5
$\begingroup$

In organic chemistry lecture, we were given a chart of approximate signal shift ranges for different carbon types in CNMR spectra. carbonyl carbons (C=O) have a range of about 155-210 ppm, but the nitrile group (C≡N) has a signal range of about 110-140 ppm. I understand that oxygen is a more electronegative atom, so it might pull electron density away from the bonded carbon (deshielding), but the C≡N is a more polar bond, which means the electron density spends a lot of time around nitrogen in this bond. Doesn't this imply the C≡N carbon would produce a signal more downfield than that of the C=O carbon?

From the given data ranges, it seems the opposite is true and the carbon of nitrile is more shielded despite the bond polarity.

$\endgroup$
3
$\begingroup$

TL;DR: It's complicated and likely has to do with paramagnetic shielding effects rather than the usual arguments based on electron density, ring currents, and so on.

Firstly, because of low-lying π* orbitals, π-bonded carbons typically have lower excitation energies than purely σ-bonded carbons and are thus more deshielded. However, due to the presence of a π-bond order term, alkynes experience less deshielding due to paramagnetic effects. So the general order of chemical shifts is: alkenes > alkynes > alkanes.

Carbonyls are more deshielded than alkenes, and nitriles more deshielded than alkynes (both due to the paramagnetic contribution), but this doesn't overturn the existing bias of alkenes > alkynes.


It is instructive to start by comparing alkenes (~120 ppm) vs alkynes (~70 ppm). There is already a large difference between those two. Carbonyls are essentially deshielded alkenes, and nitriles essentially deshielded alkynes. And so it is no longer too surprising that carbonyls come in at a higher chemical shift than nitriles.

The only question left is to ask why sp2 carbons are generally more deshielded than sp carbons. Although it's tempting to just look purely at the hybridisation and s-percentage, this doesn't explain why the pattern is generally sp2 > sp > sp3. I should first point out that the physical factors controlling 13C shifts are (usually) markedly different from those controlling 1H shifts, so any text that tries to "carry over" those ideas is likely incomplete or incorrect. The reason for that is described in further detail here, and you can see Hans Reich's excellent website for more information about the various terms that contribute to the chemical shift.

With that out of the way, I will briefly quote from Gunther's NMR Spectroscopy, 3rd ed. (p 410):

For the paramagnetic contribution, early theoretical considerations [...] led to the expression:

$$\sigma_\mathrm{p}^i \approx -\frac{1}{\Delta E} \left(\frac{1}{r_i^3}\right)_\mathrm{2p_z} \left(\color{red}{Q_{ii}} + \sum_{j \neq i} Q_{ij}\right)$$

where $\Delta E$ is a mean electronic excitation energy [...]

I have coloured $Q_{ii}$ in red because it is not in Gunther's text (I suspect this is an error). I have added it to bring it in line with an article by Pople (Mol. Phys. 1964, 7 (4), 301–306).

Note first that the paramagnetic shielding is negative, so a larger magnitude corresponds to a less shielded nucleus, or a higher chemical shift. Now, the "excitation energy" for an alkane is probably a $\sigma \to \sigma^*$ excitation, which is quite large. On the other hand, for unsaturated compounds (alkenes / alkynes), there are low-lying $\pi^*$ orbitals, so $\Delta E$ becomes smaller and $\sigma_\mathrm p$ more negative.* This accounts for why unsaturated groups have higher chemical shifts than alkanes.

It also explains why carbonyls/nitriles are more deshielded than alkenes/alkynes respectively: because a heteroatom (O/N) has a high-energy lone pair, the primary excitation becomes a $n \to \pi^*$ transition which has a smaller $\Delta E$.

[...] $Q_{ij}$ is a bond order term that originates from the presence of $\pi$-bonds. [...] The bond order term superimposes additional changes that are most pronounced for [...] the $\ce{^{13}C}$ resonances of alkynes. For [alkynes], theory yields $\sum_{j\neq i} Q_{ij} = 0$ as in alkanes, whereas for [the central carbon of] allene $\sum_{j\neq i} Q_{ij} = 0.8$.

Gunther doesn't say exactly what this term would be for a plain old alkene (not allene), but Pople writes that it is 0.44. According to Pople, the $Q_{ii}$ term (in red) is equal to 2, so the overall term in parentheses is equal to 2 for alkynes and 2.44 for alkenes. This means that an alkyne carbon experiences a smaller (less negative) $\sigma_\mathrm p$, and consequently is less deshielded than an alkene carbon.


* Note also that $\pi \to \pi^*$ excitations are symmetry-forbidden in the context of paramagnetic shielding, so the primary excitation is probably a $\sigma \to \pi^*$ or $\pi \to \sigma^*$.

| improve this answer | |
$\endgroup$
0
$\begingroup$

The nitrile is sp hybridized while the carbonyl is sp2 hybridized, meaning that the orbitals around the nitrile carbon have more s character, leading to more shielding.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I'm afraid it's not as simple as that... if this argument was all it took, then sp3 compounds like alkanes should be even more deshielded than sp2 compounds like alkenes/carbonyls. Shifts in unsaturated groups generally have more to do with other effects that aren't related to electron density. This is especially true for 13C shifts. see e.g. chem.wisc.edu/areas/reich/nmr/06-cmr-02-shifts.htm $\endgroup$ – orthocresol Jan 2 at 22:22
  • $\begingroup$ The reason for sp2 compounds being more deshielded is due to the planar node. It is an exception to the rule. $\endgroup$ – sweetandtangy May 14 at 2:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.