Provide the mechanism of the radical reactions shown.
(...

Question

Provide the mechanism of the radical reactions shown.

(b) Chemical reaction showing allylic bromination with NBS and heat, resulting in bromine addition to a cyclohexene structure.

Accepted Answer

All right. Hi, everyone. So this question is asking to provide a suitable mechanism to show the formation of the product shown below. And the transformation in question or reaction in question involves E 39 dimethyl on Decca 18 dien reacting with N BS and heat to generate 22 bromo B 10, 2 L one methyl, one vinyl heine. So recall really quickly that N BS is the primary reagent of a process known as a lili bro and in allylic, bro, right, a bromine atom is being added to the allylic position. In other words, recall that the ay position is an sp three hybridized carbon adjacent to an all not part of it, right? But next to it. So here a carbon hydrogen bond within the allylic position is being replaced with a carbon bro bond. And a lilic pro nation proceeds through a radical mechanism that has three general steps, recall that they are initiation propagation and termination. No recall that N BS also reacts with very low concentrations of BR two and HBR because if they were present in high concentrations, then addition would compete with a lilic bro. But with that being said, let's go ahead and proceed with the mechanism itself, right? The first step, like I said previously is initiation in which your first radicals are beginning to form. And when we have and BS, right, the initiation step is going to involve RN BS which stands for N bromo cide. So here, the very first step in the process is going to involve the bond between nitrogen and bromine splitting into two radicals, right? One on nitrogen and the other on Bromy. Now, the bromine radical is what's especially important for our purposes because the bromine radical is what's going to proceed with the substitution of our or our diet, right? So here our bromine radical is going to react further to generate new ones in the process and create a chain reaction known as propagation. So now I'm going to draw, are dying once again, I missed the carbon. So let me just add it. So now I'm just making sure that I'm keeping my structure consistent before I proceed with the next step. There we go. So now, now our bromine radical is going to interact with one of the allylic positions. But the tricky thing about this is that here we have two alkenes in the same molecule, which means that we actually have a few different allylic positions to consider, right, our first double bond is in between carbons, one and two of the parent chain, which means that carbon three is one allylic position that does have a hydrogen. Now, on the other hand, right, there's another double bond in between carbons eight and nine of the parent chain, which means that either carbon seven is an a lilic position and so is carbon tent. Now, what I want to bring to your attention, right is the fact that carbon seven and 10 are both secondary whereas carbon three here is tertiary. Now recall that radicals follow a similar trend as carbo ions when it comes to stability, primary radicals are less stable, followed by secondary radicals, followed by tertiary radicals which are among the most stable. So here, right, this tertiary position is going to make the better radical and therefore there is where it's going to be so always consider multiple possibilities. But now that we've cleared that up, right, our brome radical is going to go ahead and take the hydrogen from this tertiary allylic position to make its corresponding radical. So here I'm going to actually move this over to see if I can give myself some more space. There we go. So here I'm going to draw my 11 carbons once again and I ran out of space. So instead, right, I'm going to make my first step a little bit smaller. I don't want it to be too cramped, but I also need more space than I have. Here we go. So that's there we go. And now I can go ahead and proceed with drawing my now a lilic tertiary radical. So now, right, what's worth mentioning is that our final product is not a straight chain instead, right, we have the cyclohexane ring structure. So what I want to bring to your attention is the fact that if we were to count the space between carbon three, which is the radical and carbon eight, which is your second double bond there. What you'll notice is that there are 123456 carbons in between, including both the radical and carbon eight of the parent chain. So that means that the radical is going to react with the second alkene to generate a six member ring and there is your second propagation step. So here I'm just going to draw the, the starting material or intermediate a little bit differently because here you can see that the radical is going to react with the pie bond of the second alkene to form a new bond and therefore close the rank. And that as a result, right is going to form a six carbon ring like we see in the final product and also a new tertiary radical. So now are tertiary radical is going to go ahead and further react with a molecule of BR two which is in solution but in very low concentrations. So here is BR two and now here, right, our radical is going to combine with one atom of bromine, the yo bromide. Here's our alkyl bromine with HBR and another bromine radical as byproducts, right. So the importance of this step is that we've regenerated a new bromine radical. And the reason that's important is because the bromine radical is what kick started the propagation step in the first place, right. So now my bromine radical is going to react with another molecule of E 39 dimethyl on Decca 18 dien and carry this process over and over again like a chain reaction. Now, at some point, right, there will be no more molecules of starting material to react with. And at that point, the reaction ends and we can introduce our final step which is termination. Now termination like the name implies, right involves two radicals combining together to end the reaction because once there are no more radicals, then this reaction will no longer proceed. So now, for example, right, two bromine radicals can combine together and create BR two. But it's worth mentioning that there are other possible termination steps like for example, two Ocu radicals can combine or one bromine can combine with one acu radical, one bromine radical to an A radical. But here you have it right. So today, what we talked about was a lilic bro using N BS and like and the other radical mechanism we proceeded through three distinct steps. The first was initiation in which your radicals formed for the first time followed by propagation in which your first radicals combine with other materials in solution. To create new ones that will create a chain reaction. Now, once the chain reaction has been exhausted, right? And there are no more molecules to react with two radicals combined together in the termination step because once there are no more radicals in solution, then the reaction comes to an end. So with that being said, if you stuck around to the end, thank you so much for watching. And I hope you found this helpful.

Provide the mechanism of the radical reactions shown.
(b)

Key Concepts

Radical Formation

Propagation Steps

Termination Steps

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