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.

b. Toluene reacts with bromine in the presence of light to give benzyl bromide. Propose a mechanism for this reaction.

Accepted Answer

Hey everyone and welcome back in today's today, we are going to solve this problem. Been Zilla carry ions and ions and radicals are all more stable compared to simple alcohol intermediates. Two bromo two Halloween reacts with bromine in the presence of light to give a benzel bromide derivative proposed a plausible mechanism for this reaction. And if you look at this reaction, we see that the alcohol chain or in particular the metal group to be more accurate one of the hydrogen atoms and that has been replaced with bromine. And we understand that this resembles the free radical mechanism and this is what we are going to explain today. So if you recall the free radical mechanism or in this case the free radical imagination, because we're using roaming implies three steps initiation, Right? That would be our step one propagation, Our step # two and termination Step # three. So let's look at each of them and suggest a plausible mechanism. So starting with number one, initiation, we already know that during the initiation stop roaming. That is present in the mixture undergoes the hom elliptic cleavage of that roaming roaming bond. And what that means is we are using light. This is very important because life promotes that hom elliptic cleavage. We are essentially taking these romaine romaine bonding electrons and we are in pairing them. One of the electrons goes to the left roaming and another one goes to the right grooming. So because they are equally given to each of the two bromine atoms. We call this a harm elliptic bond cleavage and we are forming to bromine radicals. So that would be two bromine radicals. Each bombing has an unfair the electron. And this is where we are ready to look at the actual reaction because we have formed our radical. So now we are moving on to our second step which is propagation During the propagation is that this is where it's important to use our substrate. If we look at our substrate we can just redraw it and we will expand that carbon hydrogen bond because hydrogen is the one that is replaced with booming. Right? Notice how we have ceased to be our here instead of C3. So that means one of the hydrogen atoms is replaced with booming and we may simply draw our substrate. So that would be our benzene ring bro. Mean and here we can expand it, we can say that this is these two and then there is hydrogen bonded to it. Okay, so now what we're going to do, we're going to show how the radical form than the previous. Stop One of the bromine radicals attacks and we have another home a little cleavage over here essentially we're breaking that bond. One of the electrons from that bond will be used in the formation of hydrogen bromine bond and the other electron will be placed onto that adjacent carbon atom. Right, So we are unp airing these carbon, these carbon hydrogen electrons. Now if we look at the products formed, we first of all get our radical That would be our CH two group because we have removed hydrogen with an unpaid electron. Right. We also understand that its residents stabilized because there are adjacent pi electrons. So the radical can be de localized or basically this unpaid electron can be de localized over that ring. But we are not worried about that. We can just say that its residents stabilized. We don't have to look at the residence forms. Okay so when we have formed that we are ready to go and we are going to look at what happens to that radical. But before that we can also say that the other product formed here is H P. R. And now if you recall the radical that we have formed. Right so we are drawing out radical And now what happens here is we have to take another molecule of roaming because radicals will react with molecular species. We don't want to terminate the reaction yet. So we are going to draw a bro mean the radical essentially attacks one of the electrons is paired with it. So we are going to have that carbon booming bond formed and the remaining electron from that booming booming bond will be placed onto theater booming to form roaming radical. And this is how we get our product that would be sieged to be our that's what we are forming. We are using that unpaid electron from carbon one electron from the left blooming and we are forming carbon bromine bonds. So that will be our siege two B R. Which corresponds to the final product plus bro mean radical form. Okay, so now that we got our final product, we can essentially state that this is our final product. We are ready to think about the termination step because we are ready to stop that reaction. And there are as always multiple termination stops possible. So number three is the termination stop. And you have multiple possibilities here. The first possibility is to use two blooming radicals and essentially pair their electrons into a molecule. So that would be Bro mean molecule. Right, so this is our possibility. Number one possibility, number two is to think about the other radicals that we have in that mixture. Noticed how we had this radical. So we can essentially read through that radical. It can essentially combine with the radical of grooming and the two electrons will be paired to form the final product. Right? That's what we will get here. Siege to be our Okay. And the final possibility Thinking about the other radicals we had is just to use these two radicals combining together. So let's see how that looks like if we combine these two radicals which is our third and our final possibility because there are no more other radicals formed. We will have CH two right And we control another as a mirror image so that we can visualize it more easily stare B H two C. And now what happens is we are pairing their electrons forming a bigger compound, right? We will have two benzene rings in that molecule. So let's see how that looks like we can essentially drool one benzene ring from the left right. We are going to have roaming over here. Now here we will have siege two bonded to siege too. So that our siege to another siege to and that siege to group is bonded to benzene ring. So let's draw a benzene ring over here. Okay, and now here we will have grooming because it's the adjacent position. If you think about it, we are essentially flipping it a bit. So bro mean goes up now and we got our final possibility for determination steps. So that would be our additional product for determination stuff. So once again there's three possibilities because we have two types of radicals meaning you can always combine the radical with itself just as with it in one and three right or two radicals too different or too unique radicals into a single structure which gives us the final product. So we got our mechanism. Now looking at the mechanism, we can simply conclude that the correct answer choice to the small pool choice question is a However, if you have any questions don't have the slave, your comment down below in the comment section and thank you for watching

Key Concepts

Benzylic Position

Stability of Benzylic Intermediates

Free Radical Halogenation Mechanism

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