For the following reaction, answer questions (a)–(d).

Question

Chemical reaction: square A plus Br2 under heat yields square with two Br atoms attached.

(a) Give an arrow-pushing mechanism, including the initiation step and two propagation steps.

Accepted Answer

All right. Hi, everyone. So for this question, let's go ahead and draw the arrow pushing mechanism including the initiation and the two propagation steps. For the following reaction here we have 14 dimethyl bicycle, 220 hexane reacting with BR two in the presence of heat to produce 14 di bromo 14 dimethyl cyclohexane. So our starting material or at least one of them is an al cane and we're reacting this all cane with BR two to produce an ocho bromide. Because of this, this is going to be an example of free radical halogen nation because this follows a radical mechanism. Recall that there are three distinct steps or phases. The first of which is initiation in which the first radicals are formed. In this case, recall that the bond between the two BROM means and BR two is relatively weak. So the addition of heat is going to provide enough energy to break the bond between both bromine atoms resulting in two halogen radicals. Now, the second phase involves the propagation steps. So the propagation step or steps reference how a radical mechanism involves a chain reaction. So first, the halogen radical creates an ocho radical after interacting with the o cane and then the ocular radical is going to react with another molecule of the halogen thereby forming the cul hide. Now, our third and final step is termination and so termination involves a series of reactions where the radicals in solution combine in different ways to be consumed, right. So that there are no more radicals in solution. So even though a radical mechanism involves these three main phases, we're only going to focus on drawing the first two that's initiation and propagation. So first, let's start off with initiation. So for initiation, we're going to start off with one molecule of BR two, as stated previously, the bond between both bromine atoms is relatively weak. So because of that, the addition of heat is going to provide enough energy to break that bond. So when that bond breaks between the two bromine atoms, each bromine atom is going to receive one electron from the covalent bond itself, notice how I'm drawing fishhook arrows going towards each prom to represent the one electron going towards each of them. This concludes the initiation step because now we have two bromine radicals with one unpaired electron. So now that we've initiated this process by creating our first radicals, let's go ahead and talk about propagation one or the first propagation step. So first, let's go ahead and take one of our bromine radicals and combine it with the. So for the first propagation step. The alkane is going to interact with our bromine radical to produce an akill radical. But the question is which of the bonds in the al cane is going to break? Now, in this particular case, we do have one carbon atom creating a bridge in our structure due to the strain associated with this one carbon bridge. The molecule itself is going to be relatively happy to relieve that strain and break that bond though a slight correction because earlier I mentioned that the bridge contains one carbon. But it would be more accurate to say that the bridge is composed of one carbon carbon bond. But regardless to alleviate the strain associated with that bridge, this is going to be the carbon carbon bond that bridges. It's that bridge. So first are bromine radical is going to pair with one of the electrons from that bridge to create a new bond. Subsequently, the remaining electron of the bridge is going to go towards the opposite side to create a tertiary o radical. So what were the carbons of the bridge or the former ridge is going to have a new bond to bromine whereas the other becomes a radical. So now if I scroll down to open up some more space, the cul radical can react further in the second propagation step. So now lets talk about propagation too. So for propagation two, I'm going to start off by copying the radical that formed in the first propagation step. The cul radical is now able to interact with a second molecule of BR two. Once again, the bond between the two bromine atoms is going to break. But instead of creating two bromine radicals, we're going to end up with one bromine radical and an all kill die hali because the radical is going to pair with one of our bromine atoms using one of the electrons in the bond between the bromine atoms. Whereas the other bromine atom receives the remaining electron and becomes a radical. So that's how we get at the end, a dye bromide with an additional bromine radical. So before we go ahead and talk about our conclusion here, let me just try and make the second step smaller so I can make space or our bromine radical. There we go and there you have it. Now, once again, notice how radical mechanisms require fish hook arrows to describe the movement of one unpaired electron. At this point, our task was not to draw the termination steps, but for the termination steps, you can combine any of the radicals that appeared previously in any way. But here we have our initiation and two propagation steps. So if you watch this video all the way through to the end, thank you so very much for watching. I appreciate it. And I hope you found this helpful

For the following reaction, answer questions (a)–(d).

(a) Give an arrow-pushing mechanism, including the initiation step and two propagation steps.

Key Concepts

Radical Halogenation

Arrow-Pushing Mechanism

Cyclobutane Reactivity

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