Organic Chemistry Proton NMR

[borogirl09]

I am currently in Ochem 2, and my professor gave this set of proton NMR problems. We were asked to draw structures for the following compounds based on their NMR spectra:

1) C₉H₁₀O₂ PMR: 8.1 (2H, d); 7.5 (3H, m); 4.4 (2H, q); 1.4 (3H, t)

2) C₉H₁₀O₂ PMR: 11.6 (1H, s); 7.0-7.4 (5H, m); 2.9 (2H, t); 2.7 (2H, t)

3) C₉H₁₀O₂ PMR: 7.3 (5H, m); 5.1 (2H, s); 2.1 (3H, s)

 

For the 1st one, I am confused about the (3H, m); based on its shift I assume that it is part of an aromatic ring since the compound has 5 degrees of unsaturation, but I don't understand how it is 3H. For the 2nd one I got that it was an aromatic ring with CH₂CH₂COOH attached. For the 3rd one I got stuck with the whole thing! I don't understand how there can be a (5H, m) when both of the other H's are singlets.

 

Please can someone help me understand this.

 

Thank you.

[hypervalent_iodine]

For the 1st one, I am confused about the (3H, m); based on its shift I assume that it is part of an aromatic ring since the compound has 5 degrees of unsaturation, but I don't understand how it is 3H.

 

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.

 

For the 2nd one I got that it was an aromatic ring with CH₂CH₂COOH attached.

 

This seems reasonable to me.

 

For the 3rd one I got stuck with the whole thing! I don't understand how there can be a (5H, m) when both of the other H's are singlets.

 

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.

[borogirl09]

For the 1st one, I am confused about the (3H, m); based on its shift I assume that it is part of an aromatic ring since the compound has 5 degrees of unsaturation, but I don't understand how it is 3H.

 

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.

 

 

So does that mean that in the aromatic ring that I can have a 2H set and a 3H set. I thought that if you had a monosubstituted aromatic ring that you would have 2 sets of 2H's and 1 lone H. So in this case would the 8.1 (2H, d) and 7.5 (3H, m) both be in the aromatic ring? Can I come up with a compound and post it and get your opinion?

 

For the 3rd one I got stuck with the whole thing! I don't understand how there can be a (5H, m) when both of the other H's are singlets.

 

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?

 

 

I know that the peak at 7.3 ppm corresponds to an aromatic ring. What I am confused about is how are the protons in the aromatic ring multiplets if the other 2 peaks are singlets. Is this due to the fact that in some aromatic rings the signals of equivalent H's overlap resulting in a multiplet signal? I know that since the other 2 peaks are singlets that they do not have any neighboring H's. Again, can I work on it and post what I come up with and get your opinion.

 

BTW...Thank you SOOOOO much

 

I couldn't figure out how to draw the structures I got on the computer, so I just googled it.

Here's what I got for the 1'st one

 

And here is what I got for the 3rd one

[hypervalent_iodine]

1347033863[/url]' post='701160']

So does that mean that in the aromatic ring that I can have a 2H set and a 3H set. I thought that if you had a monosubstituted aromatic ring that you would have 2 sets of 2H's and 1 lone H. So in this case would the 8.1 (2H, d) and 7.5 (3H, m) both be in the aromatic ring? Can I come up with a compound and post it and get your opinion?

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.

 

 

I know that the peak at 7.3 ppm corresponds to an aromatic ring. What I am confused about is how are the protons in the aromatic ring multiplets if the other 2 peaks are singlets. Is this due to the fact that in some aromatic rings the signals of equivalent H's overlap resulting in a multiplet signal? I know that since the other 2 peaks are singlets that they do not have any neighboring H's. Again, can I work on it and post what I come up with and get your opinion.

 

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 couldn't figure out how to draw the structures I got on the computer, so I just googled it.

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.

Here's what I got for the 1'st one

See above for this one.

 

And here is what I got for the 3rd 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.