The Dangers of Making Too Many Assumptions. Electronegativity, Acidity, and Chemical Shift

Last month (which you can see here), we learned about how an acidic proton behaves in a 1H NMR experiment, particularly when it’s surrounded by D2O. For example, when an H+ leaves CH3COOH to join an accommodating D2O molecule, the resulting acetate (H3CCOO-) segment is reasonably comfortable bearing that negative charge. This phenomenon is the reason the solution is “acidic” in the first place. But why is acetate so capable of dealing with this negative electronic charge?

It has a lot to do with its two oxygen atoms and their electronegativities. The more electronegative an element, the more it withdraws electron density away from its surrounding atoms in the molecule. The electronegativity of oxygen is relatively high (χ = 3.44),[1] so acetate, which contains two electron-hungry oxygen atoms, is fairly comfortable as a negatively-charged species, taking good care of things while the acidic proton goes off to hang out with D2O.

And so this brings us to another point. In the same way that H3CCOO- is pretty good at being an electron sink, D2O is a pretty good host for the floating H+. Incidentally, the acidity of acetic acid is actually attributed to two things:
1) H3CCOO- is good at dealing with a scenario involving a negative charge and a missing proton; and
2) D2O is pretty good at taking care of said lost proton. D2O is a good host because its oxygen atom has two lone pairs of electrons, so there’s room for H+ to come and party.

We learned a lot about how electronegative elements affect the chemical shifts of nearby protons in a previous blog by Dr. Araneda, and last month’s blog showed us how the H and D get scrambled up in a 1H NMR experiment when an acid is dissolved in D2O. But what would happen if we dissolved CH3COOH in something that wanted nothing to do with an H+? Something like CDCl3? The only lone pairs in this solvent are on the chlorines, and like the rest of its halogen brothers, chlorine really doesn’t like making more than one bond, so all three chlorine atoms in CDCl3 have zero interest in hosting an acidic H+. The acidic proton is better off hanging around with H3CCOO-.

And that’s exactly what happens when you dissolve CH3COOH in CDCl3, we don’t see the “scrambling” we observed in the previous blog, where all the acidic protons got stashed away in the solvent residual. Instead, we can actually see the CH3COOH acidic proton (Figure 1, red square). However, its acetate host is really hogging all the electron density, so it is considerably deshielded, showing up in the downfield region of the spectrum. The methyl peak (blue circle) shows up upfield, where you expect it to be.

Figure 1

The acidic proton shows up downfield because it does not have much electron density around to “shield” it from the instrument’s magnetic field – The electronegative oxygen atoms are stealing most of it away. But, like any good scientist, we want to know if we can push these limits and make the electron sink even more electron-greedy?

i.e., How could we make acetate (CH3COO-) even better at handling the negative charge?
i.e., How could we make acetic acid even more acidic?

If we look at Figure 2, there are a few obvious targets:

Figure 2

1) Modify the chalcogen? No, oxygen is the most electronegative chalcogen, so going down the period table to something like sulphur won’t help us achieve a higher acidity.

2) Change carbon to something else? No again, because carbon is the most electronegative tetrelide, so swapping it out for Si etc. won’t help either.

3) Alter the methyl group substituents? Now we’re talkin’! We can modify the methyl group very easily by swapping out hydrogen for something else. But swap it out for what?

When you look at a periodic table, in most cases the H is placed in the top left, atop the alkali metals. Like alkali metals, it will gladly float around as a positively charged species (acids!). But, hydrogen is a funny element, because with a comparatively small valence shell, it kind of has a split personality. When it’s not being an acid, it likes to form one covalent bond, as in C–H for example, similar to halogens (C–F, C–Cl, etc.). Although C–H bonding is similar to C–X (X being a halogen), there are some remarkable electronegativity differences between H and X = Cl, F, etc. that completely change the game. Let’s look at what would happen if we replaced the hydrogens in CH3 (methyl) with chlorines and fluorines to make –CCl3 (trichloromethyl) and –CF3 (trifluoromethyl) groups.

Let’s start first with the least electronegative halogen of the two, chlorine. By replacing three hydrogens (χ = 2.20)[1] with chlorine atoms (χ = 3.16),[1] trichloroacetic acid (Cl3CCOOH) should now be even more inclined to pool electron density away from the acidic proton, making it a better acid. In addition to the electron-withdrawing effects already provided by the COO- segment, the electronegative Cl3C group is also playing a significant role. When dissolved in water, trichloroacetate (Cl3CCOO-) is really good at dealing with the negative charge, and H+ is free to wreak acidic havoc, as is evidenced by the acidity measured by a pH meter! However, Nanalysis doesn’t manufacture pH meters, we make benchtop NMRs! And we already determined that we won’t see the acidic proton peak if we dissolve an acid in D2O. So let’s take a look at the 1H NMR spectrum in CDCl3, instead (Figure 3).

Figure 3

The first thing that is easily noted is the lack of a methyl group around 2.0 ppm. This is expected because we removed all those methyl protons and replaced them with chlorines! Looking at the downfield acidic proton (10.5 ppm), it interestingly, and unexpectedly, does not show up as far downfield as we expect. In fact, it is located more upfield than regular ol’ acetic acid (11.5 ppm, Figure 1). With electron density forcefully being pulled away from the acidic proton, one would expect this peak to be located at a much higher chemical shift!! This brings us to an important point, though: Although electronegativity of neighbouring groups is one property governing chemical shift, electronegativity is not the only factor governing where that group shows up in the spectrum. Ultimately, the source of this property is based on how much the nucleus is (de)shielded from the magnetic field.

Although adding chlorine atoms to the methyl group allowed us to inductively remove electron density away from the acidic proton through the σ-bond framework, inherently making it more acidic, we have added a significant number of electrons to the system since each chlorine atoms comes with three lone pairs each. If we look at a space-filling model of trichloroacetic acid (Figure 4, Cl = green, C = black, O = red, H = white),[2] we can see that these chlorine atoms, each bearing three lone pairs each, aren’t that far away from the proton in question. As a result, these electrons are likely shielding the acidic proton, causing it to appear at a much more upfield (shielded) resonance than expected. In short, the acidic proton is being deshielded via chlorine’s electronegativity through the σ-bond framework, while also being shielded considerably by the electron cloud from the massive chlorine atoms nearby!

Figure 4[2]

As a result, one needs to be careful not to make assumptions from 1H NMR! If you looked at the 1H NMR spectrum of trichloroacetic acid (Figure 3), you might predict it to be less acidic than regular acetic acid. However, if you use a pH meter – an instrument that, unlike NMR, measures only acidity – you can clearly see that the chloro variant (pKa = 0.66)[4] is much more acidic than the acid we find in vinegar (pKa = 4.76).[3]

Let’s keep going, though. Let’s add the mother of all electronegative elements: fluorine, to make trifluoroacetic acid, or F3CCOOH. The 1H spectrum is shown in Figure 5. Recall that we’re dissolving in CDCl3, to prevent the H/D scrambling that would occur if we used D2O.

Figure 5

As before, we do not see a methyl peak because all the protons have been replaced with fluorine atoms. If we examine the acidic peak, we see a signal (11.3 ppm) that is slightly more deshielded than the proton in trichloroacetic acid (10.5 ppm), and similar to that observed in regular acetic acid (11.5 ppm). Again, although the three fluorine atoms are aggressively stealing electron density away from the acidic proton, the lone pairs on the three fluorine atoms are also imposing electron density in its vicinity, causing a shielded (upfield) effect on the chemical shift. Fluorine is much smaller than chlorine however, so they are less imposing than the scenario involving chlorine (Fluorine: 1.47 Å, Chlorine: 1.75 Å).[4]

Phew! Take a moment to let that all sink in. Because there’s more to talk about!

Last month, we also learned about 13C{1H} NMR, which allows us to look at all the carbon nuclei in the molecule. Despite the fact that we swapped the methyl hydrogens for methyl halogens, we left the carbon atoms alone. So, let’s run the 13C{1H} spectrum of each trichloroacetic acid (Figure 6) and trifluoroacetic acid (Figure 7). Recall that the carbon of the carboxylic acid functional group should show up as a small peak downfield (red square), and the methyl carbon (blue circle) should appear in the upfield region.

Figure 6

Figure 7

Comparing both spectra, we can see a clear difference in the multiplicities of the trifluoroacetic acid resonances. But, let’s ignore that for the time being. Let’s focus more on the location (chemical shifts) of these peaks in both spectra. i.e., The locations of the red and blue shapes. Notice that the difference in chemical shifts of the carboxylic acid carbon (163 ppm, 157 ppm) changes only slightly, while the chemical shifts of the trichloromethyl (88 ppm) and trifluoromethyl peaks (110 ppm) differ considerably! Electronegativity is playing a huge role in the location of these methyl peaks, with the fluoromethyl peaks being significantly more deshielded by the electron-withdrawing prowess of fluorine. The carboxylic acid carbon clearly isn’t nearly affected here.

Now let’s revisit the multiplicities observed in the trifluoroacetic acid resonances (Figure 7). From last month, you’ve probably deduced that these are quartets, rather than two groups of close-proximity singlets, considering their equal spacing and the 1:3:3:1 peak intensities. This spectrum also looks suspiciously familiar to what we saw last month in the proton-coupled spectrum of CH3COOH! But, why are we seeing coupling here? There are no protons in this molecule (except the acidic proton), and we ran this as a decoupled spectrum!

There are a lot of questions here, so let’s take it one step at a time. Last month, we saw a quartet for the CH3 peak in the proton-coupled 13C NMR spectrum because of the three directly attached protons. The reason protons caused this kind of splitting is because 1H has what is knows as a “spin”. In the NMR world, 1H is known as a “spin ½” nucleus. I won’t get into this, but spin ½ nuclei cause the peaks of neighbouring atoms to split. The spin ½ nucleus (1H) on the methyl group caused both the methyl and the carboxylic acid peaks of H3CCOOH to split into quartets. The multiplicity disappeared, and these quartets collapsed into singlets, once we ran the spectrum proton-decoupled (13C{1H}), since 1H-decoupling makes protons invisible.

As it turns out, 19F is also a spin ½ nucleus! So it will cause the peaks of neighbouring atoms to split, much like how protons do. That’s why we see two quartets again in F3CCOOH. Although this spectrum was run as proton-decoupled (13C{1H}), we are not decoupling fluorine atoms. How do you think the spectrum would look if we ran a 13C{19F} experiment? It’s also interesting to note that the coupling constants observed here are much larger than what we observed in the regular acetic acid case (1JC–F >> 1JC–H, 2JC–F >> 2JC–H). I won’t get into this today, though.

In addition to causing multiplets all over the place, another awesome feature of spin ½ nuclei is that you can observe that nucleus as well! Meaning, you can do an 19F experiment! Let’s take a look at the 19F NMR spectrum of F3CCOOH. What do you expect to see? There are three F nuclei, so three peaks, right? NO! You’re forgetting that all three F nuclei are chemically equivalent, just like how the three protons in H3CCOOH only show up as one peak! So you will see a singlet worth three F nuclei (Figure 8).

Figure 8

Thankfully, 19F is just as easy to observe as 1H, and you get a much wider range of chemical shifts to play with, as well. Thinking a little bit further about spin ½ nuclei: If both 19F and 1H are both spin ½ nuclei, and we can observe both of them in NMR, does that mean 13C is spin ½ as well? The answer is yes! But that doesn’t necessarily mean:

1) All the carbon in the universe are the super-handy spin ½ 13C, or
2) That you need to have spin ½ for it to be observable by NMR.

But, those are topics for another day. Cheers! Thanks for visiting!

[1] David R. Lide (ed), CRC Handbook of Chemistry and Physics, 84th Edition. CRC Press. Boca Raton, Florida, 2003; Section 9, Molecular Structure and Spectroscopy; Electronegativity.
[2] [Viewed Feb 26, 2017].
[3] Bordwell, F. G. Acc. Chem. Res. 1988, 21, 456-463.
[4] Rowland, R.S.; Taylor, R. J. Phys. Chem. 1996, 100, 7384-7391.