Logic Puzzle #9: Does my Unknown contain Br, Cl, S and/or Si atoms? … Solution 2

Atoms like Br, Cl, S and Si present distinct isotope patterns on a mass spectrum. The isotope pattern for a single Br or Cl atom tends to be relatively straightforward and can be viewed directly off the spectrum. In the case for S and Si atoms, a little math is generally required to reveal their presence or absence.

The careful analysis of the intensity for the A+2 signal (m/z 156.0) at 13 eV offers a good notion as to whether any of the following atoms Br, Cl, S or Si are present. The contributions of the isotopes ⁸¹Br, ³⁷Cl, ³⁴S and ³⁰Si to the A+2 signal are related to the isotope-abundance and are listed by IUPAC at ~49.3, 24.2, 4.2 and 3.1%, respectively. Please note that the contributions to the A+2 signal from ¹³C₂, ¹³C⁸⁰Br, ¹³C³⁶Cl, ¹³C³³S and ¹³C³⁰Si will be considered ~0.0% to simply the calculations.

Since the intensity of the A signal (m/z 154.0) is 78.3% at 13 eV, then the intensity of the A+2 signal will be the following if the corresponding atom(s) is present:

1 Br ~76.1% (=78.3*1*49.3/(100-49.3))

2 Br ~152.3% (=78.3*2*49.3/(100-49.3))

1 Cl ~25.0% (=78.3*1*24.2/(100-24.2))

2 Cl ~50.0 % (=78.3*2*24.2/(100-24.2))

1 S ~3.4% (=78.3*1*4.2/(100-4.2))

2 S ~6.9% (=78.3*2*4.2/(100-4.2))

1 Si ~2.5% (=78.3*1*3.1/(100-3.1))

2 Si ~5.0% (=78.3*2*3.1/(100-3.1))

According to the calculations, Br, Cl, S and Si are not present as they do not match the experimental intensity of the A+2 signal at 1.0%.

Formula:

1. % Intensity of A+2 signal = % Intensity of A signal * (Contribution of ^a+2X + Contribution of ¹³C₂ + Contribution of ¹³C^a+1X + …)

2. Contribution of Isotope = Number of Atoms * % abundance / (100 - % abundance)

Logic Puzzle #9: Does my Unknown contain Br, Cl, S and/or Si atoms? … Solution

There are two approaches to solving this problem set. The "quick" approach is to subtract the mass of 10 carbon atoms from the mass of the molecular mass and see if the difference can account for the atoms Br, Cl, S and/or Si. The "longer" approach is to examine the isotope patterns on the MS and the relative abundance of the respective isotopes.

According to the MS below, the molecular ion (M^+.) most probably corresponds to be the most intense signal at m/z 154.0. Given 10 carbons atoms, the difference is 34 Da (154.0 – 120 Da). Therefore, isotopes 79Br and 35Cl can be ruled out leaving either one atom of 32S or 28Si for the unknown. The molecular formula for the unknown could be C10 S1 H2 or C10 Si1 H6.

 
The subsequent post will examine the isotope pattern and thus examine whether the proposed molecular formulae are consistent with the MS data.

Logic Puzzle #9: Does my Unknown contain Br, Cl, S and/or Si atoms?

Characteristic isotopic patterns in MS can assist the elucidator in revealing the presence or absence of atoms. For the Br, Cl, S and Si atoms, a good approach is to examine the peak intensity of the A+2 signal. The respective contributions by the isotopes 81Br, 37Cl, 34S and 30Si are approximately 49.3, 24.3, 4.2 and 3.1%, respectively.

Below are the molecular ion regions for two EI mass spectra for the same unknown compound. The top MS was collected at 13 eV and the bottom was collected at 70 eV. A list of m/z and intensities are also provided. Given the carbon count to be 10 atoms, are any of the following atoms Br, Cl, S and/or Si present?

 
 
A special thanks goes to Scott Van Bramer for allowing me to use the data.

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.

My Column is Bleeding

Whenever a GC column is used to identify and/or quantify a sample, the column stationary phase can bleed into the MS source along with the sample. High column bleed can hinder the analysis of a sample. The resulting spectral interference typically manifests itself as discrete peaks and/or an increase in the drift of the baseline, which in turn, produces data with low signal-to-noise and poor sensitivity.  

The GC-EI mass spectra below are two examples showing high column bleed. The ion peaks at m/z 73, 133, 193, 207, 267, 281, 355 and 429 are not part of the purified sample but pertain to the column stationary phase. The major column bleed ion, m/z 207, is a result of the formation of hexamethylcyclotrisiloxane. The presence of these characteristic masses for siloxanes indicates there is a significant column bleed and that the column may need replacing.

  

My Column is Bleeding

Whenever a GC column is used to identify and/or quantify a sample, the column stationary phase can bleed into the MS source along with the sample. High column bleed can hinder the analysis of a sample. The resulting spectral interference typically manifests itself as discrete peaks and/or an increase in the drift of the baseline, which in turn, produces data with low signal-to-noise and poor sensitivity.  

The GC-EI mass spectra below are two examples showing high column bleed. The ion peaks at m/z 73, 133, 193, 207, 267, 281, 355 and 429 are not part of the purified sample but pertain to the column stationary phase. The major column bleed ion, m/z 207, is a result of the formation of hexamethylcyclotrisiloxane. The presence of these characteristic masses for siloxanes indicates there is a significant column bleed and that the column may need replacing.

  

Re-evaluating the data from MS and NMR … Part 4

With any type of data, there is an inherent risk of misinterpretation. My advice to elucidators is to consider multiple solutions and examine each one thoroughly. In the end, the answer to any problem set lies in tying together the bits of information in hopes of understanding the bigger picture.

Recap of the problem: The ESI+ MS shows a single [M+H]+ at m/z 102 allowing a maximum carbon count of 8. The 13C NMR shows there to be 12 carbons. How can the data from the MS and NMR present such different results for the same unknown?

The data from both the 13C NMR and DEPT-135 spectra are consistent with a mixture of two similar compounds at approximately a 1:1 ratio.

A mixture with an ESI+ MS exhibiting a single molecular ion indicates that the compounds in the mixture differ by a proton. For example, the mixture comprises of one compound with an R-NH2 group and the other with an R-NH3+ group. Some example amine/aminium mixtures are shown below.

Re-evaluating the data from MS and NMR … Part 4

With any type of data, there is an inherent risk of misinterpretation. My advice to elucidators is to consider multiple solutions and examine each one thoroughly. In the end, the answer to any problem set lies in tying together the bits of information in hopes of understanding the bigger picture.

Recap of the problem: The ESI+ MS shows a single [M+H]+ at m/z 102 allowing a maximum carbon count of 8. The 13C NMR shows there to be 12 carbons. How can the data from the MS and NMR present such different results for the same unknown?

The data from both the 13C NMR and DEPT-135 spectra are consistent with a mixture of two similar compounds at approximately a 1:1 ratio.

A mixture with an ESI+ MS exhibiting a single molecular ion indicates that the compounds in the mixture differ by a proton. For example, the mixture comprises of one compound with an R-NH2 group and the other with an R-NH3+ group. Some example amine/aminium mixtures are shown below.

Re-evaluating the data from MS and NMR … Part 1

Structure elucidators will routinely use data from multiple techniques such as MS and NMR to build a proposed structure(s). When dealing with data from multiple techniques, the issue may arise that the data seem to contradict each other. In these cases, it is best to step back and re-evaluate the data from a different angle.

The ESI+ MS data below shows a prominent [M+H]+ ion at m/z 102 and its sodiated adduct. The maximum number of carbons possible for the ion is 8 (= 102 / 12). The 13C NMR spectrum below shows 12 carbons signals, all aliphatic and no quaternary carbons.

 

 

Assuming no issues with the instruments, how the data was acquired or how the sample was prepared, the lingering issue is how can the data from the MS and NMR present such different results for the unknown?

Re-evaluating the data from MS and NMR … Part 1

Structure elucidators will routinely use data from multiple techniques such as MS and NMR to build a proposed structure(s). When dealing with data from multiple techniques, the issue may arise that the data seem to contradict each other. In these cases, it is best to step back and re-evaluate the data from a different angle.

The ESI+ MS data below shows a prominent [M+H]+ ion at m/z 102 and its sodiated adduct. The maximum number of carbons possible for the ion is 8 (= 102 / 12). The 13C NMR spectrum below shows 12 carbons signals, all aliphatic and no quaternary carbons.

 

 

Assuming no issues with the instruments, how the data was acquired or how the sample was prepared, the lingering issue is how can the data from the MS and NMR present such different results for the unknown?

Determining the Site of Modification … Part 3

Tandem mass spectrometry involves the process of selecting and separating a product ion(s) (or daughter ion(s)) and fragmenting it in a second mass analyzer. This is commonly referred to as MS/MS or MS2. Additional tandem processes can be applied to ions in the MS/MS data to create MS3 data, and so forth.

The metabolites A and B share the same exact mass, and as such, cannot be differentiated by the MS data alone (see Part 1 of this series). The MS/MS data, described in Part 2, offers fragment information that can assist in eliminating one of the candidates. Taking it a step further and thus verifying the candidate metabolite A, ESI+ MS3 data is presented herein.

The fragmentation scheme below shows both the Parent and Metabolite A with the fragment at 121 Da, in bold, fragmenting to create a fragment at 93 Da. This is supported by the nearly-identical MS3 data displayed below.

 

On the right-hand side of the fragmentation scheme, the fragments 106 and 108 Da are expected for the MS3 data of the Parent and Metabolite A, respectively.

Determining the Site of Modification … Part 3

Tandem mass spectrometry involves the process of selecting and separating a product ion(s) (or daughter ion(s)) and fragmenting it in a second mass analyzer. This is commonly referred to as MS/MS or MS2. Additional tandem processes can be applied to ions in the MS/MS data to create MS3 data, and so forth.

The metabolites A and B share the same exact mass, and as such, cannot be differentiated by the MS data alone (see Part 1 of this series). The MS/MS data, described in Part 2, offers fragment information that can assist in eliminating one of the candidates. Taking it a step further and thus verifying the candidate metabolite A, ESI+ MS3 data is presented herein.

The fragmentation scheme below shows both the Parent and Metabolite A with the fragment at 121 Da, in bold, fragmenting to create a fragment at 93 Da. This is supported by the nearly-identical MS3 data displayed below.

 

On the right-hand side of the fragmentation scheme, the fragments 106 and 108 Da are expected for the MS3 data of the Parent and Metabolite A, respectively.

Determining the Site of Modification … Part 2

Peak matching involves the process of comparing spectral data from a parent or starting material to an unknown compound. (The unknown compound can be referred more specifically as the product, impurity, degradant, metabolite, etc.). The similarities between the data indicate regions that have not changed while the differences indicate regions of change.

The full scan MS data in Part 1 does not offer enough spectral information to eliminate one of the candidate structures. The next step is to examine MS/MS data.

The ESI+ product ion spectra (MS/MS) for both the Parent (m/z 226) and Metabolite (m/z 228) are shown below. The spectra share a common fragment at m/z 121 and differ in the fragments at m/z 134/136 and 149/151.

Based on the fragments shown below, the suspected site of hydrogenation is on the carbonyl for metabolite A.

 

Determining the Site of Modification … Part 2

Peak matching involves the process of comparing spectral data from a parent or starting material to an unknown compound. (The unknown compound can be referred more specifically as the product, impurity, degradant, metabolite, etc.). The similarities between the data indicate regions that have not changed while the differences indicate regions of change.

The full scan MS data in Part 1 does not offer enough spectral information to eliminate one of the candidate structures. The next step is to examine MS/MS data.

The ESI+ product ion spectra (MS/MS) for both the Parent (m/z 226) and Metabolite (m/z 228) are shown below. The spectra share a common fragment at m/z 121 and differ in the fragments at m/z 134/136 and 149/151.

Based on the fragments shown below, the suspected site of hydrogenation is on the carbonyl for metabolite A.

 

Determining the Site of Modification … Part 1

In past blogs, I have ascribed the process of peak matching as an integral part of structure elucidation. In this series, peak matching is demonstrated with the use of MS data.

The scheme below shows a parent compound (2-oxo-N,2-diphenylacetamide) with two possible sites for hydrogenation. The potential modified compounds are 2-hydroxy-N,2-diphenylacetamide and 2-hydroxy-1-phenyl-2-(phenylamino)ethanone labeled A and B, respectively.

The ESI+ full scan MS below belong to the parent compound (top) and the modified compound (bottom). With the given data, can the suspected site of modification be determined?

 

Determining the Site of Modification … Part 1

In past blogs, I have ascribed the process of peak matching as an integral part of structure elucidation. In this series, peak matching is demonstrated with the use of MS data.

The scheme below shows a parent compound (2-oxo-N,2-diphenylacetamide) with two possible sites for hydrogenation. The potential modified compounds are 2-hydroxy-N,2-diphenylacetamide and 2-hydroxy-1-phenyl-2-(phenylamino)ethanone labeled A and B, respectively.

The ESI+ full scan MS below belong to the parent compound (top) and the modified compound (bottom). With the given data, can the suspected site of modification be determined?

 

Identifying the Molecular Ion using Dimer Information

A previous blog described how a sodiated ion peak can be used to locate or calculate the molecular ion for an unknown compound. In a similar fashion, the dimer ion peak can be used to identify the mass of the unknown even if the molecular ion is no visible.

The ESI+ MS data below shows 4 ion peaks at m/z 401.0, 422.9, 801.0 and 823.0. Assuming the ion peak at m/z 801.0 is 2M+H+, the weight of the unknown is (801.0-1.007) / 2 = 400.0 Da. The same holds true if the sodiated dimer ion (2M+Na+) is used: (823.0-22.989) / 2 = 400.0 Da.