Provide a mechanism for the following E1 reactions.
(c) ...

Question

Provide a mechanism for the following E1 reactions.

(c)

Accepted Answer

Welcome back, everybody. Let's take a look at our next problem. It says, what is the E one mechanism that corresponds to the reaction shown below? And we have a reaction where we have one reagent. It's a cyclops ring on one of its carbons, it has a methyl group. And what's labeled as ad with the methyl group as a dashed wedge D is a solid wedge. And on the carbon adjacent to it, we have a solid wedge leading to a chlorine. It reacts with water and the product is a cyclen ring with a double bond between those carbons that have those substituents leaving one of the carbons with just a methyl group and one carbon with ad on it. First of all, we have to remember what D equals and D equals deuterium, which is an isotope of hydrogen. That's two neutrons. It's a little bit heavier and it is often used in reactions where you want to track what is happening to a specific hydrogen, you can tell it apart from other hydrogens. So when you want to see what's happening to the stereo stereo chemistry or if you're having hydride shifts, you can use this deuterium to label a specific site. So that's what's happening here. We seem to be seeing the results of a hydride shift because the deuterium was sharing a carbon with this methyl group. It's now on the adjacent carbon where the chlorine was. So let's think about what mechanism might make this happen. Well, we have an E one mechanism. So we know we'll have the leaving group leaving first, thereby forming a Carbocaine intermediate. And that makes sense because we can see we've had a rearrangement here which we see with carbo cds, we don't see that with E two mechanisms since they have to be concerted and happen all at once. So let's think about that first step of our E one mechanism, which would be the leaving of the leaving group. So looking at this molecule, it's pretty obvious that our leaving group will be chlorine. That's the one thing here. That's a really good leaving group. So the atoms from the CCL bond will go on to the chlorine. So we will then produce our cyclops ring. It still has the methyl group and the deuterium. And then on the adjacent carbon where the chlorine was, we now have no substituent and a positive charge. So here's our formation of our carbo cat I am. So now again, whenever we form these carbo cat ions, we need to look and say, is there a possibility of forming a more stable Carbot ion? Well, we have a secondary carbo cion here. So if we can make a tertiary carbo cion, that's even more stable. Well, we do have that possibility because we've got a methyl group on the adjacent carbon. So we will undergo a 12 hydride shift here. And we see our carbon with the methyl group also as this deuterium, it can act just like a hydrogen. So instead of a hydrogen shifting over, we'll have the deuterium shifting. So the electrons from the CD bond will move on over to that positively charged carbon. So that then gives us the result, we have a cyclops ring here. It has a methyl group left behind. So we draw that dashed wedge here. However, I started to draw as a dash wedge, but we know that because we've shifted the deuterium onto our carbo C, we now have a positive charge here. We have a Carbocaine. And when there's a carbo cion in that position, we now have an sp two hybridization there. Instead of sp three, it's no longer a tetrahedral geometry. It's now a trigonal planar. So now we just draw a solid line for our methyl group since this now has a positive charge. So now we've shifted our deuterium and I'm going to draw the deuterium as a solid wedge coming forward and then don't forget there's a hydrogen there. When this was a Carbocaine, it had two bonds to carbon. So it is implied that it had one hydrogen So I'll put the hydrogen on a dashed wedge. But it's important to note when it was a carbo cion. Again, it had trigonal planar geometry. The hydrogen was within the plane of the ring. So the deuterium can add in the front or the back, it could have been arranged the other way. So that's important to note when we're indicating stereo chemistry, it's not going to matter though because we're going to be generating a double bond. And this will go back to trigonal planar geometry, the product will be the same no matter what. So just a little side note on steri chemistry there. So now we've had this 12 hydride shift and we've had the formation of a tertiary carbo cat I because now our positive charge is on this carbon that has a methyl group. So this is more stable, which is why it will be the preferred reaction process. So now our last step is the d protonation of the beta hydrogen. Well, we have two possibilities for this because we have our deuterium and hydrogen on one beta carbon and we have another beta carbon in the ring that has two hydrogens. But our problem says show the mechanism that corresponds to the reaction below. And the product we're given and told us the product is the one with a deuterium on the other side of the double bond. So we know which proton we're going to grab to correspond to the reaction that's shown. But important to note that if we had to draw all products, we'd also want to see the double bond on the other side, neither side is more substituted. So neither side would be preferred. So we have our water. So I'm going to draw oh H two and we'll draw the lone pairs on our oxygen. Now, this is an interesting one. Why is deuterium left? And why don't we show two products? One with the hydrogen of the deuterium. Well, a carbon hydrogen bond is slightly easier to break than a carbon deuterium bond. So that will actually have preferred breaking of the carbon hydrogen bond. So our loan pair from our oxygen is going to take the proton and then the electrons from the ch bond will come on over and form a carbon carbon double bond. And now we have our product here, here's our five carbon ring, which is now a cyclops. We have our methyl group up at the top. We've now formed a double bond. That's a little weird looking. So weird looking. There we go from our carbon. That was the positively charged carbon, the hydrogen has been removed and we now have a deuterium left. And again, you can see now we're back to trigonal planar geometry around that carbon. So it didn't matter whether we portrayed the deuterium on the top or deuterium on the bottom, it will end up the same way. So you can see here that this molecule corresponds to the product shown. So we have shown the mechanism for the reaction here. So we have the formation of the secondary Carbocaine when our leaving group leaves, then we have a 12 hydride shift to cause a formation of a tertiary Carbocaine that's more stable de protonation of the beta hydrogen leaving us with our alkene product with the deuterium that has shifted over from one carbon to the other. In that hydride shift. See you in the next video.

Provide a mechanism for the following E1 reactions.
(c)

Key Concepts

E1 Reaction Mechanism

Carbocation Stability

Regioselectivity in Elimination Reactions

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