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http://www.chem.ucalgary.ca/courses/351/spectroscopy/question6/allSpectra.htm

My question is specifically about the proton NMR. The structure for this IS butanal. The singlet observed at delta 9.8 belongs to the H in the aldehyde (R-COH).

But I do not understand why this is a singlet? Shouldnt it be a triplet? How does this get read as a singlet when its neighboring carbon has two hydrogens on it?

Thank you for reading!

[pfnm]

It's (that is, the protons) neighbouring carbon does not have any protons with which it can magnetically interact, and so no splitting occurs.      

  
                                                     O
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Remember, the structure is CH3CH2CH2-C-H

The carbon atom adjacent to the hydrogen in question, has no hydrogen atoms bonde to it. So no splitting occurs.

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When we talk of splitting on protons of the adjacent carbons, we mean adjacent to the proton in question (not the carbon to which it is attached).

Hydrogens that couple (split each other) are considered as two bonds away, three bonds away (and sometimes for very weak splitting) four bonds away.

In this compound, the aldehyde H is three bonds away from the hydrogens on the -CH2- (ie H-CO-C-H (three hyphens representing three bonds).
No splitting occurs - they are not neighbouring (two bonds away from each other) and magnetic interactions aren't typically carried across a carbon atom).

[sjb]

It's (that is, the protons) neighbouring carbon does not have any protons with which it can magnetically interact, and so no splitting occurs.      

  
                                                     O
                                                     ||
Remember, the structure is CH3CH2CH2-C-H

The carbon atom adjacent to the hydrogen in question, has no hydrogen atoms bonde to it. So no splitting occurs.

---

When we talk of splitting on protons of the adjacent carbons, we mean adjacent to the proton in question (not the carbon to which it is attached).

Hydrogens that couple (split each other) are considered as two bonds away, three bonds away (and sometimes for very weak splitting) four bonds away.

In this compound, the aldehyde H is three bonds away from the hydrogens on the -CH2- (ie H-CO-C-H (three hyphens representing three bonds).
No splitting occurs - they are not neighbouring (two bonds away from each other) and magnetic interactions aren't typically carried across a carbon atom).

So why does the isopropyl group, for instance show splitting? If what you say is true we'd only see H-C-H splits.

[pfnm]

take isopropyl bromide for instance:

The two terminal methyl groups appear as a doublet integrating for 6H: but the doublet is very narrow (the coupling is 3J, that is, three-bond coupling)

The -CH-Br proton is split into a pattern that appears to be a septet, with the distance between each peak in the septet being equally narrow as the distance between the peaks in the doublet (because the coupling constants for all the signals are the same, that is they are 3J coupling constants).

Orbital interactions (governing nuclear and electronic spins of the interacting protons) transmit the spin information from one proton to another - in the 3J system the hydrogens transmit spin info via overlap between adjacent C-H sigma bond orbitals which are perpendicular to the C-C sigma bond.

[orgopete]

Look at the NMR spectrum of butyraldehyde in SDBS. Click on the peak data button and look at the peaks. The CHO peak is a triplet.

[pfnm]

Yep, then coupling does occur between the aldehyde hydrogen and the hydrogens of the methylene closest to it, this would be 3J coupling. The magnitudes are so small (1.83Hz) however as to justify why the spectrum at
  http://www.chem.ucalgary.ca/courses/351/spectroscopy/question6/allSpectra.htm
appears as a singlet:

You are correct in that it was inaccurate to say that no splitting occurs: splitting does occur, but the coupling is so weak as to make the splitting not visible on a 90Mz machine.

[Cryolite]

I once asked a professor about that same question. She said that simulated spectra given to students as exercises generally show only a singlet to simplify the analysis.

Real spectra can indeed have a triplet.