Ethanol, CH₃CH₂OH, has two carbon atoms, and its ¹³C NMR (carbon-13 NMR) spectrum shows two sharp lines, one for each. Where the ¹H spectrum records a molecule’s hydrogens, this one records its carbons.
The experiment works the same way as ¹H NMR. Nuclei in a strong magnetic field absorb radio waves, and the electrons around each nucleus shift where it absorbs. This section covers only what changes when the nucleus is carbon. The payoff is a spectrum that answers one question very cleanly: how many different kinds of carbon does the molecule have?
Why the carbon signal is weak
Almost all natural carbon is ¹²C. A ¹²C nucleus has no spin, which means it is not a magnet at all, so NMR cannot see it. Only ¹³C can be detected, and it makes up just 1.1% of natural carbon.
Each ¹³C nucleus is also a weaker magnet than a hydrogen nucleus. Its gyromagnetic ratio, the number that measures how strong a nuclear magnet is, is about a quarter of hydrogen’s. So the carbon signal is weak twice over: there are few ¹³C nuclei, and each one responds weakly. A carbon spectrum needs more sample or a longer run than a proton spectrum. That makes it harder to record, but no harder to read.
Decoupling: one line for each carbon
Record ethanol’s carbon spectrum with nothing done to the hydrogens, and each carbon is split by the hydrogens bonded to it. The splitting follows the same n + 1 rule as in ¹H NMR, with n now counting the hydrogens on the carbon itself. The CH₃ carbon, with three hydrogens, gives four lines. The CH₂ carbon, with two, gives three.
A routine spectrum removes that splitting by proton decoupling. While the carbon signal is being recorded, the instrument also sends in radio waves at the hydrogen frequency. These flip the hydrogens between their two states so quickly that each carbon feels only their average, and the splitting collapses into a single line.
Carbons almost never split each other. Two carbons can only couple if both are ¹³C, and at 1.1% abundance the chance of that for a given pair is about 1.1% × 1.1%, or one molecule in ten thousand. Those rare molecules give extra lines too small to see in a routine spectrum.
Decoupling, together with the rarity of ¹³C–¹³C pairs, gives the rule the rest of this section uses: a routine ¹³C spectrum shows one unsplit line for each distinct carbon environment. A carbon environment is a set of carbons in identical surroundings, and the symmetry section below shows how to find them.
Chemical shift ranges
Ethanol’s two lines sit at 18 ppm (the CH₃) and 58 ppm (the CH₂). The CH₂ carbon is bonded to the oxygen, which pulls electron density away from it. That carbon is deshielded, so it sits further downfield, just as its hydrogens do in the ¹H spectrum.
Carbon shifts are measured in ppm from tetramethylsilane (TMS), as proton shifts are, and the same deshielding logic applies. The difference is the scale: carbon shifts spread across about 0–220 ppm, against 0–13 for protons. The table gives the ranges worth knowing.
| Carbon type | ppm |
|---|---|
| alkyl (sp³) | 0–50 |
| C next to N | 30–65 |
| C next to O | 50–90 |
| alkyne (sp) | 65–90 |
| alkene (sp²) | 100–150 |
| aromatic | 110–160 |
| nitrile | 115–125 |
| ester / acid / amide C=O | 165–185 |
| ketone / aldehyde C=O | 190–220 |
The alkyne row looks out of place. The table shows sp³ carbons at 0–50 and sp² carbons, in alkenes and aromatic rings, above 100. Going from sp³ to sp² moves a carbon far downfield, so you might expect sp carbons, the next step along, to sit further downfield still. Instead they sit between the two, at 65–90.
The reason is anisotropy, the effect that shields an alkyne hydrogen in ¹H NMR. Circulating pi electrons set up a small field of their own, and its direction depends on where you are around them. The two pi bonds of a triple bond together form a cylinder of electron density around the bond axis. When that axis lines up with the applied field, the electrons circulate around it. Inside the cylinder, where both carbons sit, the induced field points against the applied field. The carbons feel a weaker field, so they are shielded and move upfield.
Counting carbon environments: symmetry made visible
Acetone, CH₃COCH₃, has three carbons but gives only two lines: one at 206 ppm for the C=O carbon and one at 31 ppm shared by both methyls. The two methyls are equivalent. Swapping them gives back the same molecule, so they sit in identical surroundings and absorb at the same shift. This is the symmetry test from ¹H NMR, applied to carbons.
So a ¹³C spectrum counts carbon environments, not carbons. Fewer lines than carbons means the molecule has symmetry. Comparing the line count with the molecular formula is often enough to choose between isomers.
o-, m- and p-Xylene are the three dimethylbenzenes, and all have the formula C₈H₁₀. Their ¹H spectra all show a 6H methyl singlet near 2.3 ppm. p-Xylene’s four ring hydrogens are equivalent and give one singlet near 7.0 ppm. o- and m-Xylene instead give small clusters of overlapping lines near 7.0–7.2 ppm, which are harder to tell apart. ¹³C separates all three by line count alone.
The tool is the mirror plane, a plane that reflects one half of the molecule onto the other (see Chirality). Here each one stands at right angles to the ring, so the figure draws it edge-on as a dashed line. Carbons that a mirror swaps are equivalent.
How to count, on p-xylene. Number the ring 1 to 6 with the methyls at C1 and C4. One mirror runs through C1 and C4, so it swaps C2 with C6 and C3 with C5. A second mirror, at right angles to the first, swaps C1 with C4 and C2 with C3. Between them, all four ring CH carbons are equivalent. That leaves three environments: the two methyls (one line), C1 and C4 (one line), and C2, C3, C5 and C6 (one line). 3 lines for 8 carbons.
o-Xylene. The methyls are at C1 and C2, and there is one mirror. It runs between C1 and C2 and between C4 and C5, so it pairs C1 with C2, C3 with C6, and C4 with C5. With the methyls, that makes four environments. 4 lines.
m-Xylene. The methyls are at C1 and C3, and the one mirror runs through C2 and C5. It pairs C1 with C3 and C4 with C6. C2 and C5 lie on the mirror, so neither has a partner. The environments are the two methyls, the C1/C3 pair, C2, the C4/C6 pair and C5. 5 lines, the most of the three.
So the line count alone identifies the isomer: 4 for ortho, 5 for meta, 3 for para.
DEPT: recovering the hydrogen counts
Decoupling makes the spectrum easy to read, but it throws information away. A line at 29 ppm could be a CH₃, a CH₂ or a CH, and the routine spectrum cannot say which. DEPT (Distortionless Enhancement by Polarization Transfer) recovers the number of hydrogens on each carbon without bringing back any splitting. It first excites the hydrogens and then passes their signal on to the carbons they are bonded to.
The standard version, DEPT-135, sends each line up or down by hydrogen count. CH and CH₃ lines point up. CH₂ lines point down. A carbon with no hydrogens receives no signal, so it does not appear at all. NMR usage calls every such carbon “quaternary”, even a C=O carbon, although strictly that word means a carbon bonded to four other carbons.
Compare DEPT-135 with the routine spectrum to find the carbons with no hydrogens. A second experiment, DEPT-90, shows only CH carbons, so it separates CH from CH₃, which both point up in DEPT-135. Between the two, every carbon gets its hydrogen count.
Data. IR: strong at 1710 cm⁻¹, no O–H. ¹³C: four lines, at 214, 44, 26 and 25 ppm. DEPT-135: 26 and 25 point up; 214 and 44 are missing. DEPT-90: no lines.
Formula and IR. The degrees of unsaturation are (2 × 6 + 2 − 12)/2 = 1. The 1710 band is a C=O, and the line at 214 ppm puts it among ketones and aldehydes. That C=O uses the one degree, so there is no ring and no other double bond.
Line count. Four lines for six carbons, so some carbons are equivalent.
DEPT. DEPT-90 is empty, so there is no CH carbon, and the two upward lines at 26 and 25 are both CH₃. The line at 44 has no hydrogens and is far too upfield for a carbonyl. The only oxygen is already in the C=O, so this carbon must be bonded to four carbons: a true quaternary carbon. The line at 214 has no hydrogens either, which rules out an aldehyde: the C=O is a ketone.
Assemble. Only the CH₃ groups carry hydrogens, so all twelve hydrogens are in four methyls. The quaternary carbon needs four carbon neighbors and the C=O needs two. With only four methyls left, the quaternary carbon must carry three of them plus the C=O, and the C=O carries the fourth: CH₃–CO–C(CH₃)₃. The three methyls on the quaternary carbon, a tert-butyl group, are equivalent, which is why four methyls give only two lines. That is 3,3-dimethylbutan-2-one.
An unbranched isomer such as hexan-2-one has no symmetry and would give six lines. The line count alone rules it out.
How the techniques fit together
Each technique answers a different question, and a structure problem usually needs several of them. IR names the functional groups. ¹³C counts the carbon environments and so reveals symmetry, and DEPT adds the hydrogen count on each carbon. ¹H maps how the pieces connect, through shift, integration and splitting. Work in that order, groups then symmetry then connections, and structure determination becomes a procedure instead of a puzzle.
¹³C is also the quickest check on a proposed structure. Count the carbon environments your structure predicts and compare that with the spectrum. If the numbers differ, the structure is wrong.
Every problem above starts from a molecular formula. That formula usually comes from Mass spectrometry, the technique that measures a molecule’s mass.