5

31(30 points_ Compounds _ and B have the molecular fomula C,HwO. What are the structures of and B? You must identily the peaks in thc IR Hand "C Srtru that all...

Question

31(30 points_ Compounds _ and B have the molecular fomula C,HwO. What are the structures of and B? You must identily the peaks in thc IR Hand "C Srtru that allow you l0 delennine the structure. Compound Compound IR spectrum show major paks [718 cm "CompoundspectrumTnttl FlCompound "C spectrum

31(30 points_ Compounds _ and B have the molecular fomula C,HwO. What are the structures of and B? You must identily the peaks in thc IR Hand "C Srtru that allow you l0 delennine the structure. Compound Compound IR spectrum show major paks [718 cm " Compound spectrum Tnttl Fl Compound "C spectrum



Answers

An unknown compound $\textbf{A}$ (molecular formula C$_7$H$_{14}$O) was treated with NaBH$_4$ in CH$_3$OH to form compound $\textbf{B}$ (molecular formula C$_7$H$_{16}$O). Compound $\textbf{A}$ has a strong absorption in its IR spectrum at 1716 cm$^{-1}$. Compound $\textbf{B}$ has a strong absorption in its IR spectrum at 3600-3200 cm$^{-1}$. The $^1$H NMA spectra of $\textbf{A}$ and $\textbf{B}$ are given. What are the structures of $\textbf{A}$ and $\textbf{B}$?

This is the answer to Chapter 22. Problem number 82 from the Smith Organic Chemistry textbook on this problem gives us two sets of spectral information on a molecular formula and asks us to provide the structure of these two molecules. Eso as always. I think the first place to start when you're given a problem like this is to calculate the hydrogen hydrogen deficiency index. Oh, are the unsaturated number as it's sometimes called on dso. Remember to do that we do two times the number of carbons plus two minus the number of hydrogen Tze on. Then if there's nothing else in the molecule, no halogen or nitrogen Sze, it's just all of that divided by two. So for this formula, we end up with 22 minus 12 eyes 10 over two is five. So we have an HD eye of five. Um, and so right away, whenever I see an h d ie of four or more, I immediately think aromatic ring. And so for each of these, that's where I'm going to start. I'm going to draw, uh, aromatic grease. Um, okay, let me actually move. Uh, this information that I've read enough be over a little bit so that I have some space. Okay, so we'll put that right there, all right? And so for B. So the hydrogen deficiency indexes is the same for both of these molecules, since they have the same molecular formula and hydrogen deficiency index is calculated based on molecular formula. So I think that they each have an aromatic oring, okay. And then so will focus on a first eso. We were told that there is an i r. Signal at 17. 18 Wave number on DH. So that tells me that there is probably a carbon you own this molecule That is the range that we typically associate with carbon eels on dso. That's really the only piece of information that we can get from the I. R. There s o. The NMR, of course, is much more informative. And I've just written down something about each signal. This is one way to approach a problem when you're given an NMR or any more information and so we have one signal that is in the range of ah ch three group has an integration of three on DH. It is split to a triplet, so this tells me that it's a methyl group that is next to, um ah ch two of methylene and so that that makes sense. That's something that we see a lot on. The next signal is, ah, single it with an integration of three. And so this tells me that it's probably a metal group with no neighbors around it. And so I want to start thinking about what that would look like, how to add onto this molecule so that we get a metal group with no neighbors. And what occurs to me immediately is that it's probably either a methods substitute on this aromatic ring, or it's a meth a Leicester, being, as I know already, that we have a carbon you in here. So the next thing to look at is the next signal. And so it's Ah ch too well, so it's a sitting alone, has an integration of two. So it's ah, probably a CH to group. Ah, and it's split into a quartet, which means that next to it is a ch three and up here are first signal. We have a CH three that we thought was probably next to a sea age too. So this this makes sense. This confirms what we already know more what we already suspect, at least So also going back to the HD I Ah, we said we probably have a narrow matic. Great, because that accounts for four degrees oven saturation on DH then this carbon you all that we know that we have from the i r will account for our fifth degree oven saturation. So those are all the pieces there on DH. Then we see that we have Ah, this signal around seven parts per 1,000,000 7 p.m. On DH. So it's actually two signals that are each. They each integrated too. So this tells me that we have four aromatic protons and that they are arranged in two groups of two. So we can say for certain that we have a Paris substituted aromatic ring. Um and so we have this ch three, Uh, this this single it that integrates to three that we said was probably either a metal Esther or just a method group stuck onto the ring. So I'm gonna say that that this is one of these substitution is on the ring is just a metal group And that accounts for this single right here. This single it on that we said that we had. So the next thing to think about is where is our carbon, Neil? Ah. And so if we were to say that, uh, well, no. Yeah. So then the only the only other thing Only way that we can put this molecule together that makes sense, Given all the information that we have is something like this where we have our carbon deal there. We know that this has to be a nester because this's that accounts for our second oxygen. Um and so it's an Ethel Esther. And that's how all of our pieces fit together there. Um And so then if we go on, we look it b So again, we have an I. R. That's consistent with a carbon. You in this molecule on. That's all that the I R really tells us. Um and then when we look at the NMR, we see four pieces of information to the 1st 1 The one that I want to focus on first is that there are five aromatic protons. I mean, we know that because they're right around seven. That's where aromatic protons are. Nothing else is really around there. I mean, so this suggests to me immediately, eh? Mano substituted benzene room because just bending ring would have six. And so when we see that it's five, we know that there's one substitution. So this time we have a single it again that integrates to three. So, um, Ethel Group, with no neighbors on DH, then to ch to groups that because of their splitting we can tell that each of them is next to another CH to group because they're each triplets are going by the n plus one roll a triple. It means that that signal has to neighboring protons. So probably what we have here. We have a single it when the neighbors, we have to ch two groups next to each other. So let's draw to ch two groups next to each other. So 12 on DH. Then we have another Esther in this molecule because again, it's oh, too. You know that we have a carbon you. So when Esther is really what makes sense? There we go. There's that, um and then so we might be tempted Teo to draw that way. however, looking at the chemical shifts of the two methylene groups the two ch two groups, um, they are shifted fairly far downfield on the methyl group is not shifted as far downfield. And so, um, what, actually is the correct for the correct structure here is going to be like this where the Esther is this way. So it's a ch two that's directly bound, Teo the oxygen on DH, not the methyl group. Ah, and again, the only way that you can really distinguish between this structure and the first structure that I draw that a part of me that I drew eyes looking at the chemical shifts of the groups involved. Yes. So this this is the correct structure on again. The only way you can really tell is to say, Well, this one methylene group, this one ch to group is, you know, almost at four and 1/2 ppm. I mean, that is pretty far downfield. And that suggests to me that it's bound to something very Electra negative. In this case, of course, that's oxygen. So this this one that I'm drawing the Red Arrow two is going to be a our one. That's at four and 1/2. Um, yeah. And so if perhaps the metal group was further downfield on this ch to group was further up field, You could make an argument for the molecule the way that I first drew it. But since Because since this methylene group is shifted all the way to 4.5 ppm, this is really the only, uh, a molecule that are the only the only structure that that fulfills the spectral information that we were given on. So that is the answer to Chapter 22.

This is the answer to Chapter 14. Problem number 55 Fromthe Smith Organic chemistry textbook. Ah, And this problem, um, asks us to identify the structures of Ice Hammer's A and B. Ah, and we're told that the molecular formula for each of these is C nine h 10 0 on. And then we're also given one piece of eye our data in three pieces of NM or data about each, uh, A and B. And so the first thing to do in a problem like this, where we're giving a formula and asked to come up with structures, eyes to do, ah, hydrogen deficiency index eso Hopefully we remember h d. I is going to be equal to two times the number off. Sorry, I'm gonna write this little lower, so that doesn't get cut off s. So it's gonna be equal to two times the number of Corbyn's, which in this case is nine plus two minus the number of Hodgins, which in this case is 10 Ah, and then all of that over to s 0 18 plus two is 20 minus 10 is 10 over two is five. So we have five degrees of on saturation here on and so right away. What that suggests to me, especially in a molecule with so relatively few carbons. So nine is not, uh, not such a low number of carbons, but it's it's low enough that ah, lot of double bonds are unlikely. Eso remember that? Um, an aromatic ring, six members, aromatic ring of benzene ring. Whatever you want to call it, uh, this is going to account for four degrees of on saturation on dhe. So it's extremely likely that we have an aromatic ringing these molecules. Um, I would also call your attention to the fact that each of the pieces of any more data that we've been given, um, there are signals right around 7 to 8 p. P. M's, um, and with integrations of five. And so right away, I'm thinking mano substituted aromatic. Right. Um, so I think that za safe bet, um, and the hd I and those pieces of any more data back that up. Um, so then looking at each of these molecules separately. Ah, So for molecule A are we have an i r. Peek at 17. 42. Um, wave number. And so that suggests Ah, carbon oxygen, double bond. Um, okay. And so, actually, we can look at the i r for molecule be right now as well. And so that that peak get 16. 88. Wave number also suggests a carbon oxygen double bond. And so I think in each of these molecules, we have one of those as well. So now what we need to do is sort of dissect the rest of the enemy. Lord, out of that, we have. So again, we said these were going to be aromatic protons. Um, for these third signals for for each of these so aromatic, um, again, probably probably mono substituted aromatics because of the integration of five s. So then we can look at the other two signals for each. Um, so the first signal for a eyes that 2.15 ppm, and it's a single it with an integration of three s. So this is probably a metal group. Um, and so the integration of three tells me that on the fact that it's a single it, by the end, plus one role that tells us that it doesn't have any neighbors. Um, and so this, uh, second signal eyes Also a single it with an integration of two on. So this is a methylene group? Almost certainly. And so when we put all of this together we have a method group with no neighbors of methylene group with no neighbors. Ah, Corben Eel in an aromatic rain on dso Uh, a is probably going to look like this. So here is our mano substituted aromatic ring. Um, here's our methylene group with no neighbors because the carbon that it's attached to is a key tone on. Then we have a meth, a group at the end that also doesn't have any neighbors because it's also attached to Aki tone on. And so that controls all the information that we've been given for a, um and I think that's a good structure for a so then looking at B, there are some similarities. But here we see splitting. So our first signal at 1.2 ppm is split to a triplet and integrates to three. So that integration of three really really, uh, tips us off that this is ah method probably. Um and S o r. Next signal A 2.98 p. P. M. Is a quartet and integrates to two. So once again, this is probably a methylene group. Um, you know, again, the integration of two doesn't necessarily mean that, But it could mean that um and I think that it makes sense. And then if we look at these splitting patterns eso If we remember n plus one, um, the signal that we get the splitting that we get is gonna tell us how many neighbouring protons Um, each of these protons has. Eso are metal group is split to a triplet, which means that it has to neighbors, and our methylene group is split to a quartet, which means that it has three neighbors, and so these air probably next to each other. Ah, and once again, we have, um ah Carbonell. And we have an aromatic rate S O. B, I think is going to look like this. So here is our mano substituted aromatic ring. Um and then we're actually gonna put our key tone right here and then our methylene group and then our metal group. Um and so that makes sense. The methylene group on Lee has the method group protons as neighboring protons on the metal group has the methylene groups protons, neighboring protons. And so that's gonna count for the splitting patterns that we get in the 1st 2 anymore. Signals for B. Um, we have our key tone, which we know is thereby. I are on. Then we have our aromatic rank. Um and so that's Ah, that's B. And these are our good structures for and be on. They are the correct answers for A and B. And so again, the way to get there is to start by doing an HD I, um And then to just look at all the spectral information that were given and try to figure out what each piece of information is telling us on and then put all of those pieces together. And that's the answer to Chapter 14 problem.

This is the answer to chapter 20 to problem number 85 fromthe Smith Organic chemistry textbook. Ah, and this problem Ah asks us to identify the structures of D and E which your eye summers of molecular formula see six h 12 02 and were given. I are in any more data for each of them. Um okay, So the first place to start here, as always with this type of problem, is calculating in HD I. So in this case, it's going to be two time six plus to minus 12. All of that over too, um, is gonna be to over two is gonna be one s 01 degree of on saturation in each of these molecules s. So we'll start with D. The I r tells us really only that there is a carbon oxygen double bond here. Ah, the anymore. Um, we'll go piece by piece. So the 0.9 p p. M. Triplet an integration of three eyes a metal group. There's a 1.35 ppm multiple it that integrates to two. That's a methylene group. The next signal is also a methylene group. Ah, Then we have a single it at 2.1 ppm. That integrates to three. So this is a metal group. Ah, and this is pretty d shielded. Um, so maybe maybe this is Alfa two of carbon eels open now for there? Um, yes. This is possibly Alfa to a carbon you and then we have a 4.1 ppm triplet that integrates to two. So this is a methylene group. Um, but this is a pretty D shielded methylene group. Um, OK, and so I am going to say that we have. Well, sir, we have a meth, a group next to a carbon you So that would look like this. We have a second oxygen in this molecule. So what's making an Esther? Um, and we have a methylene group that is very d shielded. So we'll put that attached to the Esther Oxygen on. Then We have two more methylene groups and another method group. So 12 and the metal group. Okay. And so that's the structure for D for E v. I r. Again doesn't really tell us anything other than the fact that we have a carbon. You, um the anymore. Uh, our first signal is too. Battle groups. Then we have a 1.9 p p. M. Multiple. It that integrates to one. Um Mmm. Okay. Eso this is all right. So we'll come back to that. Um oh, no, You know what? I'm going to say that we have an isopropyl group. And so singles one into our splitting each other. Okay, Now we have a 2.1 ppm single. It that integrates to three. So this is a different method group. Um, and it is pretty d shielded. So perhaps it is Alfa to the carbon. You all that we have, uh, and then we have a 3.85 ppm doublet with an integration of two. Uh, and so this is a methylene group. Ah, and again, just like in the first part of this problem, I think that this is a methylene group that is bound to an oxygen. Um, so it's bound to an oxygen, and it is split to a double it, which means that signal too, and it are splitting each other. So we have, uh, on one side of this molecule, we have, um, oxygen methylene group isopropyl group. So, like this? Ah, and then the other side of the molecule. We have the Carbon Neil and another method group. Okay? And so that's the structure for E. Ah, and that's the answer to Chapter 20 to problem number 85.

And only by day I asked. Patrol beats each of it. Good morning to you. The first one is 4 19. Too old or 19? 21 one toe, three food fight thinking too old. 1234 I never wanted. Yeah, so if you look at the structure, there is no conjugation and soul beginning. This is an issue. Later, kiddo. Go, which appears at 1715 centimeters. Universe, that is the key to speed. And we won't know adoption this place in my breast now feel everyone over the second car bodies to die middle. Yeah, or he has threes years. Everyone's here. She was here. That is three pending. Too old He thinking too old. Yeah, Here is contribution. So the least contribution on defeat. Off softened. Relax. One thing Age. What is significant votes? The third, um, movies who do die Emulate cycle Abendanon Cyclo Pinned it all to to die Mealtime. This is a prime number. Ring he don't Andi. It appears it so plain emerging. Appears at 17.2. Sending between words. Next, my tiger. Ruben. Guilty hired Windows blown off. Absorption. Uh huh. 170 Quite 170 play. You do see everyone for stretch and then 27 to 0 and 28 to 0 27 20 and to eat do hero absorption due to leave. See it publicly to look. Thank you. What we should we do this absorption? Next come moments like low. Have you? No one? You don't hear just three supplements, you know? So there is no congregation, so it will appeal at 17 One thing Significantly, Waas and the Mexican Bowl is uh yeah, okay to Hexen are to Cinar. 12345 Fix So 284 and everyone. So this is a communicated Tito a convicted and a Alfa Beta unsaturated Randy Haiti. So see, all absorption will be I had 170 plane for contributed idea heads.


Similar Solved Questions

5 answers
Write the partial fraction decomposition for 10 (1 + )(c - 2)2
Write the partial fraction decomposition for 10 (1 + )(c - 2)2...
5 answers
Give the best reagents for the reaction(CH; CHCH;CH:C==CH(CH;A CHCH CH;CH:CHH2O, H2S04,HgSO4BH3, H2O2, NaOHK2Cr207H2, Lindlar Catalyst
Give the best reagents for the reaction (CH; CHCH;CH:C==CH (CH;A CHCH CH;CH:CH H2O, H2S04,HgSO4 BH3, H2O2, NaOH K2Cr207 H2, Lindlar Catalyst...
5 answers
12) Find the parametric equations for the line through the point P = (6, 1,2) that is perpendicular to the plane &x + 3y + 3z = 1.
12) Find the parametric equations for the line through the point P = (6, 1,2) that is perpendicular to the plane &x + 3y + 3z = 1....
5 answers
5791859 Cdoi88 8901858 ZZi~7S 081 [02 582=LE E#EL85 9E7466 58+ |8'265182*6091SL6L813 899 LSLZ4 891"L8ee56 ZLPE8L85 of 9B ucl Wbue4 L%2
57918 59 Cdoi 88 8901 8 58 ZZi~ 7S 081 [ 02 582= LE E#EL 85 9E74 66 58+ | 8 '2651 82*6091 SL6L81 3 8 99 LSLZ 4 8 91"L8ee 56 ZLPE 8 L 8 5 of 9 B ucl Wbue4 L%2...
5 answers
Sudnose Ihat & life insurance company insures 5800 40-yeat old den Vunc peop 2 andIne doaln benefitt given year (Aseirte coalh rale 0l 2pol IO0Q people 555,O00_ How much can Lhe company excec Tne cost of Ihie prenlm gamn (or lose) 5100 yeali mswance poMpully can erac Alni (Typo Ahole nurtbt )pront
Sudnose Ihat & life insurance company insures 5800 40-yeat old den Vunc peop 2 andIne doaln benefitt given year (Aseirte coalh rale 0l 2pol IO0Q people 555,O00_ How much can Lhe company excec Tne cost of Ihie prenlm gamn (or lose) 5100 yeali mswance poMpully can erac Alni (Typo Ahole nurtbt ) ...
5 answers
Give an exanpl 4 a srie ehat is Li) Convergene 8 (2) not absolutely Con Vev 'gent, ant (3) Ndl (exacllyJiaentical t0 Ere alturnatinj harmonic serics _
Give an exanpl 4 a srie ehat is Li) Convergene 8 (2) not absolutely Con Vev 'gent, ant (3) Ndl (exacllyJiaentical t0 Ere alturnatinj harmonic serics _...
5 answers
Hi-/a Points]DETAILSZILLDIFFEQ9 4.9.009.Solve the given system of differentia equations by systematic elimination Dx DZy est (Di+1)x (D =1)y BeStI(xce)y(t))lINeed Heip?RoudllIatchltMsubm Answan
Hi-/a Points] DETAILS ZILLDIFFEQ9 4.9.009. Solve the given system of differentia equations by systematic elimination Dx DZy est (Di+1)x (D =1)y BeSt I(xce)y(t))l INeed Heip? Roudll Iatchlt Msubm Answan...
1 answers
How many grams of liquid methanol must be combusted to raise the temperature of $454 \mathrm{g}$ of water from $20.0^{\circ} \mathrm{C}$ to $50.0^{\circ} \mathrm{C} ?$ Assume that the transfer of heat to the water is $100 \%$ efficient. How many grams of carbon dioxide are produced in this combustion reaction?
How many grams of liquid methanol must be combusted to raise the temperature of $454 \mathrm{g}$ of water from $20.0^{\circ} \mathrm{C}$ to $50.0^{\circ} \mathrm{C} ?$ Assume that the transfer of heat to the water is $100 \%$ efficient. How many grams of carbon dioxide are produced in this combustio...
5 answers
Jc} 267 Lw Isf Sinz; %rea" iu' @z < # Js 6 € > I3-118-! anismg)
Jc} 267 Lw Isf Sinz; %rea" iu' @z < # Js 6 € > I 3-118-! anismg)...
5 answers
ACT/SAT The graph of which of the following equations is symmetrical about the $y$ -axis?A $y=x^{2}+3 x-1$B $y=-x^{2}+x$C $y=6 x^{2}+9$D $y=3 x^{2}-3 x+1$
ACT/SAT The graph of which of the following equations is symmetrical about the $y$ -axis? A $y=x^{2}+3 x-1$ B $y=-x^{2}+x$ C $y=6 x^{2}+9$ D $y=3 x^{2}-3 x+1$...
5 answers
Decide whether each pair 2lements torned the spaces provlded_the table below will formionic compoundthey will, write the empirical formula and name of the compoundampirical formula ionic comddungFoths lonicelement #1clement #2namtionic compoundcompound?ScammCsicimmSumtutCesimncesium sulfidemagnes VMauodnePariumurygunbasium Oxide
Decide whether each pair 2lements torned the spaces provlded_ the table below will form ionic compound they will, write the empirical formula and name of the compound ampirical formula ionic comddung Foths lonic element #1 clement #2 namt ionic compound compound? Scamm Csicimm Sumtut Cesimn cesium s...
5 answers
And experiencing anet (orce that is directed to the lell The magnitude of the force is constant The Asledis moving to the lelt specdof the sledisincreasingdecreasingconstantNoanswer text provided.
and experiencing anet (orce that is directed to the lell The magnitude of the force is constant The Asledis moving to the lelt specdof the sledis increasing decreasing constant Noanswer text provided....
5 answers
J iNje 1 { 7 1 F F 6 1 LiM?
j iNje 1 { 7 1 F F 6 1 L iM?...
5 answers
8"-3'+J = 6xe72 Epinena ( Sdbtl
8"-3'+J = 6xe72 Epinena ( Sdbtl...
5 answers
(VO pUs) Let Vbo Iha space spunrvd by Ilg tko lurictions lerllo Ino 0a5 co ((Jand vinkt) {cos(t).on()} Fimd lhe malnx. Ol lh kneat tansbmuton T ( J()) f"(0 7f"() sf(t) brorn V mito dsoll mth
(VO pUs) Let Vbo Iha space spunrvd by Ilg tko lurictions lerllo Ino 0a5 co ((Jand vinkt) {cos(t).on()} Fimd lhe malnx. Ol lh kneat tansbmuton T ( J()) f"(0 7f"() sf(t) brorn V mito dsoll mth...
5 answers
These hypotheses; are There Question eg haorhakes explail an argument AGAINST it being the correct explanation: why biodiversity is higher near the Lqualon 1 State cne &f Eio
these hypotheses; are There Question eg haorhakes explail an argument AGAINST it being the correct explanation: why biodiversity is higher near the Lqualon 1 State cne &f Eio...
5 answers
Question 26Which of the following statements correct?the lattice energy 0f KCl is higher than the lattice energy of Mgothe lattice energy of LiF is higher than the lattice energy of LiBrthe lattice energy of NaCl is higher than the lattice energy of Mg3Nzthe lattice energy of Klis higher than the lattice energy Na2o
Question 26 Which of the following statements correct? the lattice energy 0f KCl is higher than the lattice energy of Mgo the lattice energy of LiF is higher than the lattice energy of LiBr the lattice energy of NaCl is higher than the lattice energy of Mg3Nz the lattice energy of Klis higher than t...

-- 0.022053--