Propose molecular formula for a compound that exhibits the following peaks in its mass spectrum by first finding the number

Identify the compound with molecular formula $C_{6} \mathrm{H}_{10} \mathrm{O}$ that gives the following DEPT $^{13} \mathrm{C}$ NMR spectrum:

In this question, they are asked to identify the structure that gives rise to the, um, given in the marsh spectrum. And they also do it at the molecule that corresponds to the given animal spectrum has a molecule formula, See, servant. Which 14? Oh, to be see that there are twice as much as hydrogen is combated the number of carbons. So we have this formula and if we only have twice as much as, um, Bajans instead off, uh, to end last two there are in is the number off carbons. It means that there is a single on saturation that caused to hide regions to go missing in the structure.

So let's keep in mind that does. Oh, that's probably a double bond or a cycle in the structure. So if you take a look at the at the NMR spectrum, we see that doesn't speak neo from so 10 bpm. So that, um, se 13 NMR spectrum peak is usually from a carbon in a group a peek me A 210 oh is observed in a C 13 NMR spectrum, some carbon air group. So it's also keep in mind that we have a carbon, her group.

So if we had down saturation, um, that, um, saturation could be coming from the carbon, his group. And then, um so it's pretty much the same thing that we could get out in the two first points that I just mentioned. So I'm also receded. There are only three peaks in the spectrum, out of which one is for the carbonate group. So that does us that there are only three types of carbons in this molecule.

So the other two bigs besides the Carbonell Big, um, much field in the spectrum. And ah, one of them is a doublet and our divine Issa corkage. So the doublet should most probably be coming from no sietch group. So I got, uh, in plus minus one last one equals two and the clock that should be coming from most probably met their group. So are three.

Passman is for corresponding to a card. It So these are the only three types of carbons presenting this molecule of carbon. It see edge, carbon and method carbons. And, uh, because there are certain conference in this structure, which we can tell from the molecule a formula they have dual own account for all of those carbons using under these three structures. And there's only von carbon Yakob because there is only one oxygen in this molecules of farming.

So the arrested the peaks should be coming from, um, or merely ch carbons and method carbons. So this molecule consists off single Carbonell Group, and the rest of the molecule has only CH groups and CH three groups. So they are also able to tell that all the CH groups and all of the material groups in this molecule are equivalent to each other because they only give rise to one beak for each kind. So do to fulfill all that cried here, we have to come up with this image extraction that has from equivalent CH and Mr Groups. So Destructor Bill.

I agree with that statement office. All those statements service through CH loops, and they are identical to each other so they'll go tab equivalent protons and all the material protons are going to be equivalent to each other because they're all in the same environment and the same follows for the carbons. Though methane garbine is identical, do the endemic dying carbon and ultimately groups have the same type of cover because of doubt. Um, I didn't good environment. So we also have seven carbons in the structure and 14 Hydra Jin's Inman oxygen.

So we have accounted for the structure that corresponds to the given Proton NMR spectrum..

The NMR spectrum consists of what appeared to be three signals, a small quartet, a large single it and aim small triplet that's being overlapped by the big single it. So a structure that would be consistent with that would be a stage three group bonded to S. H. Two group bonded to a tertiary beautiful group. The three methods of the tertiary beautiful group would have no neighbors and not be split.

So that would give a big single at that integrates 2 9 hydrogen. The uh huh CH three hydrogen is on. The other end would be split into a triplet Based on the two neighbors and then CH two hydrogen ins would be Split into a quartet by the three metal neighboring metal hydrogen. The quartet would integrate to to Because there are two hydrogen is creating that signal and those two hydrogen czar on a secondary carbon. So they would be somewhat more downfield than the hydrogen is on the primary carbon..

This question. Masto gives the structure off a molecule based on the molecular formula provided, which is C six H 14 Canada Proton NMR spectrum provide. So we see that the proton in the more spectrum has only two weeks, and those are really upfield because there's no electric negative atom in this molecule of formal either. And there's only cool types of broken. And this is why BC on little signals.

But the molecule formula tells us that there are six carbons, and for them toe reduced to two Cygnus. It means that there's a symmetry in this molecule, which will give rise to many equivalent types off protons that can be categorized into just two groups. And we're also told that the integral ratio for those two types of protons as 1 to 6. So what do fulfill on those conditions we have do pretty much come up with this structure. So we have six carbons here.

14 Hydra Jin's on me types off. I did just the meth I in addition, and there meant that hydrogen and the ratio between on those two is 12 three plus tree, which is six sorties, part 26 So this is the structure corresponding to the provided information.

From the spectrum for The compound with C9 H 12, we can see that there is a signal around seven parts per million that would indicate an aromatic rain And it appears that it would integrate to about five hydrogen. So it's a mono substituted ring. So that accounts for six carbons and five hydrogen, Which leaves three carbons and seven hydrogen. There are three other hydrogen signals. We have a triplet, uh a triplet and in between a uh sex 10 of from splitting.

So that would be the pattern for a proposal group. So the compound is propofol bending for the formula C five H 10 oh. We just have to signals a quartet and a triplet that are in a 2 to 3 ratio, respectively. So that's the tell tale sign of an ethyl group. And since we have five carbons and 10 hydrogen ins, we can accommodate two symmetrical ethyl groups, which would each group has two carbons and five hydrogen.

So that would be Four carbons and 10 total if we have two ethyl groups. So that leaves one carbon and one oxygen left over. So the simplest way to accommodate that is to make a key tone that has ethyl groups on both sides of the carbon Neil. Then finally C nine H 10 two. Again, we have an aromatic signal in the 7-8 parts per million range.

So we're going to have six carbons and five hydrogen is on the mono substituted aromatic rain. We have a quartet and a triplet in a 2-3 ratio. So it's going to be an ethyl group. The CH two part of the ethyl group is d shielded to four point um Past four ppm, About 4.2 or 4.3. So that means the uh That's CH two group has probably bonded to an oxygen and then we still have a carbon and an oxygen left over.

So the easiest way to accommodate that is to construct an ester group where the carbon eel of the ester is bonded to the aromatic ring and the oxygen Alcock, see oxygen of the ester is bonded to the ethyl group. So that would be the structure of ethyl benzoate..