Propose a mechanism for the self-condensation of methyl 3-phen...

Question

Propose a mechanism for the self-condensation of methyl 3-phenylpropionate promoted by sodium methoxide.

Accepted Answer

Welcome back everyone to another video, write a suitable mechanism showing how ethyl teeny acetate undergoes self condensation in the presence of sodium oxide. So we're going to draw our starting material Ethyl two phenyl acid and we understand that their sodium is oxide. So sodium is an spectator, meaning it is sufficient for us to draw oxide and ethyl group with oxygen that has a negative charge, which is a strong base. We are going to locate our acidic proton which will be added to the alpha position because the carbonyl group is an electron withdrawing group. So it makes the given proton acidic and therefore the base will capture that proton forming a pi bond and breaking the og and pi bond to get the ehate anion. Now, we can show the resultant in Nate and to do that, we are going to break our pipe on first of all from the original structure. And now we are going to introduce our newly formed by bond. Oxygen gets a formal negative charge because this is what the arrow tells us. So we are going to add that formal negative charge on oxygen. And this is where self condensation begins, we're going to draw another unit of ethyl teeny acetate. The negative charge on oxygen will allow us to restore the carbonyl group by breaking the adjacent pipe bond, which will act as a nuclear file, right. So what we're going to do here is just illustrate that nucleic attacks, stop the pi bond will attack the carbonel group. We are going to break the pi bond and once again, that pipe bo will allow us to get a negative charge on oxygen. But now our goal is to introduce a bond between carbons number one and two. So now let's do that carefully. This is the most important part to make sure that we combine the two fragments correctly. And the way we can see this is as follows, we can start by drawing the left side, right. Let's see what we have so far. So we have our ethyl group oxygen. Now we are going to get our carbonyl group. There will there will be no pie bond anymore. That's good for us. And we can actually change the position of that benzene ring slightly to make sure that it is easier for us to see what's happening in the structure. We can essentially throw that car, draw that fel group below the plane. OK. So now on the right side, we will be able to combine our fragments. We're also going to add the implicit hydrogen atom because it will be important. We will soon understand why? And now from here, we are ready to draw carbon number two. So let's add carbon number two, which now will have a formal negative chart. I'm sorry, which will now be bonded to oxygen. And that oxygen will have a formal negative charge just as illustrated by the arrow. So let's add that oxygen atom with the formal negative charge. And now let's see what else is bonded to that carbon number two. Well, first of all, we also have our oxy group. So we are going to draw an oxy group and actually we want to make sure that first of all, we bond our oxygen, right? And then from here, we can draw an ethyl group. OK. Well done. Now that we got these bonds, we are simply going to focus on the remaining bonds. So we have CH two and then that benzene ring, right. So we are just going to introduce that benzoyl. So first of all CH two and now that's siege too will be bonded to the benzene ring. Well done. So we got our fragment. Now from here, what we notice is that we can have the loss of the living group. We can essentially remove that oxide, right, as a catalyst. So the negative charge in oxygen will allow us to expel it. We are going to restore the carbonyl followed by the loss of the leaving group, right. So we will get that oxide, we can essentially draw an arrow, let's slightly shift everything. So we can have more space well done. And now let's draw the resultant structure from this stuff based on what the arrows are telling us, we are simply going to form carbon. So now we actually have a double bond and oxygen. Now finally, what do we want to do from here? We can actually say that we are nearly done. The main difference is that there is still oxide, right. So we still have those basic conditions. And that's why we have illustrated our proton at the alpha position. It is significantly acidic because it is adjacent to two carbo groups. Meaning the dep protonation step will take place again to form a negative charge. So let's essentially illustrate what we have so far, we are going to draw that result and negative charge between the two carbon groups. So we're going to remove hydrogen. That's our first step. It is also important to adjust the bond angles. So what we can do is just draw a straight line because that's what we are going to observe. Now, let's add the benzene ring. And from here, we simply want to understand that we are obtaining that negative charge. So let's illustrate the negative charge between the two carbonel groups. And we want to rep proton that water will essentially allow us to do that because we have a solution. So now the negative charge will attack the hydrogen atom from water. And this is how we will re proton our structure and finish our condensation to draw the final product. We are simply going to duplicate the previous structure, eliminate the negative charge because now we are adding that proton and from here we are done. So let's see if we have our complete mechanism, we can essentially decrease it in size slightly so that we can clearly see the complete mechanism. And now we can say that this is it, we have successfully obtained our mechanism for the self condensation of the starting material. Thank you for watching.

Propose a mechanism for the self-condensation of methyl 3-phenylpropionate promoted by sodium methoxide.

Key Concepts

Self-Condensation

Nucleophilic Attack

Base Catalysis

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