|Ch. 1 - A Review of General Chemistry||4hrs & 48mins||0% complete|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 19mins||0% complete|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete|
|Ch. 8 - Elimination Reactions||2hrs & 25mins||0% complete|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete|
|Ch. 10 - Addition Reactions||3hrs & 32mins||0% complete|
|Ch. 11 - Radical Reactions||1hr & 55mins||0% complete|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 42mins||0% complete|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 20mins||0% complete|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 56mins||0% complete|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete|
|Ch. 23 - Amines||1hr & 43mins||0% complete|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete|
|Ch. 25 - Phenols||15mins||0% complete|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete|
|Ch. 26 - Transition Metals||5hrs & 33mins||0% complete|
|Alkene Stability||7 mins||0 completed|
|Zaitsev Rule||25 mins||0 completed|
|Dehydrohalogenation||8 mins||0 completed|
|Double Elimination||9 mins||0 completed|
|Acetylide||15 mins||0 completed|
|Hydrogenation of Alkynes||17 mins||0 completed|
|Dehydration Reaction||27 mins||0 completed|
|POCl3 Dehydration||6 mins||0 completed|
|Alkynide Synthesis||16 mins||0 completed|
Now it’s time to introduce some of the most important nucleophiles in all of organic chemistry, alkynides.
Concept #1: 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 alkynides. So let's go ahead and talk about how we can do that.
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 H at the end are uniquely acidic due to an effect called the hybridization effect.
Now when we talked abut acids and bases we talked about that there were basically five effects that made things more acidic. There was like the element effect, inductive effect, resonance, solvation and there was hybridization. I know this might be 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 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 called bond sites and that would tell you what the hybridization is.
So, in this case, I would just have one bond over here, a second bond over here. That's two. Two bond sites equals sp. So we know that 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.
Now just so you guys know that's pretty high s character because, for example, sp3, that's a hybridization you should be really familiar with at this point. Sp3 has one 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. Like I said, hydrocarbons are really sucky acids usually. For this to be 25, that's not bad.
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 H and give a negative charge to the terminal alkyne. So we're going to use a strong base to deprotonate that H and usually the bases that we use are either NaH, sodium hydride, or NaNH4, sodium anide, 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 abut that green hydrogen above me.
That's going to create a strong nucleophile that we're going to call a sodium alkynide. What the sodium alkynide looks like is it's just going to be a negative charge on that triple bond.
So if I were just to 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+ plus H-. Now the H- is what's going to react, that's the base, it's going to reach with the H. So I would pull off the H and donate those electrons to the triple bond giving the triple bond now a negative charge.
Recall that most hydrocarbons are terrible acids, having pKas from 40-50. However, terminal alkynes (alkynes with a hydrogen) are uniquely acidic due to the hybridization effect, having a pKa of 25.
This means they can be easily deprotonated using a strong base (typically NaH or NaNH2).
The resulting conjugate base is known as a sodium alkynide, and these are often used to lengthen carbon chains to through SN2 reactions.
Let's take a look at the mechanism of akynide synthesis.
Concept #2: The mechanism of akynide synthesis.
So the interesting thing about this is that we can use these Alkynides in what's called multi-step 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 that Sodium Alkanides can do and that is to react with a primary Alkyl Halide so I'm going to 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 alkaline and a strong base? Well what I could expect would be an acid base reaction so this is just the acid base, this would be that my NH2 negative is going to react with my strong....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 a 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 Alkyne, OK? The reason we call it a sodium Alkynide is because a lot of times the sodium will associate with it so it will be close by, alright? So now that we have that we could either stop there and for some you guys your professors are just going to ask you to stop there and that's going to be it but others of you your professor is actually going to take this a step further and say OK now that we have that Sodium Alkynide what can we react to 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, OK? Now if you haven't learned the substitution reaction yet that's OK you're going to learn it soon if you guys have learned than they should be easy for you, OK? But regardless I'm 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 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 going to be that very big sodium alkynide, alright? So just you know for those of you that already have learned substitution you could totally use my flow chart to figure this out, OK? If you haven't gotten there yet that's OK but eventually you can just use a flow chart to figure out that this would be an SN2 reaction, OK? Because this is 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 that in black because that's the color that I use 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 three 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 a substitution on that Alkyl group, OK?
Beware: You can only use this reaction with primary alkyl halides. Secondary alkyl halides will favor E2!
Let's put it all together!
Concept #3: Using double dehydrohalogenation to perform alkynide synthesis.
So now what I want to do is I want to move on to one more question and we're going to do this one together again. And you guys are going to try to help me out a little bit.
So some of you guys have already learned, most of you, have already learned how to make triple bonds. If you recall, the way that we make triple bonds is by doing a double elimination of dihalides. Now for some of you, this is going to be new information, meaning you haven't learned this yet. And that's okay, you're going to get to it. I'm just going to show you how to do this here because this happens to be one of the crucial steps of making an alkynide. In fact, usually what we do is we go from a single bond, then we make a triple bond and then once we have the triple bond, we can make the alkynide with deprotonation. I'm going to show you how to do all that right here.
So here notice that we have a dihalide. So we have our dihalide and I'm reacting it with actually three equivalents of a strong base. Why would I use three? That's a very strange number. It turns out that two of those equivalents are going to be used to make the triple bond through double elimination and then one of those equivalents is going to be used to actually make the alkynide. Let me show you how that works.
So basically, we have hydrogen here, we have a hydrogen here. The red hydrogen can eliminate with the red Br. The blue hydrogen can eliminate with the blue Br. What that means is that my first equivalent of NH2- is going to pull off one of the hydrogens and make a double bond that looks like this. Is that cool so far? And I still have – well, I should have drawn the H probably going up over here. That's the blue H. I still have the blue H left and I still have the blue Br left. So that was my first equivalent, so that was times one.
Now, what about my second equivalent. Well, my second equivalent, this would be times two, can pull off the other H, so now I'm going to do a double elimination. I'm going to make my triple bond and what that's going to give me is look, a triple bond, so one – so now I have a triple bond from the first two equivalents of my base.
But now I have this third equivalent. What's the third equivalent going to do? If I add my third equivalent, so this is times three, what that's going to do is it already has a triple bond, this is the sodium alkynide part. And this is the part that all of you guys should know. Basically, what we're going to do is we're going to take the acidic H and we're going to do an acid/base reaction with it. We're going to take that off and make a negative charge and what I'm going to end up with now is a sodium alkynide.
This is really the only pathway in organic one to make sodium alkynides. We do a double elimination and then we deprotonate and then we make our sodium alkynide.
Now I want to just point out that your professor doesn't always have to write times three. In fact, what they might write instead, they might write excess. If they write one of these strong bases either NaH or NaNH2 in excess, that's really the same thing as them saying times three. They're just saying react it as far as it will go. So that just means react it all the way to the sodium alkynide.
Now that we're at the sodium alkynide, I'm reacting that with a molecule called CH3I. What the heck is that? Well, if you think about it, that is an alkyl halide. So I just showed you guys in the previous example how to nucleophile and react it with an alkyl halide and get a substitution reaction. So now I'm actually going to stop the video and let you guys try to figure out what the final product looks like. And then when you get out, when you're done, I'll go ahead and show you what it is. So get started on that.
Example #1: Predict the final product of the following reaction sequence.
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