hypervalent_iodine

Then what for?

Right. Your data makes not much sense given your spectra. The peak at 6.51 and 6.30 look to me to part of one and the same doublet, so I am unsure as to why your data lists them separately. Even more confusing are the ranges they appear to have used to calculate coupling constants (the little blue lines above the peaks). I can't really check the calculations, because your spectra only goes to 2 decimal places and I'm not sure what power the machine they used is. The peaks at 1.57 and 1.77 ppm are nice. You'll notice they appear to be pointing to each other. This would typically indicate what's called an AB system. I'm not sure if this is the case for your compound, but it may be something to consider. The other beef I have with your spectra is the integration (red numbers underneath the peaks). They don't match up at all. Do you know why this is? I don't know what 'sc' refers to, as I've never encountered it. If it were cs, you might assume it means complex singlet, but I really don't think that is the case as that particular peak looks more to be a doublet with the peak at 3.51 ppm. Neither your proton nor carbon spectra make sense for an alkyne. What makes you think it's a CH2 group? I would be more inclined to go with 2 x vinylic CH groups. This makes sense with your carbon spectra, as you'll notice that there are (not including double ups) only 5 peaks. You have to have some equivalency somewhere. That being said, I'm still at a loss for where the remaining carbon(s) is. If we assume that you have 1 peak for both of the cabronyl carbons and 1 peak for 2 vinylic CH groups, that only gets you to 7 carbons, not 8. I'll have to sit down and think about it some more. A few other pieces of information: If we assume that you have 2 x C=CH groups at 6.31 ppm, it is obvious that these two CH groups are not next door to any other protons (but they are next to each other). They have to be near quaternary carbons. This should allow you to work out the structure of a fair chunk of your molecule. You have no -CH3 groups. This is important because it tells you a lot about the ends of the molecule. You either do not have terminating groups (i.e. all the atoms are involved in a ring of some sort) OR your terminating groups have double bonds at the end. You have no alkynes, so this means your DBE's are taken up by carbonyls, double bonds and possibly a ring.

If you can post the spectra, that would be great. As for the peak at 6.31, you have to consider what kind of bonding structure you might observe given that we aren't looking at an aromatic ring. What kind of bonds will you possibly see with all those DBE's? Where do protons involved in such bonds occur?

Why do you want to synthesise them for? I should think they'd be trivially easy and inexpensive to purchase. I'm all for just making them because you want to, but my concern here is that you are making them for consumption.

Firstly, I've noticed you wrote in a few of your peaks twice? Was this supposed to mean something? Also curious as to why you've put 1 C next to your carbon peaks, given that you don't integrate carbon NMR. I'm not convinced about there being a phenyl ring in your structure. Your carbon spectra doesn't appear to have enough peaks in the aromatic region and your proton NMR doesn't really have any. The closest is one at 6.31 ppm, which is a bit too low, isn't split as you'd expet for a phenyl ring and integrates to one proton. And really, with 8 carbons and 6 hydrogens, this shouldn't be surprising. After you include the ring, you have 2 carbons and maybe 1 hydrogen left to play with, which seems quite the impossible task with your oxygens and another 2 DBE's to incorporate.

Yes. The doublet at 8.1ppm corresponds to the protons on the ring next to the substituted carbon. The multiplet is a combination of the two triplet peaks. The compound you drew seems reasonable, but you might also want to consider the possibility of having the ester around the other way. It's a multiplet for the same reason as the multiplet in question 1; you have three overlapping signals in the aromatic region. The singlet peaks aren't going to be coupled to anything in the aromatic region because they are part of the substituent and not the ring itself, so they aren't next door to any aromatic protons. I have a program called ChemDraw that I use to draw structures with. You might be able to get a free licence for it through your institution. If so, I'd very much recommend it. It makes life much easier. There is also ChemSketch, which works fine as well. See above for this one. Same deal as above. You might want to consider the possibility having the ester around the other way. I'm not saying that it is, but you have to be able to justify why it wouldn't be if you don't think it is.

What about the peak at 8.1ppm? Aromatic regions are notorious for having overlapping signals and are reported as multiplets. So you might find that your peak at 7.5ppm has more than one equivalent sets of hydrogens in it. What makes it especially obvious is that it can't possibly be a CH3 group given it's chemical shift. This leaves you with only one other DBE to assign, so do you think you can guess at what kind of functional group that might be? The arrangement of the carbons/hydrogens corresponding to the peaks at 1.4ppm and 4.4ppm should also be fairly easy to assign. Once you have your fragments worked out, putting them together should give you two possible compounds. This seems reasonable to me. Well what does the peak at 7.3 ppm correspond to? Should anything be coupled to those protons and if so, what? If the other two peaks are singlets, what does that tell you about the neighboring atoms? Do those peaks correspond to CH, CH2 or CH3 groups? The best way to tackle these questions is bit by bit. Have a look at each peak individually and write down what you know about it. As an example, I'll go through the second question (since you've already solved it). The multiplet in the 7ppm region corresponds to the aromatic region. This is acceptable given your DBE. That there are 5 protons there must mean that the ring is monosubsitituted. With that information, we have this fragment: It's not much, but it's a start. Next, we notice two things. The most telling are the two peaks at 2.9 and 2.7 ppm. They are both triplets and both CH2 groups (since we don't have enough carbons for them to be 2 x CH groups). The triplet splitting pattern means that they are both next door to another CH2 group. A quick scan of the spectrum shows that there are no other possible CH2 groups and so these peaks must correspond to neighboring methylene carbons. That gives us this fragment: So, we've worked out 8 of the carbons and 9 of the hydrogens. We have one peak left at 11.2 ppm, which corresponds to 1 very deshielded proton. Considering our two oxygens, one remaining carbon and unassigned DBE, it is reasonable to assume that we have some sort of carbonyl group. The only possible way to fit this information in with only one remaining hydrogen is if we have an aldehyde or a carboxylic acid. An aldehyde would mean that the last oxygen is involved in some kind of ether linkage. The only possible way to combine this with our above fragments is with this compound: We know this isn't likely since the CH2 next to the ether oxygen is outside of the expected range for such a proton. If this were true, you'd expect this CH2 group to be closer to 4ppm, but it isn't. The peak at 11.2 ppm is also slightly too high for an aldehyde CH group. So, this possibility is discounted and means we probably have a carboxylic acid. We now have three fragments: The only way to combine these is as follows: This sort of approach might seem long winded, but it works. You just need to take things one at a time and you'll find combining your information to be much simpler. Hope that helps.

! Moderator Note too-open-minded, Their They're There was a spelling error in the title, witch which I have now fixed. Its It's been bugging me.

! Moderator Note ACG52 and EMField, please stay on topic and stop hijacking / derailing this thread.

Ottahhh, if you still happen to be paying attention to this thread, please do not listen to rutski so far as ignoring your high school grades is concerned. Stay in school and do as best as you can. Life will be much easier for you come college application time if you do. If you wish to supplement your learning with additional study, then that's fine, but as Genecks and ajb have said, you don't want to spend so much time learning additional material at the expense of your high school grades. You also don't want to pile on too much new material at once. Picking up a general text on whatever area you're interested in might be a good start and may even help you in your understanding of the content you're learning at school.

! Moderator Note Moved to homework. Ch5112, if your intention here is to learn, you'll need to show some sort of effort. Have you worked anything out with the questions at all?

Could you be more specific?

! Moderator Note And stop hijacking other threads.

Bipedalism simply means that an animal moves about using it's two hind limbs. So yes, they are bipedal. I think they also sometimes utilise their arms to move about while grazing, but their primary form of locomotion is certainly bipedal.

It could just as easily be the carbanion.

If you're sketchy on the idea of an oral exam, our staff parties might cause you to have heart failure. It's okay, though; Klaynos really digs that sort of thing.

So moo's up to her old tricks again, I see.

Oh, *I* didn't take *your* exam? Is that how that went, was it? Is this just a cover for the ol', 'this doesn't usually happen,' shenanigans? I thought we burned those photos and swore never to speak of it again?

Because your YouTube videos don't work.

I'm unofficially asking this to get back on topic.

It's a temporary fixture. Watch while I just as easily revoke your rights to friendship. As it turns out, you're not my friend at all, you just claim me as yours. The oral exam is mostly non-negotiable.

By which I mean, I am kindly nominating Phi to be examiner. Oh, it's bad alright

Well, it's somewhat modelled off the entrance exam to moderatorship. The key component involves an oral examination. Since I live so far away though, Phi has kindly offered to examine you on my behalf.

You would have to pass my initiation test first.

What exactly are you referring to? I made but a single improvement. Also, freedom of speech on a forum is at the discretion of the people who run it. And no, I don't.