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Determine the relationship between the following sets of molecules as: enantiomers, diastereomers, identical, constitutional isomers_ or no relationship. (2 points ...

Question

Determine the relationship between the following sets of molecules as: enantiomers, diastereomers, identical, constitutional isomers_ or no relationship. (2 points each; 20 total)MeHQ Me aMeHCANHzCIH;C HOMe OHOHNH2 CHJCOzEt ~HMeQH OHMe -~HMeHO-OHOH OHHO HO~HH;C - A= OH CHzOHH=~HMeCOzEtMeMeHOOHMeMeMeMe=Me"OHOHMeMeOHMeOHMeMeMe"Me "CH;Me=Br,_OHCHjCH} Me HOOH"CHs OHCH;CHjCH;

Determine the relationship between the following sets of molecules as: enantiomers, diastereomers, identical, constitutional isomers_ or no relationship. (2 points each; 20 total) Me HQ Me aMe HC ANHz CI H;C HO Me OH OH NH2 CHJ COzEt ~H Me QH OH Me - ~H Me HO- OH OH OH HO HO ~H H;C - A= OH CHzOH H= ~H Me COzEt Me Me HO OH Me Me Me Me= Me "OH OH Me Me OH Me OH Me Me Me" Me "CH; Me= Br,_ OH CHj CH} Me HO OH "CHs OH CH; CHj CH;



Answers

What is the relationship, if any, between the molecules in each of the following pairs? The relationship may be any of identical structures, constitutional isomers, stereoisomers, or no relationship.

Let's talk a little bit about ice, summers and relationships between molecules. So there are two different types of higher classifications of ice summers. So the first type of classification of ice summers are constitutional ice tumors. Now constitutionalism er's are molecules with the same formula, but different structures. Our second type of ice summer is a stereo is, um er. Now a stereo is, um ER is part of a larger subgroup of ice MERS in which a molecule has the same formula and the same type of structure, but a different configuration or a different confirmation. So let's go ahead and look at some different structures and determine what type of ice MERS they are or if there's no relationship between the molecules. So our first structure. Here we have C H three see, which is double bonded to an oxygen, as well as a hydroxyl group, our second molecule. Here we have CH three bonded to a carbon, which is bonded to hydroxyl group as well as double wanted to and oxygen. So let's take a moment and look at these molecules here. If I were to flip this molecule around, which means we'd have a double bonded oxygen with a carbon and then a ch three and an O. H and then rotate this molecule counterclockwise. We would have the same molecule. So therefore this first one is the same for a second example. Here we have a ring. We have a five sided ring with a hydroxyl group and a double bomb. And then we have a five sided ring with a oxygen double bonded to it. So obviously these molecules are not the same, which means that we either have no relation. Constitutionalism, er's or stereo is a MERS. So in terms of determining they're types of ice MERS, let's determine their formulas here. So we know we have C s. We have HSM you owes. You're sees with ages and their past. So this has 12345 carbons. This has 12345 card. This one will have 2456 78 hydrogen zones. This one will have 246 eight hydrogen, one oxygen and one oxygen. They both have the same chemical formula. Okay, so then they must be either constitutionalism, er's or stereo. I summers. Now the definition of a constitutionalism er is a misnomer that has the same formula but a vastly different structure. Therefore, these air constitutional I summers okay for our third structure. Here we have another five cited ring, but we're going to start to get a little bit tricky here. So on our five sided ring, I'm going to redraw that because that does not look great. Okay, on our five sided ring, we have some directionality within our methyl groups. Here we have a methyl group coming out of us and a method group being sent to the back of the page. Whereas on this second one, we have a method group going into the page on a method group coming out of the page. Now, this requires some directionality thinking here, if we were to flip this molecule so I'm going to use this curly shape to say flip, would these two molecules be the same? So imagine a carbon coming out at you if you hold your hand facing your face and you were to flip your hand over. Is it the same? No, You're looking at your palm and then now you're looking at the back of your hand. If you were to stick a finger out to represent the seat, the methyl group, and then flip it out. You have a method group facing you, and then now Method Group is a way. Therefore, these are called stereo stereo ice summers for our fourth group. We have a huge aromatic molecule here that I'm going to go ahead and draw some Mexicans for. If you're hexagons, look kind of funky now, don't worry. After many years of organic chemistry, you will become a hexagon pro. I promise. Okay, so we have a double bond double bond, double bond, double bond, double monitored oxygen, oxygen. We have double bond, double bond, double one, double bond, double 12 and oxygen double wanted to an oxygen. So visually, these look like constitutionalism. Er's. However we always it's good practice to double. Check to make sure that these molecules are actually the same. So one carbon. 23456789 10. 123456789 10 Carbons. 123456789 10 Carbons, hydrogen. Okay, one hydrogen, two hydrogen, three hydrogen four, five, six, 78 nine. Hydrogen into oxygen's Here we have one. I was just kidding. It's gonna be eight. Sorry about that. Actually. Every count. 12345678 Yeah. 123 for no, no hydrogen here. Five, six, seven, 123456 six. When I was wrong here again, there's gonna be no hydrogen in these corners. Well, I'm not even looking at my notes. And you need to look at my notes when I do this six and then two. So these are, in fact, constitutional ice MERS. Okay, last structure. We have a four sided ring shaped kind of like a rhombus. And always remember, when you see rings that are less than five sides are you are getting significantly more strain on these bombs here. So if you think about this as the number of sides decreases, strain definitely increases. However, is the number of side increases the stability. We'll also decrease because if you have like a 45 sided ring, it's more likely to break. So look for that six sided sweet spot. But rings can really be any side if you have the right Kyra Ality as well as the right monastery. Okay? Besides the point. So we have a facing up roaming and facing down, bro. Mean and then we haven't up down broking. These molecules are the same, but they're shaped differently. Therefore, we're gonna have constitutional ice allies.

Starting with compounds, eh? So the easiest way to go about this problem is to try and rotate these three groups. If you change just where their positions are around the carbon hydrogen bond to match the other compound, then you can figure out their relationship. So if we're moving around this carbon hydrogen bond, we can take this siege to H move it to where? The bro meanness that pushes the bro Ming up top and the methyl group down to where this siege to a wage wass. And when you do that rotation around that access, you get the same compound. So these air identical. If you struggle to see this, I would buy or borrow a model kit to help you visualize it. Play with it until you can actually see the rotation around that bond within your head. For part B, um, it looks like potentially these two could be the same compound. They are actually an anti merce. However, because the foreign groups that are attached you cannot draw a plane of symmetry through this point. So it's not me, Zo. It is actually in an anti works. You flip this stereo center that's on ly make it a double bunk or all of them for a change

This question as us If these compounds are the same compound Ah, in Cairo in and tumors eso there's one way that we can meet. The marriage is, um, try to see the arrangement off the subsea tone on the same cab on. So keep this center Attn. The same position and we can start labeling of the rest of the molecule with the head through. Adam ah has the highest priority would. So this is one. This is, uh too. And this is three, so you can see that. One, 23 It's got a, uh and I club for twice. And this is one again too three. So 123 This is Hawk Wise. The source. You can see that the arrangement off the city wants. It's like a different, um, arrangements. And so these are in anti. Almost now. In this case, Clariant will be one A wages too. Three. So want to three. Is this one? This is again going to swan always too. This is 3123 So you can see that is in the same order. So this is the same compound in this example. Now they are both cabin, but this cab on connected the 23 other cab on this cab on connect only two. So this is one. This is two and this is 3123 It's in this direction. So again, this is one. This is two. This is three 123 As you can see, these are two opposite direction. So this is an anti almost.

No, we'Ll work on problem forty from chapter twenty one. This question gives us three different pairs of molecules and we need to determine whether they're the same identical or an anthem. So let's go ahead and serve with a and let's reproduce the structures given here. So the first one were given a methyl group CEO. Hi. I had out alcohol hydroxyl group coming out of the page. And then another method going into the page in the second part gives us, uh, something that looks like this we have here. Chlorine ch three. Once again, we have the hydra axel coming out of the page in another method going into the page. The question here Do we have, um, and Auntie Mers, or do we have the same molecules? Now? What we notice here is that we have to substitute wants, which are the same. So they cannot be an anti mers. So an anti Mars exhibit optical asama, is, um And so since we have to substitution sees two metal groups. Okay, we say that they're the same. We could superimpose these two molecules on each other. So the second set of molecules which were given is once again we have a tetra Usual carbon. Yeah. So let's go ahead and write the structures down here. You have a high fashion, bro Mean, I have a method group going into the page and then in a mean group coming out. If we write the second structure can right, bro mean metal going into the page. And I mean coming out. So we notice about this structure is that we do have four different substitute wants. And if we can think three dimensionally or as well as we can in two dimensions, we cannot find a way in which we could superimpose thes if we were to rotate this molecule here so that the hydrogen and bro Ming are in the same position is on the left the left. We would see that the method group here would be rotated, so that's coming out. And the mean group would be going in so that we say they are an anti murders because they are non super imposible mirror images waken see right from here that they are mere emerges. So if we think about rotating them, trying to get them to be superimposed, we can see that that's not possible. So the third situation third molecule which were given is a little more difficult to think about because we have two carbons in a chain here. Small train. So let's go ahead and write the two molecules. We have a C H which is coming out of the page from this carbon. And on this carbon we have, um, Ethel going into the page. We have a hydrogen there and we have a carbon try chloride substitution year, which should be coming out at the appropriate pinkel. A little bit larger oven angle Here Now, on our second carbon, we have hydrogen and we have to methyl groups. So the second molecule second structure were given. It's a little different people. So we have It's sort of in reverse. We ever h c here and then we have our carbon here and we have this now going into the page and we have our carbon try s or we have our method group coming out of the page, you and then on this carbon here we still have the two method groups. So now the challenge here is to figure out whether these are different or the same so we can say we have to be able. Teo, imagine rotating this molecule to see if we can get the structures match. No, the way. This way, we need to look at it. Here is if we rotate about this ch bond counterclockwise. So we rotate like this. This part here will be swinging around to the back. So instead of having this group here coming out of the pages now going into the page and we can see that this struck part of the structure here, this the back half is conserved from here. And then we imagine about this front carbon, which has now come over here. So we have the carbon. We have the hydrogen, which were the sea ate. This bond here is unchanged because we're rotating about this bond and we can see now the CCL three has moved to the opposite side of this carbon. And instead of going into the page, this car methyl group is not coming out of the page. So we can say that we can obtain structure too, by rotating counter clockwise around this ch bond. So they are the same because this rotation yields the same structure So in problems like these, especially when you have more than one carbon center, it takes a second, but you need to be able to think in three dimensions to see if you can have the molecules superimposed on each other.


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