[Goala]

Sorry to double post here (kind of), but I wanted to change my question without adding the information I provided in my earlier post.

For the peak seen at 7.85ppm (shown below) which is pretty "clean", it seems to be a ddd. However, the smallest coupling constants don't add up. The two outside coupling constants are equal, but the two interior coupling constants are different than the outside coupling constants, and by a fairly large amount. Shouldn't the coupling constants in the same level all be equal for a ddd? I have attached a .jpg file to show what I mean.

In final my question is... what is wrong with this NMR Spectrum? If this is indeed a ddd why aren't the coupling constants adding up?!

[Dan]

You have not identified the coupling constants correctly.

Based on the magnitude of the coupling constants (the smallest of which is 6 Hz), I think you have picked the wrong H for this signal.

The three coupling constants look like 6.0, 6.9 and 7.7. See diagram.

[Goala]

The three coupling constants look like 6.0, 6.9 and 7.7. See diagram.

Thanks Dan! Thanks A LOT for the help Yes, I accidentally mislabeled that hydrogen.

I actually think I just solved all the coupling constants in the problem. For that specific hydrogen I got coupling constants of 6.0, 7.1, and 7.9.

For the alpha hydrogen I got coupling constants of 5.9 (I got 5.9, but shouldn't it be 6.0?) , 3.7 and 2.1.

For the two diastereotopic hydrogens I got coupling constants of 10Hz between each of them. 10, 7.1 and 2.1 for one of the hydrogens and 10, 7.9, and 3.7 for the other.

I hope I finally get this ... I think I do

[Dan]

I actually think I just solved all the coupling constants in the problem. For that specific hydrogen I got coupling constants of 6.0, 7.1, and 7.9.

Looks good, I was guessing by eye as you hadn't posted all the data yet.

For the alpha hydrogen I got coupling constants of 5.9 (I got 5.9, but shouldn't it be 6.0?) , 3.7 and 2.1.

Yeah - it depends which end you measure from. If you start at the other end you get:

3105.00 - 3099.00 = 6.0
3102.80 - 3099.00 = 3.8
3101.20 - 3099.00 = 2.2

0.1 Hz discrepancy is acceptable experimental error. If you continue this process and measure all combinations for each coupling constant (4 for each) and take the average, you will get the most accurate result.

2.2, 2.2, 2.2, 2.1 => 2.2 average
3.8, 3.8, 3.8, 3.7 => 3.8
6.0, 6.0, 6.0, 5.9 => 6.0

For the two diastereotopic hydrogens I got coupling constants of 10Hz between each of them. 10, 7.1 and 2.1 for one of the hydrogens

I measure:
2.1, 2.2, 2.2, 2.3 => 2.2
7.0, 7.1, 7.1, 7.2 => 7.1
9.9, 10.0, 10.0, 10.1 => 10.0

and 10, 7.9, and 3.7 for the other.

I get

3.8, 3.8, 3.8 ,3.7 => 3.8
7.9, 7.9, 7.9, 7.8 => 7.9
10.0, 10.0, 10.0, 10.1 => 10.0

Because I was able to find the coupling constants for each hydrogen, (I hope they are correct ) would I finally be able to assign a signal to each of the diastereotopic protons because I initially wasn't able to because they were diastereotopic ?

No, in order to do that you would need to run nOe experiments and look for a nOe (nOe = nuclear Overhauser effect) between the methyl group and one of the diastereotopic protons.

[qw098]

Thank you SO much

This site and YOU are invaluable!! I am so happy I understand this!

My stress is ALL gone because of you