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Uie LeeCHSedenantlomers (d) Kenlical dlaslerepmners (e) none ol the above difierent conformnalions ol the same moleculeWhat the relationship ol the following molecu...

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Uie LeeCHSedenantlomers (d) Kenlical dlaslerepmners (e) none ol the above difierent conformnalions ol the same moleculeWhat the relationship ol the following molecules?enantiomers (d) identical diastereomers none ol the above dillerent conformations of the same moleculeHow many planes Of symmelry are present 2-methylpentane? (D) 2Of the following; which step shoula have the highest enthalpy of actrvraton? (Boiden 0r color the letter 0t your answer )HCIH,oH,o -OOH;H,oKOHBO- ra

Uie LeeCHSed enantlomers (d) Kenlical dlaslerepmners (e) none ol the above difierent conformnalions ol the same molecule What the relationship ol the following molecules? enantiomers (d) identical diastereomers none ol the above dillerent conformations of the same molecule How many planes Of symmelry are present 2-methylpentane? (D) 2 Of the following; which step shoula have the highest enthalpy of actrvraton? (Boiden 0r color the letter 0t your answer ) HCI H,o H,o - OOH; H,o KOH BO- ra



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$$ \begin{aligned} &\text { Use these bond enthalpy values to answer Question } { . }\\ &\begin{array}{lclc} \hline \text { Bond } & \begin{array}{c} \text { Bond Enthalpy } \\ (\mathrm{k}\rfloor / \mathrm{mol}) \end{array} & \text { Bond } & \begin{array}{c} \text { Bond Enthalpy } \\ (\mathrm{k} / / \mathrm{mol}) \end{array} \\ \hline \mathrm{H}-\mathrm{F} & 566 & \mathrm{~F}-\mathrm{F} & 158 \\ \mathrm{H}-\mathrm{Cl} & 431 & \mathrm{Cl}-\mathrm{Cl} & 242 \\ \mathrm{H}-\mathrm{Br} & 366 & \mathrm{Br}-\mathrm{Br} & 193 \\ \mathrm{H}-\mathrm{I} & 299 & \mathrm{I}-\mathrm{I} & 151 \\ \mathrm{H}-\mathrm{H} & 436 & & \\ \hline \end{array} \end{aligned} $$ Which molecule, HF, HCl, HBr, or HI, has the strongest chemical bond?

In each part of this problem were given three different molecules and we want to determine which one out of these three best meets the given criterion and each part of the problem. So starting in part A, we are given three molecules and we want to determine which one of these has the highest boiling point. For each part of this problem. We need to compare the the relative strengths of the Inter Molecular Forces President in each one of the given molecules in order to help us determine this. So, for a boiling point, we know that stronger inter molecular forces will require a higher boiling point. And this is because more energy is required and therefore a higher temperature in order to overcome stronger inter molecular forces. So if we compare the three compounds given in part A, we see that they're all non polar and they all differ by the Adam before of the same Adam that are bonded to the carbon atom. If we look at the periodic table, we see that rooming compared to chlorine and flooring has a higher Moeller Mass and so therefore CBR for is the largest molecule out of these three, which leads to the highest polarize ability. So even though all of these display London dispersion forces, we can conclude that CBR four has the strongest enter molecular forces out of these three and therefore the highest boiling point in part B. We want to know which one of the three molecules has the lowest freezing point. When we free something, we're going from a liquid to a solid. And as we continue to decrease the temperature, the movement of the molecules also slows down as well. And when we when we slow down the those particles, we know that they hack together more tightly and resemble a solid. And so therefore, substances with higher inter molecular forces will resemble a solid more so than than those with lower inter molecular forces. And so that means that we're looking for the substance with the smallest inter molecular force to correspond to the lowest freezing point. We know that L I f is an ionic compound, so that is the strongest type of inter molecular force. And HCL is a polar molecule that displays dipole interactions. However, F two is a dye atomic and therefore non polar, and it only displays London dispersion forces, which are the weakest types of inter molecular forces. So therefore, since F two has a weakest internally molecular forces has the lowest freezing point out of these three part C. We want to know which one of these has the lowest vapor pressure. At 25 degrees Celsius. We remember what vapor pressure is if we have a liquid, and that liquid vapour rises into a guess than that gas exerts a pressure on the inner walls of that container. So we're at the same temperature for three different molecules. Then, if if these molecules have differences in boiling points than the one with the highest boiling point, will not have a Zeman vapor in the air because this is a constant temperature of 25 degrees Celsius. So if we have a molecule with the lowest boiling point at that constant temperature, then 25 degrees Celsius would be closest to that lowest boiling point. And it would have the most vapour particles in the air and therefore the greatest vapor pressure, so vapor pressure decreases as boiling point increases. So we really want to determine which one of these three has the highest boiling point, which means the strongest enter molecular forces ch 30 ch three is a polar molecule and ch three ch 20 h has hydrogen bonding. Therefore making it have stronger inter molecular forces than ch 30 ch three and ch three ch two ch three is non polar, so it has the weakest inter molecular forces of London dispersion. So because of the hydrogen bonding in this molecule, it has the highest boiling boiling point. Therefore, the lowest favor pressure or d we want to know which one These has the greatest viscosity. Viscosity is a resistance to flow. So we're looking for molecules with the strongest inter molecular forces. H two s is a polar molecule and HF is a polar molecule of that exhibits hydrogen bonding and H 202 is also a polar molecule. It exhibits hydrogen bonding so H F and H 202 since they both have hydrogen bonding, have stronger intra molecular forces in h two n. H two s and now we can compare those two based on their relative hydrogen bonding strengths. We see that h 202 has a greater inter molecular force from those hydrogen bonds because the presence of two oxygen atoms makes it more available for more hydrogen bonding for adjacent surrounding molecules. So that is why the hydrogen bonding for H 202 means that it is the greatest inter molecular force present out of these three Bala cules and therefore has the highest viscosity report you. We want to know which one of these has the greatest heat of vaporization and so that means which one of these requires the most energy and order to to convert from a liquid to a vapor. And so that means which one of these has the strongest enter molecular forces and when we look at ch three, ch three and CH four, those air both non polar hydrocarbons, so those both have the weakest type of inter molecular forces of London dispersion H two C e O. On the other hand, is a polar molecule which exhibits die poll interactions and therefore will have the strongest inter molecular forces and therefore the greatest heat of vaporization. Finally, in part F, we want to determine which one of these has the lowest heat of fusion. The heat of fusion corresponds to the amount of energy that is required to convert a substance from a solid to a liquid. So we really want to know which one of these melts at the lowest temperature has. The lowest melting point is the easiest to convert from a solid to a liquid. Well, if we look, we can see that CSB are and CEO are both Ionic compounds and so, therefore, those have much greater inter molecular forces since Ionic forces are the strongest out of all types of inter molecular forces, compared to the diatonic I to which is a non polar that displays only London dispersion forces. So we know that I two has the weakest inter molecular forces out of these three, and that means that there is not a lot of energy required to convert it from a solid to a liquid. And so it's Delta H of fusion will be the smallest out

So, for part a, the correct enter is CBR four, as has the highest intra molecular forces because it has the highest dispersion, which is all they all, what they all have. So for part B, it's F, too, because it has the weakest intra molecular forces, which is dispersion in this case for parts season cat. It is ethanol or ch three c h 20 H. Because as the highest inter molecular forces, which are hydrogen bonds four Part D. As in dog, you have hydrogen peroxide or H 202 as it has the highest inter molecular force. Out of all the molecules present for part E. As an Edward, you have h two c o as it has the highest inter molecular force out of all the molecules present. And then for part f, you have I to, as has the lowest intra molecular force out of all the molecules present

In this question. We were arranges them, according to toe blurted. But we must have put in order, Richards and superiority is depends on the different in electoral negativity, which is that as long as a difference in electronica tive increase, we found that the alert to increase and also the direction off polarity is from no electoral negativity at Toronto. How electoral negativity at So now let's start them. That 1st 1 will be h on. Then it should be our then edge. See, uh, and it will be a juicy then actual. Then ich if that lost one will be. It's See you, then be CIA, then is on c, uh, sort of writing get wrong. But now let's see the direction, all the direction in these ways we try t must pull the row so you will boot along the and these direction. Thank you

So I have Paris structures here and I want Teoh figure out which of the and anti MIRs are the same. And which of the pairs of an anti MERS air different? So this questions kind of worded, uh, strangely. But when I say by the same in anti murder, I mean that they have Thea there's the same molecule and therefore they have the same configuration. So I'm gonna be looking for assigning configurations for each of these molecules. And if they have the same configuration than they're the same an anti mur if they have different configurations than they are different in anti MERS. So looking at Pere A, I can go ahead and, um, a sign, an absolute configuration for the barbecue on the left. Of course, I know that my bromide, my Broman group, is gonna be the highest priority because it is the largest mass. And then I'm going to go on to my scion a group since I just connect to the nitrogen and then my, um, Method group, which is connected to three different hydrogen. And then, of course, my hydrogen group, Right, so going here looks like I'm going clockwise, which would normally say are however, the hydrogen hydrogen group is coming towards me and I wanted to be going away from me, so I would just invert the absolute configuration. And therefore I would say that this one is going ah to be s in the S configuration and now going to the other pair. I'm sorry. Going to the next molecule here. I see that I have my bro my group coming towards me. That would be one Ah, Santa Group, of course, to another Group three and hydrogen for So here, Um, looks like I am going clockwise. And since my hydrogen group is going away from me just like I wanted to be, I would say that this is in the are configuration. Um, notice how the absolute configurations for these are different. And so therefore, I would say that these are different in anti MERS. Okay, now, for ah, the pair be looking at the molecule in the left. I have my bromide, my bro Mo group here, of course, is going to be my highest priority. And then I have the Ah So I have two different carbons here. I have the car pixel gas. It And then I have the Santa group, and highest part is gonna go to the carpet Selic acid, of course, because it is connected to oxygen, which has a higher mass in the nitrogen. So this will be to Santa will be three and my hydrogen will before. So here I am going what looks to be counterclockwise. And since my hydrogen group is going away for me just like I wanted to be, I would say that this is the exhibits s configuration. And then for the other molecule. Here I have my promo group, which is priority number one, and my carpets look acid to saying, um three and hydrogen for so again, we do the same thing here. Um, looks like we're going counterclockwise, and it's great that are hydrogen is going away from us that we like it to be eso This is also gonna be s. And since these this pair exhibits the same configuration, I would say that this is going to be the same in and humor. All right, and now going to see here. Example. See, I have the following pairs of molecules. And so for this particular one, I see that my, uh, hydroxy group is going to get the highest priority because it has an oxygen. Great. And then I have another group versus a method group. Of course, the other group is going to get the higher priority here side label, that is to the method group would be three, and the hydrogen would before. So here we go. Looks like we're going clockwise. And this would be are absolutely figuration since the hydrogen is going away from us like we like it to be and looking at the next mark your here, um, do the same thing as we did for the other one. So our hydroxy group is number one are ethnic group is number two. Method group is number three and hydrogen is number four. So here it looks like we are also going clockwise. And we would say this is the are absolute configuration, and this would be the case since the hydrogen is going away from us like we wanted to be. And therefore, since they have the same configuration absolute configuration, we would say that this is the same in an steamer and looking at D, we have the falling pairs of molecules here and so I can go ahead and sign priority. Of course, our looks like our nitrogen is going to be the highest priority cause it has priority over the carbon. And now, looking at CH three versus our car basilica acid, of course, Our carpets, alas, is gonna take priority because it's connected to those oxygen's and then our method groups only connected to those hydrogen. So that's gonna be number three, and then our hydrogen is number four. So for this one, looks like we're going kind of clockwise, and somebody would say s and then for the other ones are very last one here, Um, so our hydrogen is coming into the plane of the paper, but we could do a neat little trick here to make it going away from us so we can get the right configuration for this is we can hold on to that carpet cilic acid on the top, and we could pretend were rotating 120 degrees clockwise. And so what that's gonna look like when we rotate it, and when we draw it is gonna look like this. So we're gonna have our carpets look acid, which remains unchanged. But now we rotated the three other groups. Such that are, um our amino group is now coming into the plane of the paper. Our hydrogen group is now going away, and our method group is now coming towards us. So now we can assign priorities because now that we have our hydrogen group going away from us, whereas before it was in the plane of the paper. So we're actually gonna get the right Absolutely figuration. Now, when we assign priority, of course. Our nitrogen is the highest here. Carpets, Selic, acid in second highest, methyl third and hydrogen fourth. So here. Looks like we're going counterclockwise. And this is great because I oriented are hydrated to be going away from us. And therefore, we would say we have s they're in the ass configuration here. And since this is the same, um, between the pair, we would say here that we have the same in and humor. All right,


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