Signal to Noise Ratio on Our NMR Instrument
For a high-quality NMR spectrum, one goal is getting a good strong signal from the compound
against a low-noise baseline. This requires getting the right amount of compound and solvent into
the NMR sample tube. While having too much compound in the sample is a problem, it’s rarely
encountered in the undergraduate lab. The experiments and results using an organic sample are
described.
The Experiment:
The compound was 1,3,5-triethylbenzene, and the solvent was CDCl3. The
mass of the triethylbenzene was measured, and the optimal amount of CDCl3
was added (4 cm in the tube, 0.75 mL, about 1 cm above the constriction on
a large pipette). The mass of triethylbenzene varied between 1 and 30 mg.
Proton and carbon spectra were recorded, and the signal to noise ratios were
measured for each spectrum.
Why Triethylbenzene?
There is a lot of this material available in the lab, it’s of high purity, the signals are strong, firstorder and simple to understand. For the proton spectra triethylbenzene has an equal number of
methyl (CH3), methylene (CH2), and methine (CH) groups. For the carbon spectra, there are three
primary (CH3) carbons, three secondary carbons (CH2), three tertiary carbons (CH) and three
quaternary carbons (C). All of the signals in both spectra are well dispersed.
Signal to Noise measurement:
The signal to noise ratio is expressed as S/N. It is calculated by measuring (in millimeters, use a
ruler!) the height from the baseline (approximately the middle of the noise) of the strongest signal, and dividing by the maximum separation of the strongest “peaks” in the noise.
Proton Spectra:
The standard 1H program (16 scans) was used for recording the spectra, and we actually couldn’t
get a bad proton spectrum of triethylbenzene. With about 1.0 mg of the compound in the NMR
tube, the S/N was about 320, and with 30 mg of triethylbenzene the S/N was > 6000 (estimated
in SpecView).
These two spectra (printouts from the instrument) are below in figures 1 and 2. The largest difference between the two spectra is the noisier baseline, and the increased intensity of the residual
CHCl3 and H2O signals in the more dilute sample. You can see the expected triplet and quartet
pattern for the ethyl group, and the integration ratios of 1:2:3 in figure 1.
Figure 1: Triethylbenzene, 30 mg, 1H NMR spectrum
Figure 2: Triethylbenzene, 1.0 mg, 1H NMR spectrum
Carbon Spectra, part 1:
Using the standard 13C program (400 scans), the spectra here varied from really good (15 mg or
more of triethylbenzene), to acceptable (10 and 5 mg of triethylbenzene), to poor (2.5 mg or less
of triethylbenzene). The S/N decreased from 28.4 to 1.3, in going from 30 to 1.2 mg of the compound. Even with the poor S/N, only in the weakest sample did the quaternary carbon signal
disappear into the noise.
The 30 mg and the 1.2 mg spectra are shown below as figures 3 and 4. A plot of S/N vs mass is
linear with a slope close to 1, so doubling the mass of compound in the tube doubles the S/N. In
this case, it’s a coincidence that the S/N is about equal to the mass of the sample.
Figure 3: Triethylbenzene, 30 mg, 13C NMR spectrum, 400 scans
Figure 4: Triethylbenzene, 1.2 mg, 13C NMR spectrum, 400 scans
Carbon Spectra, part 2:
Using the 30 mg sample of triethylbenzene, the number of scans was varied from 3200 to just 1.
S/N went from 72.5 to 1.3. In this last experiment, the quaternary carbon signal was just visible,
if you knew where to look. These spectra are shown in Figures 5 and 6.
The math behind the theory of the signal to noise ratio states that the signal should increase as
the square root of the number of scans, and lo and behold it does—the square root of 3200 is 56.6,
and 1.3 x 56.6 is 73.5!
Figure 5: Triethylbenzene, 30 mg, 13C NMR spectrum, 3200 scans
Figure 6: Triethylbenzene, 30 mg, 13C NMR spectrum, 1 scan
The Conclusion:
Get a proper amount of compound into the NMR tube—measure the mass if you need to—and
don’t drown it in solvent. Adding more solvent than necessary wastes solvent, reduces the concentration of your sample and thus reduces the signal to noise. Adding scans can be done, but
not on a routine basis, since it takes four times as many scans to get twice the S/N. This is very
inefficent for carbon spectra when there is a lot of sample available.
Why Not Triethylbenzene?
While this compound has a nice mixture of proton and carbon types, there are only a few different signals. Triethylbenzene is also 88.9 % carbon and 11.1 % hydrogen, and is of low molecular
mass (162 g/mol). Thus there’s a lot of molecules in 30 mg of triethylbenzene.
For a contrasting example, consider 1,3,5-tris(bromomethyl)-2,4,6-triethylbenzene. This compound is 40.8% carbon, only 4.8 % hydrogen, and a molecular mass of 441 g/mol. Thus a 30 mg
sample represents a lot less carbon and hydrogen in the NMR sample than triethybenzene. Despite this disadvantage, the proton spectrum was excellent, the carbon spectrum was good with
a S/N of 8.6, and all signals were clearly visible (see Figure 7).
Poor sample preparation in this example would be disastrous. For example, using 12 mg of sample (‘cause you guessed it was OK) and 6 cm of solvent (just because) would reduce the S/N to
about 2.3—and you risk losing signals at that S/N level.
Similar examples could be done for other compounds. Many organic compounds have a low
molecular mass, and are mostly carbon, but all carbon atoms are usually different, so each signal
is relatively weaker. Inorganic compounds, where a large % of the molecular mass is due to the
metal, need even more care to get a good spectrum. The conclusion is the same—get enough compound into the tube and don’t dilute the compound with more solvent than necessary.
Figure 7: Tris(tribromo)triethylbenzene, 30 mg, 13C NMR spectrum, 400 scans
© Peter Marrs, University of Victoria Chemistry 363 NMR Notes, September 2014