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

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.

Benzylic cations, anions, and radicals are all more stable than simple alkyl intermediates.

a. Use resonance forms to show the delocalization (over four carbon atoms) of the unpaired electron of the benzyl radical.

Accepted Answer

Everyone. And welcome back in comparisons is simple alcohol radical intermediates. Ben's Ilic radicals are more stable, draw residents forms to show that the localization over four carbon atoms of the unpaid electron of the following Menzel radical. Whenever we are starting with a pencil radical, we have to keep in mind that we are going to represent our residence forms by pairing this electron with the source of electrons from pi bonds and impairing electrons from pi bonds themselves. And we will keep going until we are done with all the all the possible resonance forms. So let's see how that works. Essentially what we're going to do. We're going to just redraw that structure to make sure that our drawings are consistent and we get everything. So if we just retry the first structure, we understand that we have our siege two which contains that unpaid electron. We have our benzene ring which allows us to perform that residence. Right? Because we have that source of pi electrons. This is our conjugated system. Right. We have adjacent linked double bonds and this is how it works. So, we got our first structure, we can label this is one of our resonance forms indeed. Right. And now what we're going to do, we are always going to start with our uncared electron and essentially it will be just paired. So we can essentially use that electron inform a pi bon out of it. As long as we break the adjacent pi bon. So we notice that there is this adjacent pi bon. And we are going to unfair these two electrons, We are going to take one of them and contributed into the formation of that pipe on. So one electron comes from CH two when electron comes from this pipe on. But the remaining electron must be placed somewhere right. And we can essentially place that electron onto that adjacent carbon atom over here. So we will form our radical over here. So these three arrows indicate how we are forming our next residence form. So let's draw it. First of all, we're including our residence arrow. And now if we draw the benzene ring, we understand that. First of all, we will have our pipe on over here. So there will be sieged too. Right now here we will have no pi bon anymore. But we understand that we have used one of the electrons and um paired it, we put it over here to form a new radical. So we will have our electron, it disposition no more pi bon. And the remaining pi bonds are unaffected. So that's how our next form. We can label this as residents form number two and we can continue that process right? We can now think about the result on structure and once again we are going to use that electron and we will try to form a pi bon out of it. But if we took one electron, we need one more. So there is this adjacent pi bon. And from that pi bond we are going to take our next electron. So one electron will be taken from here to form that new pi bon, meaning the remaining electrons from the pi bond must be UNP aired and we will place it onto that adjacent carbon atom. Okay, so, well done. We can now draw our results and radical over here. Let's not forget to include that metal group over here. Okay, so that will be our metal group. And now if you notice what's happening is that we're creating a pi bond at this position. Let's throw that pi bon. And here we are breaking our pi bon were essentially adding an unpaid electron at this position. There is still that remaining double bond. That would be our structure number three. And now we can continue that process. If we use that result in unpaid electron and use one of the pi electrons into the formation of a new pi bond. Right, we need to break that pipe on to release that unpaid electron at this position. Or basically that carbon atom. Right, let's draw it more clearly to make sure you clearly see what's happening. So, if we're using this electron into the formation of a pi bond, another electron from here, right, these two electrons are combined and the remaining electron will be placed at this position. Okay, so let's see how the structure looks like we control our six member drink. We still have our CH two double bond over here. We are forming a new pi bond over here. So let's add it. And here where the methyl group is bonded, we will have that unpaid electron now. So this is where our radical is formed now. And finally, if this is structure number four, we are ready to do our final structure. You may notice that now we are approaching this double bond. So again, we can have another residence form and we can say that this electron will be used in the in the formation of this pi bond while we we will be breaking the adjacent pi bond. So if we break it, we take one of the electrons out of it to create that pi bon, right? And then the remaining electron from that pi bon will be placed onto carbon. And because we are, we have a approach that carbon atom that we started off with. If we are placing that unpaid electron on the same species, that means the final structure structure number five will be our last one. Meaning we are nearly done. We just need to draw the result on structure. And if we do so we can just draw our six members ring. So let's see that that's our 6th member dring. Double bond double bond. This is our newly formed double bond. And now here we have broken our pipe on, we have added one of its electrons onto carbon. So we can say siege two, that's our radical. And this is our structure number five. So these are our residence forms. We can essentially label them as our final answer because in this question, we were asked to draw all of the residents forms. So these are our residence forms and looking at the five residents forms that we got here, we can conclude that the correct answer choice to the small cultures question is B, However, if you have any questions, don't have States, leave a comment down below in the comment section and thank you for watching.

Key Concepts

Benzyl Group and Position

Stability of Benzylic Intermediates

Resonance and Delocalization

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