We're going to be drawing the bond line formulas of all of these molecules, so let's just jump right into it, starting with a So with a. We have a parent chain of plantain, which is five Karpin's, so that's going to be one, too. 345 I'm going to label the carbons as carbons, one to three for and five based on what you label that you can add the substitue INTs. In this case, both the situations are the same, and they're going to be on Carbons one and four and its book Going to be Koreans. So that's the final molecule for part A. So let's move on to Part B. In this case, our parent chain is going to be this beautiful group, even though it's not at the end of the name. Um, there isn't really a parent tune given at the end of the names, so this isn't an eye you pack name, but, um, the beautiful group is the only group that is going to have carbons in it. That's how we know it's the parent chain. So a beautiful group is going to have four carbons, so let's just draw that out first, so that's gonna be carbon one to three and four. So the sec insect beetle means that whatever is being attach, whether it's on the secular group for the second little group is the attached to something else. The attachment is going to be on the secondary carbon. In this case, carbons two and three are both secondary carbons. Um, and technically, you can put it on either one. Um, but if you whichever one you put it on will become carbon to based on Are you packing? Uh, so I'm just going to put it on carbon to that's where I serve. Sichuan is gonna go, and that's going to be bromide. So moving on to part C, our parent chain is going to be obtained, which is seven carbon. So 12 three, 4567 And then I'm going to name for number them. So 12 three for five, six, 17 And then we see here that on carbon number four is where substitution is going to be. And we have an isil poeple group which looks like this. So isopropyl proposal means there are three carbons and then I saw ah the attachments right in the middle. Um, so we could move on to party. So party, our parent chain is going to be plantain. So pen tain has five carbons, so that's going to be 12345 I'm going to number them 12 three for five on carbon number two. We have to of situ INTs. They're both going to be medical groups. So a catch stomach so and then on carbon three, we're going to have one methyl group, so I couldn't draw that right there. Moving on to Part E per e. We have a parent chain of Heche saying that's going to be six carbons. So 12 3456 number them again. 12 34 five and six. And then we can start adding our substitue int. So I'm just going to go in order of looks right. What the name gives us. So on our carbon number three, we're gonna have an ethnic group that's going to be to Corbyn's and then on carbon number two, we're going to have a method group, which is one carbon for a part F. Our parent chain isn't a ring. It's cycle plantain, which looks like a Pentagon, and I'm just gonna pick one carbon to net number as carbon one and then I'll go clockwise. So here are my five carbons. And then I'm told that I have both of my substitute wince on carbon number one and that these subsitute mints are Corinne's. So we'll have to Koreans on carbon number one. So moving on to part G hurt G. I have another ring for my parent chain. In this case, I have psych glow propane. So propane means we have three carbons. This is a little like a triangle, and I'll pick two carbons to be carbons 12 and degree. Um, as you can see here are substitutes are going to be Sisto one another, meaning they're going to be on the same side of the ring. So I want to give them the same type of line they're going to be on carbons 12 and they're going to be methyl groups, so I want them to be on the same side. In this case, I could just draw them with a bold wedge is my first monthly group. Then here's my second medical group again has to be on the same side. It's also gonna have a bold wedge moving on. We have a very similar looking molecule for part age. So again we have the cyclo propane. Um, so we're going to have the same triangle, and I'm going to use the same number, is there? You can see the comparison. So 123 in this case are substitue. INTs are trans to one another. That means they're going to be on opposite sides of the ring. So I have to make sure that I give them different types of lines. Um, it doesn't matter which one you give which type of line as long as they're different. So in this case, I'll just make carbon one this absurd chewing, which is metal, make it bold and carbon to I'll make it dashed. And there you go. Moving on to part I or a parent chain is gonna be Penton All so Pence and all has pent in it. It is going to be our five carbons. That's 12345 So there are five carbons and now we can utter substitue INTs. So on carbon four, we're going to have a methyl group and then on carbon to we're going to have the O H group. The reason that I know it's no age group is because this ends in all, which means an alcohol. So we must have a no age group. And since this, too is closest to this, all that means that's telling us where the O a troop is gonna be just going to be on carbon number two. So you go and so we can move on to part J Hurt J. We have another ring. So here we have a hexagonal. So ignoring the O all the ol For now we have a cyclo hexane, which is a hexagon. And then I'm just going to pick. This is carbon one to three or five and six. Um, so we can see here are substitue INTs are going to be transferred to one another. So the first substitue in that I'm going to look at is going to be four ice, a beautiful that's going to be on carbon number four and an isil. Beautiful group Looks like this where we have one carbon and then we have and, um, a branch over here, so in total, you have 1234 groups, which is where the beetle comes from. And then it's I so because we have this kind of looks like isopropyl um group. So that's what it looks like. And so are other substitute A int is going to be on carbon number one. So since their trans, they have to be on opposite sides of the ring. So I'm just gonna make this substitute bold. It doesn't matter which ones. Bolden, which ones dashed as long as its opposite or other substitue in, is going to be the O H group, which we know is on the wall is from the all, and we know that this is going to be on carbon one. So which is going to be on carbon one because it doesn't specify if it's on another carbon. And if it doesn't specify, that means that you can assume it's probably going to be in carbon one. So here have my dashed line, and then I have my witch, So that looks good. We could move on to part K parte que we have, um, um part. Okay, we have ah, vaccine. This is going to six carbons. 123456 So, in this case, um, we're going to have to substitute mints. So the first substitution is going to be on carbon one. It's going to be a cyclo purple group, which is just a triangle. So it's gonna look like this, um and then we're going to have another one of these on carbon number four. So those are die cycle purples, and then it's going to be on the hexane parent chain. So it's good. Now we can move on to part l so part l. We have neo pencil alcohol. So neo pencil, it's not until you pack name, but it looks like this. So the central carbon has for Michael grips attached to it. And since it's an alcohol, we know it's going to need to have an O age group so you can put the, oh, a troop on any of the carbons that are not the central carbon because the central carbon has its octet already filled. So it's a public with this eso. Now we can move on to part m hurt family have by Saiko 2 to 2 octane. So when we have ah by cyclo, um, molecule. That means that we have essentially two rings connected to each other. And these numbers in the middle are going to tell us how money carbons number of Corbyn's um on each bridge. So in this case, we have two carbons on each bridge of this molecule. So let's say that my rings will connect here, right? So then I'm going to have two carbons coming off of this connecting area. So 12 it's going to go that way. Then I'm going to have another two carbons going upwards. And then finally, it will have another two carbons going this way. Then I'll just draw this dash so it looks like it's behind. So, as you can see, you can kind of see the two rings where they would be fused over here. Um, and then there are three bridges, each with two carbons. So we have our you to men, too. And then if you actually count all the carbons that are in this molecule 12345678 you would see that we have eight and that fits with their octane. That's perfect. So let's do the same thing for N so for n Let's first draw our connecting bridge for our rings. So then let's start with the three carbons Love. It may be the hardest one to draw. Um, so we'll have one carbon here to three. Loan connects right there. That's three carbons. And then we'll have one carbon on the other two bridges, so one forget to color code the three and then we'll have one on this last bridge. Over here it goes downwards. So again, as you could see, we have the three carbons, the one carbon and then the one carbon. And then if we count all the carbons, that's 1234567 and we have had obtained. So that works. Finally, we come to the last part of this problem, which is going to be, Oh, so oh, we have a parent chain of cyclo plantain. So it's going to be a Pentagon. As they said before, there's only one substitue int on our Pentagon, and that is cyclo pencil. We know it's a substitute because it ended that. Why else so it's incalculable constituent, So I mean this. There's just another cyclo plantain attached to this cyclo plantain. So that's two Pentagon's connected to each other, and that's how you get the last molecule. So that is how you can draw The structure is for all of these molecules.