Organic Synthesis & Retrosynthesis Explained | OChem

[00:00:01]an introduction to organic synthesis that's going to be the topic in this first lesson on a whole chapter devoted to the subject of organic synthesis now this is something you probably didn't visit in first semester organic chemistry but it is going to be a major point of focus for most of you in second semester in fact many of you are going to consider this the hardest thing you're tasked with on most of the exams you see in second semester so organic synthesis there's no way around it it really is a pain in the button you can memorize all of the reactions and be able to predict the products for anything i throw at you and you still may struggle with synthesis and it's not just about knowing all the reactions but it's about how you organize them in your head that's going to be a big point of focus for this entire chapter is not just knowing the reactions i'm going to take that as a given that you guess you've got to know your reactions but we're going to work on really how you structure them in your head in this first lesson we're going to talk about some functional group conversions how you convert one functional group into another functional group so and then we'll move on and talk about like how you make your carbon chain bigger or how you make your carbon chain smaller or how you open up a ring because you're really limited in the number of ways you know how to do that both now and in the future as we learn more reactions we'll learn a couple more ways of doing those things but not too many but uh this is going to get more complicated as the semester goes on and so this is kind of like organic synthesis up till now and every chapter where we cover a new functional group and learn a whole set of reactions for that functional group will revisit organic synthesis in one little lesson and kind of have organic senses up till now and we'll just keep updating it throughout the second semester so what is an organic synthesis problem well we're gonna give you a starting material starting material and we're gonna give you a final product and just say how do you get from this starting material to this final product show me all the reagents and all the intermediates all along the way and generally they're going to have somewhere in the ballpark of like two to five steps is pretty typical uh and every once in a blue moon you'll see one a little bit longer than that for some harder classes and stuff like this and some professors that are more demanding but pretty typical is anywhere from two to five steps uh and again you've got to know your agents really well but you've got to have them organized in your head and this first part is going to be just knowing how you convert one functional group into another we're going to cover five that involve reactions you've already learned up until now and the first is what if you start with an alkane well you've learned very little about alkanes and you've learned that pretty much the only thing you know how to do with an alkane is free radical halogenation and turn into an alkyl halide and you'll find out that much more commonly we're going to use br2 and do free radical bromination over free radical chlorination because if you do chlorination you've got two different places where you could you know chlorinate this thing you get one chloropropane and two chloropropane you get a mixture of both but with bromination being much more selective you'll find out we use it much more commonly for synthesis purposes here because you get one major product we're gonna put a bromine on that more substituted carbon so substitution reaction placing a hydrogen with a bromine like we learned in the last chapter here so if you start with alkane this is nice because this is the only thing you know how to do with an alkane so i mean technically you can do combustion or something like that but it's not going to be helpful from a synthesis perspective so if you start with an alkane first step is going to be free radical halogenation all right so now we want to move on to a couple others and if you start with an alkyl halide now like as in like you just made one potentially but if you've got an alkyl halide you can possibly do elimination so and e1 versus e2 you're probably going to be doing e2 almost exclusively so e1 and sn1 we try to avoid those for synthesis because they almost always compete with each other and there are a couple cases where maybe it could be used and stuff like that but most of the time we're just going to rely on e2 for elimination reactions and sn2 for substitution reactions in our synthesis so in this case we're going to do elimination with this guy this is our tertiary halide here and if we use a a standard base here so not a bulky one but just a standard one like sodium hydroxide or in our case it's going to be sodium methoxide or sodium ethoxide would have worked as well so nice classic strong base here so then your z zev product is going to be the major and so in this case here's our alpha carbon so we've got beta carbons here here and here and your most substituted ones are these guys one with a fuse hydrants and that's what mr zaitsev predicts where the major product should be so there's our major product so however if we use a bulky base so and oftentimes we'll use the potassium salt of our bulky base and this is one way to write it another way you could write this is you could put ch3 3 c o minus and again you'll more commonly see the potassium salt than the sodium salt but either way so a couple different ways you could see it written here so with that bulky base now you're going to go anti-zaitsev or hoffman so and you're going to form that alkene using deprotonating a hydrogen from the least substituted beta carbon and so i'll form between alpha and beta here instead so total review of e2 elimination reactions from a few chapters ago cool now what if you have a geminal or vicinal dihalide if you've got two leaving groups here we do in this case geminal means they're on your two leaving groups your two halogens are on the same carbon vicinal means they're on two adjacent carbons well now with the appropriate strong base you can actually do e2 elimination twice and in that case you'd form an alkyne and you've got to remember that for forming alkynes since there's all kinds you'll learn in the alkyne chapter that our new favorite strong base is nanh2 and we're going to use excess of it so it turns out we use excess because we got to do elimination more than once we get it twice and then once we form a terminal outcome we find out that nanh2 promotes the formation of terminal alkynes so that it deprotonates it so actually use three equivalents of any energy along the way and so because it deprotonates that terminal alkyne you have to reprotonate it typically with weak base like water all right either way it doesn't matter if you start with a geminal dihalide or the vicinal dihalide it goes through two rounds of elimination so instead of just forming one pi bond and getting an alkene you form two pi bonds and get that alkyne cool so these are both eliminations just you have one leaving group you can do it once you can form an alkene if you got two leaving groups well then you could do elimination twice and you can get the alkyne so now we've got two more functional group conversions we've got an alkyl halide to an alcohol and truth be told we haven't actually learned a chapter on alcohols but we're about to the next chapter we'll cover we'll cover a whole new set of reactions for alcohols and so making an alcohol will be useful because after the next chapter you'll be able to use them so this is kind of giving you a little foreshadowing of some reactions that will be helpful very soon for organic synthesis so if we start with an alkyl halide and generally either a methyl or primary halide we'll learn here so then we have the option of doing sn2 so if it were a secondary halide e2 becomes more likely in most cases than sn2 and so it's not great but for a methyl or primary got a good shot at it here and if i want to make an alcohol well then i want to replace the leaving group here with an oh and so we should use a nice strong nucleophile like a metal hydroxide like naoh so in fact i meant to write that in red so let's go back and do that for the fun of it keep all my reagents in red here be consistent so in this case we'll just do backside attack the hydroxide comes out attacks kicks off the bromine classic sn2 makes an alcohol cool so that's our first way to make an alcohol and so and probably not the most common one truth be told because like i said it's not great for secondary alcohols and it doesn't work at all for tertiaries because they'll just do e2 elimination instead so but other way we can do this we can turn an alkene into an alcohol as well and you had three different hydration reactions you had acid catalyzed hydration so and that's what we do here with h2so4 and water so it adds an h and an o h markovnikov it does go through a carbocation with subject rearrangements in this case though it's not going to happen we'd have a tertiary carbocation anyway it's not going to rearrange and so our result our product in this case is this guy we had an h on that less substitute side and the o h on the more substitute side cool now turns out we have an alternative way to accomplish this that was the oxymercuration demercuration so and in this case the only difference is that this one doesn't have the option to rearrange because it doesn't go through a carbocation intermediate so in this example there is no difference they're going to lead to the same major product so but there are certain situations certain alkenes you might start with where a rearrangement might be feasible for h2so4 but wouldn't happen with the oxymercuration demercuration now your other option is to go anti-markovnikov with hydroboration oxidation cool and this would be your product here here we added the h on the more substitute side we added the o h on the left substitute side that's what makes it anti-markovnikov so and suffice it to say these are your major functional group conversions now we got five of them so we turned an alkane into an alkyl halide we now turned alkyl halides whether it be a single halide or a dihalide into either alkenes or alkynes and then we learned how to turn an alkyl halide into an alcohol or an alkene into an alcohol as well and these are kind of the most common ones that are going to be helpful to us and if you realize these functional group interconversions so this is the first step in kind of organizing the reactions and as you learn more and more reactions throughout this semester you're going to need to organize more and more and have these kind of functional group inner conversions kind of organized in your head by functional group so it is a little bit of a daunting challenge but it is the kind of the way you want to organize these reactions in your head so now we want to take a look at how we can actually change the length of your carbon chain and we'll start by increasing the length of the carbon chain and it turns out there's not a whole lot of ways we know how to do this so if you're increasing the length of your carbon chain that means you're forming a carbon-carbon bond and we really just don't have a lot of ways to do this we might have a half dozen in our arsenal by the time you finish off this entire semester so people used to get you know nobel prizes for coming up with creative ways to make carbon carbon bonds so uh but we don't have a lot of ways to do this and so if you're looking at your starting material in an organic census problem and your final product and if the carbon chain gets bigger well then you know you've got one of a few different ways to accomplish this and we'll find out that we really have one major way of doing this now one more important way anyways but we'll learn another one in the next chapter and like i said a handful others throughout the course of the second semester but really not that many ways so cool first one is actually going to be a very minor way and that's simply by an sn2 reaction with sodium cyanide so or potassium cyanide for that matter just cyanide in general if we do backside attack with a strong nucleophile cyanide here it will add one additional carbon here so in this case we started off with a four carbon chain and now we've got a one two three four five carbon chain so we're pretty limited on this we can add one additional carbon done that's all you can do so from an alkyl halide to a nitrile in this case so it is a functional inner conversion i guess as well you could look at it that way but more important is just increasing the carbon chain by one carbon we'll learn later on in the semester that these nitriles here we can do various things we can convert them into carboxylic acids and things of this sort but for now we're kind of stopped this is as far as we can take this from here but again it only makes us one additional carbon and so it's not really that useful for kind of a variety of different carbon lengths and stuff like this so if you want to actually increase the length of your carbon chain in a variety of ways then you want to start with a terminal alkyne and here i'm going to start with acetylene but we could have a carbon chain on one of the sides but in this case we've got an sp carbon on both sides both of which are bonded with hydrogen and those can be deprotonated with nanh2 we learned so some of you might also learn that you can deprotonate a terminal alkyne with a grignard reagent but most of you probably don't even know what a grignard reagent is yet we'll learn about them in the next chapter so but for those of you that did i just want to point that out real quick but for most of us it's really going to be sodium amide here nanh2 used to deprotonate this terminal alkyne you can deprotonate it on either side but you can't deprotonate both sides at the same time once you've deprotonated one side it becomes monumentally more difficult to deprotonate the other side so but we form this lovely acetylide ion and he's a strong nucleophile and being a strong nucleophile we could now react him with an electrophile of you know our choosing to make the carbon chain longer and so in this case you've really got three different options you know you learned about one of them earlier but the other two you may or may not have learned about in the alkyne chapters we're going to explicitly cover them here as well so the one you probably did learn about is just reacting this with like an alkyl halide so let's just say you want something that's either methyl or primary since we're about to do sn2 so in this case we would just do backside attack kick off the bromine and attach one two three more carbons so here's our again from our satellite and we want to attach three more carbons so one two three there's three carbons and now we want to attach them so when you're forming a carbon-carbon bond it's often easy to make carbons disappear so i highly recommend you count them so we got two from this guy one two and then one two three from this guy one two three and again count your carbons don't count your bonds cool but there's our lovely result and we now turned a two carbon chain that we started with now into five carbons and the truth is depending on the alkyl halide as long as he was methyl or primary so while methyl i guess would only have one carbon but if he's a primary halide we can make it as long as we want to and essentially we can make this as long as we want we're not restricted to just like adding one carbon and one carbon only we've got lots of options here now it turns out an alkyl halide and sn2 is not your only option you can also react this guy with like a ketone or an aldehyde as well so if we take a look at reacting this guy with an aldehyde instead we'll find out we've got to protonate it after back with some dilute h3o plus and for whatever reason we often don't write the word dilute and sometimes we just use water but we really need just a very slightly acidic solution aqueous solution to accomplish what we need to here so but what's going to happen here is we're going to do attack on the carbonyl here's our electrophile and so we don't have a leaving group per se but this carbon is partially positive to a significant degree being double bonded to an oxygen so but he's already got a filled octet so if we're gonna attach a new bond to him then we have to lose one so and that's the pi electrons push up to the oxygen become a lone pair on that oxygen and so in this case you're going to end up with a single bond to your oxygen and a negative charge and you'll now to this carbon have attached these three carbons so i'm just going to draw that extra bond out there so and that's what we get so and this thing is called an alkoxide when you have that negative charge in oxygen and that's why we add the dilute h3o plus we're just going to protonate so that lovely species turning it into an alcohol now the truth is i've drawn this h in here only to be consistent but for a hydrogen bonded to a carbon the only time we generally draw those in is when they're the hydrogen of an aldehyde but it's not an aldehyde anymore so generally you're probably not going to be drawing that h in i'm just trying to be consistent so we're not like what happened to that h chat all right so if you look here we made this bond right here that's the new bond we created right here so and if you examine it so this side over here was the nucleophile this side over here is the electrophile and you know it's the sp carbon of an alkyne that can be your nucleophile when it's an acetylide ion and so when you look at the the carbon carbon bonds you're making if the carbon you're bonding to is bonded to an oxygen of an o-h well then you know that you attacked a ketone or an aldehyde so whereas if the carbon you've bonded to so in this case it's not bonded to an oxygen there's no oxygen nearby well then you know you probably just have to now kill halide well we've got one other option here a third option so and that's going to be to attack an epoxide an epoxide is a three-membered ring here with oxygen i'm going to put one more carbon on this just for the fun of it and we'll finish this off with a little dilute h3o plus and once again you can write water here or just h3o plus you don't have to actually write dilute in some cases but the truth is we want an aqueous solution that is just slightly acidic to pull this off all right so when you've got an epoxide you're two carbons in your three-membered ring that's what an epoxide is so those two carbons are partially positive because they're both bonded to oxygen and this ring strain makes them particularly good places to attack because it's going to open up this ring we'll learn in the next chapter this is called the ring opening of an epoxide the whole class reactions with them and they can react with a variety of nucleophiles but the one of interest here is our satellite again and so it turns out when you look at the two carbons of the epoxide that are partially positive in monoxine you want to attack the less substituted one with your strong nucleophile here so we're going to come and attack over here and again this carbon has two hydrogens that are not drawn in and we already had a filled octet so if we're going to make a new bond there one of them has to break and that's the bond to the oxygen cool so that opens up that ring and so that's actually the driving force in your ring opening of an epoxide is that you're relieving the ring strain by opening up that ring so let's see what this looks like here so it turns out we are doing backside attack here and that's why we attacked the left substituted side and if we kind of draw this out here this oxygens now up here are a satellite ion attached to this carbon which this carbon is this carbon and so the acetylide is now attached to it cool and we still have this guy over here so notice this carbon doesn't get inverted it's when we attack that gets inverted but it's not a chiral center so we don't really notice the inversion so and then this auction by the h3o plus is also going to get protonated just like we did here so in fact maybe i should draw the steps to be consistent cool and so in this case we can see that final result this is the carbon-carbon bond we made this side was the nucleophile this side was the electrophile so and key thing you should recognize here is that the carbon we ended up bonded to on the electrophile is not bonded to an oxygen so we didn't attack a ketone or aldehyde like we did up here that carbon was bonded to an oxygen so but in this case there is an oxygen in the area and so it's not to the carbon we bonded to on the electrophilic side but to the next carbon over and that's what happens when you attack an epoxide because the carbon you attach to that you attack loses this bondo oxygen but the one right next to it will still have that bond oxygen and so when you're looking at a synthesis problem involved in increasing the length of your carbon chain so oftentimes what you want to look at is you you're going to look and say okay i'll probably do this during the satellite at least up through now next chapter will learn one new method as well but for now it's through an acetylide ion so here and if what you attach to there's no auctions in the neighborhood you attacked an alkyl halide if the carbon you attached to has an oxygen bonded to it you attacked either a ketone or aldehyde and if the carbon you attach to doesn't have an oxygen but the one next to it does well then you attacked an epoxide and that's one of the patterns you want to recognize when you're increasing the length of your carbon chain okay so now we want to look at decreasing the length of a carbon chain and this can happen to a couple of contexts so either with an alkyne or an alkene turns out we have a way of shortening the chain and you guys learned ozonolysis and so with a an alkyne here o analysis was carried out with ozone followed by water but with your typical alkene we got ozone and under oxidizing conditions with an alkene we use hydrogen peroxide cool now with a terminal alkyne you learned that oxidative cleavage here with ozonolysis can be that carbon-carbon bond so and you get carboxylic acid on both sides if it's internal so although if it's terminal though then the terminal side becomes just carbon dioxide instead so here we're going to have a one two three carbon carboxylic acid plus we're going to get carbon dioxide from this side now had this been a longer chain and an internal alkyne well then you would have just gotten two carboxylic acids and if it was symmetrical two two of the same carboxylic acid so but if it was asymmetrical two different carboxylic acids and with two different carboxylic acids it's not so common to kind of go that route because then you got to purify and stuff so but real common to shorten your chain by just one carbon when you have a terminal alkyne through this process same thing can work with a terminal alkene as well and once again with ozonolysis under oxidizing conditions you are going to cleave that carbon-carbon bond so and on this side we're going to get a double onto oxygen and initially this is going to form an aldehyde because we do have a hydrogen right here and that hydrogen is still there but under oxidized conditions aldehydes get oxidized to carboxylic acids and so we're doing this under oxidized conditions so this is once again going to become a carboxylic acid and just like in the case of a terminal alkyne with those analysis with the terminal keen and ozonolysis under oxidizing conditions so your single carbon on the other end also becomes carbon dioxide and so in both cases we've gone from one two three four one two three four carbon chains and turned them into three carbon chains carboxylic acids but three carbon chains we've decreased the length of the carbon chain by one in both cases now not only can we use ozonolysis here again an example of oxidative cleavage to decrease the length of our carbon chain we can also use it though to open a ring all right so we'll take a look at opening a ring here and in this case with an alkene is what we'll look at you typically not going to have an alkyne in a ring because those angles need to be 180 on either side of the sp carbon so not common so technically i guess it could happen with a very large ring but you're not going to be able to get a triple bond in a six membered ring so with an alkene here much more common to open up a ring and we're going to do ozonolysis again but in this case you might actually see it one of two ways either under reducing conditions by using like dimethyl sulfide here dms for short or zinc and water works for the second step as well both of which are reducing agents and those allow any aldehydes that are formed to stay aldehydes but once again we can also do this under oxidizing conditions which causes any aldehydes that initially form to get oxidized to carboxylic acids so but in either case you're going to do oxidative cleavage you're going to cleave that carbon-carbon double bond and so in this case perfect you know personally i just like to predict my ozonolysis products by redrawing this i'm going to draw it really huge keep the carbon chain the same but where you used to have a carbon-carbon double bond erase it and in its place put two carbon oxygen double bonds instead so and notice we could draw this out as a big long chain because it's no longer actually a ring these are not connected anymore and so in this case the top one here is a ketone and if you form a ketone ozonolysis it's going to remain a ketone but this one is actually an aldehyde there's a hydrogen right there and if you do this under reducing conditions like with dimethyl sulfides sulfur ah dimethyl sulfide it's going to stay in aldehyde and so here we got one aldehyde on one end ketone on the other so and this thing is no longer a rain we could you know draw it out as a one two three four five six seven you know length chain so one two three four five six seven and it's an aldehyde at one end and a ketone one in from the other end so you could draw it like this as well now you could also do this ozonolysis under oxidizing conditions with hydrogen peroxide here it's going to work the same way and once again i would draw this really big leave my carbon chain alone but where i had a carbon carbon double bond once again draw two carbon oxygen double bonds instead it's no longer a ring but instead of getting an aldehyde like we have right here under oxidized conditions that becomes a carboxylic acid so our ketone still becomes a ketone that's not going to change but any aldehydes that might initially form become carboxylic acids and once again it's no longer a ring and you might draw it out so we get an aldehyde at one end under reducing conditions but under oxidized conditions that's now a carboxylic acid but we still get this ketone in exactly the same location here cool so only ways you currently know how to open up a ring and probably the only ones you're probably going to know throughout this course now we'll learn how to form some rings later on in this course and like i said we're going to learn some more uh functional group reactions all throughout the semester and so the different functional group conversions and maybe how to form some carbon carbon bonds make longer chains we'll learn some additional ways for that but like decreasing the length of a chain or opening up a ring this is pretty much it i don't think we're going to learn any additional ways to accomplish this for the most part so i'll correct myself if i'm wrong somewhere along the way but i can't think of anything offhand all right in the rest of the lessons in this chapter we're going to cover what i'm going to call common patterns in synthesis so and we'll just go through some common patterns and then we'll actually work some actual examples of organic synthesis problems so you can kind of get an idea of how they're worked so but for here we've laid a pretty good foundation we've already talked about functional group conversions for the ones that you know up until now we've talked about how you make a carbon chain longer or shorter or how you open up a ring and these are the tools you really need at a foundational level before we move on and start incorporating more functional groups in the future throughout this semester now if you have found this lesson helpful would you consider giving me a like and a share a couple of the most helpful things you can do to support the channel and if you're looking for practice problems if 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