Draw a condensed structure for each of the following:
k....

Question

Draw a condensed structure for each of the following:

k. cyclopentylacetylene

l. 5,6-dimethyl-2-heptyne

Accepted Answer

Hi, everybody. Let's look at our next question. It says provide the corresponding condensed structure of the following compounds. We have two different compounds for which were just given a name. And we need to think about how would we generate a condensed structure? First off, we need to recall what characterizes condensed structures. We just have the atoms indicated by letters, all of them. So we know in a skeletal structure, for instance, we don't specifically indicate carbons or hydrogens that are bonded to each other and a can then structure they are there, we have no bonds shown. So we don't have lines or dots indicating bonds and our CH groups are shown together. So if you have a carbon with three hydrogens, you'd write it as CH three with parentheses indicating substituents or repeating CH units. So if you have say a chain where you have repeated CH two units there, you would put CH two in parentheses, we'll put using subscripts to show you how many of a particular group you have. So with that in mind, let's look at our compounds here. I'd like to start with generating a skeletal structure when I start with just a name. It really helps me to see how the groups are oriented together in space. And kind of helps me keep track as I convert to a condensed structure, I find it much more difficult to go from name to directly to condensed structure. So for compound number one, the name here is four ethanol, 12 dimethyl cyclops. So let's start with my base chain or unit here, which would be cyclops. So I have a five carbon ring. So let's just draw a Pentagon here as my skeletal structure for that. And now I need to look at the substituents I have. So I show 12 dimethyl. So I have methyl groups on carbons, one and two. So I'm just going to arbitrarily call the carbon on the upper right carbon one. So I just draw a line for a methyl group there. And then on carbon two, I have another methyl group. Then on carbon four, I have ethanol. So the or Ethan hyl, I'm not quite sure. The YL tells me it's a substituent, but the constituent off the main chain, the YN and Eine tells me we have a triple bond, you remove that e when you put it all together. And then the F part at the beginning tells me we have two carbons in the substituent and it's on carbon four. So let's find carbon four in the ring. And we know we have a two carbon substituent with a triple bond. So from the carbon four, I have a line going out to one carbon and that carbon has a triple bond, two, a second carbon. So that's my skeletal structure. Now, let's convert that to a condensed structure. Well, here's one exception to the no bond shown except rings. There is no other way to indicate that we have a ring here, we need to show the bonds. So we'll start with the top, which would be carbon five just because it's easier to draw, starting from the top. And that is a carbon with no substituents in the ring. So it's ach two group. So we'll draw C and then I'll put the H two on top of it and then we'll draw a line going to carbon one. Now, here's why you can see that I made carbon one on the right because if I had the substituent on the top, it would just be a little bit visually trickier to add that substituent in there. Whereas here I can draw it, write it out to the right of my carbon. So I've got my line going to carbon one. So that's I write in C for that carbon. Now, this carbon has one methyl group substituent. So therefore, it must have one hydrogen. So after the C, I'll write H for its attached hydrogen and now I'm going to indicate my methyl substituent. So in parentheses, I write CH three. So you can see that the next to the sea shows me that's the hydrogen bonded to that carbon. Whereas then in parentheses, ch three shows me that's a substituent. Now I make a line going down to carbon two that also has a methyl substituent. So I write that also as CH and then in parenthesis, ch three close parenthesis, now we go across the bottom, this is carbon three has no substituents. So it is just C and then to the left of it ch or H two. So it looks like H two C from left to right now, go up and to the left. I'm on carbon four. So I don't indicate bonds. So I'm not going to indicate I have a triple bond, you're going to imply it by how many hydrogens I put in. So carbon for itself, it has one substituent that vinyl. So it has one hydrogen. So I'll put H here. So it says HC. So I know that hydrogen goes with that carbon and then my first carbon on the substituent has a triple bond to one carbon and a single bond to another. It has no carbons. So I just put a letter C there and then the terminal carbon there has a triple bond and so it will have one hydrogen. So then another C to the left and then an H. So the whole group says HCCHC in that order from left to right. And then finally, I put my last bond in from carbon four to carbon five. Note that whenever you see two CS next to each other, there's a good chance you're looking at a triple bond there. So here we have our skeletal structure. I mean, excuse me, condensed structure. And again, we did have to indicate bonds because we have a ring, but we have our CH two on the top. And then ch in parenthesis, ch three, ch parenthesis, ch three, there's our two methyl groups, then CH two, then CHCCH for that ethyl substituent. So that's our first compound. Now, let's look at compound number two, which is 667 tri methyl oct three I. So our base chain here we have a three, I mean, we have eight carbons. The Inyne with a three means we have a triple bond from carbon three to carbon four in this case. So I'm going to draw my skeletal structure. So I'll draw an eight carbon chain with that triple bond. So carbon 123, triple bonds are always a little tricky to draw on a skeletal structure because we have this linear geometry. So I'd like to put a little dot that shows carbon three line to carbon four. Let's add the other two bonds and then 4 to 5. So you note that carbons 2345 are all in a straight line. And from carbon five, we go to carbon six. So we're going back to our zigzag shape. Carbon 67 eight, I always double count. It's very easy to get mixed up in skeletal structures with long chains like this. I have 12345678 carbons got my triple bond on carbon three. Now let's turn to my substituents 667 triethyl. So counting backwards, carbon eight, carbon seven gets a methyl group. Carbon six gets two methyl groups. So there's my skeletal structure. Now, let's do my condensed structure. I'm going to write the little carbon numbers here because again, with a chain like this, it's very easy to get mixed up. And I've got to keep track of all of my hydrogens and where things are going. So I'm going to go ahead and add the numbers here on my skeletal structure to help me refer back to seven is kind of small there. So let's write out the conde structure. No bonds are shown ch groups indicated together. Carbon one is just a plain old Ch three to start with. So we write CH three. I'll put a little number one up there to help me keep track. Carbon two, no substituents. And I write CH two, carbon three has a triple bond and it's internal. So it's just c by itself, no hydrogens, carbon four is internal with a triple bond. So it's another c by itself then carbon five, it has no substituents and it's internal. So that's ach two. So note that looks a little funny. Because we've got three C's in a row again, whenever you see those CS next to each other with no hydrogens, think about triple bonds. So, we're on to carbon five. You see now how the numbers are helpful, so easy to lose track when you're writing it out in a straight line like this, carbon six has two methyl group substituents. And so it has no hydrogens. Again, it's got two bonds, two carbons, five and seven and two methyl groups. So it's ac by itself that's carbon six. And then in parentheses, ch three closed parentheses, subscript two showing those two methyl groups there. Now we get to carbon seven and we could show that this has a single methyl group and then carbon eight, but it's a little Tidier in a condensed structure. Note that what this looks like is that there are two methyl groups coming off of carbon seven, the substituent and carbon eight. So it's a little easier to indicate it in the following fashion. So for carbon seven, it has one bond to carbon six and then two bonds to other carbons. So it has just one hydrogen silver white ch that C is carbon seven. And then we can show this as being two methyl groups attached to carbon seven. This makes it a little easier and Tidier. So I'll write CH three in parentheses and then subscript two. So one of those methyl groups is actually carbon eight just to keep track. So my condensed structure for number two is if I read it out left or right CH three, CH two CCCH two C and then parentheses CH three, close parenthesis, subscript two ch then in parentheses, CH three close parenthesis subscript two. That is my condensed structure for my 667 tri methyl OCT three. I see you in the next video.

Draw a condensed structure for each of the following:
k. cyclopentylacetylene
l. 5,6-dimethyl-2-heptyne

Key Concepts

Condensed Structural Formula

Alkynes

Substituents in Organic Chemistry

Watch next