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

Question

Provide the mechanism of the radical reactions shown.

(a) Chemical reaction showing anti-Markovnikov addition of HBr to an alkene, resulting in a brominated alkane.

Accepted Answer

All right. Hi, everyone. So for this question, let's draw a suitable mechanism to show the formation of the product shown below. So here, right, the fact that we're reacting in a with H pr in the presence of hydrogen peroxide is indicative of this reaction being an example of hydro halogen nation. Right? So recall it normally without peroxide hydrogenation results in the Markov Kov addition of the halogen itself. So here, the halogen which in this case, it's bromine, right? Without peroxide, the bromine would add on to the more substituted carbon of the alkene itself. However, right, because this time we do have peroxide and solution, then hydrogenation is going to be. And Tamara Niko, in which bromine is actually on the less substituted position. And the reason for that is because hydrogenation with peroxide proceeds through a radical mechanism. So recall it radical mechanisms have three general steps, right? We have initiation propagation and finally termination. So let's go ahead and see how these three steps would play out. Now, the first step, like I said previously, is initiation in which our radicals form for the first time. So the reason why the presence of peroxide is significant is because recalled peroxides are excellent initiators for radical reactions because the bond between the two oxygens is relatively weak. So here hydrogen peroxide can break apart easily in the presence of heat, right, to generate two hydroxy radicals. So now one of these hydroxy radicals is going to react with a molecule of H pr because when that happens, right, the hydroxy radical is going to take the hydrogen and it's going to yield a bromine radical in addition to a molecule of h2o of water. Right. So the formation of the bromine radical is what proceeds this reaction onto the propagation step because in propagation, right, the radicals that formed in the first step react further with your other reagents to create a chain. Right. So here my bromine radical is going to interact with my key. So here, right, the pi bond of my all keen is going to react with the bromine radical and that in turn is going to create another radical. So here right here recall that radicals follow a similar trend to carbo ions. And here our double bond is in between a secondary carbon and a primary carbon. So in this case, right, the secondary carbon is more stable or rather would make a more stable radical because it's more substituted. And as such, the primary less substituted carbon is going to attach to the Bromy. So now our bromine is attached to the primary carbon. Here is my cyclo propane ring and now we have a secondary radical. However, right radicals can also rearrange just like Carbocaine ions can. So here we're going to take advantage of the fact that our secondary radical is right next to that cyclo propane rig. So here right, this radical is going to rearrange because my cyclo propane ring here is going to break. So here, right, I'm going to create a pie bond and that's also going to break apart the cyclopropane ring and that we'll turn out a new tertiary radical. So here's my new double bond and now we have a new more stable tertiary radical which is now going to react further in my third propagation step. It's going to react further with another molecule of H PR because here my tertiary radical is going to take the hydrogen from HBR and that's going to create a brand new bromine radical. So here here is my product, my organic product and also I end up with a new bromine radical. And now that bromine radical is going to react with another equivalent of my one ethyl, one methyl, two vinyl cyclo propane until finally, there are no more left to react with. Now, at that point, right, when there's nothing else to react with, we end up at the termination step. And in the termination step, recall that two radicals are combining here to form different byproducts. So in this case, my bromine radical, as well as my second hydroxy radical from the first step are going to combine to create Broh No here. Right. It's worth mentioning that other termination steps may be possible. Like for example, two bromine radicals can combine or two hydroxy radicals can combine as well. But the idea is that two radicals are combining until finally, there are no more radicals left in solution and the reaction therefore ends and there you have it. So just as a recap right, the initiation step involved the formation of our initial radicals. And then from there, right, our initial radicals reacted further with our organic starting material which was one ethyl, one methyl, two vinyl cyclo propane to y our anti Markov Kov edition product which formed after rearrangement of the radical itself to a more stable position. Now that in turn generated another bromine radical as a byproduct which kept propagating the reaction forward. Until finally, there was nothing else to react with. At that point, the termination step took over where two radicals combined together until there are none in solution, which therefore ends the reaction itself. So with that being said, thank you so much for watching and I hope you found this helpful.

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

Key Concepts

Radical Reactions

Initiation Step

Propagation Step

Watch next