Carboxylic acid NMR

Ethylmalonic acid

R-COOH: 11.0 - 12.0. A very characteristic peak. Highly deshielded by induction and anisotropy.

HC-COOH: 2.1 - 2.5 ppm. Same argument as esters.

Proton exchange will cause the acidic proton peak to broaden and reduce in intensity - in which case the integral should not be trusted.

As expected, this peak will disappear when dissolved in a deuterium-donating solvent. But since carboxylic acids are often insoluble in these (such as CDCl₃), sometimes a piece of sodium is added, which removes the proton to make a sodium salt, the salt is then soluble in a deuterated solvent.

Ester NMR

Isobutyl acetate

O-C-H: 3.5 - 4.8 ppm. Deshielded by the O atom, but without the extra anisotropy effect.

O=C-CH: 2.1 - 2.5 ppm. This has a slightly higher range compared to the typical O=C-CH, since there is extra induction from the other oxygen atom.

A methyl ester would show a tall 3-proton singlet around 4 ppm. Induction falls rapidly with distance, so an ethyl group would show a 4 ppm quartet and a 1.3 ppm triplet - they will tend to merge into multiplets as more carbons are added.

Ketone NMR

3-methyl-2-pentanone

These show the usual H-C-C=O peak at 2.1 - 2.4 ppm. Methyl ketones are particularly noticeable since they show a 3-proton singlet in this region.

Aldehyde NMR

2-methylbutyradehyde

R-COH: 9.0 - 10 ppm. Heavily deshielded by anisotropy and induction. This is an easy way to notice an aldehyde, since no other proton environments appear in this region.

R-CH-COH: 2.1 - 2.4 ppm. Same deshielding effects, just reduced by distance. All CH-C=O environments are in this region.

CH-CO-H coupling happens with J = 3 hz. This is why the aldehyde group on the spectrum above is a finely split doublet, though it would have to be expanded to observe.

Nitrile NMR

Pentanonitrile

The only chemical shift to notice is from the CH₂ attached to the cyanide group, which is slightly deshielded at 2.1 - 3.0 ppm.

The deshielding is from anisotropy of the C≡N.

Amine NMR

Propylamine - note the broad and weak NH₂ peak, the degree of coupling, and the shift of the methylene protons due to the electronegative nitrogen

Shifts:

R-N-H: 0.5 - 4 ppm. Variable shifts/broadening for the same reason as alcohols.

Ph-N-H: 3.0 - 5.0 ppm. Deshielded by both anisotropy and resonance.

CH-N: 2.2 - 2.9 ppm

Coupling:

N-H: 50 Hz. Usually not observed. Presumably due to proton exchange.

N-C-H: About 0 Hz. Usually not observed. I'm unsure why. Perhaps due to resonance creating  (+)N=C hence proton exchange..

H-C-N-H: About 0 Hz. Usually not observed. Presumably due to proton exchange.

As with alcohols, adding D₂O to the reaction mixture will cause the NH peak to shrink or vanish.

Ether NMR

Butyl methly ether

R-O-CH: 3.2 - 3.8 ppm

In epoxides, the protons are less deshielded due to ring strain, so they are found at 2.5 - 3.5 ppm. I don't know why ring strain has this effect.

Alcohol NMR

2-methyl-1-propanol

C-OH: O.5 - 5 ppm

CH-OH: 3.2 - 3.8 ppm

The OH proton shift is very variable. It depends on the degree of both hydrogen bonding and the rate of exchange with other protons in the solution. These two factors depends on temperature, concentration, solvent, presence of water, or presence of acid impurities. These effects can also broaden the peak.

CH-OH has a coupling constant of about 5 hz. But this is rarely observed, due to rapid exchange with other protons in the solution, such as trace quantities of water. The coupling can only be seen in very pure samples. For example, freshly distilled alcohol, or alcohol from a previously unopened commercial bottle.

The rapid exchange of an alcohol proton with solvents can be used to identify an OH peak. If D₂O  is placed in the NMR tube, deuterium will exchange with the protons, and the OH peak will reduce or disappear after a few minutes.

Alkyl halide NMR

1-chlorobutane

CH-I: 2.0 - 4.0 ppm
CH-Br: 2.7 - 4.1 ppm
CH-Cl: 3.1 - 4.1 ppm
CH-F: 4.2 - 4.8 ppm

Obviously the shifts can be related to electronegativities of the halides.

Coupling can happen with ¹⁹F, which has a spin of ½:

CH-F: 50 Hz
CH-C-F: 20 Hz

Alkyne NMR

1-pentyne

C≡C-H: 1.7 - 2.7 ppm. The presence of this proton confirms that the triple bond is at the end of a chain.

C≡C-C-H: 1.6 - 2.6 ppm.

H-C≡C-C-H coupling has J of 2-3Hz.

Allylic coupling isn't very strong, but it can be noticed. For example, C in the above spectra is actually a finely split triplet. Likewise, D would show a triplet of doublets, if it were expanded.

Aromatic NMR

α-chloro-p-xylene - Note the deshielded methyl group, from e^- donation into the ring

Aromatic protons: 6.5 - 8 ppm. These are easily identified since very few environments absorb in this region. A highly deshielded vinyl proton could be found here, but it is rare.

Benzylic protons: 2.3 - 2.7 ppm

The largest shifts are found when electron withdrawing groups such as NO₂ are attached, deshielding the protons in the ring. The reverse is true for electron-donating groups like MeO.

Aromatic protons experience coupling right across the ring, but the clear differences in J values often allow substitution patterns to be determined by measuring them - described in a future post.

Alkene NMR

A vinyl hydrogen is attached directly to a double bond, an allyl hydrogen is attached to a carbon attached to the double bond.

C=C-H: 4.5-6.5 ppm, deshielded by anisotropy

C=C-C-H: 1.6-1.2 ppm, also deshielded. The anisotropy effect falls rapidly with distance.

Vinyl protons are rarely equivalent, becausce no rotation happens around the double bond. This can make the spectra of alkenes quite complicated. Another difficulty is that alkenes have noticeable ⁴J coupling - which can be even longer if the molecule is conjugated.

2-methyl-1-pentene

Alkane NMR

CH₃: 0.7-1.3 ppm

CH₂: 1.2-1.4 ppm, in long chains these may overlap into one unresolvable group.

CH: 1.4-1.7 ppm

-CH-CH-:³J = 7-8 hz

Shown above, the CH₃ region can overlap slightly with CH₂ region, but this is not the case for CH and CH₂.

The³J of aliphatic protons are 7-8 hz.

The spectrum of octane below shows CH₂ groups being blended together:

Octane

NMR 11 - Connecting peaks with coupling constants and skewing

Axiom: The coupling constants of a group of protons which split one another must be identical.

The above is extremely useful in interpreting spectra. Example:

Using J values, we can tell that C is coupled to A, and B is coupled to D.

The magnitude of the coupling constant also provides structural clues, such as whether a bond is cis or trans.

"Skewing" is another way which can be used to link multiplets. This is a tendency for the outer lines on a multiplet to lean towards the peak it is coupled with.

An example of skewing can be found in the previously-mentioned ethyl iodide:

NMR 10 - Coupling constants

The distance between peaks in a simple multiplet is called the coupling constant, J. It is a measure of how strongly a nucleus is affected by spin states of it's neighbor - the stronger the coupling the stronger the constant.

J is always expressed in hertz, irrespective of the operating frequency. Chemical shift is expressed in parts per million of the operating frequency. In a 60Mhz machine, each ppm corresponds to 60hz. In a 300Mhz machine, each corresponds to 300hz, and so on. In other words, as the operating frequency gets higher, the distance in ppm between coupled peaks will get smaller. Here is an example of nitropropane at two different operating frequencies:

The magnified peaks in the bottom spectrum are called "expansions" and are added for convenience  Many textbook examples will use low-resolution spectra to avoid having to draw expansions.

J is written with a superscript indicating the number of bonds between the coupled nuclei. So for a typical H-C-C-H coupling we would write ³J.

NMR 9 - Relative intensities

Consider methyl iodide again:

The relative intensities of the split peaks can be explained by the possible spins of coupled protons.

An easy mnemonic to describe the relative intensities, without having to draw out all the possible spins, is to use pascal's triangle:

NMR 8 - The origin of spin coupling

Consider a simple case of two CHs next to eachother:

The magnetic shielding of H_A is slightly affected by the spin of H_B. In some molecules, H_B will have a +½ spin, in others it will have -½. At room temperature, spin populations are pretty much populated 50:50. So half of H_A will be in one environment, and half of H_A will be in a slightly different environment.

Using the same logic for H_B, the spectra of these two hydrogens will be a doublet of doublets.

The n+1 rule just sums up the result of this thinking process. For example, for H_A coupled with two H_Bs, it gives a splitting of three. The three different H_A environments are both H_B at +½ spin, both at -½ spin, and one at +½ and the other at -½.

NMR 7 - Coupling

1,1,2-trichloroethane

The ratio of the integrals make sense, since there are two protons in one environment and one in the other.

The relative shifts should make sense too, since the CH is attached to two chloride atoms, deshielding it more than the CH₂ which is attached to just one.

The signal splitting can be explained by the n+1 rule. Each type of proton environment "senses" the number of equivalent protons, n, which are three bonds away. That environment is then split into n+1 peaks. This process is also called splin-splin splitting, or coupling, or spin-spin coupling.

So in the above specta, the CH is coupled to two CH₂ protons, so it split into three peaks, or a triplet. The CH₂ environment is coupled to one CH proton, so it split into two peaks, or a doublet.

Coupling does not happen between protons in the same environment. This is why the NMR of CH₄ would give a single peak.

Two more examples are below, ethyl iodide:

And 2-nitropropane:

2-nitropropane

You might wonder why coupling only works within three bonds. That isn't entirely true, it is just that the coupling effect falls rapidly with distance. So except in special cases, such as double and aromatic bonds, coupling further than three bonds will not be visible on spectra. It would be visible if we had higher resolution machines.

Consider 1-propanol:

The n+1 rule tells us that both the CH₃ and the CH₂O group will split into 3 peaks. Coupling does not happen to the OH proton since it is rapidly exchanged with the solvent, so it has an average spin of 0, and experiences an average coupling of 0.

For the central CH₂ group, there are two different sets of neighboring equivalent protons, so we have to use the n+1 rule twice, one for splitting by the CH₂O and once for splitting by the CH₃. The result is a quadruplet of triplets, or a triplet of quadruplets, giving us 12 predicted peaks in total:

In practice, high number of peaks are very hard to distinguish. A peak like the one at 1.5 ppm above would just be called a "multiplet", at least at undergrad level.

Remember that coupling only works between non equivalent protons. So H_A and H_B below would give a singlet.

NMR 6 - Magnetic anisotropy

Protons attached to simple vinyl groups resonate from 4.5 to 7 ppm. Part of the explanation is that sp² carbons are more electronegative, since the bond contains more s character (33% compared to 25%). The carbon s orbital pulls bonds in tighter, away from the attached protons.

By itself this does not explain such a large shift. A greater effect is magnetic anisotropy. Anisotropy also explains why acetylenic protons are found from 3 to 2 ppm, since we would expect them to be higher than vinyl protons based on hybridisation alone. It also explains why the chemical shift of benzylic protons is so high.

An anisotropic field is a field that is not isotropic. An isotropic field either has a uniform density, or a spherically symmetric density distribution.

When aromatic molecules are placed in a magnetic field, mobile pi electrons in it are induced to circulate around, which itself produces a new magnetic field - an anisotropic one.

The benzylic protons are in the region deshielded by the induced anisotropic field. Protons above the ring would be shielded by the induced anisotropic field, as shown empirically in these molecules:

A similar effect is found in pi bonds, with the current moving around the internuclear axis:

For double bonds, this deshields the end protons. For triple bonds it shields them:

So an aromatic molecule in an NMR machine experiences three different magnetic fields. The external field of the machine, the diamagnetic shielding from valence electrons (an isotropic field that counteracts the external field) and any anisotropic fields induced by currents induced by the external field. The chemical shift of a proton environment depends on the net effect of these.

These opens a lot of questions such as:

Why do magnetic fields induce current in some systems but not others?

Why does the moving current itself create another magnetic field?

Why are pi electrons more mobile?

Why do pi electrons in triple bonds induce in the opposite direction to double bonds?

And so on. The answers, if any, are found in physics textbooks. Spectroscopy textbooks will just summarize what happens and what the effects are.

NMR 5

NMR does not only show relative deshielding. You can use the chemical shift value itself to predict what type of enviroment it is in. Correlation charts can be used for this:

While it difficult to memorize a lot of number ranges, it is easy to get a "feel" for what regions the peaks are in, especially for the less-cluttered left half of the chart.

Notice that all protons attached to sp³ carbons are at 2-1 ppm, assuming the carbon is not attached to a heteroatom. Protons on primary carbons have a higher shift than on tertiary carbons:

This can be explained with the same reason for tertiary carbocations are more stable - alkyl group donate electrons inductively, shielding the central CH.

Sulfonates infrared spectra

Methyl p-toluene sulfonate

S=O stretch: Strong asymmetric at 1350, strong symmetric at 1175.

S-O stretch: Several strong bands at 1000-750.

Sufonyl chloride infrared spectra

Benzenesulfonyl chloride

S=O stretch: Strong asymmetric at 1375, and strong symmetric at 1185.

Thiol infrared spectra

Benzenethiol

S-H stretch: One weak peak near 2550

IR of amino acids

Leucine

Amino acids exist as zwitterions, so they can be considered both carboxylate salts and amine salts.

COO^- stretch: Frequency is lowered compared to parent acid, since resonance gives the carbonyl much more single-bond character. Strong asymmetric at 1600 and strong symmetric at 1400.

N-H⁺ stretch: Broad stretch at 3300-2600. Ammonium ions absorb to the left of this range, while tertiary amine salts absorb to the right. A broad peak often appears near 2100.

N-H⁺ bend: Occurs at 1610 to 1500. Tertiary amines absorb only weakly.

Nitro compound infrared spectra

1-nitrohexane

Nitrobenzene

Aliphatic NO₂ stretch: Strong asymmetric at 1600-1530 and medium symmetric at 1390-1300.

Conjugated NO₂ stretch: Strong asymmetric at 1550-1490 and strong symmetric at 1355-1315.

These bands may partially overlap the C=C region, but are usually easy to distinguish.