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Topic: Spectroscopy problem  (Read 4723 times)

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Offline aldoxime_amine

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Spectroscopy problem
« on: February 19, 2009, 01:16:09 PM »
The 1H-NMR spectrum of compound E (C8H8O) consists in two singlets of equal area at 5.1 ppm (sharp) and 7.2 ppm (broad). By treatment with excess HBr, compound E yields a single dibromide (C8H8Br2). The 1H-NMR spectrum of this dibromide is similar to the spectrum of E, as it also shows two singlets of equal area at 4.7 ppm (sharp) and 7.3 ppm (broad). Suggest reasonable structures for E and its dibromide.


What do the terms sharp and broad signify here?

Offline azmanam

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Re: Spectroscopy problem
« Reply #1 on: February 19, 2009, 01:55:30 PM »
broad usually signals the magnetic environment of the proton is changing on the nmr timescale.  this usually signifies slow-to-exchange protons (relatively stable alcohol-H, amine-H). 
I've also seen it with a mesityl group.  with the relatively slow rotation of the single bond connecting the mesityl group to the rest of the molecule, the 2 equivalent methyl groups on the mesityl ring broadened into one broad signal integrating to 6 protons. 
You can also measure restricted rotation about a C-N amide bond this way.  At cold temperatures, the 2 protons on the NH2 group of an amide are locked and give distinct singlets (rotation about C-N bond very restricted).  As you slowly raise the temperature (and take a new nmr every so often), you can watch the singlets get closer together, until the just coalesce into one broad signal.  You can use the coalescence temperature to calculate the energy associated with the restricted rotation about the C-N bond. Warm it even more, and the signal coalesces completely into one sharp singlet as the rotation is now fast on the nmr timescale.


What does it mean here?  Honestly... I think it means the nmr was taken on a low-field magnet and the magnet was not strong enough to separate the multiplet around 7.2...

(ps... I think the dibromide will yield one signal at 4.7, not 2 singlets...)
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