Are Proviei irhhe ststrctctee betoe . of monomers would be required to make the polymer Kevlar? Two repeat units...

What two monomers are needed to prepare each polymer?

Swore ours to monomers of Kepler. Kepler is deporting a might words a structure over here. So from here you can see that the two monomer off K flare would be the No, I mean expensing one Ford I mean and I sile new glory and together move on cable.

Okay, I'm gonna draw the monomer used to create get floor and catalogs of this. We're going to have an education attached to to Hodgins and it's attached to a ring, which means ph is attached to uh another notion attached to one hydrogen. And it's attached to a key tune. It was attached to a ring. And there he tune is attached to a nice chin, which is ha option as well as a hurry stash question. And that keeps just continuing on. But it's an example of a conversation polarization reaction takes place between a dyke. Are possible acid and a diet. I mean, it's really to carve octopuses in the same structure and to a means on the same structure. This is actually our die carb oxalic acid. It's just too Carlos of acids attached to a benzene ring. And they're in the power position and this can be our diet. Mean, So you can draw this replacing ph with a benzene ring and putting one mean, are here. Now the other parents that four carbons away. This is we have another ring and to carve outs of acids. So if you react to carbolic acid within a mean looking ahmad and this will be the monomers. This is done with heat at around 200 degrees Celsius. And this reaction, the reaction of the immune is going to deeper into the car vessel acid of hydrogen, of the car boss of acid. And in the presence of heat will join these two groups together of the loss of easter hydrogen, josie auction. Does the dehydration reaction or will also water. And if we destroy the structures together, it's gonna look like the formula here. And obviously this can react further as well. This car bostock acid over here can react with another diet, mean? And this immediately here could react with another doctor boss of acid. And it could be as simple as just like a two carbon carbolic acid Graduate with a two carbon dia Mean, or a dial. Or it can look like more complex like this. Any and this is used to create Kevlar, so it's certainly pretty useful.

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