Logic Puzzle #8: Strong Coupling or Peak Overlap? … Solution

There are two approaches to deciding which scenario best fits the NMR data in the puzzle. A structure elucidator can provide evidence that one scenario is more probable than an other and/or eliminate one scenario on the grounds of insufficient/contradictory data to support it.

For the following 1H NMR spectrum, if the integral of the doublet at 7.42 ppm equates to 1 H atom from a single structure, then the 1H signals at 7.58-7.59 ppm integrating to 2.57 (and assuming all signals are accounted for) relate to overlapping signals (probably a doublet overlapping with a singlet) from a mixture of compounds at varying concentrations.

Another approach, as Adolfo has nicely provided, the magnitude of the strong coupling demonstrated by the signals at 7.58-7.59 ppm indicates a nearby partner on the left side circa 7.70 ppm. Since this appears not to be the case, then by process of elimination, the signals at 7.58-7.59 ppm relate to two overlapping multiplets.

 

Logic Puzzle #8: Strong Coupling or Peak Overlap?

The goal of this puzzle is to distinguish between strong coupling and peak overlap.

For the following 1H NMR spectrum, are the 1H signals at 7.58-7.59 ppm exhibiting strong coupling or are they two overlapping multiplets or both?

 

Logic Puzzle #7: Almost Missed It … Solution 2

With intense solvent signals present on a spectrum, a smaller signal(s) can easily be missed. If 2D NMR data is available, then this extra information can assist in clarifying whether a small signal(s) is obscured by larger signals.

On the 1H-13C HMBC below, the correlations for CDCl3/CHCl3 (due to 1J coupling responses and more) are more intense in comparison to the weak correlation at approximately 6.9 and 77.3 ppm. In this case, the weak correlation is attributed to a quaternary carbon obscured by a set of intense solvent signals.

 

Logic Puzzle #5: Coupling + Tilting + Integration = Benzene … Solution

The goal of this puzzle is to determine the respective benzene ring systems that would exhibit the following 1H NMR spectrum.

The number of substituted benzene ring systems that would exhibit the following 1H NMR spectrum is 3 (assuming no repeating units). The keys to solving this puzzle lie with the determination of which multiplets are coupled in combination with the integral information.

 

Starting with the easy set of multiplets, M08 and M07 are coupled to each other. The second order effects (evident by tilting towards each other) and the similar coupling constants (roughly +/- 0.2 Hz) support this claim. Along side the integral ratio of 2:2, this pattern is intrinsic for a para-substituted benzene ring. Next, the multiplets M01, M03 and M05 are coupled. The 1:2:2 integral ratio and identical coupling constants support a mono-substituted benzene ring. Finally, M02, M04 and M06 are coupled. The 1:1:1 integral ratio and meta-coupling pattern (post#2 link) indicate the presence of a 1,2,4-trisubstituted benzene ring system.

Logic Puzzle #5: Coupling + Tilting + Integration = Benzene

The goal of this puzzle is to determine the respective benzene ring systems that would exhibit the following 1H NMR spectrum.

For the following aromatic region of the 1H NMR spectrum, how many benzene ring systems are present?

Logic Puzzle #5: Coupling + Tilting + Integration = Benzene

The goal of this puzzle is to determine the respective benzene ring systems that would exhibit the following 1H NMR spectrum.

For the following aromatic region of the 1H NMR spectrum, how many benzene ring systems are present?

Logic Puzzle #4: Correlating to the Correct 13C Signal … Solution

The goal of this puzzle is to resolve the ambiguity exhibited within a 2D NMR spectrum and thus provide the correct signal correlation. Although this exercise may seem a trivial one, it is important to go over the rationale when correlating one signal to another.

For the following 1H-13C HSQC-DEPT NMR spectrum it is important to note that the 2 correlations are phased positively (red) and thus represent either a CH or CH3 group and not a CH2. Next, one must ensure that the carbons at 61.5 and 62.2 ppm are 1 carbon each. Although this detail is not certain, we will assume this to be the case. Finally, it is best to start with the easy part first. The 1H signal at 3.50 ppm is correlated to the 13C signal at 62.2 ppm. By process of elimination, one can conclude that the 1H signal at 2.73 ppm is correlated to the 13C signal at 61.5 ppm (see this post for more details).

Ambiguity in correlating 1D and 2D NMR data can routinely occur. Some extra steps that can help avoid this issue are re-aligning the data and/or re-processing the ‘raw’ data with different parameters.

 

Logic Puzzle #4: Correlating to the Correct 13C Signal … Solution

The goal of this puzzle is to resolve the ambiguity exhibited within a 2D NMR spectrum and thus provide the correct signal correlation. Although this exercise may seem a trivial one, it is important to go over the rationale when correlating one signal to another.

For the following 1H-13C HSQC-DEPT NMR spectrum it is important to note that the 2 correlations are phased positively (red) and thus represent either a CH or CH3 group and not a CH2. Next, one must ensure that the carbons at 61.5 and 62.2 ppm are 1 carbon each. Although this detail is not certain, we will assume this to be the case. Finally, it is best to start with the easy part first. The 1H signal at 3.50 ppm is correlated to the 13C signal at 62.2 ppm. By process of elimination, one can conclude that the 1H signal at 2.73 ppm is correlated to the 13C signal at 61.5 ppm (see this post for more details).

Ambiguity in correlating 1D and 2D NMR data can routinely occur. Some extra steps that can help avoid this issue are re-aligning the data and/or re-processing the ‘raw’ data with different parameters.

 

Logic Puzzle #4: Correlating to the Correct 13C Signal

The goal of this puzzle is to resolve the ambiguity exhibited within a 2D NMR spectrum and thus provide the correct signal correlation.

The following 1H-13C HSQC-DEPT NMR spectrum shows two one-bond correlations linked to the 1H signals 2.75 and 3.50 ppm and two closely spaced 13C signals at 61.5 and 62.2 ppm. Does the 1H signal at 2.75 ppm correlate to the 13C signal at 61.5 ppm or the one at 62.2 ppm?

Note: the blue line was added to help align the correlation to the F1 domain.

Logic Puzzle #4: Correlating to the Correct 13C Signal

The goal of this puzzle is to resolve the ambiguity exhibited within a 2D NMR spectrum and thus provide the correct signal correlation.

The following 1H-13C HSQC-DEPT NMR spectrum shows two one-bond correlations linked to the 1H signals 2.75 and 3.50 ppm and two closely spaced 13C signals at 61.5 and 62.2 ppm. Does the 1H signal at 2.75 ppm correlate to the 13C signal at 61.5 ppm or the one at 62.2 ppm?

Note: the blue line was added to help align the correlation to the F1 domain.

Logic Puzzle #1: The Missing Link

A great skill to master is the capability to conceptualize a fragment or structure directly off a spectrum without resorting to paper-and-pen work. This skill is learnt through lots of practice. Whenever partial information is available, an elucidator can conjure up a mental image of possibilities and should it be required instinctively hunt for any missing data.

In the following example, a set of fragments including 13C and 1H chemical shifts and long-range coupling information were extracted from an HMBC experiment (not shown). The green arrows represent the 2-3J coupling responses between the 3 equivalent methyl groups and the carbonyl’s quaternary carbon. Based on these restrictions, what fragment(s) support the data and is there anything missing?

To accommodate these restrictions, three potential fragments, assigned A, B and C, are shown below. Fragment A can be disregarded on the basis of the carbon valence. Fragment B is not a good candidate because the CH3 chemical shifts do not support the presence of an adjacent heteroatom. Fragment C seems to be the most logical choice. However, there is a missing quaternary carbon. The next step is to re-evaluate the NMR data in search of a weak 13C signal at ~40 ppm.

Logic Puzzle #1: The Missing Link

A great skill to master is the capability to conceptualize a fragment or structure directly off a spectrum without resorting to paper-and-pen work. This skill is learnt through lots of practice. Whenever partial information is available, an elucidator can conjure up a mental image of possibilities and should it be required instinctively hunt for any missing data.

In the following example, a set of fragments including 13C and 1H chemical shifts and long-range coupling information were extracted from an HMBC experiment (not shown). The green arrows represent the 2-3J coupling responses between the 3 equivalent methyl groups and the carbonyl’s quaternary carbon. Based on these restrictions, what fragment(s) support the data and is there anything missing?

To accommodate these restrictions, three potential fragments, assigned A, B and C, are shown below. Fragment A can be disregarded on the basis of the carbon valence. Fragment B is not a good candidate because the CH3 chemical shifts do not support the presence of an adjacent heteroatom. Fragment C seems to be the most logical choice. However, there is a missing quaternary carbon. The next step is to re-evaluate the NMR data in search of a weak 13C signal at ~40 ppm.

When an NMR Instrument Fails

Instruments can fail mechanically and when they do fail, it is important to recognize the signs. For NMR data, any irregularities in the baseline can indicate an instrument issue. The best strategy to minimize instrument failures is to perform regular maintenance and collect data for standards with well documented results prior to any data collection.

The legs used to support the NMR magnet are cushioned by lifts to reduce excessive floor vibrations. If one of the lifts is not performing as it should, then the acquired data will exhibit some vibrational noise. The noise can affect the analysis of the data.

Below are two 1H NMR spectra for the same sample magnified by a factor of ten. The spiky baseline (somewhat symmetrical too) in the top spectrum is a result of a malfunctioning lift on one of the legs. The bottom spectrum does not exhibit these spikes; it was collected from the same NMR instrument with the same sample under identical conditions but the lift was repaired.

For the case where all the legs’ lifts are disabled, Glenn Facey’s blog shows the resulting spectrum.

I would like to give a special thanks to Kent for proposing the idea.

When an NMR Instrument Fails

Instruments can fail mechanically and when they do fail, it is important to recognize the signs. For NMR data, any irregularities in the baseline can indicate an instrument issue. The best strategy to minimize instrument failures is to perform regular maintenance and collect data for standards with well documented results prior to any data collection.

The legs used to support the NMR magnet are cushioned by lifts to reduce excessive floor vibrations. If one of the lifts is not performing as it should, then the acquired data will exhibit some vibrational noise. The noise can affect the analysis of the data.

Below are two 1H NMR spectra for the same sample magnified by a factor of ten. The spiky baseline (somewhat symmetrical too) in the top spectrum is a result of a malfunctioning lift on one of the legs. The bottom spectrum does not exhibit these spikes; it was collected from the same NMR instrument with the same sample under identical conditions but the lift was repaired.

For the case where all the legs’ lifts are disabled, Glenn Facey’s blog shows the resulting spectrum.

I would like to give a special thanks to Kent for proposing the idea.

Will the correct structure please stand up? … Part 2

Part 1 presented a challenge to determine an experiment to distinguish two very similar products from each other, namely 3-methyl-5-(pyridin-2-yloxy)pyridine and 5'-methyl-2H-1,3'-bipyridin-2-one. The products have identical formula weights and the LC/MS and 1H NMR are too similar to draw any conclusion from.

 

The first step is to determine what is different between the two products and then identify an experiment specifically designed to focus on that difference. The obvious difference between the two products is the position of the oxygen atom—an ester group verse a carbonyl group. An FT-IR experiment, as commented by the reader Felipe A., can be used to sort out the products.

 

Other experiments can include the use of reducing agents, 15N NMR, 1H -13C HMBC, 1D NOE, 1H-1H TOCSY, MS2, etc. Note free water, acids and sample concentration can inhibit the use of some of these experiments.

 

A 13C NMR experiment may appear to be another good choice when trying to identify a carbonyl group. However, the carbonyl is part of a conjugated system and so the 13C chemical shift is expected around 160 ppm, which also happens to be expected for the 13C chemical shift of the O-C=N group on the other product.

Will the correct structure please stand up? … Part 2

Part 1 presented a challenge to determine an experiment to distinguish two very similar products from each other, namely 3-methyl-5-(pyridin-2-yloxy)pyridine and 5'-methyl-2H-1,3'-bipyridin-2-one. The products have identical formula weights and the LC/MS and 1H NMR are too similar to draw any conclusion from.

 

The first step is to determine what is different between the two products and then identify an experiment specifically designed to focus on that difference. The obvious difference between the two products is the position of the oxygen atom—an ester group verse a carbonyl group. An FT-IR experiment, as commented by the reader Felipe A., can be used to sort out the products.

 

Other experiments can include the use of reducing agents, 15N NMR, 1H -13C HMBC, 1D NOE, 1H-1H TOCSY, MS2, etc. Note free water, acids and sample concentration can inhibit the use of some of these experiments.

 

A 13C NMR experiment may appear to be another good choice when trying to identify a carbonyl group. However, the carbonyl is part of a conjugated system and so the 13C chemical shift is expected around 160 ppm, which also happens to be expected for the 13C chemical shift of the O-C=N group on the other product.

Will the correct structure please stand up? … Part 1

Many organic chemists—if not all—check to see if a synthetic reaction is complete via TLC and LC/MS and/or 1H NMR. At the same time, the chemists are using the analytical data to verify that the final product is what they intended on making. In some cases, LC/MS and 1H NMR do not adequately distinguish one potential product from another. It then becomes a question of identifying a technique(s) that can clearly verify the correct product.

The chemical structures shown below (3-methyl-5-(pyridin-2-yloxy)pyridine and 5'-methyl-2H-1,3'-bipyridin-2-one) are two possible products for a synthetic reaction. They have an identical formula weight (FW) and a nearly identical MS and 1H NMR (not shown). What other experiments can a chemist/spectroscopist propose that will assist in identifying the correct structure and thus distinguish the ester from the carbonyl product?

 

I would like to give a special thanks to David C. Adams for proposing the idea.

Will the correct structure please stand up? … Part 1

Many organic chemists—if not all—check to see if a synthetic reaction is complete via TLC and LC/MS and/or 1H NMR. At the same time, the chemists are using the analytical data to verify that the final product is what they intended on making. In some cases, LC/MS and 1H NMR do not adequately distinguish one potential product from another. It then becomes a question of identifying a technique(s) that can clearly verify the correct product.

The chemical structures shown below (3-methyl-5-(pyridin-2-yloxy)pyridine and 5'-methyl-2H-1,3'-bipyridin-2-one) are two possible products for a synthetic reaction. They have an identical formula weight (FW) and a nearly identical MS and 1H NMR (not shown). What other experiments can a chemist/spectroscopist propose that will assist in identifying the correct structure and thus distinguish the ester from the carbonyl product?

 

I would like to give a special thanks to David C. Adams for proposing the idea.

Ah Sugar, Sugar … Residue

Sugar residues (saccharides) can be tough to elucidate. They tend to have 1H NMR spectrum with overlapping and sometimes poorly-resolved 1H signals, and the 2D NMR data presents lots of ambiguous assignments. With a little practice, an elucidator can quickly pick out a sugar moiety based on a minimal amount of NMR data.

For a given set of atoms with known 13C chemical shifts (shown below), a hexopyranoside sugar moiety is evident even without any long-range coupling information. The general pattern is as follows: a CH at ~100 ppm, 4 CHs between ~70 to 77 ppm, a CH2 at ~60 ppm, 6 O atoms and 4 H atoms. Note: the H atoms count may vary depending on the number of connection points.

An example sugar, phenyl hexapyranoside, with 13C chemical shifts is presented below.

Ah Sugar, Sugar … Residue

Sugar residues can be tough to elucidate. They tend to have 1H NMR spectrum with overlapping and sometimes poorly-resolved 1H signals, and the 2D NMR data presents lots of ambiguous assignments. With a little practice, an elucidator can quickly pick out a sugar moiety based on a minimal amount of NMR data.

For a given set of atoms with known 13C chemical shifts (shown below), a hexopyranoside sugar moiety is evident even without any long-range coupling information. The general pattern is as follows: a CH at ~100 ppm, 4 CHs between ~70 to 77 ppm, a CH2 at ~60 ppm, 6 O atoms and 4 H atoms. Note: the H atoms count may vary depending on the number of connection points.

An example sugar, phenyl hexapyranoside, with 13C chemical shifts is presented below.