[Marcus]

Sup,

I'm having trouble analyzing an NMR spectrum of the aforementioned compound. Namely, I'm provided a spectrum with only 5 sets of peaks, but I'm sure it has 7 groups of heterotopic hydrogens.



First, I calculated the ratio of integrals to each other, which I hoped would give me the number of hydrogens represented by each region. 2:1:1:2:3. I have no idea if that's even a correct technique, but it helped me confirm several things. The first peak, upfield at around 2.277 ppm is the methyl group; I've determined that. From there, I'm slightly lost. The singlet at 4.679 ppm is probably the AR-O-CH2, since its integrated intensity corresponds to a ratio of 2 hydrogens.

After that, I'm confused as to why there are so many split peaks. The doublet could correspond to carbon #3's hydrogen adjacent to the methyl group, since the (n+1) rule is followed. Following that, no clue. For example, how could a quintet be produced here? Is it actually a quartet with the singlet from the hydroxyl group of the acid?

Finally, how could I measure the coupling constant (J) for these peaks? A doublet is just the difference between the two peaks (ppm) multiplied by the MHz of the machine (400 MHz in this case), the triplet is the difference of the inner peak and an outer peak multiplied by MHZ (do I provide the range? or just pick one of the differences?); does the same principle apply for the quintet, i.e. find the difference between the inner peak and an outer peak? And singlets do not have a coupling constant, correct?

As you can tell, I am thoroughly confused.

It seems to me that all of the hydrogens are unique. There are no equivalent hydrogens, except those bonded to the same carbon.

[Scatter]

The peak for the H on the OH is going to be even further downfield because of dimerization of the carboxylic acid group.  It should show up around 11ppm or so.  

The far right (integration of 3) is your methyl.  

The middle corresponds to the two geminal H's on the C next to the O.

Think about your aromatic H's.  I might know what's up with them, but this is where you could think a bit more.  Think about what makes the splitting and where the H's are.  The n+1 rule corresponds to peaks, so if you have a doublet, it is one H or group of equivalent H's split by one other H.  So it couldn't be the H on the carbon adjacent to the carbon with the methyl.  You should be able to reason this out.  That closeup is very clear.  Specifically look at the quintet.  Why do you think it might look different?

Hope this helps...

[Smrt guy]

Quite simply, as was previously said, the O-H is not shown because it would be considerably further downfield and may not show up at all due to H-D exchange.  Your doublet is likely alpha to the O (the donation of the O to the ring should have a shielding effect on the 2, 4, and 6- positions).  The triplet would then be the 4-H by the same argument.  The remaining peaks you have identified as a quintet.  This is incorrect.  The 5 peaks are not evenly spaced with a 1:4:6:4:1 relative height as is the case in a true quintet.  Instead, what you are seeing is the two remaining peaks overlapping to make a non-first order region to the spectrum.  As for coupling constants, for all regions of the spectra that are first-order the coupling constants, J, are equal to the distance between the split peaks (in ppm) times the magnetic field strength (in MHz).  The coupling constants will have units of Hz.  The non first-order region coupling constants can be determined but it is much more complicated.  Singlets are not coupled and thus do not have coupling constants.

[Scatter]

Yeah, oops.  I was thinking the other two aromatic peaks were overlapping, but I still ended up referring to it as a quintet for some reason.  It was like 2:30 in the morning.  I was tired.   

[Marcus]

Ok, cool. Thanks for the help. This is what I've come to understand:

So, methyl group is most upfield peak. Next in line are two geminal H's on carbon proximal to O producing the singlet there. Following that, is the doublet corresponding to H on C2, because the O is shielding it, C4, and C6, and because there's a proton on the adjacent atom. The triplet is explained by the H of C4, due in part to lesser shielding effect and also surrounded by two adjacent protons.

Finally, it seems the "quintet" is not actually a quintet just a triplet and a doublet overlapping? The doublet corresponding to C5, and the triplet corresponding to C3?

If this is all correct, I have a couple of more conceptual questions. First, how does the O serve to shield C2 H, but also shield C4 and C6, while not affecting C3 or C5?

Secondly, which peaks of the doublet-triplet correspond to each other? It looks like the first and fourth peak might represent the doublet, while the second, third and fifth peak are those of the triplet?

[Scatter]

Uh... I was thinking C6's H was the doublet.  Then C4 triplet, and C3 & C5's were overlapping...I was thinking a doublet and a triplet.

The 2 4 and 6 H's are affected because while the ring is going to be pretty much planar, the H's are going to be alternating between slightly up and slightly down.  so the H's that are "up" along with the first substituent (with the O) will be 4 and 6 because of steric interactions.  As in, the H adjacent to the methyl (on C3) will go down also because substituent 1 is so large.

Is the attached molecule correct?  Because that's what I'm going off of.

Someone correct me if I'm wrong, but that's what I'm seeing happen here.  Otherwise, I don't understand NMR spectra and orientation of molecules as well as I think I do.     

[Marcus]

Seeing the orientation of the hydrogens really helped me understand the shielding effect of O.

Here is my visual interpretation of what's going on. You're spot on about the structure of the compound, by the way.

[stewie griffin]

the H's are going to be alternating between slightly up and slightly down.  .

No. It's an aromatic ring. It's flat. There is no "up" or "down" with those hydrogens.

Marcus, we have two short answers for almost everything in sophomore organic chemistry 1) resonance and 2) inductive effects. Those two reasons are your friends, use them... and fyi I listed them in order of importance...

[Marcus]

All right, then shouldn't the oxygen have the strongest inductive effect on the methyl group and hydrogen's D and C, since it resonates with that double bond?

That leaves hydrogen A and B to account for the doublet/triplet interaction?

Actually, where's some good literature on shielding, the inductive effect, and resonance?

[Scatter]

Well now I'm confused too.  Don't listen to me on the H orientation then.  Your labeling is correct Marcus, but I don't know why it's correct.  I know about inductive and resonance effects, but I'm unclear as to how they come into play with NMR spectra, especially in this instance.

How do resonance and/or inductive effects come into play to shift C3 and C5's H's more downfield?  I guess the resulting carbanions on the 2, 4, and 6 carbons in the resonance structures would shield the corresponding H's a bit?  Is that right?

[azmanam]

haha.  I know what school you go to

[Scatter]

Me or him?  And what is that supposed to mean?

[azmanam]

... marcus.  I recognize the scan of the nmr as the one we use in our undergrad labs   It made me chuckle when I saw it.

[Scatter]

haha nice.  It's a good one.

[Smrt guy]

The 3- and 5- positions are shifted downfield due to the inductive effect of the O.  RO- substituents are resonance donating and inductively withdrawing.  This means that the positions affected by resonance (2, 4, 6) are shielded and the other positions are withdrawn.  Because resonance almost always wins, RO- groups are activating.  (The exception to the resonance vs. induction importance is with halides).