Help me with this Organic Chemistry doozie.
June 20, 2007 2:27 PM   Subscribe

Organic Chemistryfilter: In H NMR spectroscopy, why is it that as a carbon is more substituted, it is also more shielded? Better explanation inside.

This is a bit of a long shot. My professor offered extra credit if we can figure out the reason behind the following trend. He has looked in multiple textbooks and cannot find an explanation.

In H NMR, the following trend is seen: an R3CH has a chemical shift of 1.7 ppm, R2CH2 is 1.5 ppm and RCH3 is .9 ppm. In effect, increasing the R groups apparently leads to less shielding on the proton. This seems to go against common sense because R-groups are electron-density donors, which should actually lead to a shielding effect. What is the reason for this trend?

If I am not clear enough in my question, I'll check back periodically and try to clear up any questions. Thanks, MeFi, you're my only hope!
posted by Paul KC to Science & Nature (9 answers total) 2 users marked this as a favorite
My initial thought is that although they are electron donors to the carbon, they aren't to the H, and actually let that electron be a bit more delocalized. Give me a bit to play around with some density maps to see if that helps.
posted by devilsbrigade at 3:36 PM on June 20, 2007

Response by poster: devilsbrigade Thanks for the help! Hopefully you'll be able to come up with something.
posted by Paul KC at 3:52 PM on June 20, 2007

Caveat: I don't know enough NMR for you to give my answer much weight.

But this is very intriguing; I would guess that this finding is evidence that in CH4, the bonds are actually hybrid bonds in the sense that the hydrogen atoms are bonded not only to the central carbon atom, but to each other as well.

That the hydrogens are partially bonded to each other would mean that the electrons would be hanging around the hydrogens being shared between them, because that's how bonds work, much more than you would predict if the bonds were only between the carbon and hydrogens.

If the electrons are hanging around the hydrogens more, they effectively shield the protons more.

In this view, when you substitute the R-groups the hydrogen-hydrogen bonding is interfered with and even the electron-density the R-groups donate is not enough to make up for that.

That hydrogen atoms can form multidirectional covalent bonds of the kind I'm positing here seems to be supported by the existence of diborane, B2H6, where each of the two 'bridging' hydrogens are considered to be bonded to both of the boron atoms at once.
posted by jamjam at 3:57 PM on June 20, 2007

In ethane the two carbons have equal partial charges, so neither will give anything extra to the other. Each carbon will have exactly one electron's worth of charge from the C-C bond. A C-H bond would give the carbon more charge than that, since carbon is more electronegative than hydrogen.

In propane, both end carbons will be more electronegative than the centre one, and will then donate about 1/3 the difference, giving all three electrons equal electronegativity Still, CH3 < ch4, and this will actually deshield all>
With three R groups it will be 1/4, with four 1/5. Each time the end carbons are giving up less, but they'd still be better off joined to a hydrogen.
posted by Orange Pamplemousse at 5:32 PM on June 20, 2007

Oops, forgot the closing bit.

The less electronegativity a carbon gets from its carbon neighbours, the more it's going to rip from its hydrogens. The difference in carbon shielding between methane and ethane will be slight, but ethane will be grabbing more from its hydrogens.
posted by Orange Pamplemousse at 5:43 PM on June 20, 2007

Pamplemousse probably has it, but is "R" a sigma / pi donor?
posted by wilko at 7:15 PM on June 20, 2007

For what it's worth, I emailed my orgo professor about this -- he's something of an NMR maven. He replied back as follows:

More substituted carbons are actually less shielded, or completely deshielded in 1H NMR. The electron(s) around the hydrogen atom shield it from the magnetic field. Carbon is more electronegative than hydrogen, so it pulls some electron density away from the hydrogen being observed, so it is deshielded.
posted by killdevil at 8:22 PM on June 20, 2007

Response by poster: killdevil, pamplemousse
Your answers are similar to what my professor was speculating in class. It looks like we've reached a sort of consensus.

Thanks to everyone for your help! I think I will just pass this thread on to my prof and see what he thinks.
posted by Paul KC at 10:16 PM on June 20, 2007

There's an entire branch of quantum chemistry dedicated to quantitatively calculating shielding using molecular orbital calculations.

Here's a nice powerpoint presentation I found using the google search terms 'shielding calculations nmr proton carbon' which starts basic and works its way up. There were plenty of other useful looking pages as well.

Good luck-- I never thought I would see an NMR theory question on Mefi, but there you go!
posted by ThinkNut at 2:56 AM on June 21, 2007

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