So we're going to be drawing out our bond nine formulas for these, um, names for molecules. So starting with a So we see that we have octane so talked, meaning there's gonna be It's carbons. Een means that there is going to be a double bond. It's all right, d B. And it's going to be cysts. So we want to keep in mind that when we are looking at the double bond, um, our double bond, it's substitue INTs are going to be on the same side and we can see, um, that our double on will be on the third carbon. So let's do that now. So let's start. We're gonna have one to three. So here is gonna be our third carbon. And if we know it's going to be cysts, we want to make sure drawing are substitue INTs on the same side. The 34 five 67 h Let's just double check that 12345678 or rights. And then again, we want to put our double bonds on third carbon 123 You put a double bond right there, and that is going to be our um, molecule. So let's move on to be so be, we are going to have trans to heck scene. Um, so in this case, we have hex. So we're gonna have six carbons again we have seen. Which means we're going to have that double bond. And the information we're given about this double bond is that the double bond will be trans. Um, so we don't have to worry about it being on the same side. We can destroy our normal zigzag, and then it'll be on the second carbon. So let's just draw out our six carbons. 123 456 Let's double check. 123456 And then on the second carbons, the 12 we're going to cuts our double bond. So when? Six. Okay, um, let's move on to see so again, we look at the end of the know the name to check what we're going to be dealing with. So we have pence as our prefix, So we're going to have five carbons again you haven't seen, so we know we're gonna have a double bonds. Um, and then we have closest to that. We have our double bond is going to be on the second. Carbon were also given a couple of constituents subsidence. Um, so we have two methyl groups, you know, die means to, and they're going to be on, um, carbons two and four. So let's first just draw out our parent chain here, so we're gonna have pent in. So that's one, 234 five sets are five carbons and again or double bonds gonna be in the second carbon. So, um, let's just pick, um, aside to assign carbon one, I'll pick this one. So 12 three, or sighs. Well, what are double von here? And then again, we're going to put our, um substitutes. So our dimethyl, which is gonna be on carbon too, and then carbon four. So, um, that is the molecule for C. Um, Now we can move on to D. So same thing is, before we're going to be looking at the end of the name first. So we see here we have Butte to mean. That means we're going to have four carbons with a double bond on the second carbon. And we also told that the double bonus trans So let's draw our parent chain right now. That is going to be 12 three for, um Then let's pick a side it to assign carbon number one. So I'm gonna put this is carbon number 12 three and four. Um, and then we can add our double bond on carbon number two. Ah, And as you could see, the substitue INTs of the double bonds are across from each other. So that looks good. That means they're trends. And now we can add our chlorine and chlorine is going to be on carbon number one can see here one claro Um, so we will put that over here on carbon number one. Really? It wouldn't matter which side you put it on, since it would give the same name either way. But since I picked this carbon is carbon number one all, just leave it there. So that looks good. We can now move on, Teoh the next molecules. That's gonna be molecule e so pent again. We've seen this before. It's gonna be five Corbyn's. Then again mean so again, we're gonna have that double bond that's gonna be on carbon number one. That's all the information that we were given about our parent chain, so let's just draw it out for now. So we have 12 345 Um so let's take a carbon to assign is carbon number one. So let's say I pick this one. So 12 three or five. So, um, we can now place are double bonds on carbon number one and then we can place our other substitue INTs. So we're told that we have die. Brahmos. That's to Brahmos. And they're living located on carbons four and five. That's gonna be right here. Brahmos, bro mean right here. And that is the molecule for East. So moving on, let's take a look at F. So f is a lot longer than the other molecules we're looking at. But let's break it down. So let's first look for our parent chain of the very end of our name. So we have cyclo pence and between now is going to look like a Pentagon. So here's their cycle up insane. Um, now, let's just assign what we want to be. Carbon one to three, four and five. So looking at our name. So we have one, two and one hell, that's a very long names. Let's break that down. It's gonna be on carbon one. So it's gonna be on talk here. We have pro pope, so we know that there is going to be three carbons on this Malkiel group. We know it's a local group. You can just see the Y l at the end that indicates to now kill group eso It's going to be just sees in ages. Um, so let's just draw in those the re carbons for now, said one to three. So we have that and were also given that it's two n so to e n. That means we're going to have a double bond on these second carbon. So if this is carbon number one, which is the carbon that is attached to the parent chain, we know that that's carbon number one, cause there's a one here in front of the Y el. I'll even write that in So criminal one to three. If we know that, then we know that carbon number two will be. This carbon right here in our double bond is going to be right there, So that is what this first substitution looks like. Now let's move on to the next substitue in. I'm just going to go, um, continuing to the left. We have to metal. So put a metal group here. And then finally you have our, um our first substitue int one borough, mo. And I'm going to put that right here, which is gonna be carbon number one. So that is going to be the molecule for F. So let's move on to G. So Fergie again, we're going to look at the end of our molecule zero parent chains of cyclo pence een So we know that Pantene is going to be a Pentagon. It doesn't give us a number for where the double bond is going to be. Um, so in this case, since the methyl groups of substitue INTs or carbons great and four, let's just assume that our double bond is going to be on carbon one. And now that we have that, we can add in our substitue INTs again, uh, on carbon 74 we're gonna have Dime Ethel. So carbons three and four will have these methyl groups hanging off of them. Um, so that looks good. We can now move on to H. So again four h we want to look at the end to see our parent, Jane. So we have cyclo pension again. So we're gonna have our cycle a plantain. And now we went to a sign. Um, actually, since we only have one subsidiary and we don't need to assign the numbers for the carbons Ah, that's my bed. But if we actually take a look at the beginning of the name here we have ah, vinyl. Um, and remember that vinyl means that it will be in the position that is attached to the double bonds. Um, between carbons. So, um, we would draw that onto our molecule like this says you can see our, um our substitue int here. The double bond between these carbons is attached to our parent chain, Um, in a position that is where it's directly attached to the carbon participating in the double bonds. Friends. That's the vanilla positions. That's why it's called. Um, that's why it's called Rydell. So we can now look at I ah, this time we have cyclo hexane scene, so we know that vaccine is going to look like a hexagon again similar to G. We aren't told where the double bond is going to be because we have e any, but we can assume it's going to be between carbons one and two. So 12 three, four, five, six So we'll put it right here. So I used the wrong color. There we go. So now we can add in our sub situ INTs and it's die Claro. So to Claros on carbons one and two so we can write them in right now. So we'll have our seal right here and her seal right here. And that is going to be the molecule for I. So let's finish up with molecule J. So molecule J we look at the end of the molecules were pent mean, and we know that the double bond is going to be on the second carbon and it's going to be Trans, so he could destroy our zigzags. So 123 45 Now let's just pick a Corbyn to assign carbon one. So I'll say this is one, 23 for five on that. We can put in our double bonds on carbon, too, for here. Um says you can see we have our double bond and the substitutions are across from each other. So it's Trans. That's good. Um, no. We can add our other subsequent store parent chain. So we had die Claro again, which is to Kouros Corbyn's one and force. Let's add those. And now we'll have a quarry in carbon one and then we'll have a chlorine on carbon four right here. So that is the last molecule that we had to draw in. I'll just scroll back to the top in case he wanted to look at our answers again. So again, here is the answer for molecule number one or a sorry, um, molecule B mark. You'll see molecule day molecule e mark your f molecule G mark your age molecule I and lastly molecule J and that is all.