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 #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 #3: A Play on the Molecular Formulae

The goal of this puzzle is to determine how the molecular formulae of the intermediates may assist in reasoning out the final product.

In this puzzle, let’s consider the following one-pot synthetic reaction (solvents and additional reactants are not shown). The molecular formula (MF) and the RDBE information are also presented. The chemical reaction illustrates a C11 compound reacting to form a C9 compound and subsequently a C15 compound. The reaction continues and produces an unknown compound with a MF of C24 H23 N1 O2. Based on the given information, how can the MF of the unknown compound be explained?

 The unknown comprises of 24 carbon atoms and so if you add the carbons from the 2^nd intermediate with the carbons from the 3^rd intermediate you arrive at the 24 carbons (9+15=24). This is also evident for the nitrogen atom count (0+1=1) and the RDBE count (5+9=14). This is not the case for the hydrogen (10+15=25) and oxygen (2+1=3) atoms. However, if the loss of a H2O molecule is considered, the unknown can be a combination of the 2^nd and 3^rd intermediate.

Logic Puzzle #3: A Play on the Molecular Formulae

The goal of this puzzle is to determine how the molecular formulae of the intermediates may assist in reasoning out the final product.

In this puzzle, let’s consider the following one-pot synthetic reaction (solvents and additional reactants are not shown). The molecular formula (MF) and the RDBE information are also presented. The chemical reaction illustrates a C11 compound reacting to form a C9 compound and subsequently a C15 compound. The reaction continues and produces an unknown compound with a MF of C24 H23 N1 O2. Based on the given information, how can the MF of the unknown compound be explained?

 The unknown comprises of 24 carbon atoms and so if you add the carbons from the 2^nd intermediate with the carbons from the 3^rd intermediate you arrive at the 24 carbons (9+15=24). This is also evident for the nitrogen atom count (0+1=1) and the RDBE count (5+9=14). This is not the case for the hydrogen (10+15=25) and oxygen (2+1=3) atoms. However, if the loss of a H2O molecule is considered, the unknown can be a combination of the 2^nd and 3^rd intermediate.

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.

Exchangeables Protons acquired in different Deuterated Solvents

A deuterated solvent can impact whether exchangeable protons, such as OH or NH, are visible on a 1H NMR spectrum. The advantage of eliminating any contribution from an exchangeable proton(s) is to simplify spectral interpretation. One disadvantage of not seeing exchangeable protons is the deficiency to the total proton count in establishing the molecular formula.

The diagram below compares two 1H NMR spectra for the same functional group in two different deuterated solvents. The 1H NMR spectrum in deuterated-DMSO shows two doublets whereas the spectrum acquired in deuterated-methanol shows only the singlet for the CH group.

Exchangeables Protons acquired in different Deuterated Solvents

A deuterated solvent can impact whether exchangeable protons, such as OH or NH, are visible on a 1H NMR spectrum. The advantage of eliminating any contribution from an exchangeable proton(s) is to simplify spectral interpretation. One disadvantage of not seeing exchangeable protons is the deficiency to the total proton count in establishing the molecular formula.

The diagram below compares two 1H NMR spectra for the same functional group in two different deuterated solvents. The 1H NMR spectrum in deuterated-DMSO shows two doublets whereas the spectrum acquired in deuterated-methanol shows only the singlet for the CH group.

Hampering Data Interpretation

A common misinterpretation of 2D NMR data can occur when dealing with weak correlations. Weak correlations are commonly introduced in how the sample is prepared or how the data is collected or processed. Examining the spectrum down to level of the density matrix can ensure all correlations are picked up.

The 1H-13C HMQC below shows a protonated carbon at 3.2 and 32 ppm. Since the proton multiplet at 3.6 ppm has no carbon correlation, the proton is most likely from an exchangeable group such as NH or OH group.

When the spectrum threshold is lowered to 1 % relative to the most intense correlation, a weak correlation at 2.6 and 41.9 ppm is seen. Therefore, the proton at 3.6 ppm is actually a CH group and not an exchangeable one.

Hampering Data Interpretation

A common misinterpretation of 2D NMR data can occur when dealing with weak correlations. Weak correlations are commonly introduced in how the sample is prepared or how the data is collected or processed. Examining the spectrum down to level of the density matrix can ensure all correlations are picked up.

The 1H-13C HMQC below shows a protonated carbon at 3.2 and 32 ppm. Since the proton multiplet at 3.6 ppm has no carbon correlation, the proton is most likely from an exchangeable group such as NH or OH group.

When the spectrum threshold is lowered to 1 % relative to the most intense correlation, a weak correlation at 2.6 and 41.9 ppm is seen. Therefore, the proton at 3.6 ppm is actually a CH group and not an exchangeable one.

Extracting Proton Information from 13C NMR data

In addition to using chemical shift information to ascertain a carbon’s proton count (i.e. C, CH, CH2 or CH3), 13C NMR experiments can be set up in a variety of ways to assist with this process.

The following simulated spectra compare a variety of 13C NMR experiments for aspartame. Please note that there are variations to the list below that are not being shown. These include such experiments as DEPTQ-135 and DEPT-HSQCSE (sensitivity enhanced).

 

   

 

   

 

 

Extracting Proton Information from 13C NMR data

In addition to using chemical shift information to ascertain a carbon’s proton count (i.e. C, CH, CH2 or CH3), 13C NMR experiments can be set up in a variety of ways to assist with this process.

The following simulated spectra compare a variety of 13C NMR experiments for aspartame. Please note that there are variations to the list below that are not being shown. These include such experiments as DEPTQ-135 and DEPT-HSQCSE (sensitivity enhanced).

 

   

 

   

 

 

Searching for Unknowns via the Internet … Part 2

A previous blog described the Internet as a valuable tool in searching for previous work based on a molecular formula. In addition to searching the Internet “directly”, certain online sites can assist and facilitate the search process. Some of these sites are free while others have a fee.

Be aware that the Internet is full of misleading and often times incorrect data. It is one thing to make a mistake in the elucidation process; it is another to follow someone else’s mistake. As always, the next step is to verify the data and publish once.

Searching for Unknowns via the Internet … Part 2

A previous blog described the Internet as a valuable tool in searching for previous work based on a molecular formula. In addition to searching the Internet “directly”, certain online sites can assist and facilitate the search process. Some of these sites are free while others have a fee.

Be aware that the Internet is full of misleading and often times incorrect data. It is one thing to make a mistake in the elucidation process; it is another to follow someone else’s mistake. As always, the next step is to verify the data and publish once.

Searching for Unknowns via the Internet

Searching for unknowns across an internal library or database can serve as a major time saver. A more accessible database is the Internet. Many chemists and elucidators search the World Wide Web using a mass, a molecular formula or a fragment.

Be aware that proprietary data sent over the Internet runs the risk of being intercepted.

Searching the Internet via Google™ for the molecular formula C17H11N5 produced 787 hits with Letrozole appearing in the first dozen hits or so. Using a variation of the molecular formula C17 H11 N5 with the elements spaced out produced over 34,000 hits without any clear reference to Letrozole.

Searching for Unknowns via the Internet

Searching for unknowns across an internal library or database can serve as a major time saver. A more accessible database is the Internet. Many chemists and elucidators search the World Wide Web using a mass, a molecular formula or a fragment.

Be aware that proprietary data sent over the Internet runs the risk of being intercepted.

Searching the Internet via Google™ for the molecular formula C17H11N5 produced 787 hits with Letrozole appearing in the first dozen hits or so. Using a variation of the molecular formula C17 H11 N5 with the elements spaced out produced over 34,000 hits without any clear reference to Letrozole.

Limiting the Atom Ranges

In a previous blog, the atoms C, H, O, and N were limited to a specific range, 0-50, 0-100, 0-10 and 0-10, respectively. Starting from a wide atom count range ensures molecular formulae are not overlooked. The ranges for the atom count can be restricted further using additional information such as NMR, IR, etc.

A common tactic for elucidating compounds containing nitrogen is to acquire a 15N NMR experiment to assist in setting a narrower range for nitrogen count. The 1H-15N HMBC experiment below indicates the presence of at least 2 nitrogen atoms. Therefore, the nitrogen range can be set to 2-10 atoms and thus reduce the number of candidate molecular formulae.

    

Limiting the Atom Ranges

In a previous blog, the atoms C, H, O, and N were limited to a specific range, 0-50, 0-100, 0-10 and 0-10, respectively. Starting from a wide atom count range ensures molecular formulae are not overlooked. The ranges for the atom count can be restricted further using additional information such as NMR, IR, etc.

A common tactic for elucidating compounds containing nitrogen is to acquire a 15N NMR experiment to assist in setting a narrower range for nitrogen count. The 1H-15N HMBC experiment below indicates the presence of at least 2 nitrogen atoms. Therefore, the nitrogen range can be set to 2-10 atoms and thus reduce the number of candidate molecular formulae.

    

Fitting a Molecular Formula to a Mass

With a monoisotopic mass identified from a mass spectrum, the next step is to fit a molecular formula (or elemental composition) to the unknown. For organic compounds, the starting elements of choice are Carbon, Hydrogen, Oxygen, Nitrogen and sometimes Sulfur. Information from starting material and known derivatives can also help when deciding what elements and valences to work with.

The mass of an unknown organic compound is 950.298 g/mol accurate to 0.002 Da. If the elements C(0-50), H(0-100), O(0-10) and N(0-10)—atoms ranges are listed within the brackets—are used, then no molecular formula fits this mass. For this case, expanding the elemental list to include sulfur will produce a list of 54 candidate MFs. The top 9 MFs, based on lowest error, are displayed below.

No.     Formula         Monoisotopic Mass     Error

1     C42H78OS11         950.2980             0.0000

2     C27H82N8OS13     950.2981             -0.0001

3     C28H88NO6S13     950.2981             -0.0001

4     C20H84N7O9S12     950.2979             0.0001

5     C41H58N8O8S5     950.2981             -0.0001

6     C34H74N6O4S10     950.2979             0.0001

7     C21H96N3O3S16     950.2983             -0.0003

8     C49H62N2O5S6     950.2983             -0.0003

9     C35H72N3O10S8     950.2983             -0.0003