Philosophy to Chemistry to Elucidation

Logic Puzzle #29: Adjusting the Threshold … Solution

Part of the process of interpreting data from 2D NMR spectra is to adjust the intensity threshold in order to differentiate visually the signals from the noise. A certain level of cautious peak picking is needed so not to overlook a weak signal nor confuse a noisy region for a real signal.

For an unknown compound, the 1H-13C HSQC spectrum shows several overlapping 1H multiplets equating to 4 non-exchangeable H atoms, four C atoms with one C signal more intense than the other two and two 1H-13C correlations with the third correlation in question. If all H atoms are to be accounted for, there are three scenarios to consider:

1. 1 CH at 123 ppm, 2 CH's at 120 ppm, 1 CH at 122 ppm = 4 H and 4 C atoms
2. 1 CH at 123 ppm, 2 CH's and 1 CH2 at 120 ppm, 1 C at 122 ppm, 1 artefact correlation = 4 H and 5 C atoms
3. 1 CH at 123 ppm, 3 CH's at 120 ppm, 1 C at 122 ppm, 1 artefact correlation = 4 H and 5 C atoms

Unfortunately, the data at hand may not be sufficient to rule out any of the three scenarios.

Logic Puzzle #29: Adjusting the Threshold

Spectral data that exhibit weak signals (low S/N) can easily be misinterpreted. A good approach is to check the data against other data to confirm whether the signal is real or an artefact.

For an unknown compound, the 1H -13C HSQC spectrum below is shown at two different intensity thresholds. The aromatic protons at 6.97 and 7.00 ppm equate to 4 non-exchangeable hydrogen atoms. Is the correlation between the 1H signal at 6.97 ppm and the carbon at 122 ppm real or an artefact?

Logic Puzzle #28: Interference from the Solvent … Solution

With deuterated chloroform, the signals from the solvent can overlap with the signals from the unknown of interest. The following spectra are some examples one may encounter.

In the expanded regions for each 13C NMR spectrum, Cases 1, 2, 3, 4 and 5 display a weak signal at ~76.8, 77.3, 77.5, 77.5, 77.5 ppm, respectively. The weak signals, indicated by blue arrows, are attributed to the unknown of interest.

In all five 13C NMR spectra, a clear, intense triplet is evident. The triplet, labeled with green arrows in Case 1, is a result of the solvent CDCl3. At ~77.2 ppm, a small signal is evident. This signal, labeled with a red arrow, is a result of the solvent CHCl3.

Logic Puzzle #28: Interference from the Solvent

Solvent signals that overlap with signals from the unknown can be a hurdle in an elucidation. The energy wasted when misidentifying a signal can make an elucidation attempt wearisome and frustrating to put it mildly.

Several 13C NMR spectra were collected in deuterated chloroform for a set of unknown compounds. Are any non-solvent signals evident in the expanded regions below?

Logic Puzzle #24: Conflicting Masses? … Solution

The explanation offered herein covers one possible scenario to a complicated situation. Please note that other explanations are possible.

The LC/MS data shows an unknown at 1.8 minutes exhibiting an [M+H]+ at m/z 260 and an [M-H]- at m/z 256. The difference of 4 mass units lends itself into a scenario of conflicting data.

One potential explanation for the differing LC/MS data is the unknown is a really comprised of a mixture of two compounds ionizing under separate ion modes and differing by 2 mass units. The short column run (< 2.5 minutes) done under UPLC conditions appears not to be adequate to separate out the mixture as they elute.

Thanks Graham M. for your assistance.

Logic Puzzle #24: Conflicting Masses?

The purpose of this puzzle is to examine data collected from multiple detectors with the hopes of sorting out the conflicting data surrounding the unknown.

In an attempt to identify the structure for an unknown, UPLC data was collected using a C18 column with a run time of 2.2 minutes. The UV, ESI- and ESI+ traces are presented for the unknown eluting at 1.8 minutes. What is the mass of the unknown?

Logic Puzzle #19: Can you spot the solvent? … Solution#2

In the previous post, the solution noted that the multiplets at 1.2 and 3.5 ppm were related to a single compound (labeled B on the 1H NMR spectrum below). This post will attempt to clarify this reasoning.

The relative integrals for the multiplets at 1.2 and 3.5 ppm are 1.20 and 0.81, respectively.

The ratio of the integrals = 1.20:0.81 or 1.48:1 or 2.96:2 or 4.44:3

The ratio 2.96:2 closely resembles a 3:2 ratio, or potentially a CH3:CH2. Furthermore, the coupling patterns t and q support a methyl group adjacent a methylene group.

Logic Puzzle # 19: Can you spot the solvent? … Solution

The following publication by Gottlieb et al. (linked here) offers an excellent collection of 1H and 13C chemical shifts to commonly-used NMR solvents.

From the 1H NMR spectrum below, the first step is to differentiate the signals pertaining to the unknown from those of the residual solvent. The key here lies with the integral values.

The integrals for the multiplets at 2.4, 2.7 and 3.2 ppm are very close to whole numbers and as such point to one compound (labeled A on the spectrum). The integrals for the multiplets at 1.2 and 3.5 ppm indicate a second compound (labeled B on the spectrum).

Diethyl ether exhibits a typical 1H NMR pattern, a triplet at ~1.2 ppm (CH3) and a quartet at ~3.5 ppm (CH2) in CDCl3 at room temperature. Ethanol exhibits an identical coupling pattern to diethyl ether but the chemicals shifts are slightly higher at ~1.7 (CH3) and 3.7 ppm (CH2). Based on the similarities between the chemical shifts, the residual solvent appears to be diethyl ether.

A special thanks goes to Bogdan, Carlos, R. Murugan and Djalma for their comments.

Logic Puzzle # 19: Can you spot the solvent?

Residual solvents are typically organic volatile chemicals that are not completely removed from the sample. Many residual solvents exhibit a characteristic pattern when examined via NMR. The goal of this puzzle is to identify a residual solvent by NMR.

The aliphatic region of the 1H NMR spectrum below is for an unknown acquired in CDCl3. Are any residual solvents evident?

Logic Puzzle #16: Can you spot the Impurity(s)? … Solution



If an elucidator is too successful differentiate an impurity signal from a signal belonging to a compound-of-interest, then the following may be useful: foreknowledge of the prep work, experience with similar compounds and impurities, variations within the spectral data, trial-and-error, etc.

From the 1H NMR spectrum, the integral of the 1H signal at 2.52 ppm is approximately a third less than the other integrals. This slight difference is an indication that this particular signal may be an impurity.

The 1H -13C HSQC spectrum shows all the 1H signals as protonated carbons. The integrals from the 1H NMR indicate that most of the protonated carbons are methine carbons (CH).

When attempting to solve for this unknown compound, it may be best to not use any information correlated to the 1H signal at 2.52 ppm and see what potential candidates arise without it.