Different methods for reducing alkynes to alkenes or alkynes. Created by Jay.
Want to join the conversation?
- at about6:25, why doesn't the Na bond to the C? Why instead does the C get an H from the NH3?(11 votes)
- It does bond to the C, but the bond is ionic rather than covalent. The electronegative difference between C and Na is great enough so that the bond is essentially ionic.
The carbanion is a very strong base. It can easily remove a proton from NH3 to form the much weaker base NH2-.(13 votes)
- At1:40, Jay introduces liquid ammonia as a reagent. If i remember correctly, ammonia is a gas at room temperature and at normal pressure. Does that mean that this reaction needs to take place at either a low temperature or a high pressure (or both)?
Will this reaction occur with ammonia as a gas? Perhaps the ammonia is aqueous or in some other solution? ... or am I missing something?(8 votes)
- That is correct. The boiling point of ammonia is -33°C. If the flask is immersed in a mixture of dry ice and acetone (-78°C), the ammonia will condense to a liquid. High pressure is not needed. There must not be any water present.(11 votes)
- Why does using Lindlar Pd as the catalyst reduce it only to an alkene while using Pt as the catalyst allows it to be reduced all the way to an alkane? What is the stopping mechanism with the Lindlar Pd catalyst?(7 votes)
- Lindlar’s catalyst is a Pd catalyst that is "poisoned" with traces of lead and quinoline. This reduces its activity so that it can reduce alkynes but not alkenes.(13 votes)
- In Na/NH3 system, Is there any possibility that the alkene products are further reduced to form alkane or polymerized like free radical polymerization of vinyl monomers?(7 votes)
- No, due to the radical mechanism, the pi electrons in the alkene will not split like in the alkyne. The Na/NH3 system is similar to the poisoned catalyst system where the alkyne will stop after the formation of the alkene.(4 votes)
- What happens to the sodium atoms when they donate that one electron?(3 votes)
- Hi since NH3 is a catalyst ,it should be conserved after reactions ends and product is formed but we are only left with NH2 not NH3 and in the reaction you did not mention Any role of hydrogen(1 vote)
- NH3 is not used as a catalyst, it is a reagent. Lindlar Pd is a catalyst because it is not used.(5 votes)
- I get it that you can do cis or trans hydrogenation, but why is each reaction restricted to form either form? What makes the Lindlar Pd one give out cis alkenes and what makes Na / NH3 give off trans alkenes?(1 vote)
- The hydrogen is absorbed onto the metal catalyst. When the alkyne approaches a hydrogen molecule absorbed onto the catalyst, the hydrogen atoms are both on the same side of the triple bond, leading to a cis alkene. With sodium and ammonia, as explained in the video the sodium donates an electron to one carbon which causes one of the pi bonds of the alkyne to split, with one carbon ending up as a radical (with a single unbonded electron) and the other as a carbanion (with a pair of unbonded electrons). The electrons of the radical and the carbanion repel each other by moving to opposite sides of what is now an alkene bond - this leads to a trans product.(4 votes)
- How exactly is palladised charcoal formed?
and how is quinoline a poison for alkynes?(1 vote)
- Add PdCl₂ and HCl to a warmed, aqueous suspension of activated charcoal. Then add formaldehyde (this reduces Pd²⁺ to Pd). Neutralize with NaOH. Filter off the catalyst. Wash it with distilled water, and dry it over KOH.
Quinoline is a poison for the catalyst, not the alkyne.
The nitrogen lone pair in quinolone binds chemically to the activated sites on the surface of the catalyst.
This reduces the number of active sites
Poisoned sites can no longer accelerate the reaction the catalyst was supposed to catalyze.
The activity of the catalyst is reduced.
Thus, a poisoned Pd/C catalyst will reduce an alkyne to the corresponding alkene, but it cannot reduce the alkene product to the corresponding alkane.(3 votes)
- Here NH3 is used to convert a alkene to trans alkene, I need to know that what if HCl is used instead of NH3 (HCl can serve the same function as NH3 i.e, donating a proton). Are the products formed be same in both cases?
Is this reaction an ideal one (if HCl is added)?(2 votes)
- HCl in its pure form is a gas and unable to act as a solvent. HCl as a solution will react with the sodium and prevent the reduction from occurring.(1 vote)
- Why is the Na,NH3 reduction trans? Why can it not be cis?(1 vote)
- The radical electrons want to be as far apart from each other as possible, so they arrange to be on opposite sides of the molecule. Jay mentions this at3:50.(3 votes)
In this video, we're going to take a look at two ways to reduce alkynes. The first way is a reaction we've seen before. This is the hydrogenation reaction. And we saw it before when we hydrogenated alkenes to form alkanes. Here we're going to hydrogenate an alkyne to form an alkene. And to do a hydrogenation reaction, we need some hydrogen gas, so some H2 right here. And then a metal catalyst, so we're going to use Lindlar palladium, which is a special type of catalyst. It will catalyze the reduction of the alkyne on the left, the alkene on the right. However, the reduction of the alkene to the alkane down here is slow. So, so slow that we can stop it if our goal is to just make an alkene. So this reaction will form a cis-alkene. And it was a syn edition of our hydrogens. So we're going to get the two hydrogens adding on to the same side, and this has to do with the mechanism of a hydrogenation reaction. So you can check out the earlier video on hydrogenation of alkenes to see more details. So Lindlar palladium, a poison catalyst, it will reduce an alkyne to an alkene. It will produce a cis-alkene. All right, so that's how to make a cis-alkene. Let's take a look at how to make a trans-alkene. So how do we reduce an alkyne to make a trans-alkene. So here is our alkyne So we have our triple bond like that. And we're going to add sodium metal, and we're also going to add liquid ammonia like that. So we're going to form a trans-alkene. So I'm going to put-- this time my two hydrogens are going to be on opposite sides of each other. So this is formation of a trans-alkene like that. And it does this by an anti addition of hydrogens, right? So these are adding from opposite sides like that. Let's take a look at the mechanism to form a trans-alkene. So I start with my alkyne. So I'll go ahead and put in my carbons there and put an R group on the left side. And I'll make this an R-prime group to distinguish it from the R group over there. So we start with sodium, which we know, being in Group 1, has one valence electron like that. And in the first step of the mechanism, this sodium atom is going to donate its valence electron to the alkyne. So when we're showing the movement of one electron, we use a half-headed arrow. So I'm going to show this electron moving over here, but it's only one electron so I'm only going to do a half-headed arrow like that, not a full-headed arrow. So one of these bonds here between the carbons is going to break. And one of the electrons is going to move over here to this carbon like that. And one of the electrons is going to move over to the carbon on the left. So let's go ahead and draw the result of all those electrons moving around. So we have an R group here. And we had a triple bond, but now we only have a double between our two carbons, and then we have R-prime over here. So the carbon on the right picked up an electron from sodium, and it also picked up an electron from the breaking of that one bond there. So now it has two electrons around it like that, which gives us a negative 1 formal charge on this carbon. So it's a carbanion. It's an anion here. The carbon on the left picked up one electron for the breaking of that bond like that. So that's a radical that's something we haven't talked about before. So we actually form what's called a radical anion here. So let's go ahead and write that. This is a radical anion, so radical because there's an unpaired electron there. And then it also has a carbanion in the same molecule like that. So we have these electrons that are pretty close together, at least how I've drawn them, right? So we know that electrons are all negatively charged, so all these electrons are going to repel each other. So this isn't the most stable way for this molecule to have in terms of a conformation. These electrons are going to repel, and they're going to want to try to be as far away from each other as they possibly can. So what's going to happen is, we have our two carbons right here. And let's say that these two electrons stay over here on this side. This one electron's going to go over to the opposite side. They're going to try to get as far away from each other as they possibly can. And same thing with these R group here, right? So this R group is going to try to get as far away from this R-prime group as it possibly can. So this trans conformation is the more stable one. So this is our negatively charged carbanion right here. So in the next step of the mechanism, we remember ammonia is present. So let's go ahead and draw an ammonia molecule floating around like that. So here is our ammonia molecule. And the carbanion is going to act as a base, and it's going to take a proton from the ammonia molecule. So this lone pair of electrons is going to form a new bond with this proton, and these electrons are going to kick off onto the nitrogen. So let's go ahead and draw the results of that acid-base reaction. So now we have our two carbons, with an R group right here, R-prime right here. And now this carbon on the right is bonded to a proton. It bonded to a hydrogen like that. And then we still have our radical down here, so there is one electron on that carbon as well. All right, so the next step of our mechanism? Well, there's plenty of sodium present. So here's a sodium atom with one valence electron. The sodium is going to donate this electron to this carbon. So just use a half-headed arrow to show the movement of one electron. So if that sodium atom donates that one valence electron to that carbon, let's go ahead and draw the results of that. So we have two carbons double bonded, an R group over here, a hydrogen, and an R-prime. And this carbon had one electron around it. It just picked up one more from a sodium atom. So it's like that, which would give it a negative 1 formal charge. So this carbon has a negative 1 formal charge. So let's go ahead and draw that negative 1 formal charge. It's a carbanion. And once again, ammonia is floating around, so let's go ahead and draw ammonia right here, so NH3 like that. And the same thing is going to happen as did before, right? The negative charge is going to grab a proton. It's going to act as a base. And these electrons are going to kick off onto the nitrogen here. And so we protonate our carbanion, and we have completed our mechanism because now we have our two R groups across from each other. And we added on two hydrogens across from each other as well like that. So we formed a trans-alkene. All right, so that's the mechanism to form a trans-alkene. Let's look at a few examples. Let's start with this alkene right here. OK, so carbon triple bonded to another carbon, and we'll put a methyl group on each side like that. OK, so let's do a few different reactions with the same substrate here. So our first reaction will just be a normal hydrogenation with hydrogen gas, and let's use platinum as our catalyst. So this is not a poison catalyst. This is a normal catalyst. So what's going to happen is, first you're going to reduce the alkyne to an alkene. And then since there's no way of stopping it, it's going to reduce the alkene to an alkane. So this is going to reduce the alkyne all the way to an alkane. So if we go back up here to beginning, remember, we said that a poison catalyst will stop at the alkene, but if it's not a poison catalyst, it's just going to hydrogenate your alkene to an alkane down here. So this reaction is going to produce an alkane. Let's go ahead and draw the product. So we know that there are four carbons in my starting materials. There's going to be four carbons when I'm done here, so these two carbons in the center here are going to turn into CH2's. And then on either side, we still have our CH3's. So this is going to form butane as the product. All right, this time let's use a hydrogen gas, and let's use a Lindlar palladium here. This is our poisoned catalyst. So it's going to reduce our alkyne to an alkene, and then it's going to stop. And you have to think, what kind of alkene will you get? You will get a cis-alkene. So if we draw our two hydrogens adding on to the same sides, so now we have our methyl groups going like that. So our methyl groups will be going-- this and this would be our product, a cis-alkene. All right, let's do one more, same starting material. So this one right here, except this time we're going to add sodium, and we're going to use ammonia as our solvent. And remember, this will reduce our alkyne to an alkene, but it will form a trans-alkene as your product. So when you're drawing your product down here, you want to make sure that your two hydrogens are trans to each other. So they add on the mechanism, and then your two methyl groups would also be on the opposite side like that. So look very closely as to what you are reacting things with. Is it a normal hydrogenation reaction? Is it a hydrogenation reaction with a poison catalyst, which would form a cis-alkene? Or is it reduction with sodium and ammonia, which will give you a trans-alkene.