Suggest a bromoalkane and the conditions necessary to produce ...

Question

Suggest a bromoalkane and the conditions necessary to produce the alkenes shown.

(b) Flowchart illustrating elimination reactions, showing a bromoalkane converting to an alkene with a phenyl group.

Accepted Answer

Hi, everybody. Welcome back. Here's our next problem. It says, select the correct Chloro Alcaine and condition to produce the given alkene. So we're shown a reaction where we have just a blank box for our reactant and then a blank box over the arrow representing the conditions, the Nile or base and the solvent. And then we're given a structure of an alkene product. It's an eight carbon chain. And if we look at the numbers here, we have a double bond from carbons 3 to 4, there are methyl groups on carbon three and carbon four and then it finishes up with two methyl groups on carbon seven. So we have an alkene. So we're going to be looking for an elimination reaction. So we'll be looking for either an E two or E one reaction. So we need to think about which one we would use to produce this. So let's think about what our Chloro Alkane would be. So we know there'll be a chloride group. We know that when we eliminate chlorine as the leaving group, we need to eliminate it from one of the two carbons that will form our double blonde. Now, we could have things like carbo Carbocaine rearrangements leading things to be different. But that gets really complicated and we don't need to do that unless something leads us to decide that we have to. So to be simple, let's think that we'll put our chlorine on either carbon three or four. So let's draw out our structure without the double bond. So we could put a chlorine here on carbon three and when that's eliminated, then we would remove a beta hydrogen. And those beta locations in this case would be on carbon four or carbon two. We also would have a beta location on the outside of the methyl group. So there would be three different places where we could remove a beta hydrogen and form a double bond. But we know that the most substituted a would be favored. And we look at this structure, we can see that then the most substituted alkene would be the one that our product shows because that between carbons three and four, because that gives us a doubly substituted alkene on both sides. So I'm just going to highlight the beta carbon or excuse me, beta hydrogen or beta carbon uh being carbon four. So it looks like I've hit upon this right away. I could also have the chlorine on carbon four. So let's think about that scenario. So here's my structure without the double bond. And if I stuck a chlorine on carbon four and then looked at my beta hydrogens, I'd have one on carbon 51 on carbon three and one on the methyl group. But again, that most substituted alkene would be the one where we had the C three C four bond. So the preferred beta hydrogen to remove would be the one on carbon three. So it doesn't really matter whether I put the chlorine on carbon three or carbon four. So I'll just use carbon three. So I'd say could do this way. But this is just the first one I drew. So we will use this one or the chlorine's on carbon three. So now we have our correct Chloro Alkene. So we'll just add that in the box. So I've got, again, my eight carbon chain two metals down on carbon seven on carbon three, I have a methyl group and a chlorine and then on carbon four, I have a methyl group. So there's my starting product. Now, what condition would I use? Well, that has to go back to that E one or E two question. And for now I'm going to erase my second possible chloral. Just sort of keeping in the back of my mind that that was a possibility. Well, first, let's think about what kind of leaving group I have here. I have a bulky leaving group that chlorine is attached to a tertiary carmen. So what kind of mechanism could take place? Well, we often think right away about the E one mechanism which is good for tertiary carbo cds, which would be generated in this place case, excuse me, if the chlorine left first, we would have a tertiary Carbocaine. However, what do we recall? E one and SN one compete? So if we want on purpose to produce this alkene and have that as our only product, it would be better if we could use E two. And why is that? Because for E two reactions, as long as we have a strong nucleo, they can take place with an inaccessible leaving group, which we have here because we have this bulky molecule. Whereas in essence to doesn't take place with a bulky leaving group. So if we make sure the conditions favor an E two reaction, we would have only the E two alkene produced, we wouldn't have any substitution product. So the E two mechanism would be better here. So what conditions favor E two mechanism? As we said, a strong nucleo file and particularly a strong base to deproteinate those beta hydrogens. So a good base for instance, might be sodium oxide, naeto, excuse me, not Ohnaeto minus and A plus E minus because that would be that strong nucleo file, it has a negative charge because it will dissociate in water and it will be a strong base. And then what kind of solvent do we need? Well, the E two sn two reactions prefer an a product solvent solvent because that a product solvent will hinder the nucleo file. So we don't want that here. So we might use, for instance, Acetone as the solvent there. So we'd want a strong base and we want an a product solvent again, to favor that E two reaction. So we have here and again, there's other things we could have chosen. We just want to make sure again, we're choosing a strong Nile, strong base and a product solvent. So we have our chloral canne and our proper conditions. So again, we added that chlorine to carbon three that we could have added it to carbon four. And we have a strong base here, sodium meth oxide and an ap protic solvent of acetone. See you in the next video.

Suggest a bromoalkane and the conditions necessary to produce the alkenes shown.
(b)

Key Concepts

Bromoalkanes

Elimination Reactions

Regioselectivity

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