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 #7: Almost Missed It … Solution

‘How many signals are present?’ is such a simple question, and yet, it is a fundamental question to an elucidator. Mistaken a signal or overlook one and the elucidator can run the risk of wasting time and effort.

The 13C [1H] NMR spectrum below shows 3 discernible signals that are attributed to the solvent CDCl3. In addition, one can begin to speculate on weaker signals; there might a fourth signal at ~77.2 ppm (most likely due to residual CHCl3), possibly a fifth at ~77.5 ppm, perhaps a sixth one at ~77.3 ppm and maybe more.

The next step is to examine additional data and verify whether the ‘weak’ signals are real or not. This can include:

1. comparing the weak signals to other structural signals,

2. applying deconvolution/peak fitting to this region,

3. checking 2D NMR data,

4. acquiring data in a different solvent,

5. modifying the acquisition parameters to exclude the solvent or increase S/N, etc.

Logic Puzzle #7: Almost Missed It

When dealing with small sample concentrations of an unknown compound, the NMR region where a large solvent signal appears rarely gets a second look. However, a large NMR signal can easily obscure a small signal. From one elucidator to another, check the solvent signal for any structural signals.

In the 13C [1H] NMR spectrum below, how many signals are present?

If the spectrum is zoomed in, how many signals can be seen now?

Logic Puzzle #6: Dealing with an Extra 13C Peak … Solution

Socrates is known for saying ‘Know thyself’, along the same line, chemists should ‘Know thy instrument’. The example below is one such case.

The 13C NMR [1H] spectrum below exhibits 6 signals. If an unknown compound comprises of 7 carbon atoms, then the following scenarios, or combinations thereof, are possible to account for missing or extra carbon peaks on the spectrum:

1. the 13C peaks are too weak to be clearly evident,

2. the 13C peaks are overlapping (equivalent or coincidental), and/or

3. the 13C peaks may pertain to instrument artefacts, mixtures or impurities.   

The carbon signals are 111 and 118 ppm may account for 2 carbons atoms each. This is based on the relative intensities of the peaks and this is a characteristic prevalent for aromatic carbons. The total count is now at 8 atoms. The higher atom count indicates the presence of an artefact or impurity. The signal at the exact centre of the spectrum, 100 ppm, can be attributed to a quadrature spike – an artefact produced from the instrument. If so, the total count matches the known carbon count.

Although this analysis is not conclusive, it is worth noting that there are several other possible interpretations and only additional data will help substantiate one possibility over the other.

Thank you Adolfo, Maxa and Serge for your comments.

Logic Puzzle #6: Dealing with an Extra 13C Peak

Structure elucidation by NMR involves a deep understanding of various technical aspects behind data acquisition. Being aware of how the instrument works can facilitate the process and reduce the aggravation.

For today’s puzzle, an unknown compound is known to comprise of 7 carbon atoms and exhibit the following 13C NMR spectrum with 1H decoupling. How can the 7 atoms be accounted for in the spectrum below?

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 #2: How to link 3 Fragments

The goal of this puzzle is to logically combine a set of fragments using valence and NMR information.

In this puzzle, three fragments are correlated through 2-3J coupling responses (represented by a green arrow) that were extracted from a 1H-13C HMBC data (spectrum not shown). The carbon atoms with the 13C chemical shifts displayed in blue indicate the presence of an adjacent heteroatom. Based on these criteria, what 'complete' fragment(s) supports the data and is there anything missing?

In order to accommodate these restrictions, a logical fit is to consider a trivalent atom, e.g. nitrogen.

Logic Puzzle #2: How to link 3 Fragments

The goal of this puzzle is to logically combine a set of fragments using valence and NMR information.

In this puzzle, three fragments are correlated through 2-3J coupling responses (represented by a green arrow) that were extracted from a 1H-13C HMBC data (spectrum not shown). The carbon atoms with the 13C chemical shifts displayed in blue indicate the presence of an adjacent heteroatom. Based on these criteria, what 'complete' fragment(s) supports the data and is there anything missing?

In order to accommodate these restrictions, a logical fit is to consider a trivalent atom, e.g. nitrogen.

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.

Signals can simply disappear on a DEPT-135 experiment

There are many advantages in working with a 1H-13C HSQC-DEPT spectrum over a 13C DEPT-135 and a 1H-13C HSQC (see Post 1 & Post 2). In most cases, a 1H-13C HSQC-DEPT is more valuable than either one of those experiments.

The aliphatic region of a 1H-13C HSQC-DEPT is spectrum below. Two coincidental carbon signals are overlapping at 38.51 ppm, one pertaining to a CH while the other a CH2 group. The DEPT-135, attached to the F1 domain, exhibits a weak 13C signal that can easily be misconstrued, for example as an impurity, if not for the extra information from the 2D NMR experiment.

Signals can simply disappear on a DEPT-135 experiment

There are many advantages in working with a 1H-13C HSQC-DEPT spectrum over a 13C DEPT-135 and a 1H-13C HSQC (see Post 1 & Post 2). In most cases, a 1H-13C HSQC-DEPT is more valuable than either one of those experiments.

The aliphatic region of a 1H-13C HSQC-DEPT is spectrum below. Two coincidental carbon signals are overlapping at 38.51 ppm, one pertaining to a CH while the other a CH2 group. The DEPT-135, attached to the F1 domain, exhibits a weak 13C signal that can easily be misconstrued, for example as an impurity, if not for the extra information from the 2D NMR experiment.

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.

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.

Allene versus Carbonyl

Functional groups such as allene (C=C=C) and cumulene (C=C=C=C) present an interesting challenge in an elucidation project, especially when an elucidator is not expecting them. The 13C chemical shifts for allene carbons are typically expected at 80, 200 and 100 ppm +/- ~20 ppm per shift. The confusion, or better described as bias, arises at treating a 13C chemical shift of 200 ppm as a carbonyl (C=O) rather than considering the possibility of an allene group.

In the example below, there are an odd number of sp2 carbon atoms along with 1 oxygen atom and 6 hydrogen atoms.

The example leads into two possible candidates: an allene in 1-(ethenyloxy)propadiene ethenyl propadienyl ether or an alkene plus a carbonyl group in penta-1,4-dien-3-one.