Predict the products of the following reactions.
(c) 2-...

Question

Predict the products of the following reactions.

(c) 2-methylpropene + NBS, light

Accepted Answer

Hey everyone and welcome back into this video. We are going to solve the following problem through the structures of all expected products of the following reaction in the given reaction we are treating and Elkin with NBS in the presence of flight and we may recall that this is a free radical mechanism due to the fact that we have light present here as well as N. B. S. Right. So to particular right products. We first of all want to identify the positions in our structure. In particular if you notice the position of the level bond, the adjacent carbon atoms are called a lilic carbon atoms. So we have to A little carbon atoms position # one in position # two. What we are going to do we are going to take the brahmin radical formed in the initiation stop and we will form two radicals. So if we take a look at position number one in particular we made through our six member ring and our goal is to remove that hygiene atom to produce a radical. So what happens is that the romaine radical formed in the initiation stop will attack We will have the harm elliptic cleavage of that carbon hydrogen bond. So that one of the electrons from here we will will form the hygiene roaming bond and the other electron will form the radical that we need to get. Now the radical that we're getting here. Yes, the following. So there will be an unpaid electron at that position. And what we may notice is that this radical is residents stabilized. It's the presence of the adjacent pi electrons. So we may draw another radical out of it. If we simply paired these electrons one which was unfair did with the one from the pi bond and then the remaining electrons from the pi bon will be placed at this position to form a new radical. So this would be our residence form. Right? We may say that we get our resonance stabilized radical and the second form would now have a double bond at this position and there would be an unpaid electron over here. So these are two possible radicals that we can get if we remove hydrogen from position one. But what if we remove that hydrogen from position to now? If we redraw the original structure And if we expand that implicit hydrogen at position number two. As we previously sad. The blooming radical from the initiation stop will attack we will have the harm elliptic cleavage of the carbon hydrogen bond. One of the electrons will form the hydrogen rooming bond and the other electron will form that radical. Let's draw that radical. Now now at this position we have an unpaid electron. And similarly to the previous case, we may have resonance. Right, so we are going to show that residents structure simply by pairing these electrons, one from the radical one from the pi bond and the other electron will be placed at that position. Now essentially we get our residence form with a mako substation over here. The position of the radical is here and then we have a double bond. Okay, so now we have to evaluate these structures and we have to think whether we can get four different products or do we really have some of these radicals being equivalent? And in particular, if you evaluate these four structures, you may notice that these two residents forms are equivalent to each other. These two the line of symmetry. Right? If you draw a vertical line of symmetry essentially you will notice that in one case the double bond is on the right, on the other cases on the left but we are going to produce exactly same structures out of them because they're equivalent structures. So we may label Our radicals as one T. We and this is also three as we previously said, they have an internal line of symmetry. The position of the double bond just goes on the other side of that line. But we essentially have two equivalent radicals. Now because we have three different radicals, we will produce three different products. Right? So let's see what kind of a product we would get from the first radical we mean out. We may now read troll The first radical. So that will be our 6th member ring. That's structure number one. We have an unpaid electron. The double bond, this metal group. Now the booming molecule formed in the initiation step. Right, because there's a trace of hydroponic asset which reacts with NBS to produce the BR two molecule. So we may simply indicate the presence of BR two. We will expand the booming booming bond. Now essentially we are forming one of our products by pairing this electron with the electron from the booming booming bond, forming another radical of grooming. So this would be our first product in which we get that six member ring. We will add the metal group, the double bond. And now at this position we will have roaming. Well done. So that's our first product. If we look at the second radical number two, we will just redraw it. Now the uncared electron will react with a molecule of roman. There will be an attack. These two electrons will form a bond and we will also form the booming radical. Let's draw our final product for this one. So that's our six members ring the metal group, the double bond. And now at this position we will be adding the carbon bromine bonds. So let's simply draw a booming at this position. Great job. So now let's look at structure number three. Number three, If we redraw the six member ring. Now we have a metal group, there's also a double bond that position and there's an unpaid electron. Now another molecule of roaming approaches the radical attacks, we are forming a new carbon booming bond as well as an additional grooming radical. So if we draw the main product we get our six member ring, the methyl group, the double bond and alb romaine will be bonded at that position where the unpaid electron was. So we may now state that these three are our products. Let's label them as our final answers, and we may now conclude that the correct answer to this multiple choice question is B. However, if you have any questions, don't have states, leave your comment down below in the comment section and thank you for watching.

Predict the products of the following reactions.
(c) 2-methylpropene + NBS, light

Key Concepts

Allylic Bromination

N-bromosuccinimide (NBS)

Radical Mechanism

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