrophiles because the double bond in the alkene has a high electron density we might say it is electron rich and so therefore one atom in the electrophile which is electron deficient is attracted to that electron-rich double bond and attacks it how does this happen well first of all we have got an alkene I'm going to draw ethene as our example and so if we had hydrogen bromide the bond in hydrogen bromide is going to be polar because the bromine atom is more electronegative and pulls that electron density slightly closer towards it than the hydrogen and so then the hydrogen is Delta plus that will be attracted to the electron-rich double bond now in terms of the curly arrows that we draw we draw the movements of electrons remember and so this curly Arrow needs to start at the double bond and it needs to point to the hydrogen atom this curly arrow is symbolizing the formation of a new bond between one of the carbon atoms in that double bond and the hydrogen of the hbr now if that was the only thing that happened that hydrogen atom would then have two covalent bonds and hydrogen can only have one covalent bond and so what needs to happen next is the bond between the hydrogen and the bromine breaks and that Bond electrons both go to the bromine atom and so we show the curly Arrow as starting at the bond the pair of electrons that are in that Bond and we point to the bromine because it is taking both of those electrons now in electrophilic addition we always have to have a second step to our mechanism process and in that step we have to show what has been formed so far and so one hydrogen atom has been added onto in my case here the left hand carbon that had the double bond and the right hand carbon that originally was part of the double bond has got a positive charge and so we say this whole molecule is now a carbocation because it is carbon based and it is positively charged cations are positive and this is unstable and so this doesn't end here the bromide ion that has now formed has got that lone pair of electrons that it's just taken from the bond that it broke and so it will now be attracted to the positive carbon and so the final step of this mechanism is for the lone pair to form a new Bond to the carbon that is currently positively charged and so that's what this third curly arrow means and then if you were asked to do this in an exam question this is where you're likely to end your work unless you have been asked to show clearly the structure of the organic product and if you have been asked that then obviously that's what you must do and in our case this is what the organic product would look like and if you'd been asked to name this product we would simply call this bromo ethane it's a halogenoalkane when you use bromine as an electrophile the mechanism that we draw needs to be slightly subtly different and the reason is that br2 actually doesn't have a dipole in its normal State however when a bromine molecule moves towards an alkene because the alkene is electron Rich those electrons in the double bond repel the bonding electrons between the two bromine atoms slightly away from the double bond and slightly towards one of those two bromine atoms in the molecule and for this to happen one of the bromine atoms needs to be closer to the double bond than the other and so when we draw a mechanism for br2 it's important to show the molecule in the orientation that I'm doing here and because of this repulsion this induces a dipole in the br2 molecule the bromine that is furthest away from the double bond becomes electron rich and the bromine closest to the double bond becomes electron deficient this then allows that highest up bromine atom to be attracted to the double bond and to attack it and so we draw the curly Arrow coming from the double bond and this is showing the formation of a bond between one of the carbon atoms in the double bond and that top bromine atom all the other curly arrows from here are the same as for hbr and that is the bromine to bromine Bond needs to be broken and the lower bromine in the molecule takes both electrons from that covalent bond then we form the carbocation this time there is a bromine on the left hand of the two carbon atoms and the bromide ion attacks the carbon that is positive in this carbocation and that third curly Arrow shows the formation of a new bond between the carbon and the second of the two bromines this time the organic product looks like this we have produced one two dibromoethane sometimes you're asked to draw mechanisms for the reaction of alkenes that are not symmetrical this opens up an interesting line of questioning because you can get two different products being made these are simply referred to as the major product and the minor product in exam questions and then you have to unpick this to work out which of our two possible products will be the major one and which will be the minor and so if we take a look at my propene that I've got here I'm going to react this with hbr now of course we've already seen that the hydrogen is electron deficient and the bromine will be electron rich and so the red pathway that I'm going to draw will show the formation of one of the two possible products and so we have the new Bond forming between the carbon and the hydrogen the hydrogens of bromine Bond breaks this produces my left hand carbocation intermediate you can see it's very similar to the ethene example we've got a propane chain with the carbon in the middle being positively charged this is what's referred to as a secondary carbocation because the carbon that is positively charged has got two carbon chains sometimes called alkyl groups pushing their own electron density towards that positive carbon and thereby stabilizing it a little bit because what we have is what's called a positive inductive effect carbon chains push electron density towards positively charged carbon atoms and thereby stabilizing them a bit and so this particular carbocation is referred to as a secondary carbocation because there are two alkyl groups attached to the positive carbon the bromide Ion with its lone pair will of course attack this carbon that is positively charged we'll get a new Bond forming between them and this will be the organic product that we make and we would call this two bromo pro pain alternatively we could have this purple curly Arrow occurring where it looks really similar at the beginning the hydrogen still attacks the double bond but this curly arrow is symbolizing the attack of that hydrogen on the right hand side of the double bond and so what we end up producing is a very similar looking carbocation but this time the positive charge will be on the carbon on the left of the double bond because the hydrogen joints are the one on the right this is actually only a primary carbocation because there is only one carbon chain attached to the carbon that is positively charged and the bromide ion will of course attack this positive carbon and will lead to this organic product called one bromo propane now you need to know that carbocation stability decreases in this pattern tertiary carbocations are more stable than secondary carbocations which are more stable than primary carbocations and so the more stable carbocation is the one that will lead to the major product and the less stable carbocation will lead to the minor product and so in this example the major product will be the two bromo propane because it formed from the more stable secondary carbocation and the minor product will be one bromopropane you will still get both of these products forming but because it is harder to make the less stable intermediate there will be a smaller percentage of this organic product if you do get a mixture of organic products you can separate that mixture by distillation tertiary carbocations are more stable than secondary carbocations which are more stable than primary carbocations you couldn't just accept that if you like and you can just memorize that order to justify it more closely though we can take a look at a simple energy profile diagram if we have got a alkene on the left hand side as our reactant we can produce two different carbocation intermediates a secondary and a primary in this particular example now stability is to do with how much energy something has got so if something is more stable it's more stable because it has got less energy and the opposite is true of something that is less stable it must have more energy and so if it's got more energy the energy barrier or the activation energy that goes towards producing that carbocation is going to be bigger if the cargo cation is less stable and so it follows that in a reaction mechanism where you've got two possible intermediates the one that's got the least energy is the one that is easiest to produce and so it is the one that will form the most often and so the more stable carbocation is the one that will lead to the major product and the less stable carbocation will lead to the minor product the second mechanism I want to look at is nucleophilic substitution typically you'll get nucleophilic substitution occurring when you've got something such as a halogenoalkane and it is reacting with things such as potassium hydroxide or sodium hydroxide in other words something that contains the hydroxide ion and then potassium cyanide because that will contain the cyanide ion and also ammonia these three are all potential nucleophiles nuke nuclear files are positive seeking species nucleo is like nucleus and the nucleus is positive and philic is to do with Seaking and it's called substitution because one of these positive seeking species is replaced by another why does it happen well in the halogenoalkane we have got a polar bond and the Delta plus region in this polar bond is particularly attractive to negative ions or molecules that have got a lone pair of electrons and that's a characteristic of a nucleophile it must have a lone pair of electrons because this is what it's going to use to form a new Bond to the electron deficient atom in the halogenoalkane and so when we draw the mechanism for this we don't actually need to show the Delta plus and the Delta minus in the halogenoalkane but if we take a look at the iodo ethane that we've got here the iodine is much more electronegative than the carbon so it would be Delta minus and the carbon would be Delta plus and therefore attractive to the cyanide ion that I'm going to use as my nucleophile and so the lone pair on the carbon in the cyanide ion would attack the Delta positive carbon and this curly arrow that I've drawn drawn here is showing the formation of a new bond between that carbon atom and the carbon in the halogenoalkane carbon should only have four bonds so that carbon in the halogenom alkane has got five and so what needs to happen is the carbon to iodine Bond breaks the iodine takes both of the electrons from the bond which is why the curly Arrow points at the iodine atom and then this iodine leaves as an iodide ion and in terms of the mechanism that's actually the end of the mechanism when you're using a negative ion as the nucleophile but in case you are asked to show the formation of the organic product here is what we would produce on the right hand side and we would call this propane nitrile because we've got that carbon triply bonded to the nitrogen that's the nitrile functional group but we do count that carbon atom in terms of the number of carbon atoms present in the molecule so propane nitrile that e is very important and so the chemical equation for this would be shown like so and we need to include the negative charge on the cyanide ion and on the iodide that gets produced when ammonia is the nucleophile there are a couple of really important differences firstly instead of only one step taking place we need to have two the first step begins as normal for nucleophilic substitution there is a new Bond forming between the nitrogen in the ammonia and the electron deficient carbon and again the bond between the carbon and the halogen needs to be broken leaving us with a halide ion in step two the reason that the ammonia nucleophile leads to something different is that now the nitrogen has attacked the halogenoalkane it's actually got four covalent bonds and it should only have three so its electrons have been spread into more bonds than it typically would be that leaves the nitrogen as being positively charged another reason to justify why the nitrogen should be positive is that something has left the halogenoalkane and it is negatively charged and so therefore it must be that something is positive to compensate for that and it is it is the nitrogen that is positive and so the marks for the mechanisms are going to be one mark for the first two curly arrows each so that's two marks for step one in step two we would get a mark for showing the intermediate positive molecule this needs to have its positive charge removed to do that the nitrogen needs to take back those two electrons that it used to bond the carbon but it doesn't take the two electrons actually that it literally used it takes two electrons from its bond to one of those three hydrogen atoms it doesn't matter which of those three bonds you break but what does matter is that how you break it how you show it breaking with the curly arrow is that the pair of electrons needs to point to the nitrogen to show that the nitrogen is taking both of those electrons from the covalent bond it can sometimes get quite squashed up when you draw this and so we can really curl that Arrow to show very clearly we know where it is starting and where that Arrowhead is pointing it's really important to note that the thing that takes that hydrogen away is likely to be an ammonia molecule because in the reaction between ammonia and the halogenom alkane once that halide ion has been removed there's going to be a really low concentration of halide ion and there's still going to be quite a high concentration of ammonia and so the thing that will accept that hydrogen in other words act as a base a hydrogen ion acceptor is likely to be in terms of probability an ammonia molecule and so when we draw our mechanisms we often don't actually have to show what it is that's taking that hydrogen away but it should if you want to draw it be an ammonia and what this means is we end up producing our amine in this case e-file a mean and we will also produce ammonium iodide and this means that when we write the chemical equation for what's happening we need to use two ammonia molecules overall one of those ammonia molecules acts as a nucleophile and substitutes out the halide ion and the other molecule acts as a base and accepts the proton and then forms ammonium iodide there's actually a second reason why ammonia ends up being slightly different to the other two and the reason is that what we end up producing is an amine and the amine specifically the nitrogen in the amine is able to act as a nucleophile because it has got a lone pair of electrons and so what can happen is the amine can actually attack the original halogenoalkane and we can end up getting multiple substitution occurring in order to avoid this happen opening we can use an excess of ammonia to limit this multiple substitution and not only does it do that it ensures that we get the ammonia being the molecule to remove the proton from our positively charged nitrogen in that intermediate step and so we probably will still get a mixture of organic products which we can separate using fractional distillation but we get to choose which form that we get by adjusting the conditions an excess of ammonia means that we're most likely to get mostly primary amines being produced whereas if we don't use an excess of ammonia if we use an excess of halogenoalkane we're likely to get multiple substitution occurring and not primary amines perhaps secondary or tertiary or even quaternary ammonium salts this is where this mechanism bridges over between the first year of chemistry and the means topic which is in year two of chemistry the final mechanism that I want to look at is elimination there are actually two different types of elimination reaction that you need to know about for a level chemistry one involves halogenoalkanes and potassium hydroxide dissolved in ethanol solvent the other is when you have got concentrated acid reacting with an alcohol what they both have in common is that a small molecule is removed from a larger molecule when elimination is occurring in halogenom alkanes we end up producing a hydrogen halide for instance hydrogen chloride when we do elimination of alcohols we eliminate water why does it happen well let's take a look at these one at a time when we are reacting a halogenoalkane what happens is the hydroxide ion specifically when it is dissolved in ethanol solvent will act as a base what bases do is they accept hydrogen ions and so if we take a look at chloroethane here we can see that the hydroxide ion is going to attack the hydrogen on the carbon that is adjacent to the carbon with the chlorine attached those three hydrogens are all equivalent we could attack any of them I'm choosing to attack the bottom one and so when we draw our curly Arrow we need to begin the curly arrow on the lone pair of the hydroxide ion and this is symbolizing the formation of a new bond between the oxygen of the hydroxide ion and that hydrogen atom and we're going to produce water as a result the carbon atom is going to to end up not being bonded to enough atoms currently it's only bonded to three once that hydrogen on the bottom left has been removed and so what happens is the pair of electrons that were in the bond between the carbon and that hydrogen move into the space between the two carbon atoms and this is symbolizing the breaking of the bond between the carbon to hydrogen and those bonding electrons moving over to turn the single carbon to carbon Bond into a double bond and then the carbon on the right hand side has currently got five bonds which is one too many and so what happens last of all is the chlorine atom takes both of those bonding electrons away from the carbon and it makes a chloride ion and so what we end up producing is the alkene ethene in this particular example the molecule water because the hydroxide ion has removed that proton and then we've got the chloride ion which would form a compound with the potassium from the potassium hydroxide that was dissolved in the ethanol solvent in the first place elimination from halogenoalkanes can get a bit more complicated if the halogenoalkane is not symmetrical so you can see here that I've got two chloro three methyl butane as my halogenoalkane now the hydroxide ion base has got two potential carbon atoms to take a hydrogen from if we take a look at the first path I'm going to draw it's going to remove a hydrogen from the left hand side of the carbon with the chlorine attached in other words this one here and so one step that's hydrogen forms its new Bond to the oxygen which is what this first curly arrow is showing the pair of electrons that were used to bond the carbon to the hydrogen are going to move to create that double bond and the chlorine atom is going to break its bond to the carbon and leave as a CL minus and alternatively though we could look at the right hand pathway where this lone pair of electrons is going to be used to form a bond to one of the three hydrogen atoms on the right hand most carbons could be any of these three I'm going to choose this one and so this is a new one that's forming the next curly arrow is again the breaking of the bond between carbon and hydrogen making a double bond this time in this position and the chlorine once again leaves taking the pair of electrons away with it the two possible alkenes that get produced are this one for the left hand pathway and that would be called two methyl Bute two in alternatively the right hand pathway would lead to this alkene being produced and that would be three methyl Bute one in and so we've got this mixture of alkenes being produced which will have different boiling points we can separate them using fractional distillation you've probably seen that these two mechanisms for elimination and nucleophilic substitution actually have an identical reagent the potassium hydroxide or sodium hydroxide is the source of the hydroxide ion that is either going to act as a base in the case of elimination or a nucleophile in the case of nucleophilic substitution which of these things it acts as depends on the conditions used for the reaction typically if you want elimination to happen you want the potassium hydroxide to be dissolved in ethanol solvent with specifically no water being present the mixture is then heated if you want the hydroxide ion to act as a nucleophile you can actually do this at room temperature but it is quite slow to speed it up you would warm the mixture in this case the hydroxide is going to be dissolved in water but also you might use a little bit of ethanol to help the halogenoalkane to dissolve it dissolves better than it does in just simply in water but the important distinction is that if you want to make the alkene through the process of elimination the temperature needs to be higher and the solvent must only be ethanol whereas for substitution the solvent is primarily water and it takes place at low temperature at most you would just warm it to speed up that reaction a bit Additionally the type of halogenoalkane is going to actually have an influence on whether substitution occurs or elimination occurs a primary halogenoalkane namely with only one alkyl group attached to the carbon with the halogen attached that's going to favor substitution whereas tertiary halogenoalkanes like this one shown is likely to favor elimination and secondary being between both of those two is likely to do a mixture of both of those processors alcohols can undergo the mechanism of elimination in one of two ways both of them would be classed as dehydration reactions and the dehydrating agent is what is different between the two processors first of all alcohols can be dehydrated with an excess of concentrated hot sulfuric acid or you could pass the vapors of the alcohol over a catalyst of aluminum oxide that has been heated up either way you make an alkene as your product elimination from alcohols is slightly longer as a mechanism than it is from halogenoalkanes the first step is the protonation of the oxygen in the alcohol group and so we show a curly Arrow like this the lone pair of electrons makes a new Bond to the hydrogen ion and that works because the oxygen is electron rich and it's attracted to the positively charged hydrogen region that's step one and in our intermediate you can see that that oxygen is now positively charged because it's used its electrons to form this new Bond and so it's taken on that charge of the hydrogen ion and so you'd get a mark in a mechanism for the formation of this organic intermediate obviously oxygen can't remain as positively charged so what happens is the carbon to oxygen Bond breaks the oxygen takes both of the electrons from that bond to itself thus making it neutral again and we show this by pointing the arrow from the bond to the oxygen atom that would leave the carbon that I've shown here as only having three bonds and it of course needs to have four and so what happens is a hydrogen from the adjacent carbon breaks its bond to the carbon and that bonding pair of electrons goes towards turning this single carbon to carbon Bond into a double bond and that hydrogen leaves as a positively charged ion this produces our alkene and it's also regenerates the hydrogen ion as a catalyst and we produce water which is the product of our dehydration of our elimination reaction mechanism whenever you do an elimination reaction you can end up with a mixture of alkenes being produced which would be classed as isomers for this particular example I want to use butane tool and the two mechanisms that could be used to produce two different alkenes are shown like so if we take one of the hydrogen atoms away from the right most carbon atom we will produce the alkene Bute one in if we take away one of the hydrogen atoms from the carbon to the left of the alcohol group we will ultimately end up producing Bute two in so these are two isomeric products not only that Bute two in itself has got further isomers we could have Zed Bute two in where those two methyl groups are on the same side and that's given the subscript Z which means together Zoot salmon from German and then the other isomer is when the two methyl groups are apart they're on opposite sides and that's e Bute two in so we've got in fact three different isomers produced Bute one in Zed butuene and E Bute two e okay that's the end of this summary of organic mechanisms from year one chemistry I hope it was useful I'll see you again soon