Reactions and Mechanisms of Alkynes

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 butane. What's going to happen if we react it with hydrogen gas and palladium overcarbon? Initially, when you add the first hydrogen molecule, this will turn into an alkene. And this reaction occurs with syn addition. 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 catalysts, 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 butyne 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 alkene level, it stops at the alkene level. Now you need to know that the Lindner'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 Lindler's catalyst sometimes you might see it like this as opposed to seeing the word Lindler's 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 alkaline metal such as lithium metal too. 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 alkene 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-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 vanillic 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 this 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 transalkenes. The hydrogen atoms are on opposite sides. Now let's go over some other reactions. Let's say if we have propane, 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 OH 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 alkynes it also produces alcohols. Now when dealing with alkynes instead of using BH3 typically you'll see R2BH. Sometimes you might see SIA2BH, disa-amo, borane. Now, the reason why you want to use this when dealing with terminal alkynes is because borane 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 the hydrogen there. So, I'm going to leave it like this. Now, this can react with another borane molecule, turning 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 borane 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 alkyne, 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 OH is going to be on the less substituted carbon atom. So this reaction proceeds with more carbonic acid addition. The OH group is on... the secondary carbon if you use mercury sulfate, but if you use hydroboration oxidation, the OH 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 tautomerize, 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 hydroboration oxidation reaction of alkynes is that you get an aldehyde. Using 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 2-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 going to get a mixture of products. we're going to get two pentanone and also will get three pentanone. Now here's a 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 propose 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 2 plus ion and the sulfate ion. The sulfate ion is a suspected 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 2 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 2 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 can show that the lone pair combines with this, giving 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 with 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 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, an 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 can see the enol function group. We have an alkene function group with an OH molecule. So what's our next step here? At this point, we're going to react this intermediate with H3O+. 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 carbon O 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 2, 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+. So H3O plus will be the acid, it's used for protonation, water is used for deprotonation, 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+. 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 our final product. So that's one way you can propose a mechanism for this particular reaction. What's going to happen if we mix propyne 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 VGO chemistry, which means the bromine atom is going to go on a secondary carbon as opposed to the primary carbon. Now what's going to happen if we add HBr 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 VGO chemistry, so we get this product. Now here's another example. Let's say we have two penton. 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 E and Z 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-I-SMR. 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 what we have here is a vanillic 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 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 the 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 venocadines 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 proposed, which looks like this. So the hydrogen bears a partial positive charge. And then a bromide ion comes in, attacks this carbon. causing this bond to break, and this one too, expelling 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 propyne 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 alkene. 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 HBr 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-Morkovnikov 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 propyne, 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 HBr 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 the 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 bond is 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 vicino 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 2 and 3 instead of 1 and 2. So for any one of these 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 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 vicino 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 hydroxyl 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-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 the 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-base. And keep in mind, this step is also reversible. So this is what we now have. So now we need to use NH2-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 a 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. Given us a terminal alkyne. Now, any NH2 or NH2-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, NaNH2, can form terminal alkynes, even though terminal alkynes are less stable than internal alkynes. It's because NaNH2 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 alkanide 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 we 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 kind and the reason for that is the internal kind 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 alkyne. 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 internal alkyne being more favorable, even with any NH2. So this is still more stable. NaNH2 can take off this hydrogen. So as soon as this terminal alkyne is generated, NaNH2 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 alkanide ion, and it's no longer reversible at this point, given 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 intern 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... and electrophilic carbon.

Initially, when you add the first hydrogen molecule, this will turn into an alkene. And this reaction occurs with syn addition. 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 catalysts, 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 butyne 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 alkene level, it stops at the alkene level.

Now you need to know that the Lindner'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 Lindler's catalyst sometimes you might see it like this as opposed to seeing the word Lindler's 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 alkaline metal such as lithium metal too. 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 alkene 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-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 vanillic 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 this 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 transalkenes. The hydrogen atoms are on opposite sides.

Now let's go over some other reactions. Let's say if we have propane, 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 OH 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 alkynes it also produces alcohols. Now when dealing with alkynes instead of using BH3 typically you'll see R2BH. Sometimes you might see SIA2BH, disa-amo, borane. Now, the reason why you want to use this when dealing with terminal alkynes is because borane 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 the hydrogen there. So, I'm going to leave it like this. Now, this can react with another borane molecule, turning 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 borane 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 alkyne, 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 OH is going to be on the less substituted carbon atom. So this reaction proceeds with more carbonic acid addition.

The OH group is on... the secondary carbon if you use mercury sulfate, but if you use hydroboration oxidation, the OH 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 tautomerize, 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 hydroboration oxidation reaction of alkynes is that you get an aldehyde. Using 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 2-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 going to get a mixture of products. we're going to get two pentanone and also will get three pentanone.

Now here's a 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 propose 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 2 plus ion and the sulfate ion. The sulfate ion is a suspected 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 2 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 2 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 can show that the lone pair combines with this, giving 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 with 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 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, an 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 can see the enol function group.

We have an alkene function group with an OH molecule. So what's our next step here? At this point, we're going to react this intermediate with H3O+.

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 carbon O 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 2, 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+. So H3O plus will be the acid, it's used for protonation, water is used for deprotonation, 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+. 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 our final product. So that's one way you can propose a mechanism for this particular reaction.

What's going to happen if we mix propyne 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 VGO chemistry, which means the bromine atom is going to go on a secondary carbon as opposed to the primary carbon. Now what's going to happen if we add HBr 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 VGO chemistry, so we get this product.

Now here's another example. Let's say we have two penton. 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 E and Z 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-I-SMR. 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 what we have here is a vanillic 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 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 the 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 venocadines 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 proposed, which looks like this. So the hydrogen bears a partial positive charge. And then a bromide ion comes in, attacks this carbon. causing this bond to break, and this one too, expelling 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 propyne 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 alkene. 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 HBr 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-Morkovnikov 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 propyne, 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 HBr 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 the 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 bond is 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 vicino 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 2 and 3 instead of 1 and 2. So for any one of these 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 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 vicino 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 hydroxyl 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-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 the 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-base. And keep in mind, this step is also reversible. So this is what we now have. So now we need to use NH2-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 a 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. Given us a terminal alkyne. Now, any NH2 or NH2-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, NaNH2, can form terminal alkynes, even though terminal alkynes are less stable than internal alkynes. It's because NaNH2 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 alkanide 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 we 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 kind and the reason for that is the internal kind 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 alkyne. 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 internal alkyne being more favorable, even with any NH2. So this is still more stable. NaNH2 can take off this hydrogen.

So as soon as this terminal alkyne is generated, NaNH2 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 alkanide ion, and it's no longer reversible at this point, given 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 intern 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... and electrophilic carbon.

Transcript for:Reactions and Mechanisms of Alkynes

Transcript for:
Reactions and Mechanisms of Alkynes