Synthesize the following molecules beginning with only organic...

Question

Synthesize the following molecules beginning with only organic molecules containing three carbons or fewer.

(g)

Accepted Answer

Welcome back, everyone. Let's take a look at our next problem. It says, using only molecules containing three carbons or fewer synthesize the molecule below, we have a pretty intimidating molecule here. Sort of looks like an inverted pyramid. We've got a total of a 12 carbon chain and from carbon six to carbon seven, they're part of an oxide ring. So six and seven are the base of that triangle. Then we've got two solid wedges going to the oxygen of that ring. So we've got to put this whole thing together starting from reagents that have no more than three carbons. So this is pretty intimidating, but there's some basic things and pieces of knowledge we can start with, we can go from an alkene two at a pox side. So if we had a double bond right there in the middle of our molecule, we could convert it to an epoxy. And of course, as we know, it's pretty straightforward to go from an alpine to an alkene. And when we think of all kinds and their reactions, we hopefully will think of the Alkan reaction. And that's a way to put together smaller carbon chains and put them together ending up with a triple bond in the middle. What do we want a triple bond in the middle that can go to an oxide and we need to put together smaller chains to make a bigger chain. So focusing on the idea of this alkaline reaction, we can definitely get where we want to go. So let's imagine what it would take to get to a molecule where we had a 12 carbon chain with a triple bond from C six, two C seven. So I've drawn that molecule here. Well, in an alkaline reaction, our alkaline has a terminal cc triple bond where hydrogen has been removed, giving us. So I could draw it as R and then I'll draw a triple bond and then a negative charge at the end has a terminal negative charge or where the hydrogen has been removed. So we end up with this molecule that, that terminates in a triple bond. So let's imagina break a bond between the triple bond and the rest of the chain. What does that give us? Well, that gives us a piece with seven carbons ending in a triple bond. So it will be a seven carbon alky and a piece with five carbons but five carbon Alcaine. Well, what does that make us think of? Well, if this could be an alkaline, so we could react to seven carbon alkan plus, how would we combine it with a five carbon chain? While the alkaline acts as a strong nucleo file. Strong nucleo can be paired with primary alkyl halides and perform an sn two substitution. So we could react that with a five carbon alkyl Halide. Put those two together and generate the al kind we need there. Well, we're still at seven carbons and five carbons. So we need to build up to those. Notice that the smallest alkaline we could use to add something on is two carbons. So we could have this five carbon alkyl Halide tack on two more carbons in a triple bond and get our seven carbon alkaline. So we can actually use our five carbon alkyl Halide as an intermediate on the way to that seven carbon alkaline. How would we generate our five carbon alkyl Halide? Well, we could start with a three carbon alkyl hide again, think of how we can add on at least in twos using our alkaline reaction. So we could add on a two carbon alkaline, a negative charge that will give us a five carbon alkaline or excuse me, a five carbon alkyl, excuse me. So five carbon alka al Kine. Too many similar words here that will give us a five carbon alky. Can we go from an alky to a hide? Absolutely. We could go down to an alkene and from there to an alkyl hine. So we'll have to think carefully about which reaction we'll choose for that. But we definitely have a pathway to do that. So that gives us kind of kind of our overall outline of how we start to get where we're going. So we know that we need to start with a three carbon alky Halide and our little one Ethan. So let's start there. So here we are, I've drawn one bromo propane, got three carbons ending in that Bromy and we're going to add on two extra carbons. So I will react it with one Ethan. So I have just ac C triple bond with a negative charge. That's our substitution reaction. We mentioned bromine leaves and we now have one PD. So we've got five carbons. Now, we're up to five carbons and carbons, four and five are in a triple bond. So now as we said, when we have our five carbon alkaline, we can generate a five carbon alkyl Halide. So our next step is we want this five carbon alky to become a five carbon alk. Our geometry doesn't matter. We have a terminal bond, doesn't matter how things are going on. So we can do it one of several different ways. I'm going to go ahead and use hydrogen gas, which I need. Of course. And I'll just use Lindler catalyst. Remember that Lindler catalyst ads uh or forms a cy alkin. But again, it doesn't matter here because it's going to be going away anyway, it's at the end. So that does, I don't need it to be that specific. I don't need to use that reagent, 12345 with a double bond. So now I've got my, again, five carbons still, I've got my one pine. So now I need to make this a, an alkyl Halide, but that bromine has to be added on the end I need in particular a primary alky Halide. So let's think about that. I cannot go through a Carbocaine intermediate because the positive charge on the end, a primary Carbocaine is not stable. So I need to make sure I use an anti Markov Niko reaction. They're going through a radical intermediate. So I'll put anti Markov Niko. So I'm going to need H pr because I've got to have a source of that bromine. But then I also need to add hydrogen peroxide to make sure that the bromine adds on the end where I want it to. So five carbon, I'll kill Halide. I'm up to one bromo pine, so up to five carbons there. So I've got piece one. So I'm going to sort of give it a little, I think I'll give it a blue highlight here. Oops, that's not a highlight. Give it some blue highlighting that sort of piece one there. So five carbon. So recall, I'll need this piece in and of itself to be sort of part one of the two pieces, big pieces I'll put together, but I can also use this to generate my seven carbon alkaline. So let's start again. With my one brom Entine. So again, we want to add two carbons. So we'll react that with our one Ethan. So there we go, our CC triple bond with a negative charge. And now we're going to generate our one Hep DNE. So we've added on two carbons. We started with five carbons. Now, we're up to seven carbons. We have that triple bond between six and seven and remember what we're trying to generate here. We just want our alkaline. So we just need to pull off a hydrogen so we can react this with a strong base. For instance, nanh two, get rid of hydrogen. And we have just our same molecule here still have our triple bond, but now we have a negative charge there. So there's our seven carbon oops hold on are seven carbon alkan. And I added the words Alky hide to my one bromo Penta and I just generated in the last step or not the last step. But the last set of steps, I'm going to scroll up here. So I have my two pieces to put together. I'm going to react these together. So we'll say plus here. So my two blue highlighted Alky Halle and alkaline and that will generate my big long 12 carbon chain with a triple bond in the middle. So I've got six carbons. Now, here's six and again, with the geometry of triple bonds, this gets a bit tricky. Scroll down again, I do remember the carbons 567 and eight will all be in a straight line here. So I've got 56, continue the line just putting dots to help me keep oriented here. And between six and seven, I have my triple bond and now down to carbon eight in that straight line and then 9, 1011 12. So I've got my big old molecule, my six Jode kind um here, but now I'm kind of in the home stretch just have to transfer that. And in fact, I'm going to scoot it over, see if I can fit everybody on the same line here. So reminder, I now have a 12 carbon alky. So I'm up to the right number of carbons just have to do those last steps. So now I want to end up with a cis alchem. I have an alky that is a reaction I did earlier, hydrogen gas and Lindler S catalyst. In this case, the geometry does matter. I want a cis alkene. So I'll be specifically using this catalyst and I will end up with my 12 carbon alkene. So six carbons and then between six and seven, I have that double bond and then 89, 1011, 12, it's really tiny. Now trying to make enough room for everything. So now I've got my 12 carbon alkene and finally my last reaction I have a reagent MC PB A BP A excuse me, tongue twister MC BP A that abbreviation standing for meta Chloro Paroxy Benzo acid. So MC BP A which will then convert my alkene without changing the geometry into an oxide. So I've drawn my six carbons, I have my bond from carbon 6 to 7 and then solid wedge, solid wedge, there's now an oxygen there. They're making my oxide and then finish up 89, 1011, 12. So there's my final prod product and here's my entire synthesis. It was a long journey, but we made it. So just that quick overview, I have of my, again, looked at the synthesis. I started with that three carbon alkyl Halide used an athan reaction, ended up with a five carbon alky and then went through those steps to convert it to the five carbon alkyl Halide. Then use that five carbon alky Halide that I generated. And another alkaline reaction generating my seven carbon alkaline, put the five and seven pieces together, generating my 12 carbon alkaline to my 12 carbon alkene to my final oxide product to synthesize that desired molecule. See you in the next video.

Synthesize the following molecules beginning with only organic molecules containing three carbons or fewer.
(g)

Key Concepts

Functional Groups

Reactions of Organic Molecules

Carbon Skeletons

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