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Terminal alkynes can function as weak acids if you react them with a very strong base. So something like sodium amide, so this in NH2 minus over here came from Na plus NH2 minus, so sodium amide, which is a very strong base. And so if the amide anion functions as a base, a lone pair of electrons and this nitrogen, it's going to take this proton right here. This is the acidic proton on terminal alkynes, and that leaves these electrons in here to kick off onto your carbon. So if I take NH2 minus, and that picks up an H plus, well, that would form NH3. So now I have nitrogen with three hydrogens attached to it and one lone pair of electrons. So when sodium amide functions as a base, it forms ammonia as its product. What is our other product? So we had our carbon triple bonded to another carbon with an R group here and then a lone pair of electrons on this carbon. So these electrons right here were the electrons in this bond. So those electrons, and they're moved onto that carbon. That gives this carbon a negative one formal charge, like that. So we form a carbanion here, also called an alkynide anion for this portion. So this is an alkynide anion, a carbanion in here. It's a relatively stable conjugate base, because the electrons, these two electrons here, are housed in an SP hybridized orbital, which has a lot of S character to it, so it's relatively small. So those negatively charged electrons are held a little bit more closely to the positively charged nucleus of this carbon here. So that somewhat stabilizes the conjugate base, which is the reason why a terminal alkyne can function as an acid. So once we formed our alkynide anion, we can use that alkynide anion to do an alkylation reaction. So let's go ahead and redraw that alkynide anion here. So I'm going to draw this portion, this R group over here. And then we have our carbon triple bonded to another carbon, a negative charge on this carbon on the right here. And this can now function as a nucleophile. So a negatively charged anion can function as a nucleophile. And if we react this alkyline anion with an alkyl halide-- let's go ahead and draw an alkyl halide here. So I'm going to put hydrogen there, and I'll put my halogen over here on the right, so putting my lone pairs of electrons. Let's draw in one R group right here and then a hydrogen over here. So here is my alkyl halides. And if you react a strong nucleophile with an alkyl halide that is not very sterically hindered-- this is a primary alkyl halide right here-- you're going to get an SN2 reaction. So think about this being an SN2 reaction. We have an alkyl halide, which has a polarized bond between the carbon and the halogen. So if the halogen is more electronegative, it's going to pull the electrons and the bond between it and carbon closer to itself. So this halogen ends up being partially negative. This carbon, therefore, will be partially positive, like that. So we have an electrophile. This carbon right here is partial positive charge. It wants electrons. Of course, our nucleophile has those electrons. So the lone pair of electrons on our carbon can attack our electrophile, so nucleophile attacks electrophile. And an SN2 mechanism, remember, is a concerted mechanism, meaning the nucleophile attacks the electrophile at the same time you're leaving group is leaving here. So these electrons are going to kick off onto your halogen. So let's go ahead and draw the product. So now we would have an R group, carbon triple bonded to another carbon. And now this is bonded to yet another carbon. So we formed a carbon carbon bond here in this reaction. And then this hydrogen is up here. This R group is still here, coming out at us, and then the hydrogen going away from us like that in space. And then we have our halogen over here with now four lone pairs of electrons, a negative 1 formal charge. It is stable on its own as an anion like that. So this is an alkylation reaction, right? We put an alkyl group onto our alkyne. So our alkyl group consisted of this carbon, whatever this R group is here. And we formed a new carbon carbon bond, so we? Alkylated our alkyne. Let's look at an example of an acid base reaction followed by an alkylation. So let's start with the acetylene, so the simplest alkyne. So we have carbon triple bonded to another carbon and then two hydrogens on either side and if we react that with sodium amide here. So we know sodium amide being a strong base, if we use one more equivalent, it's going to take off one of these acidic protons. So let's say it's the proton on the right here, and so we're going to lose that proton. So we're going to leave those two electrons behind on this carbon, making this carbon negatively charged. And then the positively charged sodium ion is going to interact with that negatively charged carbanion like that. So that's so that's the first reaction, Formation of your alkynide anion. And then if you want to do an alkylation, it's a separate reaction. You take this, and let's react it with the ethyl bromide. So CH3CH2Br. If you think about what's going to happen, the lone pair of electrons on the carbon is going to attack this carbon, the one that's bonded to your halogen, like that. The halogen is going to leave, and you're going to put this alkyl group onto your alkyne. So you're going to end up with an ethyl group on your alkyne. So let's go ahead and draw that. So we have hydrogen and then carbon triple bonded to another carbon, and then we have to put our alkyl group on there. So a CH2CH3, like that. So we've alkylated our alkyne. This is a very useful reaction for organic synthesis. So let's take the molecule we just made, and let's make something else with it. So if I took this-- let me go ahead and redraw it over here. So if I took this alkyne, so we just formed this. And let's react it. Two steps. Let's first react it with our base again. So let's use sodium amide right here. And in our second step, we'll react it with a primary alkyl halide. So let's go ahead and draw a primary alkyl halide here, so that is our molecule. So we think to ourselves, what happens? I have a strong base. I still have an acidic proton left on my alkyne, right? So the proton over here on the left. So that's what the base is going to do. The base is going to take that proton forming a negatively charged carbanion, an alkynide anion. And then that anion is going to be our nucleophile for an SN2 reaction. So when you're thinking about it, these electrons in here that are going to be on that carbon giving a negative 1 formal charge are going to come all the way over here and attack this carbon and attach all of this alkyl group to our carbon. So let's go ahead and draw the products of that. We're going to have our benzene ring. So let's go ahead and draw our benzene ring here, so let's put in our electrons going around my benzene ring. And then on that benzene ring is a CH2. So that CH2 is the red one that we marked right here, and this is the alkyl group that gets put onto your alkyne. So let's just go ahead and finish drawing our alkyne here. So we have now our triple bonds, right? Carbon triple bonded to another carbon. And then our ethyl group. So CH2CH3. So you'll see in later videos how we use the acidity of terminal alkynes to alkylate when we do a few different synthesis problems.