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For the polymeric substances nylon and Dacron, sketch representations of the repeating unit in each....

Question

For the polymeric substances nylon and Dacron, sketch representations of the repeating unit in each.

For the polymeric substances nylon and Dacron, sketch representations of the repeating unit in each.



Answers

For the polymeric substances nylon and Dacron, sketch representations of the repeating unit in each.

This is the answer to Chapter 15. Problem number 71 Fromthe Smith Organic chemistry. Textbook on this problem gives us two polymers. Ah, and asks us speaking what monomers needed to form each polymer. Ah, and so, um, remember that we're talking about radical polymer ization here on, and so we need double bonds and our starting material in order to form the new carbon carbon bonds between monomers to make the polymer. So these air, um, all vinyl derivatives. And so Ah, we can just sort of look at this and cut between molecules to see what what we're trying to make here and so we can cut here and here. Ah, and that will give us three pieces. Um, then so we could I guess things are cutting here. Ah, and here is well, And so when when we do that, we see that each of these pieces, uh, is is going thio look like this. So, basically, we're cutting this into pieces like this. Um, and we know that that there needs to be a double bond in our starting material. And so, based on the side chains being these, um, dime Ethel side Jane's. Ah, we should know that. Are monomers gonna look like that? Okay. And then sew for B. We can do the exact same thing so we can think about, uh, cutting this molecule into its pieces. So, like this on, then? When we do this, we should see that, um, again, our blocks sort of look like this. Ah, and of course, we need a double bond. And so what we're talking about is this monomer as our starting material. O r. If it's easier to think about, you could write it like this. See, age to double bond. C H C 00 E t. Okay. Um, yeah, and so that's the way to to do these. I think it's it's easiest to just visualize it by cutting the bonds between each of these identical repeating segments on. That gives you an idea what, um, each of the monomer should look like. And then just remember that we do need a double bond in our starting monomer in order to form the polymer. And that's the answer to Chapter 15. Problem number seven

This is the answer to Chapter 22. Problem number 76 Fromthe Smith Organic chemistry. Textbook on this is a fairly straightforward problem. It's asking us to look at two sets of monomers on. Ask what type of polymer can be made from the combination of those two monomers on DSO for a Well, so for both of them remember that these squiggly lines just represent other the polymers continuing indefinitely in either direction on DSO. Basically, what you're going to do is just attach these end to end indefinitely. And so for a we're gonna have a polyester. So, um okay, so let's do this. So first monomer is going to be our cyclo heck scene. Uh, die. All monomer. There we go. There's that land. This will make carb oxalic acid. Pardon me, Esther. Linkages. So there we go one. Yeah, who went a little too high there, So let's bring it down. There we go. All right. So there is our first linkage of the two monomers. I was straw a 2nd 1 So no. Another cyclo hack Sane unit, uh, linked with another Esther linkage two for so there we g o. There's that second piece. And then that will be Esther Linkage, too. Ah, another monomer. And so they will alternate like that indefinitely. So there's our first are polyester. So for B, we're going to be making a poly a mine, so we'll start there. There are first piece. It will be, of course, linked to the name I. Bond is the name Polly A might implies. So there's our aim. I'd linkage 1234 then linked by an ay my bond again to the next some unit. There we go! And another a my linkage. 1234 Oh, boy, the lag time on this is quite bad on DH. So again, that would be linked. That's the nitrogen of another A mile wide. And so again, the Palmer would continue on indefinitely in that direction as well on. So these are the two polymers that could be made from each of these two sets of monomers. On that is thie answer to Chapter 22. Problem number 76

So this question asks us what similar and what's different about nylon and polyethylene. So we'll start off with what's different because it looks like we can pretty easily tell that these air very different molecules. So if we look at the in just the monomer for the nylon weaken, see that it's a fairly long chain. It's got has gone double bonded, oxygen's it's got nitrogen, Zen and we compared to the polyethylene monomer. And we can see that that we don't have ah whole lot going on there. Polyethylene monomer is very simple. Uh, so as far as what is the same between these? Well, both of them are polymers. Polymers mean that we are combining multiple monomers in order to to you were stringing them together into longer and longer chains. So in the case of the polyethylene, essentially, what we're gonna do is we're gonna bring in another polyethylene, another ethylene, another another 13. We're gonna break these bonds. So we're gonna end up with just single bonds and so we're gonna string those together. So actually, once we start looking at the polymer here, it does as as it gets as we add more and more monomers together. It does get longer, but it's a very simple pattern. It's just up and down. It's just carbons and hydrogen, so they're hydrogen is attached to each of these carbons was just carbon nitrogen, the nylon. The monomer itself is much longer, but it still contains strings that are just carbons. But it's gotten nitrogen XYZ well and double bonded Oxygen's, as we pointed out before, Uh, but what is the other difference between these? Well, the other difference. The other major difference is in how they are polymer arised. What type of reaction causes them to become polymers? So the nylon monomer as it becomes a polymer, a zit problem arises. It's going to do that using a condensation reaction, meaning that it's gonna lose a water. So as we imagine, Ah, 2nd 1 of these molecules coming in on the right hand side. What we're going to see happen is this O. H. And one of the H is that is connected to the nitrogen are going to separate out as a water, and the and the the nitrogen will bond directly to that carbon There. On the other hand, the polyethylene is going to be created through an addition. Reaction and the addition reaction, as we described before, happens when we break one of the bonds of a double bond, which means that both of those carbons now have a free electron, that they can bond with something else. And, ah, they choose to bond with another carbon from a different monomers so that we can extend the chain so similar in that they are polymers different in how those polymers were formed and also a little bit now they're structured.

Shil. This problem wants to adjacent nylon 66 chains and showing where hydrogen bonding could occur between them. So I'll draw one here. So this is one nylon 66 chain. And now I'm going to draw the same chain upside down and flipped, And so we'll start like this. So this is the same Shane, but flipped. And places where hydrogen bonds could occur are right here, right here and right here. So hydrogen bonds occur between H and either n or O. And so there could also be hydrogen bonds here, here, here, um, from this and or rather from this age to either of these o's. And so when this is in three dimensions, it looks a little clear where these hydrogen bonds could exist. But it's clear that there are plenty of places where there could be hydrogen bonds.


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