Predict the product of the following rearrangement-prone E1 el...

Question

Predict the product of the following rearrangement-prone E1 eliminations.

(c) Chemical structure showing a chlorine atom on a carbon chain, with an arrow indicating heat leading to an alcohol product.

Accepted Answer

Hey, everyone. Let's take a look at our next problem. What is the product of the following E one reaction which is susceptible to rearrangement? And we're given a reactant which is a benzene ring that has one substituent, it is a carbon chain with 12345 carbons. And on the second carbon out, we have a chlorine and on the third carbon out, we have two methyl groups and then it's reacting in ethanol with heat. So we recall our E one reaction. We're going to have an elimination reaction and it's E one. So our leaving group leaves first and it produces a Carbocaine intermediate, which of course, is susceptible to rearrangement. Since our problem tells us right there and any time you see susceptible to rearrangement, you'll be thinking E one or SN one because only carbo Canons can rearrange. You wouldn't see that in an SN two or E two where everything's happening at once. So let's look at what's going on here. Well, it's pretty obvious our leaving group, we've got that chlorine there. So let's go ahead and think about the first step here. What is the Carbocaine intermediate that would be produced because then we want to think about how old we arrange and then what our product will be. So we'll have our chlorine group leave and then we'll have 12345. And then we have our benzene ring add those double bonds. So we know we have these two methyl groups, put them there. I'm not going to draw the stereo chemistry since we're not talking about um an E two or SN two where we might be worrying about which direction the Nile is coming in, what the results are going to look like. Um So, unless we see signs that we need that stereo chemistry, I'm not just gonna make my drawing a little simpler by not including that the chlorine has left. So we're left with that positive charge on the carbon where it was. So, what do we notice about this car? Well, it's a secondary Carbocaine. So can we make that more stable? Well, the benzene ring is really tempting sitting over there, we might think about stabilizing our Carbocaine by resonance. However, the carb it is not adjacent. So the benzene ring is not adjacent two r positive charge. So we can't really use that the carbon in between the benzene ring and our positive charge is just a secondary carbon, doesn't improve matters any. But when we look at the carbon on the other side, it has two methyl groups. Well, we can't do a hydride shift because there's no hydrogens there, but we could do a methyl shift. So we have ach three shift possible. So what would happen if we do that? Well, take one of our methyl groups and its electrons get moved over to the positively charged carbon. And what does that Carbocaine look like while we draw our chain as we had before? And for the benzene ring, let's put that in. There we go. And we still have one methyl group on that original carbon. But the second methyl group has shifted over to the next carbon. And then we've got a positive charge on. We'll call it carbon 123. That's a number of them. So we'll number out from the benzene ring. So carbon, well, carbon one would be the, we'll call it the carbon next to the ring and then carbon 2345. So we've moved a methyl group from carbon three to carbon two. And now our positive charge is on carbon three. So then we just need to think about, where will the double bond form, which beta hydrogen will be removed? Well, what are our beta hydrogens? Well, we've got carbon two is a beta Carmen, carbon four is abated carbon and the methyl group has abated carbon. So we need to think about which will be the most favored alchem and the most substituted alkene would be the major product. So when we look at this, we can see that if we remove the hydrogen from carbon two, then we'd have an alkene in which we have four R groups. Obviously, removing our hydrogen from the methyl group is not going to leave us a very substituted alkene. And removing a hydrogen from carbon four only gives us three substituents on our alkane. So that would be removing a hydrogen from, we'll put C two. So I'm going to let me just erase this number two, make a little more space here and we'll put in the hydrogen. So I draw my methyl a little differently there. So there's my methyl group and here's a hydrogen. So my hydrogen is going to be removed by that ethanol. So let's draw etoh, we'll put the lone pairs on that oxygen, the lone pair on ethanol comes and gets that proton and the electrons that were on that hydrogen come around and form that new double bond. So there's our dep protonation of our beta hydrogen and now we form our alkene. So draw my chain there a little longer, draw my chain, draw my benzene ring put in those double bonds. Now let's add the methyl groups where they go. So carbon three, I've got a methyl group, carbon two, I've got a methyl group and add that double bond between carbon two and three. And you can see, I didn't actually need any of those stereo chemical indicators because my methyl groups ended up being on sp two hybridized carbons which have that trigonal planar geometry. So when I go back to my final structure, I don't actually need any dashes or wedges. So just double checking that I didn't need that to show the product. So here is my product, I have a bending group with this carbon chain. And the double bond is if I start counting from the carbon chain is on 12, from carbon two to carbon three. And then each of those carbons also has a methyl group. So again, we went through a rearrangement using a methyl shift from one carbon to the next and then picked our most substituted alkene as our favorite product. See you in the next video.

Predict the product of the following rearrangement-prone E1 eliminations.
(c)

Key Concepts

E1 Mechanism

Carbocation Stability

Rearrangement in Organic Reactions

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