How would you convert the following compounds to aromatic comp...

Question

How would you convert the following compounds to aromatic compounds?

(a)

(b)

(c)

Accepted Answer

Hey everyone and welcome back in today's video. We are going to solve the following problem. The compounds listed below are not aromatic compounds provide the reaction needed to transform them into aromatic compounds. As the problem suggests structures 12 and three are not aromatic compounds. Celeste recall to you criteria for aromatic city number one, the structure must be plainer and cyclic I system and number two, there must be four and plus two pi electrons where an is an integer as defined by the Hucles rule. Okay, so let's see what we are missing in the context of structure. # one, is it plainer? We essentially can label our sp two hybridized carbon atoms which are double bonded and we notice that there is one which does not have a double bond. It actually has two hydrogen atoms that are implicit and because of that, we are seeing four electron regions and it implies that this carbon atom, it's actually sp three hybridized, recalled from Bs CpR theory for electron clouds. Sp three. Right. So this is a problem. We want to turn it into an sp two hybridized carbon atom. And in addition to that, we're only seeing 12 double bonds, which implies for by electrons. And this doesn't satisfy our equation because if you equate four equals for employers to you get an equals one half. And this is definitely not an integer. So none of the two conditions are satisfied. And we are going to think about how to solve this problem. One way to do that is just to make sure that we have a lone pair of this carbon because it will be d localized. We have an adjacent pipe bombs and these electrons might produce as a planner structure as well. This means we want to essentially eliminate one of the hydrogen atoms. So if we do so we can just through our five member ring and we are going to propose a reaction. So we are going to draw our five member ring, two methyl groups. Now let's expand just one hygiene for complete clarity. We want to make sure that we're using a strong base for this reaction recall that one possibility is to use normal beetle lithium so that be lithium bond, it's a beetle chain. So that'd be 123, four carbon atoms, right? 1234, that's normal, beautiful lithium. This is a strong base recall that carbon, lithium bond is strongly polarized towards carbon. So we can have an easy attack at our hydrogen. And then when hydrogen is removed from the structure, the original bonding electrons will be transferred in the form of a lone pair on carbon. So this is how we are getting our new structure. In this case We can throw our five member dring double bonds. Now there's a lone pair of electrons here, which implies a negative charge. And then here we have our metal groups. Now the question is, did we really solve this problem. Notice first of all that these electrons are de localized. Right? So they are part of the pi system because you can easily draw one of the residents forms creating a pi bond here and breaking the adjacent one. So you clearly see that you have residents. Let's draw the result. And residents form just for clarity, we now have a negative charge of this position, a lone pair. The double bond is over here and here. So we indeed see that the lone pair is the localized. So it's part of the pi system. And this gives us 12 electrons plus two from the double bond and another two from the double bond. That will be +65 electrons. And in addition to that, because this carbon atom contains three electron regions, 12 bonds and one lone pair. Right? But there's actually another implicit hygiene. So that would give us 1234 electron regions. We cannot really conclude that it's sp two hybridized yet because we see four electron regions here. But if we look at this residence form that same carbon contains a double bonds. We clearly see that when the long paris de localized this carbon atom actually is sp two hybridized now because it has a double bond. So that solves the problem criteria number one, the structure is indeed playing are cyclic because we now notice that this carbon atom and green that originally was S P three is now sp two hybridized. So we have created a plane are cyclic by system. And criteria number two because we got six pi electrons. We are going to grade six equals for employers to if you subtract two from both sides, you get four equals for N. Which means that an equals one. This is an integer. So it satisfies the coast rule. This works for us as well. So the reaction is feasible. In other words for the first structure, we essentially want to propose the following reaction. If we simply tour our region normal B cell lithium on a reaction arrow, that would correspond to our reaction. So let's see that instead of redrawing everything, we will just clear the partial mechanism. And we are just going to say that we want to treat the original structure with normal bees lithium. So that would be and bisa lithium B cell lithium. Okay, so that's our normal bees, A lithium And we understand that this is the reaction that we want to perform because essentially it allows us to get rid of one of the hydrogen atoms and produce our carbon ion with a D localized lone pair of electrons. Well done. So we understand our reaction. We can label this as our partial answer for number one. Now, let's move on to the second structure. And the question is, is it planner? Because that's our main requirement. We want to make sure that it's planner. We can label our sp two hybridized carbon atoms that are double bonded and this one is not okay because it has four electron regions. 1234 bonds. It's actually sp three hybridized once again. But we have to be extra careful how many pi electrons do we have? We have 123 double bonds. And because we have three double bonds, there are three pi pi bonds and three pi bonds imply that there must be six pi electrons. Because every pi bond has two electrons, we already know that six pi electrons correspond to an integer of one. So the second rule is actually satisfied. All we have to do is just to think of a reaction which would produce a planer carbon over here. Right, so to do that, we can just use an acid catalyst dehydration mechanism. Why don't we get rid of the alcohol group by protein eating it? So simply speaking, we are just going to say that we take our seven members doing. And now if we draw our seven member ring, that would be our double bond at this position. Now let's it clearly. So that would be okay. So that's our seven member grain, we can through our double bonds and then we have our method group and our alcohol group according to the asset catalyzed dehydration mechanism. We essentially need to use sulfuric acid. It would allow us to protein eight oxygen, which would act as a leaving group forming a stable tertiary carbo. Karen, that is also resonant stabilized. Okay, so this allows us to protein it. That oxygen we would form water followed by the loss of water. We are going to form our plane or carbon because that would be our carbon Caroline with only three electron regions and three electron regions imply sp two hybridization. So let's draw the result on carb. Okay, Ryan, let's make our bond angles slightly greater so that we get our nice representation of the structure. Okay, so that's our seven member ring. And then we essentially understand that we only have a metal group at that position with a positive charge. And indeed, you may notice that we have 123 electron regions. That implies this carbon is now sp two hybridized and that works for us. We still have our six by electrons and we have our planer carbon over here. Right? Essentially. You may notice that there are three substitutions sp two hybridized. So we have solved our problem. We need to use an asset capitalized dehydration to produce that car bull cast iron. And now it's it doesn't only have six pi electrons, but the ring is also playing our cyclic pi system because we have turned this final atom into an sp two hybridized atom. Finally, for the third one we have our bromo substitute print bonded to a ring. So we are going to analyze that ring and we notice that there are two double bonded carbon atoms. They're both sp two hybridized. So that implies that we're good here. But what about this carbon atom keep in mind that there's an implicit hydrogen atoms. So we have 1234 electron regions. And once again we have that problem is sp three hybridized and the number of pi electrons is to okay, because we have one pi bond, so that would be two pi electrons, meaning an must be zero and zero is indeed an integer. So two pi electrons. That's fine for us. But how do we turn this into an sp two hybridized carbon atom. Now, one of the ways is to actually attempt to remove this roaming and transfer the electrons onto this carbon to produce a carbon ion. And see if its resonance stabilized. Whenever you have a bromo substitute, recall that silver bromide is an insoluble compound. So we can try that number three, We can essentially draw our three member ring with our bromo substitution. The region that we need to remove grooming is would be silver Caroline. Right? We are going to use silver car Ryan because the promos obsession these electrons can actually be transferred to an order to a free orbital of silver and we can create that ionic silver bromide which is insoluble. And that's what drives this reaction forward. We are creating an insoluble precipitate and therefore this is favorable for us. This really possible, we're essentially going to transfer these electrons to silver forming silver bromide, which is a precipitate. Plus, if you notice we are actually forming a carbo cat ion. I think previously I mentioned that it should have been a carbon ion. But you clearly see that because we're losing these electrons. The actual correct way to describe the structure is carbon cat eye on because we are losing these electrons from our carbon. And you may already understand that we still have these two pi electrons and the original two carbon atoms were sp two hybridized and this one is also sp two hybridized because it only has three electron regions. 12 and an implicit hygiene that will be electron regions. Tell us that this carbon atom is now sp two hybridized and because it's sp two hybridized, it allowed us to achieve a planner cyclic pi system. The second requirement is already satisfied. We have our two pi electrons which satisfies the high cultural remember four and plus two. In this case that's two. And essentially you may notice that if n is zero then the equation is satisfied. And this is an integer. So that works for us. Okay, so we see all of our reactions needed for this problem and based on our answers, we may conclude that the correct answer to the multiple choice question is the However, if you have any questions don't hesitate to leave a comment down below and thank you for watching

How would you convert the following compounds to aromatic compounds?
(a)
(b)
(c)

Key Concepts

Aromaticity

Electrophilic Aromatic Substitution (EAS)

Rearrangement Reactions

Watch next