Show how Wittig reactions might be used to synthesize the foll...

Question

Show how Wittig reactions might be used to synthesize the following compounds. In each case, start with an alkyl halide and a ketone or an aldehyde.

(c) Ph–CH=CH–CH=CH–Ph

(d)

Accepted Answer

Hey everyone and welcome back devised a synthesis for the following Elkins. Using the Wittig reaction. Starting with an alcohol. Hey, light and a key tone or an alga hide. Let's start with doing the first part because this is the Wittig reaction in order to approach this problem. Using a retro synthetic analysis, we have to identify our double bond and actually we have two double bonds, meaning you can essentially devise your synthesis using any of these bonds. Let's suppose that we choose this double bond for our analysis. So what we're going to do, we're going to break it apart and we are going to throw the two results on fragments. If we start from the left, we're just going to add that final ring. Right, essentially bonded to Ch. Now we have our double bond. Ch then we have our carbon bond etc. Metal group. We still have part of that double bond. So let's leave it for now and not worry about it. And then our next fragment corresponds to the other part of that double bond that we had originally. And we have Ch ph right, so that's our next fragment. What we want to do here since we have broken it apart, we want to choose one of these structures to be our carbonell compound and the other one hour Allied. So let's suppose that we choose the first fragment to be our carbonell compound. So we're going to add oxygen at the end of the double bond even though you can choose this one as your carbonell compound as well. There is no difference. But if you choose the first fragment to be your carbonell compound then the other one must be our prosperous allied. So we are going to add try fennel fosse fine too. So that bp age three P. Right? So that's our allied. And now since you already understand what your carbon compound is, you just want to think about the synthesis of this Allied right? You want to understand how to get this Allied? Because whenever we mix this allied with this key tone we get our final product. So to get that Allied you want to perform several steps if we go from right to left, remember that the world last step in the formation of that Allied is the deep rooted nation. Stop of course to clearly see why is the D. Protein nation. Stop. Let's recall that this Allied can have another representation. What if we break this pi bon and place the electrons onto carbon. Then we have the following residence structure that be three fennel rings bonded to phosphorus. Now we have a single bond between phosphorus and carbon. And this clearly tells you that this carbon is negatively charged and phosphorus has a formal positive charge. So we're thinking about the structure. And at the very end to get this Allied we have two D. Protein it. That carbon. So first of all we are going to protein ate it right we're going to say that we have the structure so that B. P three P. Wanted to ch two ph were essentially removing that proton using normal bees a lithium right? Our strong base to get our allied. But now the question is how do we arrive at this structure? And the thing is that you want to have a good leaving group bonded to that carbon. Let's suppose that we are choosing bro. Mean, so we're going to draw a broom mean wanted to ch two P. H. So that's our benzel bromide, right? And essentially to replace that roaming group with try fennel fast fine. We're just going to use try fennel foss fine which acts in an essence to reaction. Right? We're essentially attacking our electro Felix centerfold by the loss of the leaving group. Getting our structure here. It's really important center stand that we have a formal positive charge and phosphorus. And then when normal B. Cell D protein, it's that we get our unstable allied which is in residence with the other form of it. And that would be it. We understand how our synthesis would look like. So let's combine these into our final sentences. We understand that our starting material is Benzel bromide. So we're going to throw our Benzel fragment starting with the Benzene ring adding that siege to group to it. We may draw bro mean, so that's our L kill hey light and we are happy to see this because as the problem suggests we want to make sure that the starting material is an L. Kill. Hey light actually this be an aerial Hey light right because we have our benzene ring but that's perfectly fine. No worries. Now a reaction number one we are adding try fennel fosse fine Reaction # two treating it with normal visa lithium. And when we get our allied we just want to add our carbonell compound. So that would be this structure, this key tone ph ch we're going to try that double bond ch C. O. And then we have our metal group over here. Okay, well done. We can only slightly modify the way that we rode the structure. We want to make sure that we clearly see the connectivity between our carbon atoms. We can actually change it a bit and we can say that this would be P H. H. And these two carbon atoms are double bonded. So this would be our ch we can essentially clearly show that these bonds correspond to carbon carbon bonds and we can see the same thing in our scheme. Just for complete clarity. So over here we can slightly change it that B P H H C C H. Okay, so this is the reaction sequence that allows us to get the products shown in the first part, let's label this is our final answer for the first part and let's move on to the second part. Now in the second part, if we look at the given structure, we once again have to think about the Wittig reaction. We identify our double bond, we're going to break that double bond and we are going to get two fragments. So if we redraw that structure and start analyzing it, we get the following so we can throw our five member grain, that would be our double bond. And then let's see what kind of substitution we had. So basically we had carbon hydrogen and a metal substitution. Okay, so that's the material that we want to synthesize. As we said, we are going to break that carbon carbon double bond and we need to think about our allied and our carbonell compound. So let's suppose that we choose cyclo pencil to be our carbonell compound. Let's draw the two fragments and then we will see how we get them. So first of all this fragment at the bottom and now on top we have another fragment. So we have our double bond metal group hydrogen. Okay, so let's suppose that we're choosing the bottom fragment to be our carbonell compound. So we're adding oxygen. And for the other one we are going to make sure that this is our allied so fast for us And three final rings. Now because we have our carbon compound, the only thing for us to do is just to synthesize that allied And just as we thought about it before we are just going to remember that it can undergo resonance. So we get the following fragment where we have phosphorus with a formal positive charge carbon with a negative charge here we have our implicit hygiene. Let's not worry about it. And we essentially want to remember that once again this negative charge is formed in the previous step. So if we go from right to left, we may remember that to get that negative charge. We are treating our structure with normal bees, a lithium. So if we go backwards we can just add that missing hydrogen. Okay, PPH three. And we just want to make sure that we have a good living group for our al kill. Hey light for example, right we want to have our substrate that produces our try final fast fine. So we are going to get the following let's suppose that once again we choose our bromo substitute. Print as a good leaving group. So we can say that if we have two carbon atoms, one C if we add our bromo substitute, which is a good leaving group. Whenever we treat it with try final fosse fine we will have an essence a reaction and we are going to form that fragment which later on produces our allied. So this is our starting alcohol. Hey lied as long as cyclo pencil known as our carbon compound, we already see the reaction sequence. We can just summarize everything starting with an L. Kill. Hey light, which in this case is bromo ethane. We control bromo ethane. We have several steps. Once again we standard steps. So the first step would correspond to the addition of try final fast fine just as before. The second step would correspond to the addition of our strong base normal visa lithium just as before. So the only difference is that now we have a different carbon compound which is cyclo pencil known and that would be it. We can throw our five member grain and our Carbondale group. This would give us our product let's label. This is our final answer. And we have our answer to part two as well. Based on our answers to part one and part two. We may conclude that the correct answer to the multiple choice question is D However, if you have any questions, don't have seats leave a comment down below and thank you for watching.

Show how Wittig reactions might be used to synthesize the following compounds. In each case, start with an alkyl halide and a ketone or an aldehyde.
(c) Ph–CH=CH–CH=CH–Ph
(d)

Key Concepts

Wittig Reaction

Phosphonium Ylide

Stereochemistry in Alkenes

Watch next