How do I know if my unknown contains a fluorine atom(s)? … Part 3

Experiments such as 1H-13C HMQC, HSQC, or HETCOR can be used to suspect the presence of fluorine but not necessarily rule out the absence of fluorine. The stipulation behind this assessment is that a carbon from either a CH, CH2 or CH3 group must experience 13C-19F coupling(s).

The spectrum below is a region of an 1H -13C HSQC for the CH group belonging to the steroid-type fragment. The F1 trace is a projection of all the 13C slices. A 1D 1H NMR spectrum is attached to the F2 domain, hence the better resolution. The CH carbon resonances at 71 pm are split due to 13C-19F coupling. To confirm this, the 13C resonances correlate to a 1H at 4.1 ppm, which integrates to 1. A processed 1H NMR spectrum is shown on a previous blog.

TIP: Once again, the pattern to look out for is ‘doubling up tilt’ of correlations along F1.

How do I know if my unknown contains a fluorine atom(s)? … Part 3

Experiments such as 1H-13C HMQC, HSQC, or HETCOR can be used to suspect the presence of fluorine but not necessarily rule out the absence of fluorine. The stipulation behind this assessment is that a carbon from either a CH, CH2 or CH3 group must experience 13C-19F coupling(s).

The spectrum below is a region of an 1H -13C HSQC for the CH group belonging to the steroid-type fragment. The F1 trace is a projection of all the 13C slices. A 1D 1H NMR spectrum is attached to the F2 domain, hence the better resolution. The CH carbon resonances at 71 pm are split due to 13C-19F coupling. To confirm this, the 13C resonances correlate to a 1H at 4.1 ppm, which integrates to 1. A processed 1H NMR spectrum is shown on a previous blog.

TIP: Once again, the pattern to look out for is ‘doubling up tilt’ of correlations along F1.

How do I know if my unknown contains a fluorine atom(s)? … Part 2

Without a set routine to acquire a 19F NMR spectrum for every single sample, an elucidator must resort to routine NMR experiments for clues to the presence or absence of fluorine.

For the elucidator who routinely acquires a proton decoupled 13C NMR, the spectrum can provide some clues into the presence of fluorine. Carbons in the vicinity of a fluorine atom(s) will show up as split resonances due to 13C-19F couplings.

The segmented 13C NMR spectrum below illustrates four carbons resonances with splitting patterns. The couplings constants are as follows: CF (101 ppm) at 175.4 Hz, CH (70 ppm) at 37.2 Hz, C (48 ppm) at 22.9 Hz, CH (33 ppm) at 19.2 Hz, and CH3 (23 ppm not shown) at 6.0 Hz. The coupling constants themselves are clues to the proximity of the carbon atom to fluorine.

TIP: On a first pass, there may appear to be extra carbons present in the spectrum. It is important to look for patterns such a ‘doubling up’ of carbons. Secondly, it is wise to compare the spectral carbon count to the molecular weight and see if anything is amiss.

How do I know if my unknown contains a fluorine atom(s)? … Part 2

Without a set routine to acquire a 19F NMR spectrum for every single sample, an elucidator must resort to routine NMR experiments for clues to the presence or absence of fluorine.

For the elucidator who routinely acquires a proton decoupled 13C NMR, the spectrum can provide some clues into the presence of fluorine. Carbons in the vicinity of a fluorine atom(s) will show up as split resonances due to 13C-19F couplings.

The segmented 13C NMR spectrum below illustrates four carbons resonances with splitting patterns. The couplings constants are as follows: CF (101 ppm) at 175.4 Hz, CH (70 ppm) at 37.2 Hz, C (48 ppm) at 22.9 Hz, CH (33 ppm) at 19.2 Hz, and CH3 (23 ppm not shown) at 6.0 Hz. The coupling constants themselves are clues to the proximity of the carbon atom to fluorine.

TIP: On a first pass, there may appear to be extra carbons present in the spectrum. It is important to look for patterns such a ‘doubling up’ of carbons. Secondly, it is wise to compare the spectral carbon count to the molecular weight and see if anything is amiss.

How do I know if my unknown contains a fluorine atom(s)?

Ideally, identifying whether a fluorine atom(s) is present is as simple as acquiring a 19F NMR. However, running 'uncommon' experiments is not usually my first step in an elucidation (MDE). Imagine a scenario where the elucidator is unsure whether fluorine is present. He/she acquires a 19F NMR and sees no 19F resonances. Therefore the extra experiment can be considered a waste of time especially when other experiments may provide the clues needed.

I resort to routine experiments such as MS, 1H and 13C NMR, and HSQC to gather information for the presence of fluorine. Although a mass spectrum can indicate such losses as CF3, it is usually not the best approach for inferring the presence of a fluorine atom(s). A 1H NMR spectrum may provide some clues for fluorine but on a complicated spectrum, it may not be as obvious that fluorine is present.

Shown below is a portion of a 1H NMR spectrum for a steroid-type compound with a single fluorine atom. Only the vicinal protons to the fluorine atom show any 1H-19F splitting. However, the splitting is not conclusive evidence for the presence of a fluorine. The subsequent blog will illustrate some better clues into ascertaining the presence of fluorine.

Further complicating the matter, the multiplet at 2.4 pppm (CH group) is overlapping with another multiplet and the multiplet at 4.1 ppm (CHOH group) exhibits some hard to discern couplings.

How do I know if my unknown contains a fluorine atom(s)?

Ideally, identifying whether a fluorine atom(s) is present is as simple as acquiring a 19F NMR. However, running 'uncommon' experiments is not usually my first step in an elucidation (MDE). Imagine a scenario where the elucidator is unsure whether fluorine is present. He/she acquires a 19F NMR and sees no 19F resonances. Therefore the extra experiment can be considered a waste of time especially when other experiments may provide the clues needed.

I resort to routine experiments such as MS, 1H and 13C NMR, and HSQC to gather information for the presence of fluorine. Although a mass spectrum can indicate such losses as CF3, it is usually not the best approach for inferring the presence of a fluorine atom(s). A 1H NMR spectrum may provide some clues for fluorine but on a complicated spectrum, it may not be as obvious that fluorine is present.

Shown below is a portion of a 1H NMR spectrum for a steroid-type compound with a single fluorine atom. Only the vicinal protons to the fluorine atom show any 1H-19F splitting. However, the splitting is not conclusive evidence for the presence of a fluorine. The subsequent blog will illustrate some better clues into ascertaining the presence of fluorine.

Further complicating the matter, the multiplet at 2.4 pppm (CH group) is overlapping with another multiplet and the multiplet at 4.1 ppm (CHOH group) exhibits some hard to discern couplings.

Deciphering a crowded region in a 1H NMR spectrum using 2D NMR

When faced with a crowded region in a 1H NMR spectrum, a 2D NMR experiment can assist in removing the ambiguity and in narrowing down the proton count.

For the 1H NMR spectrum below, Multiplet F (the region between 1.4 to 1.8 ppm) displays an integral of 6.71. Although several factors may contrbute to the disparity of the integral as a whole number, the question remains does the integral 6.71 represent 6 or 7 protons assuming Multiplet E corresponds to 1 proton?

A 1H-13C DEPT-HSQC, shown below, illustrates the correlation between a 1H and 13C that are separated by one bond. In addition, the correlations that are negative (commonly displayed in blue) signify a methylene (CH2) group. The 1D projection along the F1 dimension represents 4 carbons at 26, 27, 31, and 39 ppm and each carbon correlates with 2 inequivalent proton resonances.

Based on the information from the DEPT-HSQC spectrum, one can say with a higher degree of certainty that there are 7 protons within the 1H region of 1.4 to 1.8 ppm.

Deciphering a crowded region in a 1H NMR spectrum using 2D NMR

When faced with a crowded region in a 1H NMR spectrum, a 2D NMR experiment can assist in removing the ambiguity and in narrowing down the proton count.

For the 1H NMR spectrum below, Multiplet F (the region between 1.4 to 1.8 ppm) displays an integral of 6.71. Although several factors may contrbute to the disparity of the integral as a whole number, the question remains does the integral 6.71 represent 6 or 7 protons assuming Multiplet E corresponds to 1 proton?

A 1H-13C DEPT-HSQC, shown below, illustrates the correlation between a 1H and 13C that are separated by one bond. In addition, the correlations that are negative (commonly displayed in blue) signify a methylene (CH2) group. The 1D projection along the F1 dimension represents 4 carbons at 26, 27, 31, and 39 ppm and each carbon correlates with 2 inequivalent proton resonances.

Based on the information from the DEPT-HSQC spectrum, one can say with a higher degree of certainty that there are 7 protons within the 1H region of 1.4 to 1.8 ppm.

Extracting Integral information from regions with overlapping multiplets

A very useful bit of information on the proton count comes from integrating a 1H NMR spectrum. However, overlapping resonances can make integration less than straightforward and in some cases ambiguous.

Integration involves the following decisions to be made by the elucidator: which resonances to integrate, where the integral begins and ends, and what the reference integral is and what value to set it to.

The 1H NMR spectrum, illustrated below, is an example where multiplets overlap and affect the neighbouring integral values. The spectrum shows six labelled multiplets with integral information.

With the integrals in place, the elucidator needs to decide what reference integral to use. Multiplet E (or Multiplet A) is a good choice as the reference integral since it is isolated from other multiplets that would otherwise interfere with the integral. The next step is to decide what whole number to set the reference integral to; typically, it is a value of 1 or 2 or 3.

Multiplet B is a broad singlet and its integral overlaps with the integrals for Multiplets C and D. Although both Multiplets C and D appear to have an integral value of 1, one can check this by applying peak fitting (or deconvolution) to the broad singlet and subtracting out that contribution.

Multiplet F poses a different challenge. The integral shows a value of 6.71, which can be taken as 6 or 7 protons (I'd be more inclined to round up and take it as 7 protons). Peak fitting may help here but it may be more practical to resort to 2D NMR data for further analysis.

Multiplet A = CH
Multiplet B = NHn or OH
Multiplet C = CH
Multiplet D = CH
Multiplet E = CH
Multiplet F = 7 CHs (or any combination of CH, CH2, CH3 totalling 7 protons)

Extracting Integral information from regions with overlapping multiplets

A very useful bit of information on the proton count comes from integrating a 1H NMR spectrum. However, overlapping resonances can make integration less than straightforward and in some cases ambiguous.

Integration involves the following decisions to be made by the elucidator: which resonances to integrate, where the integral begins and ends, and what the reference integral is and what value to set it to.

The 1H NMR spectrum, illustrated below, is an example where multiplets overlap and affect the neighbouring integral values. The spectrum shows six labelled multiplets with integral information.

With the integrals in place, the elucidator needs to decide what reference integral to use. Multiplet E (or Multiplet A) is a good choice as the reference integral since it is isolated from other multiplets that would otherwise interfere with the integral. The next step is to decide what whole number to set the reference integral to; typically, it is a value of 1 or 2 or 3.

Multiplet B is a broad singlet and its integral overlaps with the integrals for Multiplets C and D. Although both Multiplets C and D appear to have an integral value of 1, one can check this by applying peak fitting (or deconvolution) to the broad singlet and subtracting out that contribution.

Multiplet F poses a different challenge. The integral shows a value of 6.71, which can be taken as 6 or 7 protons (I'd be more inclined to round up and take it as 7 protons). Peak fitting may help here but it may be more practical to resort to 2D NMR data for further analysis.

Multiplet A = CH
Multiplet B = NHn or OH
Multiplet C = CH
Multiplet D = CH
Multiplet E = CH
Multiplet F = 7 CHs (or any combination of CH, CH2, CH3 totalling 7 protons)

Blog Stats and Using the RSS feed

Although this blog has no scientific merit, I would like to point out two bits of information that could be of interest: 1. statistics on the P2C2E Blog and 2. how to use the newly-added RSS feed.

As of March 28, 2008, the P2C2E weblog has 42 email subscribers and 133 direct access viewers. Thank you everyone for your time.

To extend the reach of the weblog, an RSS feed was added. For those interested in the RSS feed, Ryan's Blog on NMR software explains What is RSS? and how to use it.

I would also like to take this opportunity to thank Ryan Sasaki, M.Sc., for his help in guiding me through the process of blogging.

Blog Stats and Using the RSS feed

Although this blog has no scientific merit, I would like to point out two bits of information that could be of interest: 1. statistics on the P2C2E Blog and 2. how to use the newly-added RSS feed.

As of March 28, 2008, the P2C2E weblog has 42 email subscribers and 133 direct access viewers. Thank you everyone for your time.

To extend the reach of the weblog, an RSS feed was added. For those interested in the RSS feed, Ryan's Blog on NMR software explains What is RSS? and how to use it.

I would also like to take this opportunity to thank Ryan Sasaki, M.Sc., for his help in guiding me through the process of blogging.

Maximize Data Extraction (MDE) from a 1H NMR Spectrum

Here is a lesson I learnt over time while working on small molecules—my Elucidation Evolution.

Thinking back to when I started doing elucidations of unknowns, my mindset was to collect loads of data (NMR, MS, IR, etc.) whether I needed it or not. Initially inexperienced, I was extracting bits and pieces of information from various datasets and building up a list of fragments that needed to be combined together to form a candidate structure. Although this offered a means to practice and learn how to interpret a wide array of data, it was not an efficient approach to an elucidation.

Please note that there is value in collecting additional data to confirm or verify a candidate structure. However, for elucidation purposes, one should maximize data extraction while minimizing data collection.

As my skills grew, I felt more comfortable working with less data. I began to maximize the information I could extract from a simple 1H NMR, thus avoiding the need to collect and analyze “duplicate data”. (An example of duplicate data is a 1H NMR and a 1H-1H COSY.) Although 1H-1H COSY data could be valuable in cases with complicated resonances and a high degree of signal overlap, a 1H NMR could suffice in extracting structural information.

For the purpose of an elucidation, one should extract the following bits of information from a 1H NMR spectrum:

presence of proton resonances,

spectrum purity (identifying impurities or mixtures),

identifying aromatic and/or aliphatic protons,

presence of exchangeable protons,

chemical shifts,

proton count from integrals,

presence of symmetry,

coupling patterns and constants,

second-order effects.

TIP: Every bit of information helps. A negative result is also telling you something.

Maximize Data Extraction (MDE) from a 1H NMR Spectrum

Here is a lesson I learnt over time while working on small molecules—my Elucidation Evolution.

Thinking back to when I started doing elucidations of unknowns, my mindset was to collect loads of data (NMR, MS, IR, etc.) whether I needed it or not. Initially inexperienced, I was extracting bits and pieces of information from various datasets and building up a list of fragments that needed to be combined together to form a candidate structure. Although this offered a means to practice and learn how to interpret a wide array of data, it was not an efficient approach to an elucidation.

Please note that there is value in collecting additional data to confirm or verify a candidate structure. However, for elucidation purposes, one should maximize data extraction while minimizing data collection.

As my skills grew, I felt more comfortable working with less data. I began to maximize the information I could extract from a simple 1H NMR, thus avoiding the need to collect and analyze “duplicate data”. (An example of duplicate data is a 1H NMR and a 1H-1H COSY.) Although 1H-1H COSY data could be valuable in cases with complicated resonances and a high degree of signal overlap, a 1H NMR could suffice in extracting structural information.

For the purpose of an elucidation, one should extract the following bits of information from a 1H NMR spectrum:

presence of proton resonances,

spectrum purity (identifying impurities or mixtures),

identifying aromatic and/or aliphatic protons,

presence of exchangeable protons,

chemical shifts,

proton count from integrals,

presence of symmetry,

coupling patterns and constants,

second-order effects.

TIP: Every bit of information helps. A negative result is also telling you something.

Identifying Meta coupling in a 1H NMR Spectrum

In a substituted benzene ring, aromatic protons that are in the meta position can exhibit coupling to each other. This is referred to as meta or 4J coupling. The coupling pattern is typically a doublet with a coupling constant of ~2 Hz.

On the contrary, a spectrum without any meta coupling indicates a lack of protons in the meta position. Although a 2D COSY experiment can produce the same result, one can save time by looking for this information in a 1H NMR spectrum first.

Illustrated below is a portion of a 1H NMR spectrum for a substituted benzene ring. Proton A is a doublet with a 2.3 Hz coupling and proton B is also a doublet with 2.2 Hz coupling. Proton A and B are coupled to each other due to the similar coupling constant (+/- 0.2 Hz). Another indication of coupling is the slight tilt of the multiplets to each other.

TIP: Be careful on the extent of line broadening applied to the FID, too much and the meta coupling information can be lost.

Identifying Meta coupling in a 1H NMR Spectrum

In a substituted benzene ring, aromatic protons that are in the meta position can exhibit coupling to each other. This is referred to as meta or 4J coupling. The coupling pattern is typically a doublet with a coupling constant of ~2 Hz.

On the contrary, a spectrum without any meta coupling indicates a lack of protons in the meta position. Although a 2D COSY experiment can produce the same result, one can save time by looking for this information in a 1H NMR spectrum first.

Illustrated below is a portion of a 1H NMR spectrum for a substituted benzene ring. Proton A is a doublet with a 2.3 Hz coupling and proton B is also a doublet with 2.2 Hz coupling. Proton A and B are coupled to each other due to the similar coupling constant (+/- 0.2 Hz). Another indication of coupling is the slight tilt of the multiplets to each other.

TIP: Be careful on the extent of line broadening applied to the FID, too much and the meta coupling information can be lost.

Structural bias in an elucidation

A biased elucidation is an elucidation where the chemist makes certain assumptions about the data at hand based on a previous experience(s) and not deviating from it. Depending on the elucidation, it can be a good thing or a very bad thing. As a good thing, it can speed up the time spent on an elucidation. However, an incorrect assumption(s) can mean wasting time and frustration to the point of being unable to elucidate the compound.

Shown below is a correlation map for an unknown with a molecular formula of C14H9NO2. The hybridization states of all the carbons are sp2, the ring size is restricted to 5 and 6, and there is an exchangeable proton (OH or NH). How many structures can you draw that fit these restrictions? For clarity reasons, the atoms in red are the ones to be arranged.

The answer is 28.

Two answers are shown below. Which tautomer did you draw first? A common structural bias is to take the 5 sp2 carbons and the nitrogen and draw a pyridine ring – pyridine being a commonly encountered substituent. Here, structural bias for this unknown is a bad thing especially if you overlooked a possibility.

The reference listed is for the tautomer on the right side. In hindsight, it is commonly seen when a pyridine ring is conjugated with a carbonyl group(s).

TIP: The best approach is to not to encourage an outcome over another, that is, make no assumptions. For elucidations where the starting material is known, consider not viewing the starting material so as not to cloud the mind. Remember to cover all the bases so all possibilities are taken into account.

Structural bias in an elucidation

A biased elucidation is an elucidation where the chemist makes certain assumptions about the data at hand based on a previous experience(s) and not deviating from it. Depending on the elucidation, it can be a good thing or a very bad thing. As a good thing, it can speed up the time spent on an elucidation. However, an incorrect assumption(s) can mean wasting time and frustration to the point of being unable to elucidate the compound.

Shown below is a correlation map for an unknown with a molecular formula of C14H9NO2. The hybridization states of all the carbons are sp2, the ring size is restricted to 5 and 6, and there is an exchangeable proton (OH or NH). How many structures can you draw that fit these restrictions? For clarity reasons, the atoms in red are the ones to be arranged.

The answer is 28.

Two answers are shown below. Which tautomer did you draw first? A common structural bias is to take the 5 sp2 carbons and the nitrogen and draw a pyridine ring – pyridine being a commonly encountered substituent. Here, structural bias for this unknown is a bad thing especially if you overlooked a possibility.

The reference listed is for the tautomer on the right side. In hindsight, it is commonly seen when a pyridine ring is conjugated with a carbonyl group(s).

TIP: The best approach is to not to encourage an outcome over another, that is, make no assumptions. For elucidations where the starting material is known, consider not viewing the starting material so as not to cloud the mind. Remember to cover all the bases so all possibilities are taken into account.