Show how you would accomplish the following syntheses efficien...

Question

Show how you would accomplish the following syntheses efficiently and in good yield. You may use any necessary reagents.

(b)

Accepted Answer

Hey everyone and welcome back to another video. Today we are going to show how the given conversion can be achieved with a good yield and provide any necessary regions. The starting material is psycho hex unknown. That's our key tone. And we noticed that the final product is an elk een because we have our double bond present, There's no more carbon yield group. Let's recall that one of the most efficient ways to perform that conversion is the Wittig reaction. And we are going to look at the steps of the Wittig reaction in this case. So now the Whitaker reaction essentially tells you that whenever you have a carbon compounds such as a key tone or an elk hide, you can easily convert it into an alky and what a good yield. Using this reaction and what you want to do, you first want to understand what you're Allied is. And to do that, you're going to compare the two structures. Now. Specifically, you're going to concentrate on your double bond below the double bond, you still have that ring but above your replacing oxygen with an alcohol substitute in bondage. That sp two hybridized carbon atom. Right? So we're going to see clearly LeBeau the fragment that comes from Allied. So this fragment comes from Allied and we noticed that there are two carbon atoms bonded to our final substitution. So we can easily draw the Allied required for this reaction, let's draw two carbon atoms and then let's add our final substitution. Let's recall that Allied requires us to have a negative charge. And this carbon atom that is directly bonded to our sp two hybridized carbon atom within the ring. Right? So this is our negatively charged carbon because it's part of the double bond and now adjacent to it we have our try fennel fosse finds that would be three final rings bonded to phosphorus. And because phosphorus has four bonds it would have a formal positive charge D to these charges. And especially because carbon is negatively charged, this allied is unstable. Right? So this is the ally that we are going to use for our reaction. But the question is how do we form that ally then? We may something recall that to formalize. We need two steps first of all, we are going to redraw our carbon chain. So let's draw two carbon atoms and then let's add our final substitution. And specifically we want to make sure that there is a good leaving group bonded to our carbon where we form the negative charge. Right? We are going to add a good leaving groups such as blooming. And here in particular we need to use tear agents first of all we're going to replace roaming our good leaving group with Try Fennel Fosse finds. So let's do that. We are going to use reaction number one try fennel foss fine. And then when we replace blooming with try fennel fosse fine. We also want to use normal Bethel lithium which is a strong base and allows us to D protein eight, that carbon and form a negative charge. So that's the whole idea in our sequence now based on these steps let's just look at the mechanism and understand the rationale and then we will complete our synthesis. So first of all let's see how the reaction takes place. If our starting material is this l kill hey light we are going to essentially draw it. That would be that that would actually be errol hey light. Right because we have our federal ring. So now let's throw it out. That would be our promo substitution bonded to 12 carbon atoms. Then we have our federal ring. In the first step of the reaction. We are going to replace roaming with try fennel phosphor in right. We are going to use this as our first region. So we can say that okay in this step we are adding try fennel fast fine. And now if you think about the mechanism there's a lone pair on phosphorus and we have an electra filic position. So we can think of this as nuclear Felix substitution. We have the loss of the leaving group right? And then we are getting our fragment that would be three final ranks bonded to phosphorus. And now we are going to read through our chain we have positive charge and phosphorus that's good for us. That's part of the allied and then we just want to get that negative charge over here to do that. We understand that this hydrogen can now be removed with a strong base such as normal diesel lithium is. We're going to essentially show that there is this hydrogen notice that phosphorus has a positive charge. Right? So the bond between phosphorus and carbon is polarized and in particular normal base elysium, which essentially looks like this, we have our lithium 12, 3, 4 carbon atoms. And essentially we are going to attack that hydrogen. We are going to transfer these carbon hydrogen electrons onto carbon and form a negative charge here. Right. This is how we are going to form our Allied. So in this step we form the required Allied, let's show that Step # two, we are using normal B cell lithium and let's draw the resultant Allied now we have our negative charge because we have transferred our electrons in the form of a lone parent carbon. And now that we have that Allied, we are ready to think about the Wittig reaction. Right now from here we are going to add our actual ketone cycle Jackson on leads to a cycle hex unknown in the step We can through our six member ring Carbondale group and think about the steps. So first of all in the world. First step we have two plus two cyclo edition, the negative charge on Allied would essentially attack the electrolytic position of the carbon eel group. And because we're forming a new bond, the pi bond will be broken attacking the positive phosphorus and we are going to get our ring structure. So let's see what we get in particular. We can just draw our six member ring. Now we have single bonded oxygen, oxygen is bonded to phosphorus which contains three final substitutes, who are going to say p Ph three. Now, because phosphorus will contain a total of five bonds that will have no charge. Right? And now, what we're going to notice is that that's carbon atom is bonded to that ch group over here. Right, So now this ch group is bonded to phosphorus. But in turn that ch group is bonded to cease to PHR Benzel substitutions. Let's show CH two and then our final substitution. Right? So that's basically our Benzel bonded to that carbon. And now that we have formed this structure, we will observe the reverse two plus two cyclo addition, essentially what we're going to do, we're going to draw a few arrows to see how the final L keen is formed. We are just going to exchange our bonds and essentially we are going to form that oxygen phosphorus by bond. And then looking at these bonds, we're going to swap them this way so that we form our double bond at this position and this gives us our final product. Notice that from here we get our L king. So we can show the result on structure that would be our six member ring. Notice how we are forming this double bond. We are just going to change the orientation and now we have our siege substitution. Wanted to Benzel. So that would be sieged. Two ph Now you may also notice that the by product would be oxygen double bond. It's of phosphorus, right? So we're going to have our byproduct. Three final rings bonded to phosphorus and phosphorus is bonded to the oxygen. That's our byproduct. Let's not worry about it. We got our final product. So now that we clearly see how the Wittig reaction works, let's just summarize our steps first. We start with our aerial bromide. Right? We're going to state that the starting material must be fennel rang sees too. Another siege to blooming. Right? So that's our errol bromide and that we are performing two reactions to form our Allied. These are two different steps. First of all we're adding try fennel fosse fine. So that be PPH three. Then in the second reaction we're adding normal B cell lithium to form that unstable Allied we are ready to draw the resultant Allied that would be exactly same structure. But now we will have our allies. So let's throw our carbon atoms. We have our negatively charged carbon atom and from here we're essentially just adding our carbon compound. As we said before. That's our cycle Jackson on, we are ready to throw it out as our region for the third step, the reaction. So that would correspond to cycle Jackson on. And this is how we get our final product. So we have our double bond ces two bonded to the final substitution. That's our Benzel group bonded to that carbon. Okay. And now we are ready to label this is our final answer. This is our reaction sequence. Just to be everything. We start with our errol bromide. We understand that because we want to form an elk in out of a kid's zone, the Wittig reaction would work for us. So we're forming our allied using these two common steps. Try try final phosphorus followed by addition of normal bees A lithium. This gives us our allied after we get our allies, we treated with our car carbonell compound in this case, cycle hexen on. And we get our elkin now based on the proposed synthesis scheme, we can conclude that the correct answer to the multiple choice question is a However, if you have any questions, don't have seats leave a comment down below. Thank you for watching

Show how you would accomplish the following syntheses efficiently and in good yield. You may use any necessary reagents.
(b)

Key Concepts

Imine Formation

Aldehyde Reactivity

Phenyl Group Influence

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