Predict the products of the following reactions.
(j)

Question

Predict the products of the following reactions.

(j)

(k)

(l)

Accepted Answer

Hey everyone and welcome back in today's video, we are going to tackle the fulling problem, determine the final products expected in each of the reactions given below. And if you notice there are three different reactions A. B and C. With different sets of regions. So we are just going to analyze each of them separately. If we start with reaction A. We understand that the starting material is an iP oxide because there's a three member ring with oxygen and two carbon atoms. So we are just drawing cycle pen sign and that ip oxide group bonnets it. So that's our starting material. And the first region that we're using is final lithium. You may think about it as benzene ring bonded to lithium and since lithium is electro positive, the electron density is largely focused on that carbon. So that means this bond is nuclear filic. It can participate in a nuclear attack. In addition to that. It's important to understand that this bond essentially the polarization of that bond and the possible formation of a negative charge of this position makes it basic and because it's basic, there are two positions that could be attacked here. Right? There are two electra Philip carbons. We are going to attack a less substituted position since we have basic conditions. So this position is going to be attacked for by the loss of the leaving. We may now draw our product for the first step. We understand that. Okay, we have that five member grain. Now we have one carbon atom bonded to it during the loss of the leaving group we are going to form and all negative at that position. So we may just draw draw all negative. Right? Because we have the loss of the leaving group. Oxygen gets these electrons to become negatively charged. And now at the other position. Or in particular, if you notice there's carbon number one. Carbon number two. So we're going to add that carbon number two, that carbon number two now gets the final electrons right from the from this bond. These electrons are used in the formation of a new bond. So we are essentially adding fennel to that carbon number two. We may indicate that this carbon number two has the final group attached said. And the next step is number two where we are adding H 30 plus. So before that we may simply say that in the first step we added final lithium. And now we are treating this product with H plus. That indicates acidic conditions. If those conditions are acidic, it just implies that an acidic proton will be donated to the base. So Alcock side form will be protein ated to form an alcohol. Right? So we may now draw a product after the protein nation stop. We are getting our five member ring two carbon atoms. And then the final group on it to the second carbon. The first one has in a wage group because we have that proton transfer. In addition to that, we can just expand ph as a benzene ring. So we are pretty much ready to do our first product for A. And if we drew it carefully, we may simply start by Drawing that five member drain one Carbon Atom which has that alcohol group, right? The second carbon atom which is CH two. And now that second carbon atom will have the benzene ring attached set. So we may simply draw our final group. That will be our first product for A. And we are ready to go, right, we got our product here. Now if you look at B we will not need to worry that much about the mechanism if you notice we are starting with a conjugated dying in the ring. Right? And we have an elk in bond over here and we are treating that structure with em CPB meta chloral paroxetine, Zurich acid. This is our solvent, right? That chloral methane. So you have to keep in mind that because this dying is conjugated. It's not really important to think about the oxidation step. Since here we're using oxidation of Elkins, write Em CPP is responsible for an oxidation step meaning what we're doing. We're thinking our double bond and producing an ip oxide out of it. So the step here is very simple but we're not worried about oxidizing these bonds because it's a conjugated dying. So we're only taking the ALc in part and we are only oxidizing this bond. So if we do that, we may simply withdraw our dying part. This is our ring with two double bonds. And now we are going to redraw everything that we have. How many carbons? 123. Right, So that would be one, two, three carbon atoms. And now in particular position number two and three, they are going to form any pock sides who are going to draw two new bonds, carbon oxygen bonds. That'll be our final structure. We get any Parkside. And finally, let's think about C. Regarding see. Thinking about it in more detail. We are going to draw a partial mechanism to it. Right? We are going to think what's happening in C. Regarding C. Again, we have a five member ring. That's our cyclo insane. It has another ip oxides. Once again, we are looking at an ip oxide reaction and the regions that we're using here, there's just a single region. Right? It's US HPR hi jen, bromine. Now because these are acidic conditions, right? Hydroponic acid is a strong asset. Recall that acidic conditions imply that between the two electra Philly carbon atoms, the one that is more substituted will be attacked during the nuclear attack. Stop. But before we look at that stuff, let's recall that in the mechanism the very first step is the protein ation stop. So hydroponic asset is used as a proton donor the lone pair text, the acidic hydrogen. We have the loss of the leaving group that will later act as nuclear file. We are producing our protein ated intermediate. So we may draw that protein ated intermediate. That would be oxygen with a form of positive charge. And now the brown white and are informed in the previous stuff stuff will act as a nuclear file. We already said that a more substituted position is going to be attacked. So we are going to attack that position and open the ring. So we are opening the ring. Roaming will be attached over here. We're going to release an a wage group with no formal positive charge. Meaning if we look at our product We made redraw our five member grain how many carbons do we have? There are a total of two. So let's add two carbon atoms. One to position number one now has blooming attached to it. So let's just add blooming position. Number two has O. H. Right? Because we're breaking this bond and releasing this Oh age group. So that be seized to a wage sees two. And then we're adding O. H. Well done. So let's redraw this is our product for C. Going back to our product list We essentially have a five member ring two carbon atoms. There's an a wage bonded to one of the carbon atoms. So let's do it this way. That'd be a wage. And then this would be where blooming goes. So this is our product for reaction. See let's label our final products. We can see clearly that we got everything that we needed. We wanted to get these three products for our problem, and we are nearly done right. We got our three products for a B and C. We may conclude that the correct choice. This multiple choice question is C. However, if you have any questions, don't have the slip your comment down below in the comments section and thank you for watching.

Predict the products of the following reactions.
(j)
(k)
(l)

Key Concepts

Nucleophilic Addition

Reactivity of Organolithium Reagents

Protonation of Alkoxides

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