Acetylide - Online Tutor, Practice Problems & Exam Prep

Terminal alkynes exhibit unique acidity due to their sp hybridization, allowing them to act as strong nucleophiles when deprotonated by strong bases like sodium amide. This process generates acetylide ions, which can participate in nucleophilic substitution reactions with primary alkyl halides, leading to the formation of new carbon-carbon bonds. Additionally, dehydrohalogenation of vicinal dihalides can yield alkynes through elimination reactions, further expanding synthetic pathways in organic chemistry.

Understanding how to convert terminal alkynes to alkynides.

What I want to talk about now is a way that we can make triple bonds into strong nucleophiles. Now that might sound weird because if you remember, triple bonds are hydrocarbons, and usually hydrocarbons aren't really good nucleophiles or electrophiles. They kind of just sit there. But it turns out that triple bonds are uniquely acidic because of a principle that we talked about way back when we talked about acid and base chemistry. And we can use that to our advantage to make them strong nucleophiles. And these are called alkanines. So, let's go ahead and talk about how we can do that. So basically, remember that we talked about in the acid and base chapter. So if you didn't watch that chapter, it's fine. I'm just going to remind you now that terminal alkynes, the alkynes that have one hydrogen at the end, are uniquely acidic due to an effect called the hybridization effect.

Now when we talked about acids and bases, we talked about that there were basically 5 effects that made things more acidic. There were the element effect, inductive effect, resonance effect, solvation effect, and hybridization effect. I know this might be like, bringing back some bad memories for you guys, or maybe you just completely have no clue what I'm talking about. But hybridization was one of the things that could make something more acidic. And here's an example of a terminal alkyne right there. And let's just go through the, you know, go through the hybridization. Basically, what the principle said was that the more s character that an atom has, the more acidic it's going to be.

And notice that this carbon right here has what kind of hybridization? It has \( sp \) hybridization. The reason it's \( sp \) is because remember that the way we figured out hybridization was you count up the number of groups or what we call bond sites, and that would tell you how what the hybridization is. So in this case, I would just have one bond over here, a second bond over here, that's 2. Two bond sites equal \( sp \). Alright? So we know the hybridization is \( sp \). Then out of the entire hybrid orbital, what percentage of that is s? That's actually really easy to figure out because you just say how many hybrid orbitals are there total? Two. There's the s and there's the p. S would be 50% of that. Okay?

Now just so you guys know, that's pretty high s character because for example \( sp^{3} \). That's a hybridization you should be really familiar with at this point. \( sp^{3} \) has 1 s and 3 p's, so that one has a 25% s character. So that one's a lot less, so having 50% is really good. That must mean this is pretty acidic. And it turns out that it is. Back in acids and bases, I made you guys memorize that the \( pKa \) of a triple bond is 25, which is unusually low for a hydrocarbon. As I said, hydrocarbons are usually not very good acids. For this to be 25, that's not bad. Alright?

So how are we going to use this to our advantage? Well, it turns out that if we can use a strong enough base, we can pull off this hydrogen and give a negative charge to the terminal alkyne. Okay? So we're going to use a strong base to deprotonate like NaNH₂, or sodium amide. And both of these are very strong small bases that are going to be able to pull that hydrogen off of the terminal alkyne. So I'm talking about that green hydrogen above me. Okay? So that's going to create a strong nucleophile that we're going to call a sodium alkynide. Okay? And what the sodium alkynide looks like is it's just going to be a negative charge on that triple bond.

So if I were to just erase some of this stuff and show you what the acid-base reaction would look like for this reaction, it would be that I have, let's say, NaH. Right? Well, that's going to dissociate. Right? That would dissociate into \( Na^{+} \) and \( H^{-} \). Okay? Now the \( H^{-} \) is what's going to react. That's the base. It's going to react with the hydrogen. So I would pull off the hydrogen and donate those electrons to the triple bond, giving the triple bond a negative charge. Alright?

The mechanism of akynide synthesis.

So the interesting thing about this is that we can use these alkynides in what's called multistep synthesis. Alright? Now, for some of you, you're going to have to learn that in this chapter right here. For others, we're going to talk about that more later and that's fine. But I just want to kind of introduce you to at least one of the reactions sodium alkynes can do. And that is to react with a primary alkyl halide. So I'm going to just show you guys right here. This will be a worked example where we do this one together. And actually, we'll probably do the second one together as well just because I know this is new for a lot of you.

So first of all, notice that what I have is a terminal alkyne and I'm reacting that with NaNH₂. So, what kind of reaction could I expect from a terminal alkyne and a strong base? Well, what I could expect would be an acid-base reaction. So this would just be acid-base. This would be that my NH₂^- is going to react with my terminal alkyne and produce a negative charge on it. So, if I wanted to draw that mechanism, it would just look like this, where I always start from my negative charge, I pull off the acidic H, and I give those electrons to the triple bond. So, what that's going to make now is a triple bond that looks like this. Isn't that cool? So now we have a negatively charged triple bond and this is my nucleophile that's called an alkynide. Okay? The reason we call it a sodium alkynide is because a lot of times the sodium will associate with it, so it'll be close by. Alright.

So now that we have that, we could either stop there. And for some of you guys, your professor is just going to ask you to stop there, that's going to be it. But others of you, your professor is actually going to take this a step further and say, okay, now that we have that Sodium Alkynide, what can we react that with since it's a good nucleophile? And it turns out that we can use primary alkyl halides as a good reagent for this. Why? Because there's this reaction called a substitution reaction. And substitution reactions are basically done by a negatively charged nucleophile attacking an alkyl halide. Okay? Now if you haven't learned the substitution reaction yet, that's okay. You're going to learn it soon. If you have learned it, then this should be easy for you. Okay? But regardless, I'm just going to show you right now what would happen. Basically, the negative charge would go ahead and attack the carbon that's attached to the Bromine. And it's going to kick out that Bromine.

So what we call this we call this a substitution reaction because the Br is going to be switched with the nucleophile, which in this case is that very big sodium alkynide. Alright? So, just so you know, for those of you that already have learned substitution, you could totally use my flowchart to figure this out. Okay? If you haven't gotten there yet, that's okay. But eventually you can just use my flowchart to figure out that this would be an SN2 reaction. Okay? Because it's a primary alkyl halide. So now let's go ahead and draw our final product. Our final product would be basically a triple bond. But now that triple bond is going to have a new bond, so let's draw that new bond. I'm going to draw it in black because that's the color that I used for that arrow that came down, so that's the new bond. And that needs to be attached to a 3 carbon chain. So now I'm going to attach it like this. So now what I have is a triple bond and I have a 3 carbon chain on the other end. Alright? So that's the end of that first reaction. Pretty cool, right? So we made our sodium alkynide and then we did an SN2 reaction and did a substitution on that alkyl group. Okay?

Using double dehydrohalogenation to perform alkynide synthesis.

Hey everyone. So here we're going to do dehydrohalogenation of a Vicinial Dihalide. Here we have 2 bromides that are on neighboring carbons. In step 1, we're using 3 moles of sodium amide, a strong base. Now, this strong base here wants to deprotonate this structure. It wants to remove a beta hydrogen. Because the amide ion is small, it wants to do it by Zaitsev's rule. If we were to take a look at these 2 carbons, this carbon on the left only has 1 hydrogen and this one here has 2. So initially, this amide ion wants to attack the carbon with 2 hydrogens. That's the less substituted carbon. So because of this, we'll say that this is alpha and this is beta. The first mole of the amide ion comes in, removes this hydrogen, the bond here falls to make a double bond and kicks out the Br. Initially what we'll get is this alkene. That Br got kicked out. We still have this other Br here. Now, we still have an H here and an H here. We still have more moles of sodium amide, so the second mole can come in. Remember, sodium's around as a spectator ion. Both of these carbons here have 1 hydrogen, but it's the one on the right side now that has a leaving group. So, this would mean that this is beta now and this is alpha. The alpha carbon is the one with the leaving group. This amide ion comes in, removes this hydrogen, falls here to make a triple bond, and kicks out this bromine. What we're gonna have now is we're gonna have our triple bond. Remember here that a triple-bonded carbon with a hydrogen is extra acidic than we typically see with a hydrocarbon. And because of this, the 3rd amide ion will deprotonate it. The carbon here holds onto the electrons. This is important because now, this terminal carbon, a carbon at the end, is negatively charged, and that's where step 2 comes into play. In step 2, we have a methyl iodide. Because it's a methyl halide, it can undergo SN 2. So this negatively charged carbon here comes, hits this methyl carbon, and kicks out the iodine by SN 2. So at the end, we just added a methyl carbon to that triple-bonded carbon. We've now created an internal alkyne structure as our final answer. Right? So this would be our final answer to this particular question.

Predict the final product of the following reaction sequence.

Alright guys, so I want you to predict and draw the mechanism for this second reaction. And we know this is going to be an SN2, right? So your negative charge would attack the backside, do a backside attack onto the carbon with the leaving group, right? And now, we've got a problem because carbon already had 4 bonds, that new bond is going to make 5 bonds. It doesn't want to have 5. So if we make that bond, we have to break a bond and kick out the leaving group. And what is that going to give us? Well, we're going to get the same carbon backbone as before, but now there's one difference guys. Notice that before, I had a certain number of carbons. I actually had 8 carbons. And I can count them out for you if you want. There's 6 in the ring, right? 7 and 8. So there were 8 carbons in my structure before. And now after I react with the CH₃, what do I get? An additional carbon on my structure. So now I start off with 8, I now end up with 9 carbons, plus my I^- ion. I don't care about that. Now we're done. But what's interesting about this guys is that this shows that alkanides are really good for what we call organic synthesis. One of the most important jobs of an organic chemist is to take smaller, easy to attain molecules and make them bigger. In fact, that's some people's entire job their whole life is to take small molecules and turn them into molecules that if they can sell for a billion dollars as pharmaceuticals, right? So what we just did was we took a smaller molecule and by using an alkanide we made it bigger. So this is going to be one of the main ways that we have in orgo 1 to make smaller molecules bigger. So that's something that you want to look out for in the future and you want to hold on to that thought because we're going to be using it more. Now, one other point before we stop, before we turn the page, always count your carbons, guys, when you do these reactions, when you're adding them together to make sure you drew them right. So notice that we start off with 8 here and 1 here because my CH₃I had 1. So that means that you're really just adding them together. You should literally do this math at your exam, 8 +1 equals 9. I'm good. I drew it right. You'd be shocked how many of your classmates are going to get these questions wrong because they forget to draw 1 of the carbons in the end structure. Alright, guys. So let's keep going. Turn to the next page.

Here’s what students ask on this topic:

What makes terminal alkynes uniquely acidic?

Terminal alkynes are uniquely acidic due to their sp hybridization. In sp hybridization, the carbon atom has 50% s-character, which increases the acidity of the hydrogen attached to it. This is because the s-orbital is closer to the nucleus, making the bond more polar and easier to break. The pKa of a terminal alkyne is around 25, which is unusually low for hydrocarbons, indicating higher acidity. This property allows terminal alkynes to be deprotonated by strong bases, forming acetylide ions that can act as strong nucleophiles in various organic reactions.

How are acetylide ions formed from terminal alkynes?

Acetylide ions are formed from terminal alkynes through deprotonation by a strong base. The process involves using a strong base like sodium amide (NaNH₂) to remove the acidic hydrogen from the terminal alkyne. The reaction can be represented as:

${R-C\equiv C}_{-}+{\mathrm{Na}-\mathrm{NH}}_2\to {R-C\equiv C}_{-}+{\mathrm{Na}+}_{}+{\mathrm{NH}}_3$

This reaction generates a negatively charged acetylide ion, which is a strong nucleophile.

What is the role of sodium amide in the formation of acetylide ions?

Sodium amide (NaNH₂) acts as a strong base in the formation of acetylide ions. It deprotonates the terminal alkyne by removing the acidic hydrogen atom. The reaction can be represented as:

${R-C\equiv C}_{-}+{\mathrm{Na}-\mathrm{NH}}_2\to {R-C\equiv C}_{-}+{\mathrm{Na}+}_{}+{\mathrm{NH}}_3$

The sodium amide provides the amide ion (NH₂^-), which abstracts the proton from the terminal alkyne, resulting in the formation of the acetylide ion (R-C≡C^-).

How do acetylide ions participate in nucleophilic substitution reactions?

Acetylide ions, being strong nucleophiles, can participate in nucleophilic substitution reactions, particularly with primary alkyl halides. The negatively charged acetylide ion attacks the carbon atom bonded to the halide, displacing the halide ion in an SN2 mechanism. The reaction can be represented as:

${R-C\equiv C}_{-}+{R\text{'}-\mathrm{Br}}_{}\to {R-C\equiv C}_{-}+{\mathrm{Br}}_{-}$

This reaction forms a new carbon-carbon bond, extending the carbon chain and allowing for the synthesis of more complex molecules.

What is dehydrohalogenation of vicinal dihalides and how does it lead to alkyne formation?

Dehydrohalogenation of vicinal dihalides involves the elimination of hydrogen halides (HX) from adjacent carbon atoms, leading to the formation of alkenes or alkynes. Using a strong base like sodium amide (NaNH₂), the process can be driven to form alkynes. The reaction proceeds in two steps:

1. The base removes a β-hydrogen, forming a double bond and eliminating a halide ion.

2. A second equivalent of the base removes another β-hydrogen, forming a triple bond and eliminating the second halide ion.

The overall reaction can be represented as:

${R-\mathrm{CH}\left(\mathrm{Br}\right)-\mathrm{CH}\left(\mathrm{Br}\right)-R}_{+}$

This reaction is useful for synthesizing alkynes from dihalides.