Draw structures for the following:
c. (3Z,5

Question

Draw structures for the following:

c. (3Z,5Z)-4,5-dimethyl-3,5-nonadiene

d. (3E,5E)-2,5-dibromo-3,5-octadiene

Accepted Answer

All right. Hi, everyone. For this question. Let's go ahead and draw the molecular structures of the compounds given below. Part A is one Z three Z 24 dichloralp Penta 13 di in 20 and part B is two E four, E 35 Di Chloro Hea 24 dot E 20. So first, let's go ahead and start with part A, right? And for part A and any type of this problem in general, right? I always like to start by drawing out the parent chain and the parent chain in question, here is going to be Pento 1320. So first, I'm starting with the carbon chain which is five in total. So I'm going to go ahead and draw a five carbon chain. And then from there, I have two things, right? I have two pieces of information. First and foremost is a hydroxy group on carbon one as indicated by the suffix here 10 and then I have a dye, which means that I have two double bonds in the same molecule. The first is going to be in between carbons one and two. And the second is going to be in between carbons, three and four like that. And from there, I have two other substitutes. Specifically the phrase here, 24 Chloro means that I have two atoms of chlorine in this molecule. The first one is going to be on carbon two of the parent chain and the next one is going to be on carbon four. But here I'm not done right. I've put everything together. I've added my substituent where they need to be. But here, the Iupac name in question actually specifies the configuration of the All right. So let's talk about that a little bit. The question specifically mentions carbons one and three given that they are part of a double bond, right? Carbon one is in the Z configuration and carbon three is also in the Z configuration. So recall really quickly that when it comes to deciding the configuration of an alkene, here you have to go ahead and determine the relative priorities of substituent on both sides of the alkene, meaning connected to both SV two carbons. And when an alkene is in the, in the Z configuration, what that means is that the highest priority substituent for I, for each side, the highest priority substituent are on the same side when it's in the Z configuration. So when I go ahead and I look at carbon one here, let's go ahead and look at my substituent, right. Carbon one is connected to both a hydroxy group and hydrogen So essentially right, the atoms that are making the connection to the sp two carbon of the alkene here, carbon one are oxygen and hydrogen. Now, oxygen has an atomic number of eight, whereas hydrogen has an atomic number of one. And because oxygen has a higher atomic number, it's going to be of a higher priority in this case. So the hydroxy group of carbon one is the higher priority of that side. Now let's look at carbon two, right, because carbon two has a chlorine and another carbon connecting to it, chlorine has an atomic number of 17, whereas carbon here has an atomic number of six. So once again, right, because chlorine has the higher atomic number here, it's going to be the highest priority of its respective site. So here, what you'll notice is that both my chlorine and hydroxy groups of carbons two and one respectively, they are on the same side of the double bond, meaning they are in the Z configuration already and I don't have to change them. So that's kind of nice. But now let's go ahead and do the same thing for carbon three just to make sure that we don't have to modify the structure to make sure it's accurate. So now let's talk about carbon three, right? Carbon three is connected to carbon two here and an extra hydrogen. Now between carbon and hydrogen, carbon is going to have the higher atomic number of six. So the carbon here is going to be the higher priority of this specific site. But now let's talk about carbon four here. Carbon four is connecting to chlorine and another carbon chlorine once again is going to win out in this case because chlorine has the higher atomic number than carbon. So therefore, right, this structure is accurate because both of my all keys are in fact in the Z configuration. So I'm good on this front. But now let's do part B and for part B, I'm going to do the same thing as part A and start with my parent chain exa 24 dyin 20. So let's start by drawing out our carbon chain which has six carbons. And once again, I'm going to start with my parent chain, which is six carbons long. So that's one, two, three, 45 and six. So now let's go ahead and add substitutes, right. First and foremost, I have a hydroxy group on carbon two. This time, I have a hydroxy group on carbon two and I have another dying here, right? Because here I have a double bond in between carbons two and three and in between carbons four and five. And additionally, I have two chlorine atoms on carbon three and carbon five. So now let's go ahead and check the orientation of my double bonds here just to make sure that they're accurate right here. Carbons two and four both are going to be in the E configuration, which means right, that when we compare the highest priority groups on both sides of the all keen, the highest priority groups have to be on opposite sides. So let's start with carbon two carbon two. Here we're comparing the hydroxy group and additional carbon that are bound to the SP two carbon of the all key, right, oxygen is the atom connecting to the SP two carbon in the hydroxy group and it has a higher atomic number than carbon. So because of that right, our hydroxy group is going to be the highest priority of this specific side. On the other hand, right, if we look at carbon three of the all keen here, carbon three is connecting to a chlorine and another carbon chlorine once again has a higher atomic number than carbon, meaning specifically it has an atomic number of 17, right. So here chlorine is going to be the higher priority. And when you look at the orientation of the all it is in fact, in the e configuration, the way that I drew it because my highest priority groups on both sides are on opposite sides of the alkene itself. Now, lastly, right, let's go ahead and focus our attention on the here that starts at carbon four. So if we look at carbon four specifically, right, carbon four is connecting to a hydrogen and a carbon and carbon in this case, is going to be the higher priority because it has the higher atomic number on this side of the key. But here for carbon five, we're comparing a chlorine and another carbon. So here chlorine is going to have the higher priority because it has the higher atomic number. But there's a problem in the way that I drew this structure because look chlorine and the carbon, which are the highest priority groups of both sides are actually on the same side of the alkene. But the Iupac name states that they have to be in the E configuration relative to each other. So here I'm going to make a switch right on carbon five. My chlorine is going to go where I initially drew my carbon and my carbon here is going to go up top because as soon as I do that right, I fix the configuration and now chlorine is on the opposite side as carbon in my team. And now we finally arrived at our final answers and I know it was a lot, but we went through it and if you stuck around to the end, thank you so very much for watching. And I hope you found this helpful.

Draw structures for the following:
c. (3Z,5Z)-4,5-dimethyl-3,5-nonadiene
d. (3E,5E)-2,5-dibromo-3,5-octadiene

Key Concepts

Geometric Isomerism

Diene Structure

IUPAC Nomenclature

Watch next