What orbitals contain the electrons represented as lone pairs ...

Question

What orbitals contain the electrons represented as lone pairs in the structures of quinoline, indole, imidazole?

Accepted Answer

All right. Hi, everyone. So for this question is asking us what orbital accommodates the non bonding pair of electrons in each of the structures shown below. And we are given five of them. So before we begin, let's recall really quickly that we can determine the hybridization of a given atom by considering how many groups so to speak are bonded to them. Right? So when I say groups, I'm referring to two different categories, I'm referring to electron lone pairs, but I'm also referring to atoms that are bonded to the atom that we're looking at, right. So the total number of groups you have associated to an atom is going to determine the hybridization specifically, if we have a total of four groups, we have sp three hybridization If we have a total of three groups S P two, and if we have just two groups, then we have sp hybridization. And so with that being said, let's go ahead and apply these this concept right to the non bonding pair of electrons on each nitrogen here. So for molecule, one molecule, one has just one nitrogen atom with one non bonding pair of electrons, right. So in order to figure out the orbital that it's going to occupy. Let's go ahead and figure out the hybridization of our nitrogen here. So nitrogen has a total of three groups, right? Because we have one pair of electrons plus two carbon atoms that this nitrogen is bonded to. So if we have a total of three groups, then this nitrogen on molecule one is going to be S P two hybridized. Mhm. So now we can go ahead and proceed to part two. So this um just to recap this lone pair of electrons on nitrogen is going to occupy an sp two orbital because our nitrogen is sp two hybridized. But now let's go ahead and consider this nitrogen on molecule too because this case is slightly different, right? What? So what I want to bring to your attention is that this nitrogen on the ring, we have four groups in total, right? Because we have the lone pair of electrons itself plus three carbon atoms attached to nitrogen, meaning we have a total of four, but we have to be very careful, we have to be very careful because what I want to bring to your attention is the fact that this lone pair of electrons on nitrogen is actually going to be part of the aromatic system present within molecule. To right, because if we go ahead and we count, well, if we go ahead and ignore the lone pair of electrons on nitrogen for just a second, the number of pi electrons we would have in our ring system would actually make this anti aromatic, right, because we have a total of eight pi electrons when we exclude this lone pair on nitrogen. So that would make this anti aromatic which of course is not favorable, right. So this lone pair of electrons are nitrogen is going to contribute actually to the aromatic system. And therefore, it's going to occupy a P orbital, it's gonna occupy a P orbital because it's going to participate in the conjugated pi system of this aromatic ring here. So this is the case for this nitrogen, right? And now let's move on to molecule three and molecule three, this time has two nitrogen, one associated to a method group And the other as part of this five member ring. So let's go ahead and focus on this nitrogen on the left side, which is not attached to this method group, right. So what I want to bring to your attention is that we have um we have three groups in total, right? We have three groups in total because we have two carbon atoms bonded to nitrogen plus the lone pair itself. So that means that our nitrogen is sp two hybridized. And so this lone pair of electrons on nitrogen is also going to occupy an sp two orbital. But we have to be very careful with the second nitrogen attached to our method group, right? Because if we look at our ring structure and we see how many pair of electrons, sorry, how many pi electrons are present within this ring. If we exclude these nitrogen lone pairs, then we'll notice that we have four right and having four pi electrons would make this molecule anti aromatic once again. So this nitrogen here bonded to our method group is going to once again contribute its pair of electrons towards the aromatic system. Yeah, it's going to contribute to the aromatic system and therefore just like in molecule to yeah, for the nitrogen on the right, its lone pair of electrons is going to occupy a p orbital. Alright. So now we can move on to structure for And structure four has 4 nitrogen atoms. So let's begin with these two that are on the six member ring, We have two on the six member ring and the six member ring is aromatic. So each of these nitrogen is here have two carbon atoms bonded to them and they have one lone pair of electrons each. So what that means is that both of these nitrogen have three groups, they have three groups and neither of them need to contribute to the aromatic system because the six member ring is already aromatic, right. So these two pairs of electrons are just gonna occupy sp two orbital's okay. But now let's go ahead and switch gears towards the nitrogen on this five member rate, right? Because The one on the top here, the one that has a nitrogen carbon double bond, it has three groups just like the two nitrogen on the six member ring. And it does not contribute to the aromatic system just like the other two from the six member rate, right. So this one is going to occupy an sp two orbital. I'm not sure you can really see that. So let me just write it a little. There we go. So this nitrogen on the top is not part of the aromatic system. And it is therefore, or rather its lone pair of electrons is going to occupy an sp two orbital. But the same is not true for this nitrogen attached to our method group on the bottom, right? Because this nitrogen, it's sp three hybridized because we have a total of four groups, we have three carbon atoms plus the lone pair itself. So that's a total of four, which means that we have sp three hybridization for this nitrogen. But we have to be very careful, right? Because if we go ahead and we just ignore this lone pair of electrons on this nitrogen for a second, we'll see that will have a total of eight pi electrons contributing to the aromatic city. And eight pi electrons would indicate that this molecules anti aromatic, which of course is not favorable. So that means that this nitrogen bonded to our method group would donate it's it's non bonding pair of electrons towards the aromatic system, which means that this non bonding pair of electrons is going to occupy a P orbital because it's contributing to aromatic city. And now we can focus on molecule five, right. This is the final stretch and We have two nitrogen Attached to part of a six member ring, which is already aromatic. So both of these nitrogen have three groups, right? Because we have the lone pair of electrons and two carbon atoms bonded to them. So each of them has a total of three groups, which means that we're talking about sp two hybridization. And because their lone pairs of electrons do not contribute to aromatic city, because this ring is already aromatic, those non bonding pair of electrons are just going to occupy an SP two orbital. And so now we've gone ahead and we went through everything we made it guys. And so this is going to correspond to option d within your multiple choice. And so now this is going to be pretty much it for this question, right? Depending on how many groups are associated to the atom that you're looking at. You can figure out their hybridization. But you have to be very careful because you have to ask yourself whether or not that non bonding pair of electrons is going to contribute to residents at all. So with that being said, if you made it to the end, thank you so much for watching. And I do hope you found this helpful.

What orbitals contain the electrons represented as lone pairs in the structures of quinoline, indole, imidazole?

Key Concepts

Lone Pairs

Hybridization

Molecular Structure and Resonance

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