In the following molecules, identify the carbon where the radi...

Question

In the following molecules, identify the carbon where the radical is most likely to form in the first propagation step.

(d) Chemical structure of a cyclopentene ring with a methyl group attached to the allylic position.

Accepted Answer

All right. Hi, everyone. So this question is asking to identify the carbon in the structure where the radical is most likely to form in the initial propagation step of free radical halogen nation. And here our starting material is one tuttle six methyl cyclo he one in reacting with N bromo icin amide or N BS for short in the presence of heat with peroxide. Now, the question mentions that this is free radical halogen nation, right? But we can name this reaction a bit more specifically because recall that the reagent N BS is crucial in what's known as a lick Broin nation, right. So in aic brominated, we're talking about the addition of a bromine onto an aic carbon atom via a radical mechanism. But what is the aly position? Right. What is aly carbon referred to? Well, if we have an all keen right, recall that the aly carbon or the aly position is an sp three hybridized carbon that is adjacent to an alkene but not part of it. So what were looking for or possible AIC positions that might receive a bromine atom substituting it, right. So the reason why the initial propagation step is important is due to the fact that we have to recall that every radical mechanism has three distinct steps. We have initiation, we have at least one propagation step. And finally, we have the termination step. Now, the initiation step in the context of a lilic brominated involves the formation of our first radicals that will then react further with our organic starting material. So in this case, right, the initiation step is going to feature the formation of a bromine radical which will then react with our starting all keen because in the first propagation step, right, if I take any starting material that has an aic carbon and that a lilic carbon has a hydrogen, right, our bromine radical is going to go ahead and react with or a little like carbon and it's going to remove that aly hydrogen to yield an aic radical at the very end. Now, I'm going to move this over actually because you won't be able to see the product that I need to draw. So after that, right, the aic carbon has lost a hydrogen and it now has a radical in its place. Now, eventually, right, the termination step involves existing radicals colliding together so that there are no more left in solution and therefore the reaction ceases to occur, right, the reaction ends. But the main focus here is on this propagation step because what the question is asking us about is to identify the carbon in the structure that will become the aic radical after the first propagation step. So that's what we're looking for here. And in order to find that information, we have to look for what aly positions do we have to choose from. And if we consider the location of the double bond itself, we end up with three aly positions, we have one on the top or rather on the top of the ring, we have the carbon that has our methyl substituent and we have the carbon of our tutto substituent our ordinary carbon. So if we have multiple options to choose from, right, the answer to this question is going to be to identify the carbon that would make the most stable radical because they're all alike positions, right? But what I want to bring to your attention is the degree of each one, right? Because all of them are sp three hybridized and we have one secondary carbon, one tertiary carbon and the ordinary carbon of our tuttle substituent. Now, the problem with our ordinary carbon is based on the fact that there has to be a hydrogen on our lytic position because our bromine radical has to remove the hydrogen in order to yield the aly radical. So because the ordinary carbon has no hydrogen, it cannot become a radical. So based on that alone, we can eliminate it. And that leaves us to choose between our secondary and our tertiary carbons here inside the six member ring well, recall it, the more substituted, the more our groups are attaching to a radical, the more stable that radical becomes. So the aic tertiary radical would be more stable than an aic secondary radical. And therefore, it's the tertiary aic carbon that becomes our radical at the end. So here after the first propagation step, we're going to end up with a radical on that tertiary aly position. And there you have it, here is our final answer. And just as a quick summary, right, the first propagation step of any aly brominated reaction involves the formation of the AIC radical. So in order to determine which carbon forms the AIC radical, if you have more than one aly position, you must choose between the most substituted because that would make the most stable radical. So that's why the tertiary aly carbon won over the secondary aly carbon. So with that being said, thank you so much for watching and I hope you found this helpful.

In the following molecules, identify the carbon where the radical is most likely to form in the first propagation step.
(d)

Key Concepts

Radical Stability

Propagation Step in Radical Reactions

Hyperconjugation

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