Identify the two compounds that display stereoisomerism and draw their structures. e. 2,4-dimethyl-3-pentanol

Video transcript

Hello. Everyone in this video we're given four different molecules And we want to identify. Well first straw all the molecules and then identify which one exhibits stereo terrorism. All right. So let's go ahead and start drawing out our molecules in. So for our first one let's go ahead and first draw the parent structure. We can see in the name that will all of these actually have the prefix of heP. So that is going to be meaning that our parent chain has seven carbons. So we can go ahead and add that or put that in. So for first one we have C. So that's three wow sort of a linear chain for our parent chain. So we can start numbering this but we have 123456 and seven. So that fulfills the hep part of this there we see a four and then a anal. So that is on our fourth carbon. We have an O. H. Group. So our fourth carbon is located right over here. We'll just add A O. H. Substitue ints all right then we have a dimethyl so we have two metals because of the prefix di and our local number is two and six. So basically on our second and six carbon we have two methyl groups. So this is where the positions are let's go ahead. Our our methyl groups we extend the methyl groups a little bit more. Okay, so that's going to be our structure for number one. Let's do it for a number two again we have our Leonard chain. We have seven carbons. So do the exact same thing. Number them 123456 and seven. So we have the hips of the way. And for all of these we can see actually that therefore help channel. So always we have an H. Substitue ints hanging off. Alright, So for this one, what's different is again we have two methyl groups because dimethyl but now it's on our 2nd and 5th carbon. So right here and right here let's go ahead and add those in. Okay, Now, for our 3rd molecule, Arthur molecule again has seven carbons. Okay, so we have 123456 and seven. So for 1/4 carbon we have an O. H. Then we have a 35 die couple instead of a dimethyl this time. So our 2nd and 5th has Our 3rd and 5th, I apologize has an ethyl group. So we're gonna draw that in. Right this. Alright, so that's going to fulfill our molecule name for that. And lastly our fourth molecule let's put this in red again, of course we have our seven carbon parent chain. So we have that and then our fourth carbon in the middle we have an O. H. Group 456 and seven. Perfect. All right. Now what's different about the fourth one is that we have again this thigh methods we have two method groups on carbons three and three. So basically with that 33 means is that we have two Method groups on the same carbon, which is the third carbon. So let's go ahead and add that in. All right, So that's what we have. And we can see once we have drawn everything out that the only summers are stereo cameras that we have is going to be carbons two and four. So this one right here and this one right here, which is right next to each other coincidentally. So, I can see or identify that these to our customers. And that's because they have actual stereo centers. These two molecules and other hand, are very symmetric on both sides, no matter where we look at it, and they do not exhibit any sort of stereo terrorism. But let's take into account for the fourth molecule. We can have a stereo center right here and then for our second molecule that we've joined over here again, we can have a stereo camera right over here. Alright. And that's going to be my final answer for this problem. Thank you all so much for watching