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Studying for a test? Prepare with these 5 lessons on Alkanes, cycloalkanes, and functional groups.
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Video transcript
We've done a bunch of examples looking at the actual molecular structure and trying to come up with the name. In this video, we'll go the other way around. We'll start with the name and see if we can actually draw the molecular structure of whatever this might be. So let's start. When you first look at these, it's very daunting, but you always want to start really at the end, so you know what kind of the core of the structure is going to be. So if you look at the end of this, you have an -ane, so there's not going to be any double or triple bonds here. It's all single bonds. It's a hexadecane, so let's think about that. Hexadec-, that's six and ten. Hexadec- was a prefix for 16, so this is 16 single-bonded carbons. And cyclohexadecane, so 16 single-bonded carbons in a ring, so this part of it right here. So let me just do this. So cyclohexadecane, let's draw that part first. It's not easy to draw even a 16-carbon ring, so let's start here. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, and then sixteen. I think I got it. Let me count them again and then I can connect them up into cycles. It's one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen. All right, I drew the cyclohexadecane part. Now, if we go back a little, let's see, we have a 2,9-diisopropyl. What does this mean? This means that we have an isopropyl at the two and the nine spot. Now when you are drawing the structure from the name, you could just arbitrarily on someplace on this ring pick what your one through your sixteen spots are. I'll just arbitrarily pick them, because I could have drawn this ring any which way. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen. All right, so this next piece right here, let me do this in magenta. 2,9-diisopropyl, this is telling us that at spot number two and at spot number nine, I have isopropyl. The di- is just saying I have two isopropyls at two and nine, so you can kind of ignore the di-. I have an isopropyl, and you may or may not remember that an isopropyl looks like this. It's three carbons, so it's going to be one, two, three, and the connection point to the main ring in this case is going to be in the middle carbon, so it kind of forms a Y. All of the isos, the isopropyl, isobutyl, they all look like Y's, so it's going to be linked right over here. That's also going to happen at the ninth carbon, so at the ninth carbon we're going to have another isopropyl. We're going to have another isopropyl at the ninth carbon. All right, we've taken care of the 2,9-isopropyl. Then we have the 6-ethyl, which is just a two carbon. Remember, meth- is one, eth- is two, prop- is three. Let me write this down. So this is going to be prop- is equal to three. Isoprop- is equal to that type of shape right over there. In this case, eth- is equal to two, so it's a 6-ethyl group. So at six we have an ethyl group, so one, two, carbons, and it's connected at the six carbon on the main ring. And then finally we have a cyclopentyl. So if we look at-- let me find a color I haven't used yet-- cyclopentyl. so pent- is five, but it's five in a cycle, so this is a five-carbon ring that's branching off of the main ring. It's at the first spot. Let me draw a five-carbon rings, so pent- is equal to five, so it would look like this, one: two, three, four, five. It looks just like a pentagon. That's a cyclopentyl group and it's attached to the one carbon on my cyclohexadecane, so it is attached just like that. We're done. We've drawn 1-cyclopentyl-6- ethyl-2,9-diisop ropylcyclohexadecane. Let's do another one. I think we're getting the hang of it. So here, maybe we can do this one a little bit faster. Let's see, we have a tetramethyldodecane, so the main root here is the dodecane, do- for two, dec- for ten. This is a 12-carbon chain. It's not in a cycle, so let me just draw it out. We have one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, and so we can just number them arbitrarily, just because I could have drawn this any which way. So it's one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. That's the dodecane, all single bonds. Then we have a 2,6,9,9-tetramethyl. All this is telling us-- remember, meth- is one carbon, so all this is telling us is at the three, the six, the nine, so at the three, the six and twice at the nine spot, we have methyl groups, and we have four methyl groups. That's all the tetramethyl is saying so it's a little bit redundant. We know we have four of them here: 3,6,9,9. We have methyl groups at each of those places. We have one methyl group at three, and then that is bonded with the third carbon on the dodecane chain. We have one at six bonded to the six carbon on the dodecane chain. We have two at nine, so that's one at nine and then we have another one at nine bonded to the nine carbon on the dodecane chain. And we're done. That's it. That's 3,6,9,9- tetramethyldodecane. Let's do another one. 1,3-bis(1,1-dime thylethyl)cyclopentane. So once again, just kind of breathe slowly. It's very daunting right when you look at it, but just start with the core: cyclopentane. That's just a simple five-carbon ring. A five carbon ring that looks like a pentagon: one, two, three, four, five. There you go. That is a five-carbon ring. We can number it however we want, so one, two, three, four, and five. This is telling us at the one and the three position we have-- and the bis- is kind of redundant. This is saying we have two of these things. Obviously, we have two. We have one at the one and one at the three. So you can kind of ignore the bis-. That's just the convention and we've seen that multiple times. But at each of those positions, we have a 1,1-dimethylethyl. So what's a dimethylethyl look like? So let's think about it a little bit. Let's think about it and let me do it orange. They obviously named it using systematic naming and what we have here, we have an ethyl as kind of the core of this side chain. So if an ethyl is equal to two carbons, so this is two carbons right there. So let me draw a two carbon: one, two. That is two carbons right over there. I'm just drawing it at the three spot. I'll draw it also at the one spot, actually. So that is two carbons right there. That's the ethyl part. And then on 1,1, so if we number them, we number where it's connected, so it's one, two. This is saying 1,1-dimethyl. So on this ethyl chain, you have two methyls. Remember, methyl is equal to one, so this is one carbon. You have one carbon. That's what methyl is, but you have two of them. You have dimethyl. You have it twice at the one spot. So you have one methyl here and then you have another methyl there. Same thing over here. You have 1-methyl on the one spot and then you have another 1-methyl on the one spot. And then you are connected at positions one and positions three, so you're connected there and you are connected right over there. And you're done, That's it. That is our structure. Now, if you did this with common naming, instead of this group being a 1,1-dimethylethyl, you might see that we're connected to a group that has one, two, three, four carbons in it. The carbon that we're connected to branches off to three other carbons. It is a tert-butyl. So you can also call this a 1,3-- let me just write it down. So another name for this would be 1,3-tert-- or sometimes people just write a t there-- t-butylcyclo-- no, actually I should say di-t-butyl, because we have two of them. 1,3-di-t-butylcyclopentane.