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- In the previous video we started with the molecular formula C3H8O and we looked at one of the possible Lewis dot structures that you can draw that has that molecular formula. >From this Lewis dot structure we looked at other ways to represent the same molecule. However we didn't have time to talk about bond line structure. So, let's start this video by taking this Lewis dot structure and turning into a bond line structure. If you look at the drawing on the left it implies that these three carbons are in a perfectly straight line but the drawing on the right does a little bit better job of showing what the molecule looks like in reality. Those carbons are not in a perfectly straight line. You can see there's a bend to them like that. So, when you're drawing a bond line structure and you have a carbon chain you wanna show that carbon chain in a zig zag pattern. Next, let's think about the carbon hydrogen bonds. If you were to draw every carbon hydrogen bond in organic chemistry class it would take you forever. So, we leave those out in bond line structures. Carbon is still bonded to these hydrogens but we're going to ignore them for our bond line structure. Just to simplify things. So, I'm gonna draw this around the carbon hydrogen bonds so we're going to ignore them for the time being. And now we have our three carbons drawn like that. And this carbon is bonded to an oxygen, and this oxygen is bonded to a hydrogen. I'll put in low-end pairs of electrons on that oxygen. Next, we can simplify this even further. We can leave out those carbons, right? So, the carbons are still there. I'm just talking about not drawing the Cs in here because it can get kinda confusing. So, we take out those Cs and I'll leave off the lone pairs of electrons on the oxygen and we have our bond line structure. So, this is our bond line structure. It contains the same information as our Lewis dot structure does. But it's obviously much easier to draw. It takes less time. It's gonna help you out throughout your course when you're looking chemical reactions. So, let's focus in on some carbons here. So, this carbon in red, right, that's this carbon. So, the carbon's still there. We're just not drawing in the C. And let's look at our other carbon. So, the carbon on the right is the one in magenta so that's this carbon right here. And the carbon on the left is in blue. So, that's this carbon. The carbon in blue is still bonded to three hydrogens, right? This carbon in blue is still bonded to three hydrogens. We just leave them off in our bond line structure. The carbon on the right is still bonded to three hydrogens, all right. But again, we leave those off when we're drawing a bond line structure. And the carbon in the middle, this red carbon here, is bonded to a OH, right? That's already shown in our bond line structure and it's bonded to one more hydrogen. So, there still is a hydrogen bonded to that carbon. All right, we just leave them off to make things easier to see. So, let me go ahead and erase what I just did here. All right, so let's just take off those, let's take off those hydrogens. So, those hydrogens are still there. We just know that they are there. So, let's just take some practice. All right, let's just take some practice to figure out what these bond-line structures mean. So, let's do several examples of understanding bond line structures and the information that they contain. Let's start by analyzing this bond line structure. And we'll start with this carbon right here in magenta. So, that carbon in magenta all represented over here is bonded to another carbon, and I'll use light blue for that. So, that's this carbon right here. So, there's a bond between those two carbons, and let me draw in that bond. Now we have another carbon, I'll use red, this one right here so the carbon in red is up here. I'll show the bond between those two carbons. And then let's use green for the next carbon so we have a carbon right here in green. And finally, there's one more carbon to think about so let me, let's see, what color do we need to use here? Let's use dark blue. So, we have one more carbon right here in dark blue and I'll show that bond. So, we have five carbons in this molecule, right? So, five carbons. So, let's write the molecular formula. So, it'd be C5. Next, let's figure out how many hydrogens. Now, to do that you need to remember that a neutral carbon atom forms four bonds. So, let's see how many bonds we already have. We'll start with the carbon in magenta. The carbon in magenta already has one bond. And a neutral carbon atom forms four bonds. So, if that carbon already has one bond it needs three bonds to hydrogen. So, one bond to hydrogen, two bonds to hydrogen, and three bonds to hydrogen. Next, let's go with this top carbon here. So, the one in red. This carbon already has one bond. So, it needs a total of four. So, it needs three more bonds and those bonds are to hydrogen, right? So, it's implied that those bonds are to hydrogen. Next, we'll go for the light blue carbon in here. So, how many bonds does this carbon already have? Well, here's one, here's two, and here's three. That carbon already has three bonds. So, it only needs one more. So, we can draw in one hydrogen. We know that carbon is bonded to only one hydrogen. Next, we'll do the green carbon. So, the green carbon right here already has two bonds. Here's one and here's another one. So, if we think about a neutral carbon atom forming for bonds that carbon needs two more bonds and those bonds are two hydrogens. So, we draw in those hydrogens there. And finally, the carbon in dark blue. The carbon in dark blue already has one bond so it needs three more. So, there's one, there's two, and there's three. So, now we've drawn out the complete Lewis dot structure for this bond-line structure over here. And how many total hydrogens do we have? Well, if you count those up you'll get 12. So, the molecular formula is C5H12. Let's do another one. So, let's look at this next bond line structure here, and let's focus in on our carbon. So, we'll start with this carbon right here in the magenta. So, I'll draw that in right here. The carbon in magenta is bonded to two other carbons. I'll make this top carbon here red. So, there's a bond to the carbon in red and there's a bond to this carbon here in light blue. So, let's draw in those bonds. So, we draw in those bonds here. And let's just keep going with our carbons. Let's assign our carbons first and we'll come back to our hydrogens. So, next let's make this carbon right here in green. So, we have another bond of a carbon to a carbon, and then let's go with dark blue. So, we have dark blue over here for this carbon. So, let's show that bond, and then we have another carbon over here. So, that carbon is right here. And we can show, we can show our last bond. The carbon in red is bonded to a chlorine. So, let me go ahead and show that. So, there's our chlorine. And now let's think about hydrogens, and let's start with the, I'll start with the carbon in red. All right, so this carbon in red, how many bonds does it already have? Well, one, two, and three. It already has three bonds. So, it needs one more and so it's implied that that bond is to a hydrogen. So, that carbon is bonded to one hydrogen. Next, let's do the carbon in magenta. So, over here, how many bonds does that carbon in magenta already have? Well, here's one and here's two. So, it already has two. So, that carbon needs two more. So, we can draw in a hydrogen here and a hydrogen here. It's the same situation for all of the carbons around our ring. Now, if we go to this carbon here in light blue it already has two bonds. So, it needs two more bonds and that must mean that two bonds to hydrogen. So, we go around the entire ring and add in two hydrogens to all of these carbon. So, let me make sure I use the correct colors here. So, in blue, and then we have this one here. So, what's the total molecular formula for this compound? Well, we have a total of six carbons, right? One, two, three, four, five, six. So, C6, and how many total hydrogens? We have two on five carbons and then we have another one here. So, two times five is 10 plus one is 11. So, H11, and then we have a chlorine as well. So, C6H11Cl would be the molecular formula for this compound. Let's look at two more examples and we'll start with this carbon right here in magenta. So, I'll draw in that carbon. That carbon in magenta is bonded to this carbon in blue but notice there are two bonds between those two carbons. So, the carbon in magenta is bonded to the carbon in blue but there's a double bond between our carbons this time, and the carbon on the right here in red, there's a single bond between the carbon in blue and the carbon in red. So, for the molecular formula so far we know there're a total of three carbons in this compound. Next, we need to think about hydrogen. So, we know a neutral carbon atom forms four bonds. So, carbon forms four bonds. How many bonds does a carbon in magenta already have? Well, here's one bond and here's another bond. So, the carbon in magenta needs two more bonds. And those bonds must be two hydrogen. So, let me draw in those carbon hydrogen bonds like that. Next, we think about the carbon in blue. How many bonds does the carbon in blue already have? Well, here's one, here's two, and here's three. So, the carbon in blue already has three bonds. It needs one more. So, we show one carbon hydrogen bond. And finally, the carbon in red already has one bond so it needs three more. So, we draw in three carbon hydrogen bonds. So, how many total hydrogens do we have? That would six hydrogens. So, the molecular formula is C3H6. But you can start to think about hybridization states here too because if you look at this carbon and this carbon, you know both of those carbons are SP2 hybridized and if those carbons are SP2 hybridized we're talking about trigonal planar geometry around those atoms and we try to show that in our dot structure as best we can. All right, approximately, approximately 120 degree bond angles around here. So, hybridization can come in to it as well. Next, let's look at this one right here which has a triple bond, and triple bonds often confuse students on bond line structures. So, let's assign our carbons again. Let's start with this one right here in magenta. So, let me draw in that carbon in magenta. The carbon in magenta's bonded to this carbon in blue and there's a single bond between those two carbons. Next, there's a bond between the carbon in blue and this carbon right here in red. So, that carbon in red. There's a single bond between those. The carbon in red is bonded to one more carbon in the opposite side of our triple bond. So, that carbon in blue is right there. There's a triple bond between the carbon in red and the carbon in blue. So, now we have our carbons drawn out. That's four carbons. So, this would be C4 so far for the molecular formula. Next, we need to think about hydrogens. So, the carbon in magenta already has one bond. So, it needs three more bonds. So, we draw in three bonds of carbon to hydrogen. The carbon in blue here already has two bonds. There's one and there's two. So, the carbon in blue needs two more. What about the carbon in red? Well, the carbon in red has one bond, two, three, and four. The carbon in red already has four bonds. So, the carbon in red doesn't have any hydrogens on it at all. And finally, the carbon in blue, the carbon in blue has three bonds, one, two, three. So, three bonds already which means the carbon in blue needs one more bond and that bond is to hydrogen. So, now we have all of our hydrogens. That's a total of six hydrogens. So, we can complete the molecular formula. C4H6. And once again, thinking about hybridization, this carbon and this carbon, all right, there're both SP hybridized, and so we know the geometry is linear around those carbons. And so, that's why we draw this as being a straight line on our bond line structures. It's because of the geometry. We're trying to reflect the structure of the molecule the best that we can. So, practice your bond line structures because they're extremely important for everything that you will do in organic chemistry.