[intrusionv2]

Hello, I'm studying for an upcoming test and struggling pretty badly with some (easy?) NMR problems.

I was wondering if someone can just walk me through a problem step by step and show me exactly what I need to do for every problem? When my teacher does these types of problems, I have no idea where he gets some of the information from. I know that it's all contained in the NMR problem, but it would be nice if I knew where.

So anyways here's one problem (no chart, just tabulated NMR data):

C₇H₆Br₂. IR: 3300, 2980, 2180 cm^-1.  ¹HNMR: 3.6 (s, 2H), 2.1 (s, 1H); ¹³CNMR: 78,72,52,31.

I know the answer just don't know how to get there using this data. Thanks for any help.

[Jorriss]

A useful technique right off the bat is determining the index of hydrogen deficiency. Here we have, C7H6Br2 - 7 + 1 - .5(6+2) = 4

What does an IHD of four suggest? To me it suggest an aromatic ring, but is this consistent with the IR and NMR data provided?



Can you give some more of your thoughts? When you see all that data, what do you think?

[intrusionv2]

Thanks for your response. I do know how to calculate IDHs.

I guess I see that it can't be a ring because there are only singlets. But what do I do next? 

[Jorriss]

Right. And let's look at those carbon-13 NMR and the IR spec.

We know a peak in the 3300 range for an IR spec involve something bonded to a H. Either a N, O or C. It's not N or O, of course so it is likely an alkyne hydrogen single bond. Then, looking at the 13-C NMR, we see there is no carbon peaks in the double bond range.

It's probably an alkyne.

[ardbeg]

there are four distinct carbon signals, so from 7 atoms, you would expect something with symmetry i.e. two halves the same (3) and a middle carbon.  also the integral for three H atoms from 6 in a molecule = suggests twofold symmetry.  the ppm shift of the signals means there is no aromatic ring.  i think i know what it could be but i'll let you try first   

[orgopete]

Since the question is about how to do these, let describe how I do them. I use what I call the "Tinker-Toy" method.

I agree with calculating the IHD first, done.

• Since the MF was given, we also know the values in the NMR should be doubled and because there are two signals in the proton spectrum and four in the carbon spectrum, there must be symmetry (or there would be seven signals) and two (or more) carbons do not have hydrogens attached.
• From the adjacency data (singlets), the carbons bearing hydrogens do not have a carbon with a hydrogen attached in an adjacent position.
• The peak at 3.6 ppm must be two CH₂ groups because 1xCH and 1xCH₃ is not symmetrical and a CH₄ is not possible.
• The peak at 2.1 ppm must be 1xCH₂ group or 2xCH group. We should exclude the 1xCH₂ group in this case it becomes impossible to create a structure with 3xCH₂ groups and four degrees of unsaturation. If we add the IR data, it indicates a triple bond by the CH-stretch and the C-triple-C stretch.
• That leaves piecing together 2xCH₂ groups, 2xCH groups, 2xC-triple, 1xC-no hydrogens attached (the two missing Br?)

I argue that if you had molecular models (or Tinker-Toys), there is only one way you could put this together (and I think this will give the correct structure).

[intrusionv2]

Awesome, thanks everyone.

[willug]

A really nice way of getting a feel for NMR spectra is to use something like chemdraw ultra and it's NMR prediction feature. I've found that sitting and playing with it has helped me quite a bit. Although don't take the chemical shift values it gives you too seriously!