The benzene ring alters the reactivity of a neighboring group
in...

Question

The benzene ring alters the reactivity of a neighboring group in the benzylic position much as a double bond alters the reactivity of groups in the allylic position.

Structure of a benzyl anion showing delocalization over four carbon atoms.

Benzylic cations, anions, and radicals are all more stable than simple alkyl intermediates. (iii) Use resonance forms to show the delocalization (over four carbon atoms) of the negative charge of the benzyl anion.

Illustration comparing allylic position, benzyl group, benzyl position, and benzyl radical.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem which states and comparison to simple alcohol and I and and immediate Ben's silicon ions are more stable, draw resonance forms to show that the localization over four carbon atoms of the negative charge of the following benzel and ion. If you look at the given structure, we understand that if we want to draw restaurants, we first of all, we have to keep in mind that the starting point would be the species that has negative charge or an excess of lone pairs because we can de localize them. Right? So first of all, we notice that there is this lone pair which can be d localized over the ring, right? Because we have these pi bonds over here and that will allow us to perform residents. So let's see how that happens. If we redraw the starting and I on because we have a negative charge, right? We control our Benzene ring. Siege to group. We already know that it has negative charge. There's a lone pair and there is that metal group. So the world first step we're going to perform, we are going to take that lone pair. It might seem that it's localized, meaning it only it only belongs to carbon atom, but it's not localized. It's actually de localized. And the reason why is because if you look at this lone pair, it can actually be contributed into the formation of the pi bond here. But recall that if we're forming a pi bond, we have to break an adjacent pi bon. So the adjacent pi bonds over here. And what we're going to do, we're going to break it. And by breaking it, we are going to transfer these electrons in the form of a lone pair on that carbon. So that's what the two arrows. 2%. And we can draw residents arrows and draw the result on farm. So we are starting with our six member dring. Now if you notice here, we have used our lone pair and we have formed the double bond. So let's draw that double bond that we are siege to group that we had originally no negative charge because we have used up our lone pair. Right? But now if you look at our ring, there will be a negative charge here because we have transferred a lone pair of electrons from the original pi bon. So there is this lone pair at that carbon. And we can also indicate the presence of a negative charge. Nothing changes regarding the remaining pi bonds, they're still there and we are ready to go. This is our second residence form. You can think of. This form is our first one, so we can label this is our first one, the initial one. And this is one the result on one that we got so far. Now we will get our third residence form if we keep the localizing this lone pair. Now, what happens is you will essentially keep going counterclockwise, right because there are still adjacent pi bonds, so they allow you to de localize that lone pair even more. So you're going to start with the lone pair again and you're going to form a pi bond out of it just as with it before. Right? So by forming a pipe on, we are going to break the adjacent pipe on and we indeed have that adjacent pipe on. If we break it, we will essentially be able to transfer the pi electrons onto that carbon atom in the form of a lone pair and we will have negative charge here. Now. So let's draw the result on form. So that's our speech to you, that's our muscle group. Now, if you notice we have created a pipe on at this position. At this position, we have broken our pi bon and we have a lone pair at this carbon with a negative charge. And now here we still have that double bond which hasn't been affected yet. Okay, so we got our third form and we can keep going. There is another possibility to de localize this result in lone pair and we are going to take that lone pair, right? And we are going to create a pi bond out of it and we will break the adjacent pi bond by placing the pie electrons in the form of a lone pair on to this carbon atom. So let's see what we get. Now, that will be our six member ring, our siege to group, our metal group. Now let's see first of all we have that vibe on over here we have formed another pi bond over here, there will be no more negative charge because we have used our lone pair. So there is this pi bon and now here we have broken our pi bon and we understand that the lone pair will be over here and this is where our negative charges on that carbon atom. Right? So we have got our 4th strike chair and let's think about it. Do we have anymore? And the answer is yes, noticed that we still have an adjacent pi bond, so we can keep the localizing and we will basically we will basically get back to our negative charge on that carbon. And if you notice we can essentially say that the slow lone pair can be de localized and we can essentially form a pi bond over here by breaking that adjacent pi bon and transferring the pi electrons in the form of a lone pair on carbon. Getting a similar structure to structure number one. But the positions of the double bonds within the ring will be different. So that would be our fifth form. And let's just throw it out. So now you may notice you have a double bond here, another here and you have formed a double bond at this position. Now for Siege two, you will have a single bond because you have broken it. That would be seized to as we previously said there is a lone pair now meaning there will be negative charge and there is still this metal group over here. So that's our fifth form. And we have successfully drawn all of the possible resonance forms, right? If you essentially continue your D localization you and if you go in any direction you will just get one of the five forms that we already got. So those would be, that would be it right. As long as we get back to our original negative charge at the same at um in this case carbon carbon with a negative charge in the lone parent, you're good. You understand that you are done. We went counterclockwise, we saw all of the possibilities. We got five forms. We ended up at the same position, right? Our negative charge in structures one and five it belongs to the same carbon atom meaning we are done. So we can label these five structures as our final answer to this problem. Okay and looking at these structures and at our residence arrows, we can conclude that the correct answer choice. The small cultures question is a however, if you have any questions, don't hesitate to leave your comment down below in the comment section and thank you for watching

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