Identify A through O:

Question

Diagram illustrating a series of radical reactions with labeled steps A through O, showing chemical transformations and reagents.

Accepted Answer

All right. Hi, everyone. So for this question, let's go ahead and identify the missing products. And here, methyl cyclo pentane is reacting twice to yield two different products labeled A and labeled B. So the product or the missing product labeled A is the result of subjecting methyl cyclo pentane to a reaction with elemental bromine or BR two in the presence of light. So recall that when you react in can like methyl cyclo pentane here with bromine in the presence of light, you get a free radical halogen nation, but specifically a free radical, bro, right? And in free radical bro Nation, you're replacing a carbon hydrogen bond of your Alcaine to in this case, a carbon bromine bond. And like the question implies or rather like the name implies, this proceeds through a radical mechanism. So when it comes to drawing the missing product A, that results from this free radical, bro nation, we have to consider which is going to be the major product or the product that's going to form preferably, right. So here's the thing about free radical B domination, right? Recall that it is highly selective specifically towards the carbon that makes the most stable radical intermediate and radicals followed the same stability trend as carbo ions, right, with tertiary radicals being the most staple, followed by secondary, followed by primary radicals. So in this case, we'll notice that we have only one tertiary carbon in the structure of methyl cyclo ate. So therefore, right, our product a our major product of this free radical bro reaction is going to be an oyl bromide with the bromine at that one tertiary carbon. So product A is my methyl cyclo pentin but with bromine on that tertiary carbon and now we're subjecting product a here or a bromine and we're reacting it with oxide. Now, here's the thing, right? A fox side oxide is a pretty good nuclear file because notice how it has a negative charge indicating that it's quite strong. But the ocho bromide in question is tertiary. So an sn two reaction is not favorable because that tertiary oyl hali creates a STAIC hindrance problem for the nuclei oxide. So instead, right, a oxide is going to behave as a base in an elimination reaction and recalls an in an elimination reaction. Bromine here is a living group and bromine alongside a beta hydrogen gets removed to create a pipe bond between those two carbons. So to review some terminology, right, the carbon that bears the bromine in my ocho bromide is the alpha position or the alpha carbon and adjacent adjacent to the alpha position are my beta carbons and my beta hydrogens are bound to those beta carbons. So if I consider this the symmetry of this ocho bromide here, I'll notice that I have two non equivalent beta positions because I have two carbons inside the ring here, two beta carbons. But they're essentially the same due to this element of symmetry. And then the methyl group here is a second beta carbon. So to understand what product is going to form from this elimination reaction, I have to consider a nifty thing called Zit's rule, right. So if I take my arch bromide here, and I should actually scroll down to give myself some space. But if I consider my ocho bromide here, I can draw two distinct products depending on which beta hydrogen gets removed. Right? So if I start off with the beta hydrogen inside of the ring here, as soon as oxide detonates it, right, the new pipe on is going to form in between the alpha position and the beta position inside the ring, which is going to subsequently remove bromine and have it leave, right. So here's one product. But my other product favors or rather features the same exact mechanism, but with a methyl hydrogen being removed instead of this one inside of the ring. So my second product, my second product would yield the double bond outside of the ring in between the alpha carbon and the carbon that used to be the methyl group. So a fox side a oxide is not that bulky of a base when you consider its size, right? So it's going to favor an alkene product according to Zeit says, rule which states that the more substituted alkene is going to be the favorite product. So when we consider the degree of substitution, right, we have to consider how many R groups are bound to the SP two carbons of each. And when we consider the alkene inside of the ring, we have three R groups bound to both of my sp two carbons. So it is tri substituted. And by contrast, my terminal all keen here is only dice substituted. So because my tri substituted alkene is more substituted, right? It is considered the Zeev product and it is going to be the major product of this elimination reaction. So my missing product be here. My missing product B is going to feature in all Keen inside of the ring after elimination according to Zit's rule. And there you have it. I know it was a lot, but just to recap, right, the first reaction to yield product A was a free radical bro nation that resulted in a tertiary ocho bromite and that tertiary ocho bromine was subjected to elimination with the oxide to yield the zit product, which was the tris subbed. So with that being said, thank you so much for watching and I hope you found this helpful

Identify A through O:

Key Concepts

Radical Reactions

Electrophilic Addition

Nucleophilic Substitution

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