Determine which of the heterocyclic amines just shown are arom...

Question

Determine which of the heterocyclic amines just shown are aromatic. Give the reasons for your conclusions.

Accepted Answer

All right. Hi everyone. So for this question it's asking us which of these heterocyclic compounds are aromatic. And to explain our answer briefly. So before we do anything, let's go ahead and talk about if you recall the four criteria for aromatic city in a given molecule. So the first one is that the molecule has to be cyclic, which all of them are in this case. So we are good on that front criteria. Two is that the system has to be conjugated. In other words we have pi electrons separated by a single bond. Number three, is that all the atoms of your ring have to be plainer? Or SP two hybridized and number four, the amount of pi electrons in your ring has to obey the fore end plus to rule four pi electrons. So in other words, um you must be able to solve for n with your number of pi electrons. So before we begin, like I said before, all of these molecules are plainer. So we don't have to worry about that respect. But there are a couple we can eliminate based on the fact that they are not conjugated nor planer because they have no pi electrons. Because notice how molecule A here has no pi electrons. So our atoms are all sp three hybridized. So not only is it not conjugated, it's not planer. So that one's getting crossed out. And the same is true for molecule C. Right, because we have no pi bonds and therefore everything is going to be sp three hybridized. So that doesn't make the cut. Either molecule D is the same story because once again no pi electrons, everything is sp three hybridized. So it's not plainer. And the final molecule that has this problem is molecule I here because there are cannot stress this enough, no pi electrons, therefore everything's sp three hybridized. So it's not plainer. So we've crossed out everything that doesn't qualify now, we can talk about which of the following. Do qualify. So first we have to check if we focus our attention to molecule B here. We have to check whether or not it's conjugated and cleaner. And the answer is actually going to be yes. Because if you recall um nitrogen here has a pair of loan electrons that can participate in residence because there's a single bond followed by a double bond. So that combination of an electron pair and a double bond separated by a sigma bond that can generate resonance. So this nitrogen is actually going to be sp two hybridized. So molecule B is both conjugated and plainer. So now let's go ahead and count our pi electrons. So for molecule B I count six pi electrons because we have four coming from our two double bonds here and we have an additional two coming from our nitrogen. So for molecule B we have six pi electrons. And if I do some math off to the side here, if I equal six to R four and plus two, then I can solve for one, which is a whole number. So we know that molecule B is in fact aromatic. So I'm gonna go ahead and highlight it. There we go. Molecule B is aromatic. So now let's move on to part C. And part C is very similar in appearance to molecule B. And the same discussion that we had earlier about resonance is true for molecule E. Right because nitrogen still has a lone pair of electrons and there's a double bond separated by a single bond. So we have another lone pair of electrons followed by a single bond followed by a double bond. So um so it is in fact sp two hybridized this nitrogen attached to our method group because it does participate in residence. So it is conjugated. It is plainer. And like molecule B, it also has six pi electrons which like we established before does obey the fore end plus two rule. So we can go ahead and highlight that one as well. So there we go. And now for molecule F our nitrogen can once again participate in a resonance because notice how the lone pair of electrons that I just drew here, there are single bonds on either side of it followed by double bonds on either side of it. So like we mentioned before resonance conjugation And it's sp two hybridized. So it's plainer. So now let's count pi electrons right now pi electrons. We have Um six in this bending ring because we have three double bonds. We have an additional double bond in our five member ring here, so that eight and we have an additional two from nitrogen. So we have 10 pi electrons. And if we sold for 10 off of the side here for N plus two, we are going to get and equals two. So we do get a whole number, which means that four N plus two rule is in fact obeyed. So molecule F is in fact a romantic. So I can once again highlight it. All right, so now let's move on two, molecule G. So molecule G is a little bit more straightforward, right? Because we do in fact have three pi bonds um separated by single bonds in between. So this is a conjugated system and therefore all of our atoms in the ring are sp two hybridized, they're cleaner. So now if we count our pi electrons stemming from Our our double bonds within the ring, we get six pi electrons because we have two times three double bonds, two electrons times three double bonds. That is in fact six. So molecule G is aromatic. So now let's move on to molecule H. And molecule H like molecule G has Um three double bonds alternating each other. So it is in fact conjugated. It is in fact planer and if we count pi electrons we get six pi electrons because once again we have three double bonds each with two electrons. So molecule H is also in fact aromatic. It is in fact aromatic as well. So now we have two more and molecule J just like molecules G and H has three double bonds in a six member ring. And they're each separated from each other by a single bond. So that means they are conjugated. And if they're conjugated then each atom your ring is plainer. So now let's go ahead and count pi electrons. And we have two electrons times three double bonds. Which means we have six pi electrons once again. And therefore molecule J is aromatic. Alright, so now last but not least. Right, molecule K now, molecule K um our nitrogen here bonded to our method group has a lone pair and we have a lone pair of electrons with single bonds on either side of it and then adjacent to those single bonds or double bonds. So residents is possible. Which means that resonance is possible. And therefore this nitrogen and all members of your ring system here are sp two hybridized, They are plainer and they are in fact conjugated. So with that being said, let's go ahead and count pi electrons. We have three from our six member ring here. Or sorry, we have six in total. Right, because we have three double ones times two electrons each. And we have an additional four from our five member ring here, we have one we have to from a double bond and two more from the lone pair of your nitrogen. So six plus four, it's gonna be 10 pi electrons and therefore molecule J is also aromatic. And now we did it. We made it to the end. Guys, molecules B e, f, g h j, and K are all aromatic because they are cyclic, conjugated planer, and the amount of pi electrons obeys the four N plus two rule. So if you made it to the end, thank you so much for watching. And I hope this helped you.

Determine which of the heterocyclic amines just shown are aromatic. Give the reasons for your conclusions.

Key Concepts

Aromaticity

Heterocyclic Compounds

Conjugation and Resonance

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