Show a reaction coordinate diagram for the two processes in Fi...

Question

Show a reaction coordinate diagram for the two processes in Figure 13.41 that rationalizes pathway B as the one that gives the major product.

Accepted Answer

Welcome back, everybody. Let's take a look at our next problem. And you can see by the fact there's a little image at the bottom of the screen. This is a really big problem physically, it doesn't fit all the way on the screen. So we're going to have to look at the first part and kind of scroll down, scroll back up. So a little bit challenging here. So our problem says, consider the two pathways for the reaction and we start with one reagent, which is a cyclohexane ring, it has an alcohol group on it and then a ethyl group with another alcohol group. So we essentially have a, what's called a vicinal diol A dial. So two alcohol groups on adjacent carbons, one that's on the ring and one that's on this branch. And we show this reagent being pronated with two possible pathways. Either the top alcohol on the chain can be proton or the bottom alcohol that's on the ring can be proton. So pathway A has the one on the chain being proton. So we see in step two, it's been converted into 02 plus and then that becomes the leaving group of water forming a carbo cion where the positive charge is on the first carbon. In that two carbon chain. In pathway B, the alcohol group on the ring is proton, we see it leaving as water and forming a Carbocaine where the positive charge is on the ring. And then in our box, it says of the Carbocaine and pathway A that is secondary Carbocaine is less stable in pathway B under the carbo cannon. It says tertiary carbon is carbo cion is more stable. And then we see equilibrium arrows between the two carbo CS that says hypothetical equilibrium. Then down at the bottom on pathway B on that picture of the tertiary Carbocaine, we see a hydride shift where hydrogen is shifting from the first carbon on the chain to the positive carbon on the ring and that leads to a positive charge. So another Carbocaine rearrangement that hydride shift, the positive charge is now on the first carbon on that chain. But because it comes from that tertiary Carbocaine, we still have an alcohol group there. So it's still a tertiary Carbocaine. And then we see it stabilized by resonance. This is a resonance structure where some one of the lone pairs from oxygen comes down and makes ac o double bond. And then that second resonance structure we see gets deproteinate by water and we have the formation of a ketone. So this is the reaction that is a P A call rearrangement where we see a dial becoming a ketone or an aldehyde depending on its position in the molecule. But essentially the dial is converted, one of the alcohol groups leaves and the second one ends up forming a carbon group. So we have these two pathways, let's say, look at these two pathways, the formation of these two different carbo canons. And then the question says, does the following reaction coordinate for the boxed portion of the two pathways rationalize pathway B as the one that produces the major product. So then we will scroll down to see our reaction coordinate. And we see it's an energy diagram with the vertical axis as increasing energy, horizontal axis as the reaction progress over time, we see we start at the same point with the reactant and we have a purple pathway that's pathway A with the secondary carbon carbon Carbocaine intermediate and a green pathway that's pathway b our tertiary Carbocaine intermediate. And we notice our purple pathway reaches its first peak a little sooner. So that would be sort of the first energy barrier and then it dips down a little bit. They're at the same energy level though just pathway A reaches that point a little sooner, then they have the same point of lower energy and then they again build to a second peak with pathway A, the secondary Carbocaine being higher in energy than pathway B. And then they come down at the same time to their lower energy state with again, A being a little higher than B. And then our answer choices are a, the reaction coordinate diagram does rationalize pathway B as the one that produces the major product or choice B. The reaction coordinate diagram does not rationalize pathway B as the one that produces the major product. So let's think about, we know this diagram is just talking about the boxed portion of these two pathways. I'm going to scroll back up. We're going to look at the box portion for a little bit before we go back to our diagram. So what are the steps that are encompassed in this box portion? Well, of course, we have initially the protonation of one of our alcohol groups. That's step one. So will there be a difference in how this step takes place? We start with the same reagent? So actually, I'll put that up here first, same reagent. And we noted that in our diagram that is represented, we start at the exact same point. So is there a difference in the protonation of one of these alcohols? Well, energetically, it will be the same but protonation of the one on the chain. So the protonation in pathway A actually, I'm just going to write A, that'll save me some space there perton in A will be faster. Well, why is that? Well, A is out on this chain, but the alcohol group in pathway B is on the ring, it's got this giant six carbon ring next to it. So there will be less, we'll say due to less stric hindrance, it will be easier for this reaction to take place. Due to the molecules the being able to get in to that alcohol better, those electrons to be able to move over, you just have less stuff in the way fewer electron clouds. So the protonation a will be faster but the intermediate in this case. So the proton alcohol will have the same energy level. And the second step of the leaving of the water group, which should be the same, there shouldn't be a difference there. So step two, water leaving should proceed summly for both. So water leaving now we have the formation of the carbo cion. But what do we know about that carbo Canion that the secondary Carbocaine is less stable? So the A Carbocaine has a higher energy level then the B Carbocaine. So that is all that's going on in that boxed region. So does our energy diagram reflect that? So we'll scroll down again, trying to leave my little steps in the screen. So we can see that again, we start with the same reactant and we do see that step one proceeds faster on pathway A, we see that that purple line reaches its peak a little earlier than the green line for pathway B. So that does reflect what's going on. Then we see that that intermediate is at the same energy level. And that, then they reach that point where they form the Carbocaine at the same time again, makes sense that that leaving of water, there's no difference between the two. But then at the end, we see that the a secondary carbo cion is indeed at a higher energy level than the pathway B Carbocaine. So yes, our reaction coordinate does rationalize pathway B since that carbo cion is overall at a lower energy level. So the more stable Carbocaine will be preferred and therefore lead two more product. So our answer here will be that choice. A the reaction coordinate diagram does rationalize pathway B as the one that produces the major product because it shows that with a more stable intermediate that will be the more favored pathway and therefore will generate most of the product. See you in the next video.

Show a reaction coordinate diagram for the two processes in Figure 13.41 that rationalizes pathway B as the one that gives the major product.
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Key Concepts

Reaction Coordinate Diagram

Transition State Theory

Reaction Mechanism

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