Concept: Practice: Intro

Video Transcript

Alright guys these next few questions are cumulative retro synthesis based on the past few chapters of reactions so you do need to still know those and I have made them so that they do represent common types of synthesis, OK? But I've also made them all little bit harder than what I've seen on most tests, OK? That's a good thing what that means is that I'm giving you some hard stuff so that by the time you get to your exam it doesn't feel that bad, OK? You're like "Oh Wow I did harder things with clutch, I did harder things with Johnny online", OK? For all for all these retro syntheses this is that I give you the synthetic cheat sheet is very hopeful so you should be looking at it or referencing it when you're stuck think OK What....Are there any clues here? Are there any ways that I can get unstuck that I can figure out where we're going, OK? So for all of these I would really encourage you to try to do it in your own as best as you can they're all doable none of them are impossible but they are tricky they are hard but they're doable and then of course I'll answer it as always so best of luck guys let's go ahead and get started.

Concept: Practice 1: Supply the necessary reagents.

Video Transcript

So the very first thing you should be looking at here is that I'm starting off with four carbons and I'm ending up with five, OK? That means that I must be using an organometallic here particularly I'm probably using an Alkynide, OK? So that means that I'm going to convert this alkane to an Alkynide react it with how many carbons? 1, add it and then figure out how to transform it into an Epoxide OK? So like I said lots of stuff going on here but it's all stuff you can manage so the first step should be BR over heat what that's going to give me is a secondary alkyl halide that's important because I need it to reliably form there my second step is that I need to form a double bond because I can't make a triple bond out of that yet I need a double bond first so in this case I'll use and I want it to be terminal double bond I want it to face in this direction in the Hoffman direction not in the Zaitsev direction, Why? because if I form it in the middle I'll never be able to form an Alkynide out of it I'll be stuck so I should form it on the edge so I should use LDA, OK? So what that's going to give me is a double bond that looks like that so now for my third step I'm going to do a halogenation so I'm going to use let's just use CL2, OK? So CL2 what that's going to do is that's going to give me a CL-CL, OK? So now I have those vicinal Di-halides, why is that cool? Because now I can do a double elimination, now I can do NaNH2 excess, OK? And what that will give me is that's going to give me in three equivalents that's going to give me a four carbon Alkynide that's right over my head, OK? So now I've got a four carbon Alkynide now what we're wondering is what do we react that with to get it to be a five carbon chain? Well once again a one carbon Alkyl Halide so I'm going to take for five I'm going to say CH3 CL, OK? And I'll react that here and what I'm going to wind up getting is backside attack so I get that, OK? So now at least we have the right number of carbons but now we still have a problem what do we have at the end? We have an epoxide, OK? On top of that this Epoxide has interesting stereo chemistry it's actually CIS, it's a CIS Epoxide because I've got both of these groups facing the same side, OK? So how can we do this? Well are there any ways do you guys remember? Are there any ways to make an Epoxide out of a triple bond directly, do you guys remember that? No there are no ways, OK? But there is something that you can make an epoxide out of and that's a double bond, OK? Remember that we can use Peroxiacids on double bonds to make epoxides, OK? can I make a double bond out of a triple bond? Yes, I can get so this is giving us a big hint we want to turn this into a double bond, well what kind of a double bond? We want to turn it into a Cis or a trans double bond because that's also important and it turns out we want to turn this into a Cis double bond so that I'm going to get a Cis epoxide so from my sixth step I would get H2 over Lynn Lars and what that's going to give me is something that looks like this so now I have five carbon chain with a double bond that is now Cis, OK? So now I'm actually just one step away from getting my final product all I need to do is doing an epoxidation and that will give me in an epoxide that's Cis and the has a five carbon chain and just so you guys know the common reagents for Epoxidations the general formula was R-CO3H, OK? So even if you forget all the reagents for epoxidation just think it's like a Carboxylic acid with one more O, OK? Carboxylic acid with once more O is an epoxiacid, OK? But in case you guys might have put a specific reagent down just so you guys know the really common proxiacids were MMPP and MCPBA maybe these are ringing a bell for you maybe not, OK? But these are the two reagents so whatever I'm just going to go ahead and just use my general formula RCO3H and what that's going to give me is it's going to give me an epoxide that has both of the groups facing the same direction, why? Because they started off in the same direction and that's it that's my final product, OK? So obviously that was tough like I don't think we're going to get anything that hard in your test because I gave you the worst starting product and I gave you one of the worst ending products, I could have made it worse I could have made you like do a base catalyzed ring opening on this thing but I'm not going to torture you, OK? I'm just trying to show you guys how you're very responsible to make Alkynides react them to have the right number of carbons and then transform them I think that I went a little bit overboard with the making, you're probably going to start from an Alkane and the transforming you're probably not going to make a Cis Epoxide but who knows maybe you will and that's one from show you that these things do happen and so let's go ahead let me know if you have questions but if not let's go ahead and do another.

Problem: Propose a synthesis


Concept: Practice 3: Supply the necessary reagents.

Video Transcript

Alright so for this one we know what the first step has to be and we actually kind of what the last step has to be it's just the middle steps that we're confused on so let's actually draw both of those so first of all we know what the first step has to be a radical halogenation right there so I'm just going to put it heretical bromination same for BR to overheat, OK? What we also know is that well how do you add 2 Chlorines next to each other? 2 halogens right next to a vicinal, do you guys know of a way to do that? Well the way to do that is to add them to a double bond, OK? So what that means is that before the step immediately before the step I must have had a double bond that looked like this, OK? And in that and I added Cl2, OK? So really this is actually getting a lot easier, the question is just not how we get those two chlorines because we know we can do that the question is how do you get that bromine over there to turn into a double bond here? After everything we've been through this should be a breeze for you guys the only thing you need to figure out is are we going Markovnikov or anti-markovnikov? more substituted less substituted, right? So first of all we're moving a functional group so that means that we're going to use the whole additional elimination thing, right? So in the first step I have an Alkyl Halide should I use an addition or should I use an elimination, which one? Elimination, right? So I have a... IÕm trying to go in this direction, OK? Is that the more substituted or the less substituted direction? Well let's look at our options but sometimes it's hard to answer that question without looking at your options, I could either dehydration in this direction, OK? Or not dehydrate I could eliminate in that direction or I could eliminate in one of these directions, OK? Which of them is going to give me or get me closer to the double bond by blue or green? Green which was more substituted after it forms a double bond? Green, OK? because I'm going to get a double bond right inside the ring instead of outside of the ring so what that means is that in this first step I actually want to do is Zaitsev elimination since I want to do a Zaitsev elimination I can use any small strong base and I have a good leaving group so let's just go ahead and use, what do you want? You could use anything so let's just go ahead and use NaOET again, OK? Pretty common, alright? So what NaOET is going to give us for this second step is I'm going to wind up getting a double bond like that, OK? Now keep in mind Ooops, OK now keep in mind that...What was I going to say? So I kind of blanked out just a second, OK? Holding my train of thought whatever so we have this double bond here we have this double bond here we're getting really close, OK? But now we have to do another addition where we basically need to keep going in this direction so now let's go ahead and do another addition do you want to go in the markovnikov direction or anti-markovnikov direction what do you guys think? So I could either add something here to the blue or something here to the Red I'm going to choose the blue, OK? So I'm going to choose the blue direction and that means that I want to do an anti-markovnikov addition so let's go ahead and do HBR over peroxides, OK? What that's going to give me that's going to give me a BR right here, OK? So I have a BR there is there any way that I can turn that BR and eliminate it so that it becomes that double bond, interesting, right? OK so is there it is that one of the possible beta carbons that I could eliminate? Actually yes if you'll notice that if I do an elimination here I could either do this as a beta or if they eliminate this as a beta, OK? So I have a blue beta I've got a green beta which one will actually make that double bond that I need so desperately? The green, OK? Is the green in the more substituted position or the less substituted in terms the double bond that I'll get? Less, OK? So what that means is that I want to do a Hoffman elimination here so I should use a bulky base like LDA and there you have it we have retro synthesis, OK? So we're going to do you BR2 NaOET then three was HBR over peroxides then four was LDA than five was CL2 and there you have it, OK? So once again I'm usually taking this one step further usually about one step further than I think your professor would, OK? That's because if you're just prepared for the extra step and then on your test that extra step just isn't there, perfect that just means you breeze through it and you get on faster, OK? Because I know time is a big issue on these exams so anyway I hope that that made sense guys let me know if you have any questions I hope that you're starting to understand synthesis a little better so let's go ahead and wrap this question up.