The following molecular model is a representation of caffeine. Identify the position(s) of multiple bonds in caffeine, and tell the hybridization of each carbon atom. 1Red = O, blue = N, gray = C, ivory = H.2

Video transcript

Hello everyone. So in this video we're given this incomplete structure here. We're trying to do in this video is complete this electron dot structure of uric acid by adding necessary lone pairs and bonds. And the other thing we have to do in this problem is by providing the hybridization of each carbon atom. So we're given a little legend here for what these circular spheres represent. So first are gray spheres are going to be carbon. So we'll go ahead, add R C. Our white spheres are going to be our hydra gents, our red spirits here, oxygen. Unless we we have our blue spheres, these are nitrogen. Alright, so first thing I'm gonna do is just directly translate what we have here into what we would regular draw as a lewis dot structure. So we see here to the left of this structure, we have a six member ring. So we have a nitrogen here being one of the members. Then we have carbons all around. So maybe we'll put these carbon items as well and then we have one more nitrogen here. So each nitrogen in the six member ring here has a hydrogen attached to it. We see that this carbon here has a oxygen. So we'll add that as well just with a single bond for now as well as this one on top. We'll go ahead and add this five memory in right next to it. So we have these two central carbon atoms already we have another nitrogen here, the carbon, Another nitrogen and that connects and finishes off this 55 member drink again with the six memory ring. We have hired an attached to each nitrogen atom to go ahead and add that as well. Okay, and then for our last carbon all the way to the right here, we'll go ahead and add this auction atom. So this is just directly translating this structure here into this two D model. So now what we can do is adding necessary lone pairs as well as the multiple bonds. So for carbon carbon likes to have four total bonds and we see here for the carbons inside of our um cyclic all canes, we can go ahead and add our double bond. So we see here in the middle here this is only one or the on carbon that has a missing bond. So we'll go ahead and add this double bond here to complete that. So it seems like every other carbon in their structure has r four different bonds. Alright, so moving on to auction auction likes to have two bonds and two lone pairs to complete its octet. We see here in oxygen, we only have what we have no long pairs and then we have one bond. So what I'm going to go ahead and do is add a double bond and we'll add those to necessary lone pairs. So do that for all of the auction present in our molecule. And lastly we have to deal with our nitrogen. So nitrogen likes to have three bonds and one lone pair. So we see here that all the nitrogen atoms in our molecule have three bonds. So, for example, this nitrogen is attached to one hydrogen and two carbons. So what we need to do is just add that one more lone pair to fulfill its octet. So do it for all the nitrogen atoms. All right. So, we can see here now that we've completed by adding every all the lone pairs and all the bonds necessary to complete this electron dot structure. So now we have to go ahead and determine the hybridization of the carbon atom. So, again, carbon is just see here so we can see that all the carbons do have a double bond as part of their bonding. So it has three electron groups. So let's go actually write this out. So, you can see that all carbon Adams Have three electron groups. All right. So just for example, we're gonna take a look at this one right here. So, we have one bonding to the nitrogen, a double bond to the auction, which we just count as one electronic group still because it's just too one oxygen another to the carbon. So we have three electron groups. So we have 12 and three. Alright, so continue on. Has three electron groups. And therefore the hybridization hybridization is going to be S. P two. Again, we have three electron groups. So one of the orbital's is from the US. And then to from the P which totals to three total electron groups. So our final answer is not only this completed structure here, but it's that our carbon atoms have a sP two hybridization.