Two of the carbocations in Problem 6-42 are prone to rearrange...

Question

Two of the carbocations in Problem 6-42 are prone to rearrangement. Show how they might rearrange to more stable carbocations.

Accepted Answer

Everyone. And welcome back in today's video, we are going to solve the following problem. So how the following carbon Caroline's rearranged to more stable carbon Caroline's? Of course, we have two problems, but let's just discuss rearrangement. In general, you generally expect two types of rearrangement. That would be one comma two hydride shift, right? So this requires us to have a source of hydrogen atoms. And number 21 comma ti al kill shift generally that requires us to have a muscle group. But variations are possible right now, let's think about these and let's apply our knowledge to problems one and see starting with number one, we are going to draw that one comma four di substituted car broke Orion. So this is how it looks like as the name implies. If we look at the T concept one comet two implies that we find the position of the carbo Karen. This is a secondary carb, okay Ryan, because it has two adjacent neighbors over here and here. And we are going to look at the nature of those neighbors if we look at the two substitution. So we're here, we have a metal group and hygiene and if we look at the other adjacent position, there are two high jeans, we are going to think about the shift of each. What if we shift this hygiene over here, we will then get another secondary carbo Karen, which will be exactly the same carbo Karen as before. So definitely it doesn't make it any more stable and even not a unique car Bocaranda be exactly the same Carbo Karen. So we are definitely not going to use any of these high jeans. However, if we look at this position over here because it has a source of hydrogen, we are definitely going to shift Haijun towards here. We're not performing a metal shift because shifting a metal would form a secondary carbo Karen. So we don't want that we want to form a more stable carbo Karen. So assuming that this is true, let's see why that works for us. We have one comment to hydrate shift. And if you look at the cardboard, can I informed, you may simply draw that structure once again because you're shifting Haijun, the metal group stays hydrant goes here. So the charge is neutralized. But due to the lack of these bonding electrons, we will now have a positive charge at this position. And you may simply notice that this car bull can Ryan. Yes, terrace Sherry because this carbon Karen has three neighbors, 123, right? So it's more stable and it's favorable recall that we always want to form more stable Carbo Karen. So if we start with a secondary one, we want to form a third share, we want out of it. So that's the only possibility for us after evaluating the two positions. OK. Well done, we can say this is the rearrangement scheme from the original Carbo Karen one to a more stable tertiary Carbo Karan. If we look at number two, we have a positive charge at a different position. The Kabul Karan looks a bit similar, but it's actually at the siege to that methylene group over here, right? We're going to thank in terms of hydrate and alcohol shifts. So let's find those adjacent positions. Okay. So one of them is basically over here, right? We have hydrogen over here. So let's expand. That can essentially say that there's that methylene with positive charge hydrogen. Okay. So now here's what we're going to do. We can either think about this hygiene or an L kill shift. That's really important when you think about this carbon, you can also try opening the ring to see if you can form a more stable carbon Karen. So there are two pathways here. Let's start with an easier one. Previously, we said this one was one comment to hydrate shift. Okay. So we will make it clearer for us. For this one, we will look at that high tried shift as well, right? We have that adjacent carbon atom that contains hygiene. So if we look at one comet to hydrate shift. Again. Chef tried basically what we're going to do, we are going to transfer. That's hi Jen to our Carbo Karen to form a neutral metal group, right based, it will have no charge. And then if we're transferring hydrogen out of this carbon, we're removing these bonding electrons and the positive charge will appear over here. Now, start having started with a primary Carbo Karen because there's only one neighbor, we are now forming a tertiary carbon Karen, which looks exactly the same as the previous swan. So we get this car Bull Canton, which is their chivalry. So it's definitely more stable, that's favorable. But that's just part of the story. Let's look at the other possibility. If we have a ring nearby, we can definitely open that rain and see if a more stable carbon Karen can be formed. So we are going to perform another shift, which is called the al kill shift. And we are going to open the ring at that position from that carbon atom. That would be one comment to al Kill shift because we can actually open the ring and change the position of that alcohol group. We can essentially say that this alcohol group goes here. So this carbon gets incorporated into that six member dring to create a seven member ring, right? We are going to break that carbon carbon bond and we are going to bond carbon if we label them we can say that's carbon 123456. For complete clarity, we're going to form a bond between carbon number two and carbon number suppose seven. Okay. So carbon two and carbon number seven will now make a bond to give us our seven member rings. So let's label this is our second pathway one. Comma two. I'll kill Chef tribe. Now we went to draw our seven member dring. Let's see that. So we can essentially draw, let's make it a bit clear. Okay. Okay. Let's count the number of carbon atoms. 1234567. Well done. And now we notice that during this shift, we will still have a method group at that position number four. So let's suppose we label our carbon atoms 1234567. Now if we look at our positions, position four must have a metal group, right? Position number two forms a bond with position number seven. So actually for complete clarity, we can just label it a bit differently. Let's change our labels to make sure that we are consistent, right? So if we just start with the metal group, suppose is basically at this position, that's our metal group. And we understand that this must be positioned for if we go counterclockwise, that's position three and that's position to which is now bonded to position seven. Okay. So to avoid confusion, we can just do this way. Now position seven is bounded to position one, then we have six five and that's it, right. So we now clearly see that carbo cat ion will now be formed at position number one, because we have a loss of these carbon carbon bonding electrons. So that's where the carbon Karen will be observed. We can essentially add that positive charge at position number one. And this is a secondary carbo cast iron, which is also favorable because it's more stable than the primary one, even though tertiary one is more stable. So we can say that this is the predominant form. That's why I'm adding this inequality sign to say that we expect to see more of that one because it's more stable than the secondary one. But of course, if we can get a secondary carbo Karen out of a primary one, we can do that that's favorable as well. So we can simply say that this is our complete rearrangement scheme and the correct answer choice. This multiple choice question is B however, if he and this problem challenging, don't worry and leave your comment down below. In the comment section, we will try to help you out as much as possible. Thank you for watching this video.

Two of the carbocations in Problem 6-42 are prone to rearrangement. Show how they might rearrange to more stable carbocations.

Key Concepts

Carbocation Stability

Rearrangement Mechanisms

Reaction Pathways

Watch next

Two of the carbocations in Problem 6-42 are prone to rearrangement. Show how they might rearrange to more stable ­carbocations.

Key Concepts

Carbocation Stability

Rearrangement Mechanisms

Reaction Pathways

Watch next

Two of the carbocations in Problem 6-42 are prone to rearrangement. Show how they might rearrange to more stable carbocations.