what-molecular-ions-would-you-expect-for-the-compound-depicted-in-the-ball-and-stick-model

Name: what molecular ions would you expect for the compound depicted in the ball and stick model
Uploaded: 2019-05-18
Duration: 2 min 23 s
Channel: Numerade
Description: What molecular ions would you expect for the compound depicted in the ball-and-stick model?

Question

What molecular ions would you expect for the compound depicted in the ball-and-stick model?

Ian Kaigh · Accepted Answer

This is the answer to Chapter thirteen, problem number four from the Smith Organic Chemistry textbook. Um, and I would like to begin by saying I&#x27;ve ah, I&#x27;m working from a pdf copy of this book, and I&#x27;m having a lot of trouble seeing this ball and stick model that they have drawn Ah, in the pdf copy. So I believe that the molecule that I&#x27;ve drawn at the top here the cycle of butane with an alcohol, is the correct molecule. And if it&#x27;s not, I do apologize. But even if it&#x27;s not, uh, this solution should give you the ability to find the correct answer, even if I&#x27;ve drawn their own molecule here. And so basically they&#x27;re asking us what the molecular ion peak is going. They and so remember the molecular ion peak corresponds to the weight of the molecule, so all that you need to do is count the different types of atoms in the molecule and then multiply that number of each Adam by each Adam&#x27;s atomic weight. So, for example, in the molecule of drawn here, we have six carbons carbons each way twelve, So we have a total of seventy two. We have twelve hydrogen CZ at one a. M. U apiece for a total of twelve and one oxygen with a weight of sixteen for a total of sixteen. When we add thes three totals together, we get one hundred total on. So we would expect the molecular ion for this molecule to show on the mass spec at one hundred. Um, And again, if if I&#x27;m not seeing this molecule correctly and I&#x27;ve drawn it wrong here, this explanation should still allow you to Okay, find where the molecular ion should be. Really? For any molecule, just count the carbons. Multiply that by twelve. You know, Hodgins, multiply that by one count, the oxygen&#x27;s multiply that by sixteen and then add those numbers together on. So that is the answer to Chapter thirteen. Problem number four.

Ian Kaigh · Answer

This is the answer to Chapter thirteen. Problem number twenty four, Fromthe Smith Organic chemistry Textbook on. So this problem is giving us five molecules and asking us where the molecular ion peak is going to be by mass spec. And so the way to approach each of these is too. Write the formula out and then do the math to figure out how much each molecule ways and that weight is where your M to Z should be. And so the first molecule is benzene is benzene on? We know benzene is C six age six. So six times twelve plus six times one I was going to add up to seventy eight. And that&#x27;s where our molecular ion peak should show s o. Similarly, for be the formula is C ten h sixteen. That adds up to one hundred thirty six for sea C five h ten Oh adds up to eighty six s o for D N E. It is a little more complicated because remember, chlorine and bro mean occur in two isotopes so that the two most common isotopes are both pretty abundant. And so we&#x27;LL see the Emma Peak as well as the M plus to peek s O these molecules add up. Uh, well, so you approach them the same way. So you would write out their formula, uh, add them up, and then also add two and note that you would see both peaks. So for D, you would see a peek at one of six and a peak at Wanna wait. Similarly, for a you would see a peek at one ninety two and one ninety four, and that is the solution to prom.

Ly Tran · Answer

All right. So the question as us to name this compound, these compounds. So, uh, us, we&#x27;re too late, boy. 12 3456 Because this way, the substitute and has the lowest number. So the structure will be 33 die. Mazel side go hexane. Or in this one, we can name the two group. This group is, uh, as Bhutto. And this covers either. So it will be Aysel Beautiful easel either. Ah, this case we&#x27;re gonna name using the boxy nomenclature so you can get 12345 So one s o one to a boxy, hinting cycle. Benton. Hey.

Andrew Jensen · Answer

All right. So for this problem, we&#x27;ve been given a few ah balling stick models for several molecules, and we&#x27;ve been given the task to determine the molecular formulas from these violent stick models. So the first step is gonna be identifying which Adams are present in these molecules. So for the 1st 1 we can see there are two light blue spheres which are indicative of nitrogen atoms here. And you can confirm this by looking at how they have three bonds and then the little white spheres were going to be hydrogen atoms, which you can confirm by seeing how they only have one body eso because we have to act regions like this, we&#x27;re gonna do end. Then we&#x27;re gonna mike there two of them use your subscript to here and then because there were four hydrogen, we&#x27;re gonna put the hydrogen next, and we&#x27;re gonna put H and a subscript for Did you know that there are four hydrogen is in this group? Um, another way to right. This would be in h. So we take each group here and we say that each half of this molecule is his own little subgroup and was going to say there two of them. So NH two and then another sub to that&#x27;s our first molecule for our second molecule. We&#x27;re gonna do a similar process. We&#x27;re going to start in the left here. So the little white ball, us fears are still hundreds, and the black ones are carbon. The red one is oxygen. So you could run this by looking at all their bond numbers, and they matched what we would expect. Um, so if we&#x27;re gonna start here on the left side, we&#x27;re gonna start with the carbon, which is the black one. We&#x27;re gonna start with the whole group, so we&#x27;re gonna add it three hundreds. Then we&#x27;re gonna move on to the next carbon, so there&#x27;s gonna be a carbon here which has two hydrogen is attached. And then there&#x27;s an O age group right there. So we&#x27;re gonna add that a wage right here, and this is convenient to write it this way where we have three distinct groups, like here, here and here because we could see this kind of the structure reflected in the formula. Whereas if we just wrote it is if we wrote it here on the top with the NH four originally up here after just two C two h 60 That&#x27;s actually kind of ambiguous because that could be the structure for other ethanol or, ah, dimethyl ether. So this gives us a little bit of, ah, clue that it Oh, it&#x27;s ethanol and that can help us in future analyses. We need that from here moving onto the next one. We have the structure here. So these these slightly darker spheres were gonna be phosphorus, actually, so they also form three bonds here, and the red ones are still out our oxygen. So but this one, we&#x27;re going to count that we have four Bosphorous Adams, and we&#x27;re gonna put him right here, so it&#x27;s for phosphorus, and then we have six oxygen&#x27;s, so we&#x27;re gonna put our oxygen here. 06 writes. We get p 406 The next one we have here is going to be a similar case to the one that we did second. So we&#x27;re gonna start by looking at the color. So these roll Adams we&#x27;ve seen before, You know, no new fancy ones. We&#x27;re gonna start with the far right side here, so we&#x27;re gonna start with our carbon with our three hydrogen zahn it so ch three Then we&#x27;re going to do another three ch two. So what we&#x27;re gonna do is we&#x27;re going to something is gonna be a short cut instead of writing ch two out. You know, three times you&#x27;re gonna do it one time, so we&#x27;re going to see if it&#x27;ll catch up to me. Sorry. When sickness is lacking still again, that&#x27;s going to see each to cause there to hydrants attached to each carbon. And because there are three of these groups were going to do a three right there. It&#x27;s gonna go ch three from here. What we&#x27;re gonna do is we&#x27;re gonna go to the end of that whole thing, and we&#x27;re gonna put the carbon, the oxygen, but because they&#x27;re too oxygen&#x27;s, we&#x27;re gonna put an 02 and then we&#x27;re gonna put that last hydra around here and note that this ending is also commonly written a c 00 h. And that&#x27;s ah, fairly common Mallika. The subgroup you&#x27;ll have on various molecules those air to equivalent rates of ways of writing it, and they&#x27;re both equally valid. The last one we&#x27;re going to looking at here is the eyes molecule where you have to kind of start the middle here. So we&#x27;re going to start with our carbon against we&#x27;re gonna see each to and then I&#x27;m gonna go. Oops, issue. Now I&#x27;m gonna go, and I&#x27;m gonna add this this o h group right here. So oxygen and hydrogen noting that they&#x27;re together. And now we&#x27;re gonna go back to our next carbon C H two with their hydrogen is on it. And then we&#x27;re gonna add a chlorine because that green is a chlorine, which you can check by only having one. But so see, it&#x27;s juicy, and this skin gives us a little bit more insight into the structure of this molecule.

Natalie Almond · Answer

So for this question, we want to refer to the diagrams in the book and let&#x27;s start with Ace because he will dio so first what we should dio to say how many of each element we have. So for carping for this one, we have just one we have for hydrogen that then we have one oxygen. So let&#x27;s start with our Mitchell cap Wishes are ch three and then we see the O H group. So this is one that we know it&#x27;s math. That&#x27;s our first because of that siege three group and then o h. We know this is gonna be knoll. So it&#x27;s methanol, so we can go ahead and find how much this ways we have 12 for our one carbon. Then we have four for hydrogen. Now, then 16 for our oxygen. And we find that this is 32 0.4 okay, before two p going to do the same thing instead of oxygen. For this one, we have Corinne. So we have six carbons for hydrogen. Zwei, have 10 and then we have that. We have 13. My bed. We have one chlorine, and we see that this one chlorine is on our fifth carbon. So let&#x27;s go ahead and do our first part. So we see first our metal and C h three. Next, we&#x27;re going to see how many ch twos we have. We can count them and see that we have three. And then we have just a C h and that&#x27;s Onley one hydrogen. Because the other thing, this carbon is bonded to its R C l. And then we have our waas ch three. So naming this we know that on the second when we number them officially because we see that it was are one three is 4/5 carbon in this formula. But when we&#x27;re numbering carbons based on their priority, we see that it is actually the second that has the chlorine. So it&#x27;s too chloral. And since it&#x27;s six, we&#x27;re gonna have hex and then just came because we have no no or anything like that. We have no oxygen&#x27;s involved. So then we can find that we have six times 12 plus 13 times 1.1. Then we have our plus 35.45 the Corinne. You find this to be hired and 20 0.6 then we confined. See, have carbon hydrogen oxygen for carbons. We have five. Hydrogen is we have 10 in Austin&#x27;s. We have to. So we again are starting with that C H three camp You can&#x27;t How maney siege twos we have. We have three once again, But then we&#x27;re ending in a c o h because of our c o h. Here we know that our name is gonna end in Oh, I see. And since we know we have five carbons, we need a Penta. So we can name this pento no IQ acid. And we can find the weight just like any other. Got five carbons, 10 hydrogen ins. But then two auctions e and we find that this is equal to 102 0.1. And it&#x27;s you looking at D carbon. Hydrogen oxygen. We have four 10 him one. So we&#x27;re starting with that C H three at one end, just two ch Because attached to the ch we actually have a ch three or a methyl group. Then we have seen h two. Ohh. Then we look at the numbering of the carbons will see that the metal is on the second core in. We have to Bethel. And then here we have a one appropriate all and our to that arm Ethel&#x27;s actually attached. We see here, here&#x27;s her metal. And then if we want to find Dean molecular weight, we have four times 12 waas 10 times, 1.1 waas. One times that, 16 for the auction and we get 74.1. That&#x27;s it for this problem.

Natalie Almond · Answer

So for this question, we&#x27;re gonna wanna look at the diagrams in the book E and we&#x27;re going to start with a So first we should count how many carbons we have and looking at diarrhea, we see that we have five. Next issue County hydrogen is we have looking at the different hydrogen configurations and where they are. We see that we have 12 then Lastly, we see that we have just one oxygen and that should take care of all the elements so we can go ahead and look at where it would be. So we see a c H three on the end Then when we look at how many ch twos we have? Yeah, to that we have just a ch because this fourth carbon in this line not in the diagram myself. But here we have 1234 is attached to the O H group that we have it with another methyl group. So what? We&#x27;re naming this when we look at how are carbon zehr numbered? We find that this O age group is on the second carbon and we have five carbons total Wishing that means that we know is going to be a two pent, Uh, because it&#x27;s five within the knoll because that oh, Austin based group So now if we want to find how much this waste you can dio, we have five carbons. We have 12 higher shins and then just one oxygen and we get 88.12 So now we can move on to be We&#x27;re going to do the same. Think, let&#x27;s start first with how maney carbons we have. We only have three for hydrogen zins look like we only have six. And then for oxygen&#x27;s we have to Let&#x27;s go ahead and we see I started this each three one c h two of them for our last carbon. You have a c 00 h group. So when we&#x27;re looking at giving this a name, we see that we have only three carbons. And because we have this C o age group, we want end in Oh, I see. So we have a probe or a number of apartments and then our and no wick because of that C O h. And now we can go ahead and find how vicious ways three carbons, six high intrusions. Oh, there&#x27;s no one here. And then we have two questions. So this gives us 58.6 so we can move on to see Harvin&#x27;s. We see that we have four for hydrogen is we have nine. We don&#x27;t have any auction here. We do have a borough mean, though, just one. So we&#x27;re gonna put that bro me first. They won&#x27;t look at how many seats tools we have in between. The next method group. You see that we have three, and then our last carving to make four Is that cap C H three group. So when we&#x27;re giving this a name, we see that the roaming is on the first Kurban. So we do one, bro mo. We have four carbons, which means that we need a butte. And there&#x27;s no a sh reporting things. We could just do a. Then we know that we have four carbons, nine hide regions and then one bro mean, you know, we get 13699 and then we move on to do our very last, which is deep. We have four carbons, seven hydrogen Zinn&#x27;s auctions, looks that we have just two. But then we also have one Korean group. So let&#x27;s start with Darcy. Age three, Cab that we have a just ch. Because this one is attached to our chlorine. Then we have one C h two before our C 00 h. Okay, Now we can go ahead and name this. You see, the chlorine is on the third carbon. So let&#x27;s make sure we do that just like seaward Duke Wawrow. And then we have four. So it&#x27;s gonna be Butte again. But since we have the c o h we&#x27;re gonna do are spree are Oh, I see. And this isn&#x27;t acid. This one is also an acid up here for Be sorry about that. And we know it&#x27;s ask. Is this hydrogen this acidic? And now we can go ahead and find how much this ways. 12 times four in 35.45 So we find this to be 122 0.52 and that&#x27;s it for this problem.

Problem 4 Easy Difficulty

What molecular ions would you expect for the compound depicted in the ball-and-stick model?

Answer

$m / z=118$ and $m / z=120$

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