VISUAL SKILLS Visualize the structural formula of each of the fol...

Question

VISUAL SKILLS Visualize the structural formula of each of the following hydrocarbons. Which hydrocarbon has a double bond in its carbon skeleton?
a. C3H8
b. C2H6
c. C2H4
d. C2H2

Accepted Answer

welcome back everyone in this example, we need to determine which of the below hydrocarbons is not straight chained. So we should first recall what a hydrocarbon actually is. And recall that this is an organic compound with a series of carbon atoms bonded to hydrogen atoms. We next want to recognize that there are two types of categories of hydrocarbons. We have a lymphatic hydrocarbons versus what we would recall as aromatic hydrocarbons. Now going to aleph attic hydrocarbons, we have certain criteria. This can be either open or cyclic structural molecules whereby open. What we're referring to is having either branches or just being a straight chain. The second criteria to be an al emphatic hydrocarbon is to either be unsaturated or saturated and by unsaturated. We are referring to our hydrocarbon structure having PI bonds. So we're called that pI bonds are double bonds and that would correspond to being an alkaline or we can have triple bonds. So we have three lines which would correspond to being an alkaline. Whereas on the converse we have saturated hydrocarbons which only consist of sigma bonds only recall that sigma bonds are single bonds. Now we have to consider the criteria for aromatic hydrocarbons which are always going to be unsaturated, meaning they will normally contain either pi bonds or triple bonds. Whereas they also should have benzene like properties and we will get to that in one moment. But what that means here is that it would either be molecules that are S. P two hybridized, which we should recall means we have three bonding regions With zero long pairs and another. Benzene like property would be to have planer geometry Because of this sp two hybridization, specifically molecular geometry. So going back to our examples, we can begin with example A. Which gives us ethane. So we want to pay attention to our prefix and or suffix looking at our prefix. We can see that we recall means we have a two carbon structure where the suffix ane tells us that this is an al caine, meaning we should only have sigma bonds in our structure. And so drawing this out, we would have two carbon atoms next to each other where each carbon atom to fulfill its octet. Because we recall carbon has a bonding preference of having four bonds should have a total of three bonds to hydrogen all singly bonded. Because again this is an al caine. And as we can see we have an open system here because this is just a straight chain of two carbon atoms. And so this specifically is a a lymphatic saturated hydrocarbon because it's a open system with sigma or single bonds making up the structure. And so this would be because it's straight chained. A incorrect answer. Choice. So we can roll choice out. So let's label that as straight chained. Moving on to choice B. We have beauty which we should recognize. This prefix tells us we have a four carbon chain. And again the N ending means we have an L. Kane. So drawing out our carbons, we have 1234 carbon atoms bonded to one another where the outer carbons have three bonds to hydrogen all singly bonded. Because again this is a al cane. So it should only have sigma bonds and the inner carbon atoms should have just two bonds to hydrogen because that's all it needs to complete its octet. And so yet again, we have an open structure. Since this is just a straight chain here where this is specifically an al emphatic saturated hydrocarbon. Yet again because it's just single bonds making up our structure. So again because it's straight change, we would rule choice be out and let's move on to choice D. Since we can clearly see it's yet again another al caine with the prefix probe which tells us that we have three carbons. So we have three carbon atoms bonded to one another where each of the outer carbon atoms again have three bonds to hydrogen to complete their octet. The middle one has just two bonds to hydrogen to complete its octet. And the outer carbon has three single bonds to hydrogen to complete its octet. Since this is an al caine. And yet again we have another aleph attic saturated hydrocarbon in an open chain system where this is again a straight chain. So we would rule out choice D. This leaves us with considering choice C. As benzene. So what we should recognize when it comes to Benzene. We can't tell how many carbon atoms it has from the prefix or anything here. So we want to recall from lecture that benzene is an aromatic functional group. And recall that benzene follows two, Formula C six H six. Now, we also want to recognize that benzene is also characterized by having a fully, sorry, that says fully conjugated ring system. And so what this means is that when we draw a structure for Benzie, we want to recall that each corner represents a carbon atom. So we would have two carbon atoms here, a third here, Fourth carbon atom here. And sorry, let's just make this a bit more even so we have 12345 and then connecting these ones here, we have six. Now, when we normally draw benzene in a structure, we have the hydrogen atoms be implied. So recall that each of these corners here or carbon atoms have two bonds to hydrogen to complete their octet and these are implied hydrogen. So we don't have to draw them in. Now the next characterization as we stated, is to have a fully conjugated ring system, meaning we must have three pi bonds within our structure of benzene. Now you can place them in any location and that is important because it doesn't matter technically because this is a fully conjugated ring system. So this bond can move to this location. This bond can move to this location or this bond can be conjugated to this location. And so recall that this is described as the localization of these pi bonds in our structure here. Now, because of these pi bonds and because also of the suffix that we see we have in our name in this tells us that we have an L. Keen, which makes sense as to why we have these double bonds or these pi bonds characterizing our benzene structure. So as we can see, benzene is a ring or cyclic structure. So let's write that out. This is cyclic and we'll use the color black and this would correspond to being not straight change. And that means that C. Is our correct answer choice to complete this example as an example of a hydrocarbon that is not straight chained. I hope everything I reviewed was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video.

Watch next