hello class our discussion today is going to be on alkynes so we recall that alkynes look like this and so we need to understand some special features about this molecule so we can gain a better appreciation for why it's reactive because this what we're going to be discussing is how do alkynes react with other reagents so the first thing that we see is if I we need to know what these where these electrons are coming from so one of those bonds right there is going to be our Sigma Bond and then we have are these going to be color-coded drastically different I put a blue Bond up there and a pink one down here so in an alkyne we have a green bond in middle which is the sigma Bond and then we have two Pi Bonds on top and bottom and so what does that do when you look at the orbitals of this molecule here when we take the green marker here let's put it right here what does that Sigma Bond look like well it's going to be a what type of orbital is it going to be remember the hybridization state of this guy so this carbon right here has one electron group two electron groups so that's SP hybridized so we're going to have the other carbon over here and it's going to have the P orbitals sorry the SP Sigma Bond right there and then you'll have the two electrons right there so I'm not going to represent the electrons because it's just going to messy up the image here okay so this Bond right here is the sigma head up head to head overlap which is the green one and then what we have is the blue one is going to form a pi bond from P orbitals so I'll draw p orbital there the orbital there HP and these have phases huh and so we have overlap of electron density above and below but what's really really important to understand is that these Pi orbitals are perpendicular to this axis right here all right so if this molecule in green is in the plane the glass then what do we have then we have the P orbitals well let's take the alkane alkyne sorry right here and represent it by my hand all right so my hand represents the green atoms there the blue P orbitals are going to be above and below my hand okay so perpendicular but then what we have is the pink Pi electrons here or Pi orbitals and they're going to be perpendicular also but in and these are a little harder to draw okay because if my hand represents the green molecule the blue is above and below but the pink is going to be out and or into the glass and out of the glass all right and it's going to have and those are going to also have phases and there's going to be overlap right there and that's going to cause us to have a triple bond now because these two carbons are SP hybridized you see the bond angle between the two when we look at the central carbon here that Bond angle right there is what 180 degrees and at 180 degrees it is impossible nearly impossible no I don't want to say this there's very few exceptions to this rule you cannot incorporate an alkyne into a ring structure so if we wanted to have a five-membered ring but turn that guy into a alkyne that's not going to happen why because look at that Bond angle right there that is deviating drastically from this 180 degree and so that just does not happen okay some other features of an alkyne are the bond links here the bond length between the triple bond and the bond lengths between the r and the carbon and so what I'm going to do is uh erase the sport and draw three molecules that we're going to look at their bond lengths and make some comparisons so what we have here is ethane Ethylene and acetylene you all see that they have two carbons each right so what I want to First focus on is the bond lengths in ethane the bond length between the two carbons that's the largest or the longest and the distance between these two carbons right here is going to be the smallest now why is that the case well one reason is it's hybridization state these carbons right here are what sp3 hybridized this one's st2 and that is SP so it's hybridized right so what are we doing we are taking our orbitals here let's see here 1s so there are that's our 2s and that's our 2p orbitals right likes that and when we hybridize it to a sp3 what we're doing is we're going to take one s orbital and three p orbitals and hybridize it to the same energy so that's our sp3 and so then that's what the electron configuration would look like for carbon here it's sp3 hybridized we have those four Atomic orbitals now what's interesting about the sp3 is it is only 20 it has only 25 s character why is that well we have one s orbital and three p orbitals and we mix them together so when we come over here the shape is technically 25 percent s remember in s orbital is spherical and then a p orbital kind of looks like dumbbells so when we hybridize these two to get these guys it's going to look 25 percent like an S and 75 percent like a p not a full p and not a a full s but a hybrid of the two now if we could contrast that to the alkyne over here what's that what's that hybridization State going to look like well we said it's only an SP so what we're doing is we're only going to hybridize take 1s and one p orbital and hybridize it together so we're going to get two of those both of those are SP we took one here and one here and then left the other two P orbitals alone so that's a p and that's a p and then when we put our electrons in there now why am I only putting 4 right here is because carbon has four valence electrons so we are trying to understand here how carbon is bonding and it bonds uh like this in a SP hybridized format okay so when we look at an SP what do we have we have 50 s and 50 p so that means that this hybridized orbitals right here looks basically 50 50 of this and we have learned from General chemistry that if you have a complete s orbitals the electrons are going to be closer to the nucleus whereas in a p orbital the electrons are further away from the nucleus right and a p orbital is longer than a s and so when we look at the ethane sp3 hybridized it's going to have a longer Bond because it has more P character or other the way the textbook says it has only 25 percent s character in contrast the acetylene or the alkyne is a 50 s so that means those electrons right there are going to be or that bond is going to be shorter because it has more s character also why does this molecule have a shorter bond is because it has more s character so the electrons are closer to the nucleus and so it sucks the bond in because the electrons are closer to the nucleus it's going to shrink the length of that Bond there's more attraction for the electrons to the nucleus and then here our ethylene right here is just smack dab in the middle of that Spectrum there but what was to finish this out what would the electron configuration look like for us Ethylene well we already know the hybridization state right SP2 so that simply means we're taking 1s and two P's and we're going to hybridize those so those are SP2 and then we're going to just not touch one of the P's and so now we have the four valence electrons there and you can see now what how much s character do we have right here one out of three is 33 percent so we have 33 s character for SP2 hybridized carbons now this hybridization state all right not only affects the bond length between the two carbons here but the hybridization state is also going to affect the bond length of the hydrogens so I'm just going to draw one yeah let's draw them all right there okay so we have those three hydrogens and we can draw those out there and that one's already drawn out there for the alkyne we can also say that this carbon hydrogen bond length versus that one and that one follows the same Trend this one is the longest this one's the shortest why is the carbon hydrogen bonds shorter for a alkyne it all comes back to the shape of those orbitals more s character so it the electrons are closer to the nucleus so those bonds just shrink in and help are held tighter to each other the electrons in this hydrogen and in this carbon they're just held together more closely so that Bond length shrinks let's see another interesting feature about alkynes is the acidity of the terminal hydrogens here when we compare an alkane versus an owl kind and an alkene the alkanes are the least acidic this carbon hydrogen right here or that hydrogen right there it's the least acidic how acidic is it well it has a pka of around 50. not acidic at all and alkene has a pka of around 44. so not I mean that six orders of magnitude difference so what's that a million times so an alkene is a million times more acidic than that guy but look at an alkyne it has a pka of about 25. and let's see if we have another marker let's see if this one's a better one that one looks like there's more ink in it so we need to understand why that's the case why isn't alkyne more acidic so let's just go back to acid-base chemistry to re uh look at this again because what's going to happen if we have an acid and we have a base how does it work well the base is going to come and grab the proton and give us our conjugate base plus her conjugate acid all right that's what's happening so when we take a look at an alkyne what's happening we're going to have some kind of Base and this base has to be a strong base and I'm going to show you one in a moment that's going to come grab that proton and then give us our conjugate base that looks like this plus that our conjugate acid so that's what we want to do we want to compare why is this guy not acidic and this one is orders of magnitude more acidic than alkane which is kind of interesting well it comes down to the hybridization States again well hybridization state of this carbon sp3 SP2 and SP and we've learned that uh SP as well 50 s so when we take and form this species right here and this negatively charged species has a name and it's called an acetylite ion and you guys know I'm the world's worst speller so let's make sure I get this right settle the satellite ion is when you start with this and turn it into that okay now why is that well how do we compare acidities like what are we actually doing well we're looking at the conjugate base if the conjugate base is stable then the acid is acidic so we have to rationalize and understand why is it when we take a alkane like this have our base come and Abstract that proton and get something that looks like this negatively charged so that means that the lone pair we have a lone pair there and there well that that lone pair has to be in a orbital which orbital is that uh low pair going to be in well if we draw it we'll draw a what is it going to be an sp3 hybridized orbital and we can put the electrons there but with the satellite ion what's the hybridization State we started at asp so we have a SP hybridized orbital here and it has more s character so I'm going to kind of just draw up for drastic effects here okay to see now that that lone pair if we put them in that orbital are closer to the nucleus it has more s character and so this conjugate base right here is more stable than this conjugate base because the electrons are being stabilized because they can get closer to the positively charged nucleus there's a stabilizing effect and so what we're seeing here is another idea called effective electronegativity foreign it's mostly on the board so effective electronegativity is well let me tell you what it's not for a moment when we look at electronegativity we go to our periodic table and we see trends like okay an auction atom is further right than a carbon so it's going to be more electronegative right but when each of these atoms right here is a carbon we can't go to the periodic table and look at for Trends right because it's just carbon carbon carbon so there's that term effective electronegativity which is saying that a SP hybridized carbon is going to be more electronegative because it can stabilize that negative charge better electronegativity is wanting electrons right and it can suck those electrons closer so SP hybridized carbon it has more it's more electronegative than a sp3 hybridized carbon uh really cool stuff let's see and and this right here the whole concept now because this is so much so much more acidic than alkanes we can do reactions with alkynes because of the acidity difference here now you can have alkynes that don't have two hydrogens on either end like this here would be a terminal alkyne and you could have and form this anion right here and this has a name alkynide ion the one when you have two hydrogens it's the acetylide am I saying that right there acetylite ion versus a alkanide ion okay same mechanism whatnot what I want to talk about is the base that is used is really really important it has to be a strong base and some good bases would be sodium amide now why would that be such a strong base because look at it we have a metal here that is not a covalent bond so what we actually have is something that looks like this you see that nitrogen atom very basic has a negative charge that's a strong base we also have uh let's see sodium hydride another sodium right so what does that actually look like here it's ionic right so we have a hydride when we have just one hydrogen and it's negatively charged we call it a hydride and that is very very basic and it's a strong strong base and then you can even have some bases where it's a carbon based like this where you have a carboanion and that species is very very very basic but would hydroxide or sodium hydroxide be a strong enough base to do this reaction that's what I want to demonstrate to to you right now which ones because that's a strong base is it not by definition it is but is it strong enough to take off that proton all right so just so we don't get confused here the answer is this one is not strong enough but the other three that I talked about sodium amide sodium hydride and these carbon ion species those three are strong enough to take off that proton this one is not why well we have to look at equilibrium for that so we've done acid-base chemistry so we we're going to do this something that we've already done before so if we put our sodium amide in here negatively charged here so that's the base so that's going to turn that into ammonia right so what we have to do is look at the pka values of each acid what's the pka of this guy approximately around 25 look at the conjugate acid over here with the pka of an amine approximately 38. so at equilibrium which side is going to be favored well it's always pointed in the direction of what the weaker acid so this is weaker than that acid so equilibrium is going to favor to the products so that base is strong enough to deprotonate this alkyne but if you change that up a bit to let's say the hydroxide like so and then what would we get over here we would get water what's the pka of water about 16. now when we look where's the weaker acid this one so it shifts equilibrium to this side so it's equilibrium favors reactants so we are never going to get enough of our alkynite ion here to do the reactions because now when we have this species right here what is that well that's a negatively charged species right so that could act as a nucleophile or a base we can we've converted this alkyne into a a reactive species when we have deprotonated it so the moral of the story yes hydroxide is a strong base but not strong enough do you recall if we took a alkyl halide like so and let's make those R groups actually and oh so our another set of R groups right there remember we could take a strong base right here and do a E2 reaction where we're going to eliminate the bromine which is a good leaving group and so we would generate a alkene remember that reaction that we've gone over well we can do the same thing but when we have two halogens on the carbon so we could have something like this hydrogen another hydrogen R Group now we have two halogens like this now that guy is an alkyl halide alkyl halide di alkyl halide because we have two halites right makes sense we can do the exact same mechanism above as above now we can add a strong base we do an E2 mechanism and we are going to now generate our alkyne and just now we're going to do this reaction actually twice we're going to do one uh E1 mechanism or reaction and we will generate an alkene and then we'll do it again to form the alkyne and I'm going to show you the mechanism of how that works but when we have a di alkyl halide where the halides are attached to the same carbon we call that geminal so we could also say this is a geminal DI alkyl halide if we change it up a bit well let's let's work with the geminal first okay now let's look at the reaction of a geminal DI alkyl halide so when you have your di alkyl halide what you have to make sure that you have is on the adjacent carbon in which the halogens are attached so you see these halogens are attached to this carbon you have to go to the adjacent carbon and there has to be two hydrogens okay there has to be at least two now What's Happening Here is we add some sodium amide and that's going to act as a very strong base you need a very strong base so it's going to abstract one of these two protons then it's going to break that carbon hydrogen bond and then eliminate one of the halogens so that's going to generate a carbon like this okay and we'll have our R Group our br double bond C r h plus our bromide okay and then we're going to take a second equivalent so that's the first equivalent first equivalent then we will have a second equivalent of sodium amide in our reaction and so we're going to take that second one I'm going to come in and subtract that proton break the carbon hydrogen bond go like that and then kick off the bromine and now what have we generated we have generated our molecule like that all right and so there's our alkyne just like that pretty slick all right now what happens now what we can do is let's see if we can keep some arrows here let's replace now we can have a DI alkyl halide like this where you can see we have two carbons with a halogen all right that's interesting and so what's going to happen here is this guy is also a dialgio halide but it's called a vicinal it's a vicinal DI alkyl halide and that's when the halides are on separate carbons now the halites have to be on two carbons that are adjacent to one another okay that's really important you can't just have a a long molecule like this okay a bromine over there and a bromine over here and call those vicinal di alkyl halides no vicinal means that they have to be right next to each other so that is vicinal that is not and if I change this one get rid of that for a second and if I put a bromine there then that would be geminal okay so here's our vicinal so how would that mechanism work well we'd have our first equivalent of sodium am I doing the same thing perfect but what's different on the product side here all right we said this bromine is right here all right so now I have a hydrogen there so when we tap our second equivalent of sodium amide coming in to react what's it going to do it's going to come in and grab that proton go like that like that to give us what the exact same product so it doesn't matter if you had a vicinal diacyl halide or a geminal when you treat it with a very very strong base like sodium amide then you're going to get to a alkyne same product is that pretty cool but when we do this reaction two equivalents of sodium amide is not enough not enough at all right it has to have three equivalents three three equivalents okay that is well let me let me make a distinction here um how do I want to say this here okay let's let's write this out here so if you have a DI alkyl halide existinal or geminal all right let's draw our geminal one okay like that and we add two equivalents of our sodium amide what are we going to get we're going to generate this product right and that is what a internal alkyne huh but if you are making a terminal alkyne so if our starting materials would look like this all right what you're going to have to do is you have to add three equivalents of sodium amide because what's going to happen you're going to generate a terminal alkyne right so when the end product is a terminal alkyne you have to add three equivalents of sodium amide all right but when you're just making an internal you only need to add 2. we're going to see Y in a moment before we get into why we need three equivalents I want to make sure we understand the reagents here the standard protocol here is sodium amide in ammonia so we've got to add that little other piece to the reaction scheme here so we have sodium amide in ammonia to get this reaction to proceed now why do you need three equivalents when you have a terminal alkyne well it's this is really cool because what happens after you generate after you do the first elimination all right so let's look at this mechanistically here I probably didn't need to erase all that Oops I did so if we I'm just going to represent it with a B for strong bass right now that's going to represent the sodium amide okay so what's what's the first step in the mechanism here well the base is going to come in and grab a proton right and give us lots of horrible we're missing one of the mechanisms huh all right here so there we go that's a better mechanism and then we're going to have R carbon br like that right and a double bond and then we can do the second reaction right so that's our first equivalent then we can do it again with another base do another E2 so that's going to come in and grab that and give us our alkynide species right there all right no no no no it's going to give us that guy right there okay so now we've already used two equivalents right now if we could just visualize this reaction in a beaker and in the beaker we tell the molecules only one at a time okay only one at a time only one molecule reacts with this and then this molecule reacts with this to get that okay if that's the case look at what's going to happen here but we know it doesn't happen right if we form one out terminal alkyne right here okay and after we've used these bases they're still going to be a lot of Base left over right that's acidic and so once you form this all the leftover the two equivalents of Base that are meant to react with all the other Di alkyl halides are going to be used up because of this guy so let's just look at it in rounds so this reaction is around one all right so now let's do round two so we're going to have a second diacyl halide okay so this is round two we still have our base present all right solve our base present but now in round two what does round two have that round one did not this base is now conflicted if you come and grab this guy and start the process over again or this base is like hey I see a very acidic proton over here and so you can't stop that and so what's going to happen is you've all if you've only added two equivalents of Base you have two equivalents or you have all of your di alkyl halide that you want the two equivalents to react with so you've added okay if you've added only two equivalents of Base those two equivalents had to go all to the DI alkyl halide but now this reaction isn't going to go to completion because if you only put in two equivalents of Base you did not account for that you have this as competition and you're going to run out of Base very very fast so what you have to do it is to add three equivalents of Base and what's three equivalents of Base going to do it's going to do deprotonate all of your product right here okay basically when you look at this reaction you have to have an equivalent of base for this proton that proton and that proton so you need three equivalents of sodium amide or the strong base that you're choosing to use for your reaction when you have a terminal alkyne now if that is no longer a terminal alkyne you only need two equivalents because there's no acidic hydrogen that I hope that makes sense okay let's see here what do we want to talk about next so after you add your three equivalents here okay that's R that's b r okay so how would we do this reaction we would write it out like this we would do three equivalents of our sodium amide in ammonia and what is that going to actually give us it's going to give us this species right here the acetylene no no no no no what is this could be this guy right there there's our alkynide ion that's what this reaction is going to do so what we have to do is what's called an aqueous workout where you can just add some water hence aqueous workup and what we're doing is we're working up the reaction we're going to take this piece add it to some water to do a proton transfer step and that proton transfer step is going to put the hydrogen back onto the alkyne so that's a simple proton transfer right so that would just be that negative charge we have some water auction and we could have a proton transfer that would give us our product markers are extra squeaky today right so that's all and we've already determined that the hydroxide right cannot deprotonate that so that's good so really when you want to do this reaction you have this step and then you have a aqueous workout and sometimes you could see the reaction written like numerated like we have reaction One which is all that and then you would have reactions to shown like that and then it would just go right to the product all right things to be aware of the next reaction that we want to talk about is we can also take alkynes and alkenes and reduce them to an altane so if we have a an alkene we could treat that with some hydrogen gas and some kind of metal catalyst so it could use nickel or we could use platinum or we could use Palladium you don't have to use all three at the same time one you just pick one it does not matter hydrogen gas and what are you going to generate you're going to generate an alkane so you can see that there's one hydrogen there one hydrogen there and if we label those carbons one and two what has happened we have added a second hydrogen to both of those carbons and this process of reducing this double bond because reduction also means when you are adding hydrogens that's a reduction but this reaction is also called the hydrogenation hydrogenation reaction have you ever heard of hydrogenated oils well that's what's happening is we're taking oils that have alkenes in them and we are adding hydrogens to them like that but we can also take a uh alkyne as well we can take an internal alkyne or a terminal alkyne it does not matter if you treat it with the same reagents so let's actually do a better example here foreign with let's see one two there we go one two three four four carbons and if we treat this terminal or internal alkyne with the same reagent we're going to get the same exact product but now what has happened if we call that carbon one and that carbon two here they have zero hydrogens but now we've added two of them like that so we can hydrogenate an Alka alkene or a alkyne but one cool thing about alkynes is that if we use this condition right here and pick one of these Metals will go all the way to the alkane but if we take one of these metals and we poison it by adding poison okay we're going to get what's called it what's called a poisoned catalyst and a poison catalyst is going to take an alkyne all right like this and not go all the way to a alkane but it will stop at the alkene and alkene that it stops that is very very important it's going to stop at the CIS it does not make any of the trans alkene it's predominantly CIS so what type of poison Catalyst could we do well there's one that's a very famous one called The lindler's Catalyst I want to spell the name right here if we do lidlers lindler's catalyst so if you take an alkyne treated with hydrogen gas lindler's Catalyst you will generate a CIS alkene and lindler's catalyst is simply you're going to use Palladium all right so you take Palladiums and you mix into that Palladium the poison which is going to be calcium carbonate right there and Trace Amounts of lead oxide right so what that when you take if it's just the pure Palladium and hydrogen it's a very active and just of an amazing Catalyst it's just going to take the alkyne or alkene all the way to the alkane awesome but if you can make it an okay Catalyst but not as good we can slow it down and so it's going to stop at the alkene and so lindler's catalyst is simply a Palladium uh Catalyst that's been poisoned with calcium carbonate and Lead oxide okay that's very very very important to remember the stereochemistry because what we can also now do is we could take the same alkane kind right here and treat it with a different metal so we could treat this guy right here we could treat that with sodium metal solid sodium metal okay in liquid ammonia it's and in order to do liquid ammonia you have to get this pretty cold so it's going to be a cold reaction here and look at what's this is so cool this is so fun it is going to stop at the alkene as well but where does it stop or how what's the stereochemistry it is going to give us trans so now we have the ability to select for what type of alkene that we want in a particular product do we want an alkene that's CIS if we do then we use lindler's catalyst if we want our alkene to have a trans Bond here then we are going to use sodium metal dissolved in liquid ammonia and that reaction is called a metal dissolving metal dissolving reduction okay I think that's that is just fun and so interesting to me that you can control the stereochemical outcome just based off of the reagents that you're using now why is this the case well the mechanisms can help us figure that out okay but that's where I'm going to end this video for today