in this video we're going to go over reactions associated with alkynes so let's start with this example let's say we have two butine what's going to happen if we reacted with hydrogen gas and palladium over carbon initially when you add the first hydrogen molecule this will turn into an alkene and this reaction occurs with syndication the catalyst puts the two hydrogen atoms on the same side now it doesn't stop there another hydrogen gas molecule will add to the alkene turning it into an alkane so in the end you're adding four hydrogen atoms across the triple bond so you can simply write your final answer like this so for example here's another problem let's say we have this molecule so there's a total of five carbon atoms if you react it with hydrogen gas using palladium over carbon catalyst this will give you just an alkane that's all you need to do for this example now let's look at another example so let's react to butane with hydrogen gas but this time we're going to use something called the lindler's catalyst what do you think is going to happen this is basically a poisoned catalyst and so this is going to stop at the alkene level so it's going to give us a cis alkene like the other reaction it adds the two hydrogen atoms on the same side but it doesn't stop at the alkane level it stops at the alkene level now you need to know that the landlord's catalyst is composed of palladium mixed with barium sulfate and it's in quinoline which looks like this and you also have methanol mixed with this compound so that's the landlord's catalyst sometimes you might see it like this as opposed to seeing the word liners catalyst but the effect is the same it converts an alkyne into a cis alkene now the next reaction that you need to be familiar with is having an alkyne with sodium metal and liquid ammonia now you can use another alkali metal such as lithium metal two so it would yield the same result now this is going to stop at the alkene level it's going to add two hydrogen atoms but you're going to get the trans alkane as opposed to the cis alkene so let's go over the mechanism for this reaction so the first thing we're going to do is react with sodium metal sodium has one valence electron and so this electron interacts with the triple bond and when you see a half arrow it represents the flow of one electron so we're going to get a carbon-carbon double bond the bulky r groups will prefer to be away from each other favoring the trans-alkene and on one of these carbons we're going to have two electrons or a lone pair and on the other one just one electron so this is called what do you think this is called this is called a radical anion here we have a radical and here is the anion which is basically a carbon with a negative charge so collectively it's a radical anion now in the next step of this mechanism we're going to react the radical anion with ammonia so the carbon with the negative charge is a strong base it's going to take off a hydrogen from nh3 giving us the nh2 minus ion and so at this point we're going to have this molecule so we still have the two r groups but now we have a hydrogen atom and we still have a radical so what do you think this is called this intermediate this intermediate is known as a vanillic radical the radical is directly on the double bond so it's a vanilla group now at this point we're going to react with another sodium atom with its one valence electron and so that electron will add to this carbon converting it into an anion so let me continue this on the next page so now we have two electrons on its carbon which means it carries a negative charge so this is called a vanillic anion now at this point we're going to react it with ammonia again and so it's going to pick up another hydrogen atom from nh3 and so this will give us our final product which as you can see it's a trans alkene so that's the mechanism for the metal ammonia reduction of alkynes into trans alkanes the hydrogen atoms are on opposite sides now let's go over some other reactions let's say if we have propine and we're going to react it with mercury sulfate with water and sulfuric acid what do you think is going to happen here now this reaction is similar to the oxymercuration demercuration reaction of alkenes which would give us alcohols but for this reaction we're still going to get an oh group but the triple bond will go down to a double bond so whenever you have a double bond with an o h group this is called an enol now the enol is not stable it can tautomerize into a ketone now you're going to have a small amount of this molecule in the solution but the majority will be in this form because these two they exist in equilibrium and so we have a longer arrow pointing towards the ketone so this reaction converts alkynes into ketones now there's another reaction that you need to be familiar with it's the hydroboration oxidation reaction because with alkenes it also produces alcohols now when dealing with alkynes instead of using bh3 typically you'll see r2bh sometimes you might see sia2bh disenamo borine now the reason why you want to use this when dealing with terminal alkynes is because boring can react with a terminal alkyne twice so let's say if we react this with bh3 so first it's going to go down to a double bond and we're going to add a hydrogen and a bh2 group here now i'm going to write bh2 like this because depending on where you are in the course of the reaction this can be an h or it can be an r group where it reacts with another molecule such as that so i'm going to put that i won't put the hydrogen there so i'm going to leave it like this now this can react with another boring molecule turn it into an alkane and so you get a lot of side products here if you use bh3 with a terminal alkyne now if you use it with an internal alkyne you usually don't have this issue because the internal alkyne is more sterically hindered and so typically one boring molecule will add to an internal alkyne but for terminal alkynes you definitely want to use r2bh so typically you'll see this a lot with alkynes so this is the preferred reagent to use so we're going to use r2bh in thf followed by hydrogen peroxide hydroxide and water so it's going to be very similar to what we have here we're going to get an enol but the o h is going to be on the less substituted carbon atom so this reaction proceeds with more karnikov addition the o h group is on the secondary carbon if you use mercury sulfate but if you use hydroboration oxidation the o h group is on the primary carbon as opposed to the secondary carbon so make sure you can distinguish the results of these two reactions now just like before the enol will tosmarize but if you have a primary enol it's going to convert into an aldehyde a secondary enol will convert into a ketone and so the end result with the hydration oxidation reaction of alkynes is that you get an aldehyde use the mercury sulfate with alkynes you'll get a ketone now what do you think will be the product of this reaction so let's say we have two pentine and we're going to react it with mercury sulfate in water and sulfuric acid so this carbon is secondary and this carbon is secondary so we have an unsymmetrical alkyne so because it's equally substituted we can get a ketone at any one of these two carbons so we're gonna get a mixture of products we're gonna get two pentanone and also we'll get three pentanone now here is the question for you how can we propose a mechanism for the conversion of this alkyne into a ketone using mercury sulfate with water and sulfuric acid feel free to pause the video and propose a mechanism for this now you might see different mechanisms depending on which textbook you use but proposing an acceptable mechanism for this reaction so here's one for you so let's start with the alkyne now mercury sulfate contains the mercury two plus ion and the sulfate ion the sulfate ion is respected ion so we're not going to worry about that so here's one proposed mechanism the mercury atom has a lone pair which i'm going to put right here and that's going to attack the alkyne and the alkyne will attack the mercury simultaneously and so you're going to get an intermediate that looks like this so now we have a double bond we have the mercury atom and it has a a two plus charge and at this point we also have a hydrogen and an r group as well now what do you think is going to happen next it's important to understand that this structure has a resonance form if we break this bond this is one resonance form of this structure now we don't want to break this bond because if we do we're going to have a plus charge on a primary carbon which is less stable than having a plus charge on a secondary carbon so this structure is in resonance with this structure which tells us that this carbon has some partial positive charge now some textbooks may show the first step like this they may show the alkyne attacking the mercury ion and the mercury ion will add to the less substituted carbon atom so now instead of having a plus two charge it has a positive charge it still has the lone pair and now we have a positive charge on this carbon and then we could show that the lone pair combines with this given us this intermediate if you do it this way you can show that there's some partial positive charge on this carbon atom if you draw the resonance hybrid between these two so you might see a mechanism something to that effect but it's important to understand that those two structures are in resonance with each other so the actual intermediate is somewhere between the two i'm going to use this structure for the mechanism now water is going to come in and water is going to preferentially attack this carbon as opposed to this one and so that explains why we get markovnikov addition with this uh reaction why the oh group goes on the secondary carbon as opposed to the primary carbon it's because the secondary carbon has more partial positive charge than the primary carbon and so when it attacks that carbon this bond is going to break so we have a double bond with a mercury ion attached to it a hydrogen and r group and we have an oxygen attached to this carbon now in the next step we're going to use another water molecule to remove this hydrogen and so we have this intermediate as of now now what do you think is going to happen next so this is called a mercuric enol you could see the enol function group we have an alkene function group with an o h molecule so what's our next step here at this point we're going to react this intermediate with h3o plus so this oxygen will use this lone pair to form a pi bond causing this pi bond to break which will mean that this carbon will grab a hydrogen and then the oh bond will break giving us h2o again so now this is a single bond we still have the mercury attached to it now we have two hydrogen atoms on this carbon we still have an r group and now we have a carbonyl group with a hydrogen attached to it and it has a positive charge in the next step the mercury atom can spontaneously expel itself by doing this the molecule will become more stable notice that the total charge of the molecule right now is positive two and so it's not very stable but in this step it's going to become neutral so now we have a double bond two hydrogen atoms the mercury atom is gone and now we just have an oh group with two lone pairs so now we have a neutral enol molecule so what's going to happen next what can we do with this enol how can we show its conversion into a ketone under acidic conditions so under acidic conditions you can only use water and h3o plus so h2o plus will be the acid it's used for pronation water is used for depronation it's going to behave as the base under basic conditions water behaves as the acid hydroxide behaves as the base so under acidic conditions the enol will react with the acid in this case h3o plus so this lone pair is going to form a pi bond causing a double bond to abstract a hydrogen and so now we have a ch3 group on a primary carbon and we have a protonated ketone at this point so now this molecule or this ion rather it lost the hydrogen so now it's h2o so it's going to regain that hydrogen again when it takes it from the carbonyl group and so now we have the ketone as a final product so that's one way you can propose a mechanism for this particular reaction what's going to happen if we mix propine with hydrobromic acid what's the major product of this reaction if we add hbr to an alkyne it's initially going to go down to an alkene and this reaction proceeds with markovnikov regiochemistry which means the bromine atom is going to go on the secondary carbon as opposed to the primary carbon now what's going to happen if we add hpr again the double bond will go to a single bond now the second bromine atom will still go on the secondary carbon the reaction still proceeds with markovnikov geochemistry so we get this product now here's another example let's say we have 2-pentine what's going to happen if we react it with hbr in this case both of the carbon atoms of the alkyne they're both secondary so the bromine atom can go on either one of these carbon atoms let's put it on this one first so the triple bond is going to go down to a double bond and we're going to have the bromine atom there now in addition to getting the z isomer we can also get the e isomer so the bromine atom is still on the second carbon and so this right here is the e isomer and this is the z isomer they're both 2-bromo 2-pentene now if we add hbr again the second bromine atom will still go on carbon 2 and so we'll get one final product which looks like this but it's important to understand that you can get a mixture of enz isomers at that point now these are not the only products that we can get because we can put the bromide atom on carbon 3 as opposed to carbon 2. so it can be here and this is the z isomer again so we can get the e isomer which we can draw it this way so this is the e isomer the highest priority groups are on opposite sides of the double bond and then once we add hbr we can get this product with both bromine atoms being on the third carbon now wherever the first bromine atom is the second bromine atom is going to go on the same carbon and we're going to talk about why that's the case now let's propose a mechanism between the reaction of a terminal alkyne with hydrobromic acid so the alkyne will behave as the nucleophile going for the hydrogen expelling the bromide ion so now we're going to have a double bond and we're going to add the hydrogen to the primary carbon because it's better to have a positive charge on a secondary carbon instead of a primary carbon and then in the second step the bromide ion could attack the carbon with a plus charge and so we have here is the vanilla cation and this is going to give us this product now let's react this with hbr again so the double bond is going to attack the hydrogen expelling the bromide ion once again now let's talk about where to put the hydrogen atom so if we place it on the secondary carbon the plus charge will be on the primary carbon which is not good but if we place it if we place the hydrogen on the primary carbon the plus charge will be on a secondary carbon which is better and not only that but this carbocation is stabilized because of the bromine atom this bromine atom can donate a pair of electrons creating this resonance structure which looks like this and so now the bromine atom bears the positive charge and so that's why the second bromine atom goes on the same carbon as the first bromine atom because that first bromine atom can stabilize the positive charge on a carbon so now the bromide ion is going to attack this carbon breaking this pipe on and so in the end we're going to have two bromine atoms on this carbon so i'm going to redraw it like this this is called a geminal dihalide now let's move on to our next example by the way going back to the mechanism it's important to understand that veno cations are not very stable in fact their stability is comparable to a primary carbocation and so as a result some chemists have proposed a different mechanism for this reaction so in the first step where the alkyne reacts with hbr a pi complex is composed which looks like this so the hydrogen bear is a partial positive charge and then a bromide ion comes in attacks this carbon causing this bond to break and this one too expel in this bromine atom and so it gives you this intermediate with the bromine atom being on this carbon so that's another mechanism that's proposed for that first step so just keep that in mind different textbooks may use different mechanisms i've seen that let's react propine with br2 in dichloromethane so this reaction is very similar to reacting an alkene with bromine the alkene will go to an alkane with two bromine atoms added anti-addition in this case the alkyne is going to go down to an alkene and we're still going to add two bromine atoms with anti-addition so that's what's going to happen if we add br2 to an alkyne now if we repeat the process the alkene will go to an alkane so we're going to have four bromine atoms added in total so that's the end result of reacting an alkyne with bromine using two equivalents for each equivalent of bromine that you add you're going to lose a pi bond now what's going to happen if we take this particular product and react with hbr the bromine atom is going to preferentially add to the secondary carbon as opposed to the primary carbon so we're going to have two bromine atoms on a secondary carbon and only one on a primary carbon so that's going to be the end result if we add hpr to this particular substance now let's start with the same alkyne but this time what's going to happen if we add hbr with peroxides so in the first step we're going to get a double bond but this time the bromine atom will go on the primary carbon as opposed to the secondary carbon so we're going to have anti-morkondikov regiochemistry now we can get a mixture of isomers we can get both the e and the z isomer so this is the z isomer and this is the e isomer now let's work on some practice problems starting with propine what reagents can we use in order to make this product go ahead and try that problem so one way we can do this is we can start by adding hbr and so this will give us initially a double bond with a bromine atom and then in the second step we could simply add cl2 with dichloromethane and that will give us this final product or we could add cl2 to begin with and so we're going to get this double bond with two chlorine atoms and then we can add hbr which will give us the same final product so both methods can lead to the same product now it's important to understand that this particular intermediate is less reactive than this one and the reason for that is there's two electron withdrawing groups that pull away electron density from the double bond making the double bond less nucleophilic whereas this one only has one and so whenever you have these electron withdrawing groups that pull electron density by means of the inductive effect it weakens the nucleophile so this particular molecule is more reactive than this one so therefore put in hbr first and then add in cl2 later it's better than adding cl2 first and then hvr later but nevertheless both reactions can still lead to the same product let's try another example so let's say if we want to make this particular product what reagents should we use one way we could do this is we can add hbr with peroxides and so this will put the bromine atom on a primary carbon and then we can add br2 with ch2cl2 and that's going to put a bromine atom here and here giving us this product now let's go over one more example with this topic so starting with this alkyne how can we produce a vicinal dihalide where we have two halogens but they're separated by a carbon atom one way you can do this is we can use h2 and the lindler's catalyst to turn this into a double bond now typically we would get a cis alkene but you can't have a cis alkene when the double bonds at the end of a chain and then after that we can add just br2 with dichloromethane and so that's going to put the bromine atom the two bromine atoms across this double bond giving us that answer now what do you think is going to happen if we have a vicinal dihalide and if we react it with sodium amide the end result is that we're going to get a triple bond if we were to use let's say potassium hydroxide and it has to be heated we can get an alkyne too but it's going to give us an internal alkyne whereas this one ultimately will give us a terminal alkyne now we need to protonate it at the end and we also need three equivalents of sodium amide to get the terminal alkyne and technically it really doesn't matter what type of dihalide we have sodium amide will favor the formation of terminal alkynes and potassium hydroxide at very very high temperatures will favor the formation of internal alkynes in fact if you take let's say a terminal alkyne and if you react it with potassium hydroxide at high temperatures this will isomerize into the internal alkyne these two actually exist in equilibrium these reactions are reversible but at high temperatures the more stable alkyne will be the major product and internal alkynes are more stable than terminal alkynes so that's why potassium hydroxide at high temperatures will favor the internal alkyne now if you have the internal alkyne and if you want to get the terminal alkyne you need to use sodium amide followed by water and this will isomerize to the terminal alkyne so you can have any one of these four dihalides so we can have a vicinal dihalide we can have a geminal dihalide at carbon 2 or we can have a geminal dihalide at carbon 1. or we can have a vicinal dihalide at carbons two and three instead of one and two so for any one of these uh four compounds if we react it with let's say i need a bigger arrow than that sodium amide in the first step followed by water the end result will be the terminal alkyne so any one of these molecules will give us the terminal alkyne with sodium amide now if we use potassium hydroxide at high temperatures with any one of these terminal i mean any one of these dihalides we're going to get the internal alkyne so just as long as you understand that the end result for sodium amide will be the terminal alkyne and the end result with koh at high temperatures will be the internal alkyne then you should be okay with these problems now let's work on a mechanism problem so consider this vicinal dihalide show a mechanism for the conversion of this molecule using potassium hydroxide at 200 degrees celsius show how we can get the internal alkyne so the first thing i'll do is draw the hydrogen atoms hydroxide is a strong base and through an e2 elimination reaction is going to abstract a proton form a double bond and expel a bromine atom at the same time and so we're going to get an alkene with a bromine atom attached to it now another hydroxide ion will follow the same pattern it's going to grab a hydrogen form a triple bond and kick out the bromine atom and so we're going to get an alkyne so that's how you can convert a dihalide into an alkyne now here's another mechanism problem for you how can we convert this internal alkyne into a terminal alkyne using sodium amide and ammonia followed by water go ahead and propose a mechanism so i'm going to draw two butane like this and i'm going to show these carbon hydrogen bonds so in nanh2 the nh2 minus ion is a strong base it has two lone pairs and the negative charge so it's strong enough to take off this hydrogen put in a negative charge on a carbon now this step is reversible so it can go backwards so right now we have this intermediate so we have a negative charge on this carbon and it's stabilized by resonance that lone pair can be used to form a pi bond causing this pi bond to break so now we can put the negative charge on this carbon atom and so this is what we now have at this point so that carbon with a negative charge can react with ammonia it can take off a hydrogen regenerating the nh2 minus base and keep in mind this step is also reversible so this is what we now have so now we need to use nh2 minus again to take off this hydrogen put in a negative charge on this carbon now the reason why these hydrogens are relatively acidic is because the carbon with the negative charge or the conjugate base that forms it's stabilized by resonance and so we could use this lone period to form a triple bond put in a negative charge on that carbon and so now we have this so now we need to react this with ammonia again so by the way the previous step that led to this is still reversible so this will take off a hydrogen giving us a terminal alkyne now nanh2 or nh2 minus will still react at this point everything is reversible up to this point now this is a strong enough base to take off the hydrogen of a terminal alkyne potassium hydroxide is not strong enough to remove this hydrogen from the terminal alkyne once you get the conjugate base it's not stabilized by resonance and so koh can't remove that hydrogen and this is why sodium amide and a nh2 can form terminal alkynes even though terminal alkynes are less stable than internal alkynes it's because any nh2 is a stronger base than koh and so it can take off this hydrogen in a non-reversible step so everything is reversible up to this point but because this step is not reversible once you get this product it just remains in the solution and it's going to stay like this until you add water so once you add water this alkynide ion will take a hydrogen generating hydroxide and so that's how we can form the terminal alkyne and get it out of the solution so it's important to understand that koh or hydroxide even at high temperatures cannot take off this hydrogen it's not strong enough however it could take off an adjacent hydrogen on a carbon atom if it's heated because this hydrogen is more acidic as you mentioned before if it takes off this hydrogen the conjugate base is stabilized by resonance with the triple bond and so that makes this hydrogen more acidic towards deprotonation by hydroxide but hydroxide is not strong enough to take off the hydrogen of a terminal alkyne because once you put that negative charge on the alkyne it's not stabilized by resonance so this hydrogen is more acidic due to the stabilization of the conjugate base so let's see if i can put this all together here is the internal alkyne and here we're going to use potassium hydroxide at 200 degrees celsius and so this reaction is reversible so we have a bigger arrow towards the internal alkyne and the reason for that is the internal alkyne is more stable and so potassium hydroxide can take this internal alkyne and convert it to this it could also take this internal alkyne and convert it to that so these two are reversible however internal alkynes are more stable so potassium hydroxide will preferentially make the internal alkyl but both reactions are going backwards and forwards so it can basically go both ways however this one is more favorable when using koh now when using sodium amide the same thing is going to happen these two reactions are still reversible with the terminal i mean with the internal alkyne being more favorable even with any nh2 so this is still more stable however nanh2 can take off this hydrogen so as soon as this terminal alkyne is generated any nh2 will take it a step further to deprotonate that alkyne and this step is not reversible so as soon as this forms it doesn't go back into this form and so that's why we can get this thing out of the solution once we add water thus stopping the reaction so regardless of which base you use these two can interconvert but when you use this base sodium amide you're going to get the alkynide ion and it's no longer reversible at this point giving us the term alkyne koh can't take off this hydrogen and so that's why koh will not make this product koh will preferentially form the internal alkyne so hopefully this gives you an overview of these reactions now let's move on to our next topic so let's say if we have an acetylide ion and we're going to react it with sodium amide followed by ethyl bromide what is the major product of this reaction and propose a mechanism as well so we know that the nh2 ion it's a strong enough base to deprotonate the terminal alkyne giving us an acetylide ion so this ion is a very strong nucleophile and it's going to react with alkyl halides now we know that bromine has a partial negative charge and the carbon that bears the bromine is partially positive and so a nucleophilic carbon will attack an electrophilic carbon creating a carbon carbon bond and so the end result of this reaction will be this product so here we have the triple bond and we've added an ethyl group to it so that's one way in which you can make carbon-carbon bonds it's by combining a nucleophilic carbon with an electrophilic carbon you