Propose a mechanism for the bromination of ethylbenzene shown ...

Question

Propose a mechanism for the bromination of ethylbenzene shown below.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to draw the mechanism for the following reaction. Looking at the reaction, we understand that we are starting with isopropyl benzene and we are using one equivalent of roaming in the presence of light. We call that this implies light. This is an equation for an energy of a photon. And we are essentially replacing this hydrogen bonded to carbon with broom in. So that's the net change of this reaction. And the fact that the bombing is used as a halogen in the presence of light. We are forming a radical of blooming. So we may simply recall that this reaction is the free radical halogen ation. Right, because this is a free radical hallucination. We are going to use the basic steps or any free radical elimination recalled that the world first step in a free radical halogen nation is the initiation step. So we are going to call it initiation within the initiation stop. We want to form our radical that starts the reaction bromine is a halogen which contains that roaming roaming single bond. And in the presence of light we can observe that roaming roaming com elliptic bond cleavage, in which the two electrons within the bonds are now on paired. And we are forming two equivalent bromine radicals. In other words, we are forming to most of bromine radicals from one mole of a molecule of grooming. So this is where we have initiated our reaction because we have formed a radical out of our reactant Now in step number two which we call the propagation step, our radical reacts with the starting material. So we may simply draw our starting material. That would be our six member ring. Let's add benzene. And now we have our carbon atom With two methyl groups and hydrogen. Now, even though we are not asked to predict the product, we understand that because roaming is a selective radical in free radical imagination, it always wants to react at the Benz Illich position. That's why we're not actually replacing any of these hodgins within the metal groups because a Benz Illich radical is resonance stabilized so it's more stable and booming prefers to react with the most stable possible radical. Right in this case, that would be the Benz Illich position. We are going to replace this hydrogen in particular and we are going to take the radical form, then the initiation stop, which will essentially attack and simultaneously we're going to break this carbon hydrogen bonds. We're going to pair these two electrons in hydrogen bromide. However, the other electron from that, that carbon hydrogen bond is going to be placed on carbon to form a new radical. You may notice that in this reaction, the world first product that we're forming is hydrogen bromide because the electrons from hydrogen is paired with the election of grooming. So that will be our first product and the second product would correspond to our radical. So we can throw a benzene. Let's do it a bit more clearly so that we don't get confused. That would be our six member ring, our double bonds. And now you may notice that this carbon only will contain three bonds. So we will have carbon with a muscle group on top and on the left hand side. So we are going to add that unpaid electron to our structure. And now it's really easy to see how this radical. It's actually resonance stabilized right? Because it's basically at the Benz Illich position and adjacent to it. We have our pi system. That means it would make sense to draw our resident structures. Let's begin with the first resident structure. And let's see how that works. So first of all, we are going to say that this electron, this unpaid electron is going to be used in the formation of a pi bond because we have an adjacent pi bonds. We're going to use that lone pair. My apologies, that paired electron. And we're going to place it over here and we're going to compare our pi electrons using one of the electrons in the formation of the new pipe on over here. And because we're breaking this pipe on, we are going to place the other electron at theater position. So this is how we get our first resident structure. But that would actually correspond to the second one because the initial one is also considered the resident structures. We can label the initial one is the first one and the next one would be the second one. So let's draw it out. Our six members ring. Now you may notice how we have broken this pipe on. It now appears at the very top carbon, Which now has a metal group. So that's our c. three. And another siege three, another metal group. Now because we have broken the spy bond, we can essentially erase the bond from the ring. And we can clearly show how we have that um paired electron at this position, just as indicated by the arrow. And now this is our second resident structure. Let's think about the production of the next one. So for the next structure, we may simply show that this electron can now form a new pi bon because we have an adjacent pi bon. The electron from that pi bond can be used in the information while the other electron will be added over here to form a new radical. So this gives us our third resident structure. We control our six members ring. We may now notice how we are forming a double bond at this position. Then we are adding the original double bond. Let's add C with two metal groups. And now here we have our unparalled electron. Well done. So this is our structure number three. Let's label it it. Now. Let's think about residents. So we can still keep going because we have that additional double bond. Right? So if we essentially draw the residents form, we may understand that this electron can be de localized over here by pairing it with one of the electrons from the pi bon. The other electron is placed over here to form a new radical. So let's draw the result on structure, our six member ring, two metal groups bonded to that carbon. And now clearly we are forming a double bond at this position so that we have an unpaid electron adjacent to it. That will be our structure number four. And we can still draw one more, notice how we still have that adjacent double bond. So we are going to make another double bond by de localizing the the unpaid electron. Then we are going to take another electron from the pi bond and the remaining electron from that pi bond will go onto carbon to form our fifth structure. So here we are going to draw the result on resonance form Our six member ring. Now we may notice how we have our three double bonds within the ring. And then here we have carbon with an unparalleled tron and two methyl groups. So that would be CH three C three. Okay, so that will be our fifth residence form. And now from here we can essentially think about our next step, let's suppose that we simply analyze this resonance structure because elimination occurs at the Benz Illich position. So we are going to use either this form or the first one whether you use the first residence form or the fifth one. That's up to you because the only difference is that we have a different position of double bonds within the benzene ring and that's perfectly fine. So now what happens in this step is that you need to make sure that you use a booming molecule because you're not yet terminating your reaction. So because you have a radical you want reacted with a molecule, meaning we're just observing the formation of our carbon booming bond by the attack by the radical. Then we are impairing our booming booming electrons. One of them is used in the formation of carbon booming bond and then we are forming our booming radical and this gives us our final product. So we can throw a Benzene and then we may notice that we get carbon, we are creating that carbon roaming bond And let's add our metal groups C3 C. Swing. And then our next product is brought me in radical. Now because we have got our final product, we simply need to terminate the reaction because we obtain the product that we needed. And there are multiple termination steps. We can combine any two radicals. So there will be several opportunities, at least three of them. Right? We can either combine these two radicals together or to be more blooming radicals together or a booming radical. And this radical actually let's suppose that this one because those were just residence forms of course it's possible to bond them to booming. But generally we want to make sure that we're thinking about the residents form that actually participates in the reaction based on our mechanism. So we have at least three possibilities we can go with any of them. And let's suppose that we are just using the combination of two bromine radicals into a molecule. So that would be our step number three, the termination step. And just as we said, so you brought me in, the radicals will essentially combine into a molecule that would form B R. T roaming. Now the reaction is over right. We have our termination stop. We may notice that we indeed got our found product in the final step of propagation. And from here we have seen all of the steps of the mechanism based on the mechanism. We may conclude that the correct answer to this multiple choice question is d However, if you have any questions, don't hesitate to leave your comment down below. Thank you for watching

Propose a mechanism for the bromination of ethylbenzene shown below.

Key Concepts

Electrophilic Aromatic Substitution

Radical Mechanism

Regioselectivity

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