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.

(c) Chemical structure of a cyclohexene ring with methyl groups at positions 1 and 3, and a double bond between positions 2 and 3.

Accepted Answer

Alright. Hi, everyone. So this question is asking us to identify the carbon in the given compound where the radical is most likely to form in the initial propagation step of free radical halogen nation. So the starting material in question is one isopropyl 33 dimethyl cyclen, one in reacting with N BS or and Chromos Exide with heat in the presence of peroxide. No, we do know that this is free radical halogen nation but we can be a little bit more specific because recall that N BS is crucial for a lilic brominated specifically. Now, a lilic brominated refers to the addition of a bromine atom onto an aic carbon via a radical mechanism. But what does that mean? Right. What is the aic position or the lyric carbon? Well, the lytic position refers to any sp three hybridized carbon that is adjacent to an ache but not part of the ache. That is the lytic position. Now, just like with all radical mechanisms, there are three distinct steps, right? We've got the initiation step, we have at least one propagation step. And at the very end of the process, we have a termination step. Now, the initiation step involves the formation of the first radicals that will then react further with our organic starting material. So in the context of aic brominated, the initiation step involves the formation of our bromine radical. And then right, the first propagation step becomes important because that is where the ali radical actually forms. Because here, right, in the first propagation step, our bromine radical is going to react with an aic hydrogen and it's going to remove the aic hydrogen thus generating an aic radical at the very end. Right. So here is the aic radical that results. And then from here, right, the aic radical reacts further with elemental bromine to yield our actual brominated product. And then the termination step is where all existing radicals and solution combined together so that there are no more radicals in solution, meaning the reaction eventually comes to a stop. But the reason why I'm only highlighting the first propagation step is because we're interested in understanding which carbon of our starting material actually becomes the aic radical that we discussed in the first propagation step. So the question then becomes first of all, how many aly positions are available to us? And if we consider the location of the double bond itself, right, we end up with a total of three possibilities because we have two aly positions inside the ring. And we've got another from the Isocal substituent attached. So if we compare the degrees of all of our possibilities. Here inside the ring, we have a ordinary and a secondary position. And in the isopropyl group here, we've got a tertiary position. Now, if you're in a situation like this, where you have to decide or pick one carbon to create the radical from, your answer is always going to be the carbon that results in the most stable radical possible. However, there is one more thing to consider when it comes to our ordinary carbon here because like we illustrated in the initial propagation step, we drew the aly position, the aic carbon has to have at least one hydrogen because it's the hydrogen that gets removed after interacting with the bromine radical in order to create the aic radical as a consequence. So when it comes to a ordinary carbon, like this one where there are no hydrogens, a radical cannot actually form at that position. So we can go ahead and actually eliminate all ordinary carbons from possibly becoming a radical because we know that cannot happen. So now the tie is in between our secondary and our tertiary positions. Well, recall once again, that radicals follow a very similar stability trend to carbo ions in that primary radicals aren't the most ideal secondary radicals are more stable and tertiary radicals are the most stable among these three. So the more substituted that position is except of course ordinary carbons, the more stable that radical is going to be. So therefore, right, the tertiary aly radical is more stable than the secondary aly radical. So therefore, right, the tertiary hydrogen of the isopropyl group here is the one that's going to be removed. And that is where the radical is therefore going to be. So after the initial propagation step just going to redraw our product here or are studying material actually. But the bottom line is that the radical is going to be on that tertiary aly carbon. That is where our proton is going to be removed once it interacts with our bromine radical, and there you have it here is your final answer. So once again, right, the bottom line is to always ensure that you're identifying the carbon that will yield the most stable radical if you have more than one option to choose from. But 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.
(c)

Key Concepts

Radical Stability

Propagation Step in Radical Reactions

Hyperconjugation

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