Predict the product(s) that would result when the alkenes are ...

Question

Predict the product(s) that would result when the alkenes are allowed to react under the following conditions: (iii) Br₂, H₂O ; (iv) Cl₂, CH₃OH;

(l)

Accepted Answer

Welcome back, everyone. Let's look at our next problem. Draw the products that will form in the following reaction conditions. And we have two different reactions with the same m starting molecule, an alkene, a branch chain alkene that we'll talk about in more detail in a moment in reaction one, it's reacting with cl two. So diatomic chlorine in water in reaction two, it's reacting with BR two or diatomic bromine in ethyl alcohol ch three C h2o H. So let's think about what's happening here. We have the presence of our halogen and we have it in a solvent that is capable of being a nuclear file and performing nucleic attack. So the reaction that will take place is we're going to have the formation of a hallow hydrant. So a bridged intermediate where the two carbons that were in the double bond are both bonded to one of our halogen atoms. So to the chloride or the bromide, and then followed by a nucleic attack by the solvent molecule. And then we'll have just a final dep protonation step. And we need to recall two things in step one because we're going to form that bridge intermediate it's a concerted attack. So the stereo chemistry of the other groups are not affected, including the arrangement around the double bond. So if it's either E or Z that will be maintained, and the other thing is that, that bridge can be formed with the chlorine on either side of the double bond. So the product will be a reca mixture. So once we have our bridged intermediate, we then have this nucleic attack. So the second thing to remember is that this nucleic attack will occur on the most substituted carbon. So a Markov Niko edition, so I'm just going to draw a quick kind of focusing in on our intermediate and how that will, how that will look. And then we can think about how the water will attack here, we'll draw our products down below. So when we have our double bond, so we'll draw it as a double bond. In this case, we can see right away this is going to be um the e arrangement because our double bond has a hydrogen on either side and then a portion of the chain on either side and they, the, the two chain portions are opposite each other on the other side of the double bond. So I'm just going to label those as R and our prime on either side of the double bond. So for the intermediate, the double bond will become a single bond, my arrangement around the of the R and R prime will remain the same in that ee arrangement. And then I have this dashed line going to a chlorine with a positive charge in the middle. Now, when I think about where the water is going to attack, it will attack the most substituted carbon. But in this case, the carbons are the same, they both have one hydrogen and one R group. So this means that the nucleic attack can happen on either side. So we have two possibilities after the first step, since we can, we'll have that reca mixture there. And then we'll also have a mixture after the second step, since our water can break the ring open by adding onto either side of that bond. So let's look at our original molecule and think about what this will look like. So I'm going to look at the basic name of this. I'm not going to worry about naming the RS of the different stereo centers that will take a while to figure out, I just want to be able to talk about this molecule and where the groups are. So I find my longest carbon chain which is highlighted in blue 1234, carbon four to carbon five is my carbon carbon double blond. So that was 456789. So I've got a non in a non four here. So I've added numbers to the carbons here and I have two methyl groups on carbon two and carbon three, then isopropyl group on carbon six and without any stereo chemistry indicated I have six isopropyl 23 dimethyl non four. The S chemistry I will describe but we're not going to spend time labeling it. Rs we have the methyl group on carbon three is a dashed wedge coming out. So projecting down behind the plane of the screen, as I said, around our double bond, we have the e arrangement, the two portions of the chain are opposite each other. And then on carbon six, the isopropyl group has a solid wedge, it's projecting forward towards us out of the screen. So again, the stereo chemistry of those groups will remain unchanged even with the addition of our chlorine and our water molecule. So let's think about those products. I'm going to have to scroll downward to make more room and not to be confusing here. So I will start with reaction one, go ahead and paste up the basic structure of that molecule without my groups added yet. So I'll just highlight here the where the double bond originally was from carbons 4 to 5 just to keep ourselves oriented here. So I know I have this reca mixture of the direction I add the chlorine. So I'll start with adding the chlorine to carbon four. Again, I know I can open my ring in either direction. So I'm going to just add the chlorine here on carbon four because I know that that water molecule could have added on either side. So if I put my chlorine on a solid wedge on carbon four, I know that my ring opening must happen in an anti orientation. We'll add that up at the top. We should have mentioned that the nucleic attack can be on either side and then an anti addition to open the ring. So since it's an anti addition, the oh will be on carbon five and in a dashed wedge again, there will be a dep protein nation 02 is going to add and then there'll be a dep protein nation. So you'll end up just with an alcohol group there. So that will be product one. And let's look at product two. So you can see we've maintained that e arrangement around the double bond. So we recall we can have a reca mixture based on the first step. So now I'll draw it with the chlorine on a dashed wedge added to the same carbon four, but just where the chlorine would have formed a bridge to intermediate going the other direction. And that means the alcohol group will be on a solid wedge on carbon five. So that reflects the ramic mixture on the of the first step with the water molecule having attacked carbon five. But then I have two more products with the water molecule attack carbon four. So again, put up my basic structure here. And in this case, we would have the O Groupon carbon four to highlight that carbon 4 to 5 bond again, and the oh group could have been on a solid wedge. It was at an anti fashion. So that would mean that the chlorine on carbon five will be a dashed wedge. And then we have our very final product scroll down for that with our oh having added to carbon five or carbon four, excuse me, but the opposite steri chemistry there to reflect the reca mixture in the first step. So now the alcohol group on a dashed wedge on carbon four and the chlorine on a solid wedge on carbon five. So there's our four possible products from reaction number one. And now let's scroll down to look at the second reaction actually, before we do that, we see we start with the same exact reactant there reacting with BR two. So we're going to add a bromine and set up a chlorine and then we're in an alcohol solvent rather than water. So that ethyl alcohol is going to add as a group rather than, oh, so we'll scroll down. So we're going to end up forming an ether rather than alcohol. So since I already have my basic structure saved, I don't need to refer back to that drawing. So there's that basic structure going to highlight that C four C five bond. So same exact mechanism and type of addition as number one. But instead I'll start with bromine or bromide on a solid wedge on carbon four and then on a dash twitch on carbon five, I have och two CH three. So you can see we formed a Hallow ether here. So now let's look at the second possible product reflecting that reca mixture at the beginning. So I still have bromium carbon four. It's hard to highlight this 45 bond, but in this case, it's on a dashed wedge and then my alcohol has added on a solid wedge och two ch 31 carbon five. And we have our third product where the alcohol will have attacked on carbon four instead. So then we have a solid wedge with O CH two ch three on carbon four, then a dashed wedge with the bromine on carbon five. And the last product here again, reflecting that ramic mixture from step one, we have our OC two C three on a dashed wedge on carbon four and our bromide on a solid wedge on carbon five. So there we have those, all those reactions or all those products, excuse me from my second reaction. And again, just reflecting that in this case, the alcohol can attack either carbon as equally substituted. And that that first step of that bridged intermediate would have formed a aic mixture. So same process except for the addition of the different molecules. Number one po four possibilities. Number 24 possibilities. See you in the next video.

Predict the product(s) that would result when the alkenes are allowed to react under the following conditions: (iii) Br₂, H₂O ; (iv) Cl₂, CH₃OH;
(l)

Key Concepts

Electrophilic Addition Reactions

Regioselectivity

Solvent Effects

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