hello class I want to take some time in showing you the mechanism of when you take sodium metal and dissolve it in liquid ammonia then you treat that with an alkyne and in under those conditions what do we get we get a trans alkene and then in contrast if you take the same alkyne and treat it with different reagents let's say lindler's catalyst then you're going to get a CIS alkene and the mechanism is going to help us explain that so I'm going to show you the mechanism of the dissolving metal reduction and then that will help us to understand why it gives us trans alkenes but before we jump into the mechanism we need to understand something else about mechanism arrows so let's just talk about something that we are familiar with what if I take this tertiary alkyl halide and I do let's say start off with an sn1 type of thing here the two electrons in this Bond when I draw an arrow like that we see that we are going to get a tertiary carbocation because it only has six electrons around that carbon but the two electrons all went to the bromide so these two electrons right there came from that Bond and so it's the arrow that is showing you what's going on here the arrow is saying hey since that's a double-headed Arrow it's saying both electrons in that Bond are going to the bromine like that and so what we have is What's called the heterolytic cleavage both electrons go to the species and so that would have to be negatively charged but then we could have something called a radical mechanism in which we use what's called Fish Hook arrows and Fish Hook arrows only have one hook you see the difference between the two when we do this in a fish hooked Arrow drawn like that what are we saying we are saying what one electron goes to the carbon and one electron goes to the bromine and so we would have species that look like this with an electron there plus our bromine right here so we would have a carbon radical and a bromine radical because it's called The Radical because we have a electron that's not paired with another electron you can see these six right here are paired it's a lone pair this is not it's unpaired so whenever you have a electron or a species that has an unpaired electron you have a radical and so it's this fish hook Arrow that you need to get used to in in order to understand radical mechanisms and radical mechanisms are when you formaticals and so this dissolving the metal dissolving metal reduction reactions proceeds by a radical mechanism so we have to use these Fish Hook arrows so now let's look at the reaction that we're going to talk about oh I just realized I didn't tell you the name of this cleavage this would be called a homocle homolytic cleavage the one above was heterolytic this is homolytic so when you take a alkyne and treat it with dissolved sodium metal we have learned from General chemistry that Metals want to give up their electrons so they can obtain their nearest uh what's the nearest noble gas electron configuration so it has one electron that sodium wants to get rid of and so what it does is it notice the fish hook Arrow it's going to donate its one electron to the carbon there and then this uh Pi electrons in this the triple bond here we're going to break one of those Sigma Bonds in a homolytic fashion we'll put one electron on this carbon and another electron on that carbon so what we have going for us now is we have something that looks like this we still have our double bond and if I number these carbons carbon one carbon two carbon one and carbon two we see that carbon one is going to have a a p orbital with one electron in it shown right here by that one fish hook Arrow and then on carbon two we'll have another p orbital but what's happening here this electron came and went to carbon two and then this one electron broke off so now we have two right there so once we get to this step we can see that this is going to be negatively charged right here all right well this is going to be put it on the carbon I guess so that's going to be negatively charged let's double check the formal charges here one two three four five yes so that would be negatively charged and so when we have sodium metal dissolved in liquid ammonia are liquid ammonia is going to be our proton source all right it's going to practice the asses at the Lewis acid here so this lone pair right here is going to come and grab our hydrogen atom so that's going to give us something that looks like this boom and then H right there and then it does the same thing over again is going to react with another sodium metal now we're going to say see the same fish hook arrows in this step now all right so what's going to happen is this is still here it's still a radical there and so this sodium is going to donate its one electron there and why is it going to do that well it's going to give it a full-blown negative charge right here so that's going to be negative if the carbon is going to be negatively charged and then you can do another proton transfer with liquid ammonia so we can see this is going to come grab that go like that and then that's going to put the proton on this side right here and where do we want to draw that product we're going to go down here okay just to draw the final product and so you're going to see that we have you can't see the bottom one right here can you so this one right here is a hydrogen right here let's just let's draw the final product over here so we can see it there we go and that's going to be our trans alkene and that's the mechanism but now let's figure out using these Concepts here why does the trans alkene form right well it all has to do with repulsion let's take a look at two scenarios and you'll you'll see very clearly why the this reaction gives us a trans alkene so let's take a look at this intermediate first this is after the you have the alkyne and the sodium metal donates an electron and you get this species right here and you can see that these orbitals right here are on opposite sides of that double bond but what if and this what if it actually added something like this all right negatively charged what about this Why didn't it add like that because this is the low energy State and this is the high energy state and molecules are going to take the easy easy way lowest energy so why is this the higher energy state well I didn't draw it too accurately because what if those are bigger all right do you can you now see that there's going to be some uh Clash right there there's going to be some electronic repulsion because we have electrons in those orbitals getting really close to each other and so that's going to increase the energy it's going to this in this confirmation here if the molecules could talk they'd be like hey this is really uncomfortable why not look like this but this is way more comfortable and so that explains why when you are using sodium metal and liquid ammonia you are going to get a trans alkene because it proceeds by a radical mechanism and this Electronics uh the electronic flashing here explains why it's going to be trans but when you compare this to using like blindler's catalyst to the same alkyne you're going to get the CIS so if we have the same alkyne and we treat it with lindler's catalyst then lures and add some hydrogen gas here what we're gonna get we will get this as the dominant product the CIS so what's going on here well it just has it has to do with the Slender's Catalyst which is what um is it platinum or Palladium it's Palladium it's a poisoned Palladium catalyst and so you have basically have chunks of metal in the reaction the metal does not dissolve so you'll throw it in there and you'll just see it floating around so you have these metal chunks and so if I take that metal chunk and just zoom in really really close Okay so this is the Palladium metal chunk and on its surface the hydrogen gas or the hydrogen atoms okay are going to basically stick to that metal surface and so when the alkyne comes getting close what is it going to do just kind of picture it kind of like this let's see boom boom boom like this these hydrogens are just gonna plop right in there from the same face so they're adding sin when they're adding in a sin fashion which what I mean by that they're coming from the same side you will get a CIS alkene all right pretty slick so our next reaction that we want to look at is Hydro halogenation of alkynes and we've seen this type of reaction with a alkene before we've seen that so I'm going to pause this and clear this board so the next reaction that we want to look at is hydrohalogenation of alkynes but I do want to remind you that we've seen something similar when we have an alkene when we treat that with HX like that what are we going to get we are going to get the marconikoff product which means that the halogen is going to go to the more substituted carbon and the hydrogen is going to go to the carbon with the most hydrogens that's what we see and we've seen that before but if you take a alkyne like this and treat it with equal molar amount of halogen or the HX sorry so if you have one mole of this and one mole of that all right you're going to get a very similar marconikoff product but it's going to look like this one two three four five six one two three four five six one two three four we're going to have a double bond here and then we're going to add our halogen there like that but if you have excess amount of your acid so that could be what it could be hydrochloric acid hydrobromic acid right when you have excess of this it's going to add twice and it's going to add both in a marconikov fashion and add the halogen there so you have to pay close attention to the concentrations right here when you're dealing with an alkyne if it's equal molar like a one to one that only adds once but if there's excess it's going to add twice like that and what kind of halide is this going to be called this is going to be called a geminal dye alkyl halide because we have two halogens they're on the same carbon now do you recall when we're doing this reaction right here the mechanism so we call that the first step in the mechanism is electron Rich Pi electrons attacking the electron for hydrogen in this fashion and that's going to add the hydrogen there and give us a secondary carbocation which then allows the halogen that has just broke off to come in and attack the carbocation to give us our product right remember that look kind of the same we invoke kind of the same mechanism when we're dealing with alkynes because what do we have here we can have a electron Rich Pi electrons Tom and grab this like so and then we would have two three four five six and so what we would have our double bottom there a hydrogen would have added now we have two hydrogens there we already started with one that they added there and then what we would have a carbocation right there and then our halogen whatever it may be sees that positive charge there comes in and attacks to give us our product now you do the same mechanism over again if you had excess acid but there's a lot of controversy in with this mechanism because we know from studies a vinylic carbocation is very very unstable but we're invoking a vinelik carbocation here to explain this mechanism so it's not entirely true there's debate on how this mechanism actually proceeds and so I don't want to get into the nitty-gritty details of of this mechanism and which one's the right one which one's the wrong one well the jury's still out on that and depending on who you talk to they could argue their point so what I just want you to realize is follow this mechanism the way I've shown you but in the back of your mind just realize hey that's not entirely correct because vinyl final sorry vinyl at carbocations are very unstable okay so I'm not too worried about saying that this mechanism is 100 correct because we know it's not but we're going to use it we're going to use it because it helps us figure out where the hydrogens are going to go it's going to follow markonikoff all right so why did the this hydrogen not add at the other carbon so let me give another hypothesis here one two three four five so why did the hydrogen not add there on this first step and give us that well because compare the two this is a primary vinyl carbocation and this is a secondary vinyl carbocation when you compare these two this one is way more stable than that one and so this helps us to understand why this reaction is adding in a marconica fashion it's all has to do with that carbocation intermediate which we know is not particularly true but we use it to explain the regiochemistry of these reactions okay so let's take a look at another radical reaction we've seen this already with alkenes but now we're going to apply it to alkynes as well recall from previous chapters that if you had a alkene and you treated it with hbr and only hbr and we'll talk about the reasons why only hvr works in upcoming chapters and we add a peroxide of sorts in the reaction right there let's make that a r a peroxide remember that it's going to add anti-markconic off so it's going to add the halogen there just by adding that peroxide pretty cool so we can extend this idea to alkynes as well so if we treat it with only hbr we can't use a hydro hydroiodic acid or hydrochloric acid for reasons which we will talk about later but what we're going to get is we're going to get a mixture we are going to get a mixture uh one two three four five like that I'm going to get a mixture of E so this would be e and that would be Z which is pretty slick okay so I'm just oh wait what am I forgetting with the reagents here we need our peroxide in order to do a radical mechanism now we will cover the radical mechanisms uh later on I just want you to be aware that we've seen this before and it does work with the alkynes as well all right and you also see the correlation here anti-markconikoff right okay so the next reaction that we're going to talk about is the hydration of alkynes hydration what's that what's that word mean we're adding some water okay the hydration of alkynes you can take an alkyne you treat it with sulfuric acid and some water and you need to add some of this mercury sulfate to get the reaction going now what it turns into is this molecule right there okay and that molecule has a very special name and that's called an enol the in is for the alkene and then the all is for the alcohol so when you have a double bond right here and an alcohol directly attached we call that an enol but enols are not going to be isolated what's going to happen is once the enol is formed there's an equilibrium process which says this molecule is going to tautomerize into the keto form which looks like this all right and it's cutting off a little bit so maybe let's shrink it down right now to the keto form so and this is called the keto form and we call it the keto form because what do we have we have a ketone so what we can say is this enol keto form is a equilibrium process but once this reaction comes to equilibrium almost all of the enol is going to go to the keto form generally speaking we can always find exceptions right but generally speaking the keto form is the more stable form and the most and The prominent product so what we can say is when we take an alkyne and we hydrate it with some acid and mercury sulfate we are going to generate a ketone so put that in your head how do I make ketones this reaction right here hydration of alkynes so I want to go through the mechanism with you to show you how this all works okay so let's uh clean this up so we have all the space because this mechanism there's quite a bit of steps so the first thing that's going to happen is we're going to have our Mercury two plus ion floating around in solution and what's that's going to do is it's going to form this three the three-membered ring like we've seen before like this it's going to generate something that looks like this Mercury boom like that just like that [Music] now what's going to happen is that we had sulfuric acid and water present so we're going to take water and that's going to act as our nucleophile and so this water has options now is it going to attack this carbon or that carbon well it's going to attack the more substituted carbon based off of the same reasons that we've seen in previous examples this carbon Mercury bond is relatively weak and it's more weak because it's going to have a a stronger partial positive over here and this is going to be a a smaller partial positive and why is this going to have a larger partial positive because we have an alkyl group that can donate electron density in to stabilize that on this side there's no alkyl groups that can don't come in and stabilize it because what would we have if these bonds totally broke look at it this way if that mercury atom if this Bond broke we would have a carbocation there versus at the Mercury stayed on that carbon then we would have a primary carbocation so what we're looking at these partial positives here this one is a secondary carbocation that's a primary we know primaries don't form so when it's in this cyclic structure right here this was going to have a a more partial positive because it's going to be more stable due to these alkyl groups donating electron density in and so when we have our electron Rich species right here this is electron Rich electron pore it's going to what gravitate towards the carbon that has a higher the more intense partial positive but when it does that that Bond will break like that to give us our intermediate here that looks like this and Ultra like that okay and then we still have our Mercury hanging on there just like so that'll be our positive charge there and then what's going to happen here we'll do a proton transfer we have water present so that can act as our base let's use this color here we can do a simple proton transfer and that's going to give us Mercury our alcohol plus we just generated some hydronium from this spot right so when we have this species under acidic conditions okay it's going to react so let's draw out the hydronium like this and so what's going to happen you could look at it like these electron-rich Pi electrons are going to come in all right go like that and it's going to add the hydrogen onto the carbon that is attached to the Mercury so if we come down here it's going to look like this Mercury right and then we're going to have a partial positive or a partial charge on that carbon the hydrogen added here and the reason why we want the hydrogen adding on the carbon that's connected to the Mercury is because we need to kick that mercury off okay we need to kick that thing off and so what's going to happen here let's see here what we're going to do now is because we've generated a generated a positive charge here this Mercury Bond right here it's going to break and give its electrons to that species and so that's going to give us what like this we don't need to draw on that hydrogen so now what did we just generate you see that we have now generated our enol and so the Mercury fell off so that's going to be plus our two Mercury right there okay the Mercury falls off the generate our enol and this happens when you are under acidic conditions the miracree is going to fall off so now that we form the enol we've learned that the enols are not favored at equilibrium that they're going to favor the keto for so now we're going to take the enol and tautomerize it into the keto form and I'm going to show you how to do that let's take a look at tautomerization of the enol under acidic conditions that's really important to understand is that part because we can have tautomerization under basic conditions as well and the mechanism is going to be a little bit different so under acidic conditions we're going to have we can represent the acid that's hydronium it's not really too critical for me to know exactly which one it is okay and so the first step here we could now what's the best way here I'm going to show it like this okay what can happen is that this alkene right here is the electron rich but it's also extra electronics because we have these lone pairs right here that can come down and then it can come and grab it like so and when you do that you're going to get what you have a double bond you have that the oxygen has three bonds to it so that has to be positive positively charged and since these electrons came down we have four bonds there so the hydrogen that was abstracted wet here there's already two hydrogens here that we don't need to even show but this hydrogen right here has now been added so what do we have we have a total of three hydrogens right there and then we generated what the conjugate base of our hydronium which is just water and then we are going to have a proton transfer step where the water is going to uh come and deprotonate that to get rid of the positive charge and there we have we have our ketone what is this step what was this step called that is also a proton transfer so tautomerization under acidic conditions is just two proton transfers when you think about it uh based off of those 10 mechanistic steps that I've been teaching you it's really not that hard right it's just a simple proton transfer two of them just in a row to get you your ketone all right now you have to understand though that this is at equilibrium and so technically we should have some equilibrium arrows here because we can in fact go from the keto or the Ketone all the way back to the enol so mechanistically how do you do that will you do the exact opposite of what I just did in the green arrows do the exact opposite to go backwards and that is a very good exercise to do because you need to know this in the forward Direction and in the reverse Direction so I'm just going to show it to you in the forward Direction and good exercise try it in the reverse Direction okay let's see what else needs to be said now let's take a look at uh adding some water across a triple bond but do it in a anti-markconica Flay all right so in this reaction hydration of alkynes what did we do we have we had let's draw like this we had an alkyne we treated it with sulfuric acid some water and some mercury sulfate right and where did the water go or where did the let's say the alcohol let's do it that way where did the alcohol go carbon one or carbon two it went to carbon two because it was and so it would form this guy all right that water molecule went to the more substituted carbon right for and the reasons why we explained that was due to the the three-membered ring with the Mercury right okay but now we can take that same exact molecule and add the alcohol on the other carbon what if we wanted it to add there do you see the difference between the two both are enols it's just where did the hydroxyl get added the more substituted carbon or the less substituted carbon and so that's the next reaction that we're going to look at and it's called hydroboration oxidation of alkynes now if we take the same alkyne as we've been working with and then treat it with borane and thf and then do a second step the oxidation step so Step One is the hydroboration step two is the oxidation and the product that we're going to get would be an enol but now you see that the hydroxyl is added to a anti-marcoticoff fashion that is an enol so it is going to tautomerize into a different product here so what do you what are we what is it going to tautomerize into well that really I could draw in a hydrogen if I want we've went to the aldehyde but we still call this the keto form this is the enol this is the keto form but what functional group did we make well we made a aldehyde okay let's see here now when you do this reaction you can use boring okay but you can also replace the boring with this die cyanoborane or you could replace the borane with the 9bbn you don't need a rim remember the structures of these guys all you have to understand and remember is that this species right here looks different you can see a image of it in your textbook but all you have to do is say hey that's going to behave the same thing behave as the same way as boring and same thing for 9 BBN that's basically going to behave like boring okay now why do people want to use 9bbn over boring sometimes well it's really really big so it's ster it's really bulky and so 9 BBN is really really going to favor the anti-marconic cough uh product and so sometimes you just want to force it like all the way no chances of any side reactions but for our purposes of our class all three can work all three or fine but now that we have this enol we need to show you how to tautomerize it to the keto form and we have to show you how to do it under what basic conditions so it's different than tautomerization under acidic conditions so you have to make sure when you have to do a tautomerization you take a look at what are the reaction uh what's the reaction under or what are the reaction conditions and if you see acid then you tautomerize based off of acid and if it's under basic conditions you use the basic mechanism okay so what we have here now I'm going to extend out that hydrogen because that's important in the first part of the mechanism so the first part of the mechanism is that you are adding base you're under basic conditions so under basic conditions you have to do a proton transfer and so when you look at this molecule you're like okay base so there has to be an acid over here somewhere what's the most acidic proton well an alkyne hydrogen or alkene hydrogen is what has a pka of around 43 but an alcohol hydrogen has a pka of around 16. so that's the one that it's going to go for and so what uh does that why is that so significant well now we have it negatively charged and now we generated what the conjugate acid which is water so now we have this uh charge species now that can bring a lone pair down and collapse down and so that means this double bond has to break and it's going to come and grab another proton Now where's the hydrogen going to go well it's going to go to the carbon that's attached to the R Group so we have a double bond like no no no this and we still have that hydrogen there we can draw that in if we'd like okay and then what did we do we added there's already hydrogen there so that one's that one but then this hydrogen right there I'm going to now add it and there we go and so there's our uh aldehyde so Hydro boration oxidation of alkynes can give us a aldehyde but remember our starting material very very important to see this what type of alkyne was it that was a terminal so a terminal alkyne is going to generate a aldehyde under these reaction reaction conditions well with the time that's remaining I think it'd be a good idea to uh call it a day for this content and then we will pick up the remainder of the reactions that we need to learn in the next video okay as as always reach out if you need anything