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Draw the structure of an unsaturated hydrocarbon that has a molecular ion with an $m / z$ value of 128 ....

Question

Draw the structure of an unsaturated hydrocarbon that has a molecular ion with an $m / z$ value of 128 .

Draw the structure of an unsaturated hydrocarbon that has a molecular ion with an $m / z$ value of 128 .



Answers

Draw the structure of an alkane with molecular formula C$_7$H$_{16}$ that contains (a) one 4$^\circ$ carbon; (b) only 1$^\circ$ and 2$^\circ$ carbons; (c) 1$^\circ$, 2$^\circ$, and 3$^\circ$ hydrogens.

This is the answer to Chapter 13 problem number 21 from the Smith Organic Chemistry textbook. Um, and in this problem were given two molecules while we're going, two sets of information about molecules on were asked to determine a structure for each of these molecules based on the information that we've been given. And to me, this is where organic chemistry, like, really gets fun and interesting, because it's it's really just an exercise in problem solving, which I like a lot. So hopefully you do, too. So basically, we just need to use the information that we've been given, um, and determine, uh, different things about the molecules based on that information on, then we can propose a valid structure. So for a we're told first that a is a hydrocarbon. So this is good because it tells us that there is on li carbon and hydrogen in this molecule s. So then were given an mtz a molecular ion of 68. So our formula is goingto have to equal 68. Um, And again, we have on li carbon and hydrogen here. Um and so I think that the way to approach this is probably just to think, Um okay, well, we only have carbon and hydrogen. So how many carbons can can an M to Z of 68 accommodate? And so the answer is 55 carbons will account for 60 of that 68. Ah, And so we can then fill in what's left with eight hydrogen sze. Um and so if we, uh, then do, Ah, hydrogen deficiency index based off of that C five h eight. Um, so it's gonna be two times the number of carbons plus to minus the number of hydrogen cz all of that over to s. So we get a hydrogen deficiency index of two. Um, And so if this is indeed, um, our correct molecular formula, then we're gonna have, umm to pie bonds to account for or two rings or pi bond and a rain. Okay, um and so we can now look at the i r. Data and see what that is going to tell us. So our first piece of I r data, um, is, ah, peek at 33 10 wave number. And so that is going to correspond to an SP hybridized carbon bound to a hydrogen. So we have a C S P h. Bond. So that tells me that we probably have a triple bond, a carbon carbon triple bond in this molecule. And that would be good, because that by itself would account for the two degrees of on saturation that we determined when we did the hydrogen deficiency index. So you know, that's great. Um, so then we have our next peak at 3000 to 28. 50 s o by now. Hopefully, um, you haven't committed to memory that that is gonna be a C S p three hybridize carbon bound to hydrogen. I mean, so lastly, we have a peek at 2120. Um, and that Peter 2120 is also great because it confirms the presence of this triple bond. So we have a, uh ah. Carbon carbon triple bond. Um, Right. Okay. So that carbon carbon triple bond taken in conjunction with the peak it 33 10 which was on SP hybridize carbon bound to a hydrogen. Uh, you know, further confirms the presence of a carbon carbon triple bond in this molecule. And so, given all of those pieces of information, I think the easiest way to fulfill all of that is to say OK, so here is our carbon carbon triple bond will make a terminal one so that the carbon has a hydrogen bound to it. Um, again, this piece of information tells us that it has to be a terminal, uh, triple bond. And so then we can just make the rest of the molecule linear. Why not? It's simple enough. So ch two ch two and that's 1234 carbons on then ch three. And so, um, we can quickly count. So there's five carbons in 1358 So eight hydrogen sze So it fits with our molecular eye on it fits with our pieces of i r data. Ah, and it is a hydrocarbon, as the problem stipulates. So I think that that's ah, that's a good answer to a I'm so then looking at being weaken, do exactly the same thing. So we'll follow the same steps in the same order here. So ch and oh, and this molecule. Okay, so it's not a hydrocarbon. It's an alcohol or a key tone or an ether, something with an oxygen in it. So we have an M to Z of 60 here. Um and so this time it's not as straightforward as just saying, you know how, how many Hydra or how many carbons could this accommodate? Because we could say it's five carbons and that would get us exactly 60. But we have hydrogen and oxygen as well. Im so four carbons would be 48 but oxygen is 16. So four carbons isn't going to work either, because we know of at least one oxygen. I'm in 48 plus 16 is 64 that's too much. So let's go with three carbons. Two starts of C three, um, and then if we assume that we have only one oxygen so three carbons is gonna be 36 on oxygen is gonna be another 16 so that'll get us to 52. So that leaves room for eight Hydrogen sze I'm so our formula could be C three h 80 and I'm gonna go ahead and just assume that that's the correct formula. Um, and do the hd I So again, hd, I's gonna be two times the number of carbons plus to minus the number of hydrogen. Is all of that over two on. It ends up being eight minus eight. Um, over to so zero over to so zero s ono on saturation here. Um, so that's good. We don't have any double bonds. We don't have any rings. So we can look now at our i r. Data. And so this first piece of I R data, this broad peak at 363,200 is going to be from a no h stretch. So that tells me that we have an alcohol here. Um, and then our other piece of eye, our data is just a C S p three hybridized carbon bound to hydrogen. So C S P three h. So, um, this is a pretty straightforward molecule. Then we have some carbons bound to each other and to Hydra Jin's, and we have an alcohol. Um, and so the easiest way to draw that is just as, ah, straight proven all ch three ch two c h 20 h. Okay. And so if we want to count, we have three carbons. We have eight. Hydrogen is and we have one oxygen. So that fits our formula. So it fits our molecular eye on that we were given. And we have bonds that would account for the I R peaks that we have and you know, no other bombs that would account for peaks that we haven't been given. And so I think that this is a good structure, that for an answer to Port Bay, and so that's the way to approach this problem. We just need to individually consider each piece of information that we've been given. Ah, and use whatever it's telling us, um, and then take all of those pieces of information in aggregate to determine a formula. Um, yeah, and so I think I think the way to do it is to use, you know, they tell us what type of molecule artists, so start there, then look at the molecular eye on and try to determine a formula. Having determined a good formula, calculate an HD eye on, then used the eye. Are each piece individually to determine what type of bond that I our peak is describing on and then put all that together. And so that's the answer to Chapter 13. Problem number 21

This is the answer to Chapter four. Problem number 37 fromthe Smith Organic chemistry. Textbook on this problem asks us to draw four Al canes and for each one were given a specific parameter to meet. Um And so for a let's see, So for a were asked to draw a now cane that contains only primary and co ordinary carbons. And so the easiest way to do this, I think, is just too Draw one ordinary carbon and surround it with primary carbons. Okay, And so here we go. Um, all the methyl groups are primary carbons, and central carbon is co ordinary. Ah, and that fulfills what we're asked to draw for a So then for be, um we're asked to draw a now cane that contains on Lee secondary carbons. Um, and so the easiest way to draw on al cane that contains only secondary carbons. Um, at first you might think, you know how. How can I do this? Any terminal carbon and a chain is gonna be a primary, you know, a metal group. Um, and so the answer is, don't make a chain make a ring. Um, so you could do a five member during six member during. So this is a six member during a cyclone back saying that I've drawn here. I'm in. All six of these carbons are secondary carpets there, each bound to two other carbons into two hydrogen sze Um And so that's a good solution to that problem on again. You could draw a different sized ring on and for a you could have drawn something else too. There are multiple possible solutions to each of these. Um I'm just giving you the ones that I think are the easiest to see. Asks us to draw um, an Al cane that contains only primary and secondary hydrogen Sze. So remember, we're gonna, um ah uh characterize a hydrogen based on the carbon to which it's attached. So if we want an owl cane with primary and secondary hydrogen sze were were drawing and cocaine with only primary and secondary carbons on dso again, the easiest way to do that eyes to just drop roping. So this is gonna be two primary carbons with a secondary carbon in between them. And so, um, each of the terminal carbons air gonna have primary hydrogen tze and the carbon in the middle will have secondary hydrogen Tze And so that's ah, simple solution there. I mean, really, you could draw any, um, saturated out cane and that would fulfill See, so you could draw non in. You could draw dough deck ain, um you know, you could draw anything as long as it's just ah, straight chain on the end, pieces will be primary and everything in the middle will be secondary. So then, for d were asked to draw um, an Al cane that contains only primary inter. She ery hydra jin's on dso. This is gonna be similar to a but the difference is gonna be We want a primary hydrogen in there. So instead of putting four methyl groups on this central carbon, uh, we will put one hydrogen and three methyl groups. And so again, arm ethel groups are gonna be primary carbons, and so have primary Hydra Jin's I'm and then the central carbon here is gonna be tertiary, and it's gonna have a tertiary hydrogen on it. On dso That's ah, that's ah, possible solution to D. Um yeah, and so that's that's how to approach this problem again. There are other structures. You could draw for each of these, Um, these air just, in my opinion, that the easiest structures that fulfill other requirements for each part of this problem and that's the answer to Chapter four problem number 37.

This is the answer to Chapter 13. Problem number 27 from the Smith Organic Chemistry textbook. Ah, And in this problem, we're asked to propose four possible structures for a hydrocarbon with a molecular ion at M to Z 1 12 So, um, this problem tells us this has to be a hydrocarbon. So we're talking only carbon and hydrogen. Um, and so the first thing to do is to look at this mtz 1 12 and see what type of molecular formulas could possibly fit here. Um, and so, um, so, let's see. So, nine Corbin's ah would be, um, 108. Um and so that's probably wrong, because then that only leaves us four. Hydrogen is and nothing carbons. Um, so let's try eight carbons. Uh, eight carbons would be 96 and so that's gonna give us 16. Hydrogen is to work with. And so I think that that is probably the correct formula here. So c eight, age 16 um, seems reasonable. So the next thing I'm going to do is going to be to calculate the HD eye for this formula. So the hydrogen deficiency index for C eight age 16 is gonna be two times the number of carbons plus two minus the number of hydrogen sze all of that over too. So we get 18. Minus 16 is to over two is gonna be one. So this tells us that we have one ring or one double bond in each of these structures that were going to propose. And so I think, um, obviously the easiest structure is gonna be just a straight chain. So once you 345678 um, and then we can put a double bond somewhere. Um, okay. And so another simple structure would be to go 1234 56 on and then maybe put ah and Ethel group here. Um, so 78 And then again, we can put a double bond pretty much wherever. Um okay, um and so we could continue moving our substitutes around and continue doing straight chains. But just to spice it up a little bit, let's start some rings in so we could do a six member dring with an ethel substitue int. Or we could do a five member dring Ah, with a propulsive situated. So there we go. Um, we could also do Ah Di substituted five member dring. We could do a six member during with two metal substitutes. So there are. There are many, many structures that we could propose that would fit our parameters here. But here are four simple and straightforward ones. Um, again, though, you know, if you come up with other ones, that's perfectly fine a cz long as you're sure that they fit the requirements. And so I think, um, I think ch 16 is definitely the way to go. Um, because otherwise we would have too many carbons or too few carbons. So this makes the most sense. Um, yeah. And so once, once you determine what your formula is gonna be from the M to Z ratio, then get ah, hydrogen deficiency index. So you know whether you need to incorporate a double bond or ring and then with those parameters in place, just draw structures that I said the bill, so to speak. Um And so that's the answer to Chapter 13. Problem number 20

This is the answer to Chapter 13. Problem number 33 from the Smith Organic Chemistry Textbook. Yeah, and in this problem were given four sets of data and asked to draw a molecule consistent with each set of data. A So for a were asked to draw the molecule that contains a benzene ring and has a molecular ion at M Dizzy 107 And so that one of seven is a clue. Um, odd number for a molecular ion often indicates the presence of a nitrogen. I'm and so I'm gonna move forward with the assumption here that we have a nitrogen in this molecule on dso. Benzene is worth 78. But if we make it a mano substituted benzene, that's gonna be worth 77. And so if we, um so 77. So we need to account for 30 more. Um, and one of them has to be in nitrogen. So let's add a methylene group. So that's gonna add 14 for us. I'm in. So that will take us to 91. Then we need to add 16 more, which is easily accomplished with an N H two group primary mean, um, so that will add 16 for us. Um and so that equals one of seven. Um okay, so we've proposed Ah, consistent structure for a And so that's sort of Ah, the methodology that I'm gonna use here. Um, whatever. Whatever the problem gives us, we'll just work with and sort of fill in what we need to hit the target number. Um, so for B b is a hydrocarbon that contains only SP three hybridized carbons and a molecular ion at M to Z 84. Okay, So hydrocarbon only SP three hybridized carbons in a molecular eye on it. M to Z 84. So let's see, eh? So we're told hydrocarbon, and we're told M to Z equals 84. So that's not a lot to go on. But we can start guessing at a molecular formula from this, So MTZ equals 84 s 08 carbons would be 96 7 carbons would be, uh, 84. Exactly. We're not gonna have a molecule. It's just carbon a select ical. One less. Let's go. Six carbons. Uh, which is gonna be worse? 72. So then we will have to have 12 hydrogen sze to make up? Um, the remaining 12 that we need and so we could do a quick HD I hear for C six age 12 hd. I's gonna be two times six plus two minus 12. All of that over to so 14 minus 12 is to over two equals one. Okay, um, so the only way that I can see to do this uh, well, no, not the only way I'm. But a simple way to do this is to just make cycle vaccine. So that's gonna give us one degree oven saturation. It's gonna be six carbons and 12 hydrogen tze and it's gonna give us ah molecular ion at 84. So that's Ah, simple solution there. Um okay. And so moving on to see ah, compound that contains a carbon your group and gives a molecular ion at MT. Z equals 1 14 So, for C, we know we need a carbon. You and we know we need an empty Z of 1 14 Okay, um so again, this isn't, uh, too terribly much to work with, but we know that we're gonna have one oxygen. So one oxygen at while at least one oxygen. Right? But so one oxygen would be worth 16 which is going to give us, you know? So let me do it this way, then. So we have 1 14 We can get rid of 16 for the oxygen. That's gonna leave us with 98. So then we can think about numbers of carbons. Eso Let's do seven carbons. So seven carbons is gonna be 84. Um, and so that's gonna leave us with 14 Hydra jin's. So the formula that I'm gonna propose here is gonna be C seven h 14. Oh, um, okay. And so again, we can calculate a quick HD, I hear. Um, so it's gonna be two times seven plus to minus 14 over to it's gonna be ah, 16 minus 14 is to over two equals one. So the only double bond that we're gonna have here is our carbon eel. So let's draw, um, obtain chain 1234567 And let's throw a key tone in there pretty much wherever we want. We could do an album hide. We could put the key tone elsewhere. I'm just doing it there because it's nice and in the middle. Um, yeah, And so that's gonna fulfill all of our requirements for C. Um, okay. And then so lastly, D. So for D. We're told to make a compound that contains carbon, hydrogen, nitrogen and oxygen and has an exact mass for the molecular ion at 101.841 Um, Okay, so let's see here. Um Okay. So what's right? Our information as I have before. So carbon, hydrogen, oxygen, nitrogen. Um, exact mess. It's gonna be 101 0.841 point 08 for one. Okay. Um Mmm. Okay. So we can, uh, really approach this similarly toe how we've approached this before. Um, so just give me a second, because I'm gonna have to get back to that table and the chapter that lists of the exact weights of things, because that will come in very handy here. Okay. All right. There we go. Okay. So table 13.1, um, I have a known page for 92 of the textbook, So table 13.1. Lists of the exact masses of common isotopes of carbon, hydrogen, oxygen and nitrogen. Nitrogen is gonna be worth 14.31 Um And so we we, uh we have an odd number here when I won, so we know that we're probably gonna have one nitrogen. Eso won nitrogen is gonna be worth What did I say? 14.31 Okay. And so that's gonna leave us with, um, 87 0.810 Okay, um, so let's see what else we might have here. Um, so let's ah, assume that we have a single oxygen. So that's gonna be worth 15.9949 0.99 for nine. Uh, give me one moment, Thio, actually do the math here. Okay, So, uh, 87.810 minus 15.9949 is gonna be 71.861 Um, okay. And so now we should think about, um, carbons and hydrogen sze. So 71 point 08 six point. Um, Yeah. Okay. So carbon is worth 12. Exactly. Um, so let's say, for argument's sake that we had 60 so five carbons, which would be worth 60 So let's go see five. Um, that's gonna leave roughly 11. H is Ah, And then we said one nitrogen and one oxygen. Um, OK. And so, um, we could do the math here, So nitrogen also Okay, s O hydrogen is worth 1.78300 73 times 11 is gonna be 11.8613 Um, And then once we have left, we had OK, good. That's that's perfect. So then, plus 60 for carbon. Um, and then plus 15.9949 If the oxygen and plus 14.31 for the nitrogen is gonna be 101.8413 So that is perfect. So this formula is perfect. We now just need to draw something that's going to incorporate all of these pieces. Ah, and we should probably start again with getting an HD eye because it tells us exactly how many double bonds or rings we're gonna have. Um and so this is gonna be two times the number of carbons hoops. So this is gonna be two times the number of carbons. Uh, plus two minus the number of hydrogen sze. Um, so in this case, because we have one night's Jin. We need to add one. So it's plus the number of nitrogen Sze. Um and so that is going to get us to 12. Minus 11. Plus one is two over two is one. We have one ring or one double bond here. Um and so to make this easy and incorporate the oxygen and the nitrogen, I'm just going to draw a name ID. And so that's gonna look like this s o. We can just put it right here. So an h on. And then how many more carbons to account for? Three more carbons s. So 123 So there we go. There's a name. I mean, that certainly fits the data that we were given for party. Um, yeah. And so that's the answer to this problem. That's the way that I would approach it. Just use what you're given to try to derive a formula. Once you've derived a formula, find the HD I, um and then draw something that fits your formula. Fits your HD. I I mean, that will give you the molecular eye on that you're looking for. And that's the answer to Chapter 13. Problem number 30


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