Propose a mechanism for each of the following reactions:

Question

b.

Accepted Answer

Hi, everybody. Let's take a look. Look at the next problem. It says, draw the mechanism for the reaction below. And we have a rather complex molecule here we have a be all. So a benzene ring with an oh and then in the metas position from that alcohol group, we have a substituent that's an amine. So it's ch two ch two NH then ch two ch two that's reacting with formaldehyde. So just a single carbonel group, two hydrogens on that carbon, it's reacting in the presence of acid. So in the form of an hydro num ion H 30 plus and the final product involves now are phenol has a ring attached to it. So the nitrogen has come around and is now attached to the benzene ring with ach two and that carbon has a bond with the benzene ring and then it still has the ethyl group attached. So there's been a formation of a new carbon carbon bond and the addition of another carbon. So let's think about how this will take place. Well, we kind of look overall at our reactants, we have an amine, we have a carbon and we have acid and those three things together can react to form an IMing. So those three things together make us think of an IMing formation. Let's recall that it means are the sort of nitrogen analogs of Carbonis. So it would be ac double bonded to a nitrogen. We know that our benzene group must be involved here because we have an addition of AC C bond to the benzene ring. So when we think about adding to a benzene ring, we will think about the eas reaction, electro pilic aromatic substitution because we still have our benzene ring, it's still aromatic, but we could have added that bond to the nitrogen, the additional carbon with an eas type reaction. So let's think about how this would go. So let's start with amine formation. So we have our carbonel group and it's going to have a resonance structure. So let's do an arrow there. That resonance structure results from the protonation of the carbon group. So our carbonel has two lone pairs on the oxygen. One of those lone pairs is going to grab a proton from our hydro M ion. And we'll end up with these two resonance forms. The first one, the direct result of the Perotin nation, we'll see we saw that carbonyl co double bond there's now hydrogen on that oxygen. So the oxygen is a positive charge and the other resonance form has a single bond between the carbon and oxygen. So it's now an oh group and then it has a positive charge. Oops, I accidentally drew two carbons there. I want to make sure that we have two hydrogens attached to my carbon carbon. So add those in there accidentally left those looking like metal groups. So now we have a single bond with an oh and the positive charges move to our carbon. So we've generated a carbo cion as one of our resonance forms here. So that carbo cat I am is then capable of attacking or rather I should say of being attacked by the lone pair on our amine. So recall that our amine has a lone pair on the nitrogen and those electrons now will come on down and attack that positively charged carbon forming a new CN bond. So we see the main body of our molecule there and our nitrogen now has a new bond to carbon. And then, oh notice that this nitrogen still has a hydrogen on it. Actually, we're going to draw the oh in another direction. On the other side, nitrogen still has a hydrogen. So it now has a positively charged nitrogen which leads us to our next step, which is going to be a dep protonation of that nitrogen by a water mole mole molecule. So draw our water electrons coming over and grab the hydrogen. Then those electrons go on to the nitrogen, removing its negative charge. I'm going to scroll up here to make some more room. Now we know, to form the Iine, we're going to have to get rid of that alcohol group. So we've still got acid around so we can protein that alcohol. So we'll draw hydro in there. The lone pair from the oxygen from the alcohol group comes on over and grabs a proton which converts that alcohol to a good leaving group. So then the lone pair on our nitrogen can come up and create this NC double bond while our 02 plus group leaves as water. So now we have a nitrogen double bonded to a carbon note that there is a positive charge on this nitrogen. So now we have our, I mean, we just need our last eas part. So I'm gonna scroll down, we're going to draw this just rotated around a little bit to make our next mechanism a little easier. So we've got this molecule this, I mean, we're just going to redraw it, going to draw a little wiggle across here and then redo this. So now I have Roman nitrogen, I'm going to put the double bond going down towards that benzene ring. And then I'll put the ethyl group up top the same molecule. I've just redrawn in, in a different direction. They are rotated around that single bond. My nitrogen installs a positive charge. So now we'll have this eas reaction. So it's going to involve the alcohol group on the phenol part. So the electrons from that alcohol from that oxygen, excuse me, were to come down and form ac o double bond that's going to kick out one of the double bonds in the benzene ring. So its electrons will now come up and bond to the carbon, the terminal carbon at the end of my ime. And then the electrons from that double bond will hop on to that positively charged nitrogen. So there's the formation of that new ring note that we have now have a temporary disruption of the aromatic of the molecule, but it will be regenerated. So now we have ac o double bond with an oh group still on there, that's going to be positively charged on the oxygen. It it's not gonna be super happy like that. We still have two of the double bonds in our benzene ring. But the one adjacent to this new carbon bond has now come up and formed this new link with the carbon that was part of the CN double bond. So we have this bond to ch two and then to the nitrogen. And they remember our nitrogen comes back around and bonds to the ring, the nitrogen no longer has a positive charge, but it still has that ethyl group on top. So that's kind of tiny and hard to see we're running out of space up there, but that's an ethyl group on top there. Now, our last step is we need to regenerate that aromatic ring and we're going to do that by detonating, I'm going to draw on where the two rings join the new joining there, there is a hydrogen there. So we can deproteinate that hydrogen. So let's put in a water molecule with its long pairs. One of those lone pairs is going to grab that hydrogen, its electrons can come down and regenerate the final double bond in the benzene ring, which then kicks back the electrons from the co double bond on to that positively charged oxygen regenerating the aromatic ring and the alcohol group. So we've regenerated that final product with the alcohol group back from the phenol, no longer having that co double bond. We've put the third double bond back in the benzene ring and of course, we've added on and formed this new ring with our amy there. So there's our mechanism again, we went through the stages of an immune formation which allowed us to add on another carbon there to the nitrogen. So, I mean formation and then an eas reaction which temporarily disrupted the aromatic city of the ring, allowing the addition of that bond to the carbon and then the nitrogen from the amine. And then we regenerated the aromatic ring. See you in the next video.

Propose a mechanism for each of the following reactions:
b.

Key Concepts

Reaction Mechanism

Nucleophiles and Electrophiles

Catalysis

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