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The images shows 4 compounds; labeled A through D. Many of the structures are missing lone pairs, but they are also "understood" t0 exist whcre needed.Com...

Question

The images shows 4 compounds; labeled A through D. Many of the structures are missing lone pairs, but they are also "understood" t0 exist whcre needed.CompoundCompound BH H H H H H H-c-c-0-C-C-H HCC-H H H H H H HCompound CCompound DH H H HHH-c-HH H-C-C-C-C-O-H H_C C-H H H H HHSelect one statement that most accurately describes mixture of Compound A and Compound D.Compound A and Compound D will hydrogen bond with one anotherCompound A and Compound D will not hydrogen bond with one anoth

The images shows 4 compounds; labeled A through D. Many of the structures are missing lone pairs, but they are also "understood" t0 exist whcre needed. Compound Compound B H H H H H H H-c-c-0-C-C-H HCC-H H H H H H H Compound C Compound D H H H H HH-c-HH H-C-C-C-C-O-H H_C C-H H H H H H Select one statement that most accurately describes mixture of Compound A and Compound D. Compound A and Compound D will hydrogen bond with one another Compound A and Compound D will not hydrogen bond with one anothcr Compound A willact Js hydrogen bond acceptor while Compound D will both act as both hydrogen bond acceptor and donor Compound A will act as a hydrogen bond donor whilc Compound D will act as hydrogen bond acceptor:



Answers

Four pairs of compounds (A) - (D) are given below. Indicate, in each pair, which compound can be more effectively hydrogen bonded. (a) $\mathrm{CH}_{3} \underset{(\mathrm{I})}{\mathrm{CH}_{2}} \mathrm{OH}$ and $\mathrm{CH}_{3}-\underset{(\mathrm{IH})}{\mathrm{O}-\mathrm{CH}_{3}}$ (b) $\underset{\text { (I) }}{\left(\mathrm{CH}_{3}\right)}_{3} \mathrm{~N}$ and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$ (c) $\mathrm{HOCH}_{2}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CH}_{2} \mathrm{OH}$ and $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}_{\text {(II) }}{\mathrm{C}} \mathrm{HOHCH}_{3}$ (d) $\mathrm{CH}_{3} \underset{\text { (I) }}{\mathrm{CH}_{2}} \mathrm{OH}$ and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}$ (a) $(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{I})$ (b) (A) - (I), (B) - (II), (C) - (I), (D) - (II) (c) $(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{II})$ (d) (A) - (I), (B) - (II), (C) - (I), (D) - (I)

So here we're just looking R H two c O Lewis structure and bond angles. So, firstly, let's draw it when we have our central carbon with our carbon I'll component, which is our carbon doubly bound to an oxygen in our Lewis structures. We need to include our loan pass. So Oxygen has two lone pairs. It's Group six. And so now that accounts for all of our electrons. We also have two protons on that central carbon, and so we have an SPD to hybridize carbon. So this structure is a tribunal plane are so all of the molecules are in the same plane, and so as a result, we have bonding angles off 120 degrees.

This question is asking about hydrogen bonding for us to determine which molecules or substances if hydrogen bonding. So remember before we answer this question, we have to appreciate that for hydrogen funding a trojan pointing two conditions must be met Number one, there should be a hydrogen atom that is bonded or attracted to a highly electro naked chief at tom in another. My little So for us to justify or to conclude that there is hydrogen bonding, there should be a hydrogen atom that is bonded to an election negative atom. And when we are talking about highly electron negative items are talking about oxygen, Floren and nitrogen. So if a substance has hydrogen and any of these atoms or elements, then we can conclude to say there is hydrogen bonding or hydrogen bonding exists in that substance. So Looking at the 1st set, We've got an s. three and P. H three. If we are to draw these, We've got NH three and H three and P and N. These are all in group five. So when they born with three, this God alone, there's a lone pair of elections that is left. The same applies for p three H edge and edge. But here we'll say hydrogen pointing only exists In NH three because we have got a hydrogen atom that is pointed to a highly election negative atom first for us is not highly electron negative. So we can talk about hydrogen bonding in here. So applying the same principle to the next set, we have seen two H four and high translating into edge form. If we are to draw this, this is what C two H four then for high treason we've got and then we have a church inch hedge hedge. But we also have lone pairs of elections here. If we notice here, we've got hydrogen in both. But the hydrogen atoms in ethylene are not bonded to a highly electron negative atom or element happen is not highly electro negative, but nitrogen is. So as a result, we can talk about hydrogen bonding in hydrogen but not in italian. So moving on to another set, Yeah, we've got hydrogen fluoride H. F against hydrogen chloride. It's real. So for fluoride, hydrogen fluoride, We have got 121212. The same applies here. HNCL 121212 1 2. This is ceo so the same principle remember for hydrogen bonding the hydrogen atom should be bonded to a highly election negative atom if we look at this flooring is a highly election negative items. So we can talk about hydrogen bonding in HF. But when we look at this, chlorine is not really small and electro negative. So we can't really talk about hydrogen bonding here. So in all cases, to answer all, to answer the question At the end of the day, we've got hydrogen bonding in NH three in attrition, fluoride and also in Hi Tricia Yeah.

Using the information provided. Let's determine the chemical formulas of each of the compounds and then draw their Lewis structure. So we have five compounds which are listed A, B, C, de and E. So let's first cell for our compound d. Um, because we're told that Compound D is 43.7% nitrogen and 50 percent oxygen. So for the mass of hydrogen, it will be 100% minus 50 percent minus 43.7% which would be equal to 6.3% hydrogen. So we're also told that the density of Compound D is equal to 2.86 grounds per leader at STP. So no, let's go ahead and find the formula for de. We can find the moles of n which is going to be, um, before we do that, let's assume that we have 100 g of D. And so we have 43.7 grams of N 50 g of oh, and 6.3 g of hydrogen. Let's convert each of these two moles, uh, 14 g per mole, smaller amounts of nitrogen. This would give us all right? Uh, yeah, 3.12 malls on 3 16 g per mole. So 55 about 16 gives me 3.12 malls and times 1 g per mole. That's equal to 6.3 moles we're going to divide by have 3.12 This would give me 11 and two. So my empirical formula for Compound B would be called to, uh and oh h two and we can solve for the empirical mouse for D, which would be equal to. And that's very each to their N O H. Two. So 32.0 g per mole. We're told that the density is 2.86 grounds for leader. So, uh, to let's hope for the molar mass of D. We have 2.86 g per liter times 22.4 liters per mole and this would work out to leaders would counsel 2.86 telestrate. You play for 64.1 pants per mole eso for n which is our whole number ratio. Mm. Over e m 64.1 grounds per mole, divided by 32 g per mole. And so we killed the two. So the molecular formula for D would be equal to two times the n o H. Two, which should be equal the end to O to h two. We rearrange this, we can get a compound which is equal the n h four n 02 So there is Compound de and H four to. Okay, so now there's d. We can identify compound. See, we're told compound. See, it has one more oxygen full, then compounds mhm d. So this is equal to N H four and 03 for the formula of C. Um, Now let's move on to compound A. We're told that compound C and a half one iron in common eso has one I am in common with compounds. See on when is acting is a strong electrolyte. The solution is strongly acidic. Eso it's strongly acidic. If it is strongly acidic, a strong acid would be nitric acid. The ironing common would be the nitrate iron. So there's compound a Andi, we've got Compound B, which we can do next. And here we have the tie. Trish, in of Compound B, requires 20. So whatever compound B is, um, it's gonna be tight treated with hydrochloric acid for complete neutralization. So I know that this has to be a base so tight, traded with HCL and obtain The Moller mouse s 07 to 6 g of compound B is tight, treated with 21.98 There's a neutralization, their middle leaders of mhm one Moeller hcl The moles of HCL is equal to one Moeller times 0.2198 leaders and this is equal to a point 0 to 1 98 moles of HCL and what we're going to assume here is we're going to assume a We're gonna assume a 1 to 1 more ratio, so we're going to assume 1 to 1 more ratio. So we have 0.2198 moles of compounds be and there for the Moeller mass of be would be equal 2.726 g over 0.2198 moles And this would work out to 30 1976 33.0 grounds per mole is the Moller massive beat a compound Be, you know must be basic since there's a neutralization. So it has to contain an O H group and ohh the O. H. Group is 17 g per mole, so 33 grounds per mol minus 17 g per mole will leave us with 16 grounds for more and so on a compound B. Uh, since it must be basic and it has to be 16 g from all the 16 g per mole must be comprised of N H two. Therefore, Compound B is an H two oh each and there's the identity of Compound B. And lastly, Compound E. It's as commercially available concentration solutions of air normally 16 Mueller Um, yeah. So, uh, compound A, we're told, is h N 03 and commercially available A 16 Moeller the solution of Compound E, which is 15 Moeller Um Hmm. Which company? E. Which is 15 Moeller I mean on this Must be but be solution with commercially available. Um, all the comrades are strong electrolytes. If it is a strong electrolyte 15 Mueller, This would be n h four Oh h Now let's draw our Lewis structures for compound A component E was H and 03 and it's Louis structure for nitric acid is and plus double bond. Oh, oh, minus Yeah, yeah. Oh, and H There's a Louis structure for a for B way have N h 20 h, which is hydraulic saw aiming on Lewis structure here O h and lone pairs here or compound. See, we have ammonium they treats, which would be ammonium, the ammonium iron and nitrates, which is in Yeah. Oh, this would be a minus there on the O double bond. Oh, lone pairs ceremony. Nitrates, compound D or Sorry, ammonium nitrate in each four. No end to their Oh, no. Sorry for and through. They get it wrong. NH 403 Let's make this correction here. This is N H four n 02 Let's fix. Come. Um, this should be nitrates, not nitrate. I hear So nitrate would be and plus couple bond. Oh, lone pairs. Oh, whole minus minus And ammonium nitrate B n plus H h h h. And things should be and plus oh, minus on pairs Double bond. Oh, so ammonium nitrate and compound E waas ammonium hydroxide So and plus and a drug side Oh h minus sell ammonium hydroxide there. So those are the five compounds and their Lewis structures. Yeah,

We're just considering some more NMR scenarios here, and we're just gonna be drawing structures based on formulas that we're given, as well as the peaks that may be seen on the spectrum. So fastly I have C two h 60 here where we have one single it present. So we have one peaks, meaning that all of my protons are in the same environment. So given I've six protons, I might want to consider an element of symmetry that might be present. That will cause all of these protons to be chemically equivalent because given that we've got one single it, that means that all six protons are in the same environment. So I would start off with my central oxygen, and then we can just attach our to me 1000 groups either side. And that gives us an element of symmetry that allows us to group both of these groups of protons into the same environment. That gives us one large single it that will have an integration of six there. It has a multiplicity off, a singular as it's stated in the question, and that is because is split by no other protons that are in different environments. So given that Russian out, you can continue with the following examples that we do have and I've drawn up the structures. But in the order that we have chemical connections up on the screen and just for this last example, we do have two possibilities there. Just be mindful of that that both are correct, given the information in the spectrum.


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