Choose the more basic member of each pair of isomers, and show...

Question

Choose the more basic member of each pair of isomers, and show why the base you chose is more basic.

e.

Accepted Answer

All right. Hello everyone. So this question is asking us to choose the more basic of the two structures below. And why now, notably, we have two benzene rings here with methyl substituents as well as a pin oxide group and a nitrile. The first structure has the nitrile meta to the phenox ion and the other had the nitrile Ortho to the oxide ion. And I've redrawn the structures on the bottom of the screen here but recall that there's a correlation between stability and reactivity when deciding between the more basic of the two structures. We're asking ourselves which of the two is more reactive as a base. And generally speaking, the more reactive something happens to be then the less stable it's going to be therefore the more stable something is the less reactive it's going to be. But what determines the stability of something at least compared to something else because both of these compounds happen to contain benzene rings, we can first and foremost talk about the role of residents in increasing or decreasing stability. Now, resonance stabilize the molecule or an ion for that. But because both of these species have negative charges resonance is going to delocalized that negative charge and spread it across a larger area. So on that note, the more resonance structures are possible, the more delocalized a charge happens to be and therefore the more stable it becomes. So with this in mind, let's go ahead and draw out some resumed structures. For both starting compounds. Recall that when it comes to drawing out resonance structures, the movement of electrons is kept from areas of high electron density towards regions of lower electron density. Now, because we have oxygen atoms containing negative charges, that negative charge indicates highest electron density and similar to a standard reaction mechanism. Double headed arrows showcase the movement of two electrons and single headed or fishhook arrows showcase the movement of only one. So to start off the resonant structures of the first compound, I'm going to have one of the lone pairs of electrons on oxygen move to the single bond in between oxygen and one of the pi bonds of the benzene ring. Ultimately, what this does is push the pi electrons of that benzene ring away from this movement. Now, what this does is ensure that the negative charge has been moved onto carbon because the charges of each resonance structure should remain the same. Keep in mind also that resonance structures should be enclosed in brackets with these double ended arrows separating them from each other. So now that the negative charge has moved inside of the ring, it's going to go ahead and proceed in a circle delocalized itself throughout the ring. Now, as I'm drawing this next resonance structure, I want to point out here that in this case, the negative charge is not going to be delocalized through the nitro group or excuse me, the nitrile group. Because recall that four residents to take place, there has to be one single bond separating both groups that are being delocalized. When we consider the distance between the lone pair of electrons on carbon. And the nitrile group itself, there are actually two single bonds in between them. So because they're too far away, they're not going to participate in resonance with each other. So the lone pair of electrons is going to be delocalized throughout the ring only and the oxygen, of course. So at this point, the lone pair of electrons has made its way around the ring. So now it's going to go ahead and return to oxygen, which means that no more resonant structures are possible at this point. Keep in mind that I went ahead and moved the second base a little bit off screen just to make some more space. So here we have all resonance structures for the first base and there are quite a few. So this base is very much stabilized by resonance. But let's go ahead and compare the second compound just scroll down a bit more just to open up some more space. All right. So now lets go ahead and start off the resonance structures for the second base in a similar way. So oxygen is going to donate a long pair of electrons in between itself and carbon, which results in a carbon ion inside of the six member ring. So just like we saw in the first compound resonance structure, the negative charge is going to go ahead and make its way in a circle. The route the ring itself almost forgot my brackets. Here, there is one more pie bond in the benzene ring. So one more carbon ion can be created. And at this point is where we can go ahead and see the main difference between these bases and their resonance structures. Because now notice how there's a carb an ion on the position in which the night trial group is installed. Now, this is important because this time it is in conjugation with the nitrile group itself, which means that the negative charge can also be delocalized throughout the night trial substituent in addition to the carbons of the ring and oxygen. So in this case, there are two different possibilities, the loan pair of electrons on carbon can move towards a nitrile substituent which would result in a negative charge on nitrogen. This is something that was not possible in the previous compound due to the positioning of that negative charge. Now, the other possibility like we've seen before is where the negative charge can move back inside of the ring, which puts the negative charge back on oxygen like we saw in the beginning. So because of this, we can see why the second base is made more stable from resonance. Because notice how in addition to the negative charge being delocalized on the carbons inside of the ring and oxygen, it's also delocalized on nitrogen which is relatively electron. So because of this, it's made more stable from residents, which makes it less reactive. This means that the first compound, the first base is more reactive since it is relatively less stable. And with this in mind, we've arrived at our final answer. So I'm going to go ahead and scroll up here so I can write out the final answer. So I'm going to temporarily pause the recording. So now I've resumed the recording and the final answer reads as follows. The first structure is more basic because it is less stable than the second structure and is therefore, or and is thus more reactive. So if you watch this video all the way through, thank you so very much. I appreciate it. And I hope you found this helpful.

Choose the more basic member of each pair of isomers, and show why the base you chose is more basic.
e.

Key Concepts

Basicity in Organic Chemistry

Resonance Effects on Basicity

Inductive Effects on Basicity

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