13C NMR and Natural products

Looking for the next big pharmaceutical drug among natural products is a growing endeavour. However, a common drawback to working with natural products is the limited amount of sample.

With a sample limitation, acquiring a 13C NMR spectrum within a reasonable time can be an issue. Rather than wasting instrument time on a 13C NMR, one can save time by extracting the sum (or projection) of slices across a 2D NMR.

Using strychnine as an example, the diagram below is a comparison of two datasets. The top trace (green) is the sum of all the F1 slices from an HMBC and the bottom spectrum (blue) is an acquired 13C NMR. From the top trace, one can gather a good approximation on the minimum number of carbons for strychnine without resorting to acquiring a 13C NMR.

TIP: Be especially careful with quaternary and overlapping carbons as they may not be evident on an HMBC.

13C NMR and Natural products

Looking for the next big pharmaceutical drug among natural products is a growing endeavour. However, a common drawback to working with natural products is the limited amount of sample.

With a sample limitation, acquiring a 13C NMR spectrum within a reasonable time can be an issue. Rather than wasting instrument time on a 13C NMR, one can save time by extracting the sum (or projection) of slices across a 2D NMR.

Using strychnine as an example, the diagram below is a comparison of two datasets. The top trace (green) is the sum of all the F1 slices from an HMBC and the bottom spectrum (blue) is an acquired 13C NMR. From the top trace, one can gather a good approximation on the minimum number of carbons for strychnine without resorting to acquiring a 13C NMR.

TIP: Be especially careful with quaternary and overlapping carbons as they may not be evident on an HMBC.

What is the ring count for the structure Cubane?

Following the RDBE blogs, I decided to run a little test pointed out to me by Robin Martin, Ph.D. Without using the RDBE formula, what is the RDBE (number of rings) for cubane (structure shown below)?

Starting with the 2D representation of cubane, it comes down to counting how many bonds are removed to displace a ring.

According to the diagram above, the RDBE of cubane is 5 and not 6 according to the 3D representation.

TIP: The number of faces on a 3D structure does not dictate the RDBE value.

What is the ring count for the structure Cubane?

Following the RDBE blogs, I decided to run a little test pointed out to me by Robin Martin, Ph.D. Without using the RDBE formula, what is the RDBE (number of rings) for cubane (structure shown below)?

Starting with the 2D representation of cubane, it comes down to counting how many bonds are removed to displace a ring.

According to the diagram above, the RDBE of cubane is 5 and not 6 according to the 3D representation.

TIP: The number of faces on a 3D structure does not dictate the RDBE value.

Interpreting strong coupling information from a 1H NMR spectrum

The second-order effects (also known as tilting or roofing) exhibited by multiplets can be used to identify multiplets coupled to each other.

The 1H NMR spectrum below illustrates an example of strong coupling among three multiplets. The peak intensities across Multiplets A and B are different, that is, the peak on the right side of the multiplet is higher in intensity than the peak on the left side. The purple arrow illustrates a tilt towards the right side for both multiplets. Multiplet C shows an opposite tilt, i.e. to the left side. Multiplets that tilt to form a roof are most likely related protons, and thus are in proximity of each other. Therefore, one can say that Multiplets A and B are coupled to Multiplet C.

Although the field strength and coupling constants impact the amount of tilting, generally the closer the multiplets are the more pronounced is the tilting.

Interpreting strong coupling information from a 1H NMR spectrum

The second-order effects (also known as tilting or roofing) exhibited by multiplets can be used to identify multiplets coupled to each other.

The 1H NMR spectrum below illustrates an example of strong coupling among three multiplets. The peak intensities across Multiplets A and B are different, that is, the peak on the right side of the multiplet is higher in intensity than the peak on the left side. The purple arrow illustrates a tilt towards the right side for both multiplets. Multiplet C shows an opposite tilt, i.e. to the left side. Multiplets that tilt to form a roof are most likely related protons, and thus are in proximity of each other. Therefore, one can say that Multiplets A and B are coupled to Multiplet C.

Although the field strength and coupling constants impact the amount of tilting, generally the closer the multiplets are the more pronounced is the tilting.

What Window Function do I use for an FID?

A window function (also known as apodization function or convolution) is applied to an FID to emphasize a region of the FID over another. The general goal is to improve the SNR (Signal-to-noise) ratio. The early stages of processing such as applying the “correct” window function can affect the time spent on an elucidation especially when dealing with peaks having a low SNR ratio.

I use typically the Exponential window function for both 1H and 13C NMR data. Recently, I have noticed that the TRAF function for resolution enhancement to be doing a better job. The TRAF function comes from Daniel Traficante (Concepts in Magnetic Resonance 2000, 12, 83-101).

The four 13C NMR spectra, shown below, are from the same FID of cholestenone in CDCl3. The functions applied to the Top (purple), Middle Top (red), Middle Bottom (green) and Bottom (blue) spectra are no window function, TRAF 8 Hz, Exponential 2Hz, and Exponential 8 Hz, respectively. The 13C signal for C=O at 206 ppm is clearly visible over the noise for the TRAF function. In addition, there is a significant improvement in the baseline for the TRAF spectrum.

TIP: Processing of an FID is just as important as peak picking. Try different window functions to see which function works best for your data.

What Window Function do I use for an FID?

A window function (also known as apodization function or convolution) is applied to an FID to emphasize a region of the FID over another. The general goal is to improve the SNR (Signal-to-noise) ratio. The early stages of processing such as applying the “correct” window function can affect the time spent on an elucidation especially when dealing with peaks having a low SNR ratio.

I use typically the Exponential window function for both 1H and 13C NMR data. Recently, I have noticed that the TRAF function for resolution enhancement to be doing a better job. The TRAF function comes from Daniel Traficante (Concepts in Magnetic Resonance 2000, 12, 83-101).

The four 13C NMR spectra, shown below, are from the same FID of cholestenone in CDCl3. The functions applied to the Top (purple), Middle Top (red), Middle Bottom (green) and Bottom (blue) spectra are no window function, TRAF 8 Hz, Exponential 2Hz, and Exponential 8 Hz, respectively. The 13C signal for C=O at 206 ppm is clearly visible over the noise for the TRAF function. In addition, there is a significant improvement in the baseline for the TRAF spectrum.

TIP: Processing of an FID is just as important as peak picking. Try different window functions to see which function works best for your data.

Valences for Sulfur, Phosphorus and Nitrogen

In light of the posting by Oliver Fiehn’s group (http://fiehnlab.ucdavis.edu/projects/Seven_Golden_Rules/Ring-Double-Bonds/), I’ve decided to blog some compounds I’ve encountered in the lab. Using the 2nd formula from the previous blog (http://acdlabs.typepad.com/elucidation/2008/03/rings-double-bo.html), for some cases the RDBE calculation will work if the correct atom valence is known.

Note: the nitrochloroform compound can be drawn more than one way and thus the observed RDBE can be either 1 or 2.

Elucidating a structure containing sulfur can be tricky sometimes. My approach is to consider the likely valences starting with S(II) first, then S(IV), and finally S(VI). If I have prior knowledge of the starting material or a derivative(s), I would start with that valence. If multiple O atoms are present in the Molecular Formula and the NMR data shows no OH groups, then I might favour starting with a S(IV) first.

TIP: The best approach is to cover all your bases to avoid missing something. I strongly recommend trying all valences for P, N and S, thus ruling out any other possible structure that might have otherwise been overlooked.

Valences for Sulfur, Phosphorus and Nitrogen

In light of the posting by Oliver Fiehn’s group (http://fiehnlab.ucdavis.edu/projects/Seven_Golden_Rules/Ring-Double-Bonds/), I’ve decided to blog some compounds I’ve encountered in the lab. Using the 2nd formula from the previous blog (http://acdlabs.typepad.com/elucidation/2008/03/rings-double-bo.html), for some cases the RDBE calculation will work if the correct atom valence is known.

Note: the nitrochloroform compound can be drawn more than one way and thus the observed RDBE can be either 1 or 2.

Elucidating a structure containing sulfur can be tricky sometimes. My approach is to consider the likely valences starting with S(II) first, then S(IV), and finally S(VI). If I have prior knowledge of the starting material or a derivative(s), I would start with that valence. If multiple O atoms are present in the Molecular Formula and the NMR data shows no OH groups, then I might favour starting with a S(IV) first.

TIP: The best approach is to cover all your bases to avoid missing something. I strongly recommend trying all valences for P, N and S, thus ruling out any other possible structure that might have otherwise been overlooked.

Rings, Double bonds and triple bonds - clues about an unknown structure

Rings plus Double Bonds Equivalent (RDBE) (also known as the degree/element of unsaturation or index of hydrogen deficiency) can be determined from a Molecular Formula (MF). The calculation provides insight into the number of rings and/or double bonds and/or triple bonds to expect from an unknown structure.

RDBE = (# of C) + 1 - (# of H)/2 + (# of N(III))/2

For example C5 H9 N1 O1: RDBE = 5 + 1 - 4.5 + 0.5 = 2

Since bivalent atoms such as O and S are not counted in the formula, another way to write it is as follows:

RDBE = (# of Valence IV) + 1 - (# of Valence I)/2 + (# of Valence III)/2

Note: Since this formula will fail for MFs with higher valence states such as N(V), P(V), S(IV) or S(VI), the rest of this blog will focus on the lowest valence state for these elements. For more information on how the formula fails, I recommend reading the following blog by Oliver Fiehn's group: http://fiehnlab.ucdavis.edu/projects/Seven_Golden_Rules/Ring-Double-Bonds/

An RDBE of 1 indicates that the MF has either 1 ring or 1 double bond present. If NMR data is available and it shows one or two sp2 carbons are present, then one can rule out the ring possibility and therefore only a double bond(s) is present. An odd value for the RDBE with an odd count of sp2 carbons can indicate the presence of an allene (C=C=C) group with 3 sp2's or a carbonyl group (C=O) with 1 sp2.

Tip: A benzene ring, a carbonyl and a triple bond have an RDBE of 4, 1 and 2, respectively.

Rings, Double bonds and triple bonds - clues about an unknown structure

Rings plus Double Bonds Equivalent (RDBE) (also known as the degree/element of unsaturation or index of hydrogen deficiency) can be determined from a Molecular Formula (MF). The calculation provides insight into the number of rings and/or double bonds and/or triple bonds to expect from an unknown structure.

RDBE = (# of C) + 1 - (# of H)/2 + (# of N(III))/2

For example C5 H9 N1 O1: RDBE = 5 + 1 - 4.5 + 0.5 = 2

Since bivalent atoms such as O and S are not counted in the formula, another way to write it is as follows:

RDBE = (# of Valence IV) + 1 - (# of Valence I)/2 + (# of Valence III)/2

Note: Since this formula will fail for MFs with higher valence states such as N(V), P(V), S(IV) or S(VI), the rest of this blog will focus on the lowest valence state for these elements. For more information on how the formula fails, I recommend reading the following blog by Oliver Fiehn's group: http://fiehnlab.ucdavis.edu/projects/Seven_Golden_Rules/Ring-Double-Bonds/

An RDBE of 1 indicates that the MF has either 1 ring or 1 double bond present. If NMR data is available and it shows one or two sp2 carbons are present, then one can rule out the ring possibility and therefore only a double bond(s) is present. An odd value for the RDBE with an odd count of sp2 carbons can indicate the presence of an allene (C=C=C) group with 3 sp2's or a carbonyl group (C=O) with 1 sp2.

Tip: A benzene ring, a carbonyl and a triple bond have an RDBE of 4, 1 and 2, respectively.

Recognizing an exchangeable proton

Depending on the experimental conditions, signals from exchangeable protons such as NH and OH can be present on a 1H NMR spectrum. A key to interpreting a 1H NMR is distinguishing between CHn protons (where n = 1, 2 or 3) and exchangeable protons. Acquiring an 1H-13C 2D NMR experiment can assist in this process, specifically, an HMQC, HSQC or HETCOR.

A partial 1H NMR spectrum for the carbamate compound is shown below. The aromatic region shows 5 multiplets with a relative integral of 2:2:1:2:2 ratio. The multiplets with integrals of 2 correspond to the aromatic CHs from the 3-ring system. The doublet (d) at 7.78 ppm corresponds to a CH or NH group. The chemical shift and coupling pattern (d) provide clues to the assignment, however, the coupling constant is very close to more than 1 multiplet. Supporting data is needed to rule out a CH assignment.

The HSQC, shown below, correlates H-C groups separated by a single bond. Since a carbon correlation to the doublet at 7.78 ppm is absent from the spectrum, one can say with a high degree of certainity that the doublet at 7.78 ppm does not belong to a CH and therefore by the process of elimination the doublet belongs to the NH group.

Tip: There may be additional value to use a database or library to compare 1H NMR spectra of similar compounds.

Tip: Start with the easy stuff to simplify data analysis.

Recognizing an exchangeable proton

Depending on the experimental conditions, signals from exchangeable protons such as NH and OH can be present on a 1H NMR spectrum. A key to interpreting a 1H NMR is distinguishing between CHn protons (where n = 1, 2 or 3) and exchangeable protons. Acquiring an 1H-13C 2D NMR experiment can assist in this process, specifically, an HMQC, HSQC or HETCOR.

A partial 1H NMR spectrum for the carbamate compound is shown below. The aromatic region shows 5 multiplets with a relative integral of 2:2:1:2:2 ratio. The multiplets with integrals of 2 correspond to the aromatic CHs from the 3-ring system. The doublet (d) at 7.78 ppm corresponds to a CH or NH group. The chemical shift and coupling pattern (d) provide clues to the assignment, however, the coupling constant is very close to more than 1 multiplet. Supporting data is needed to rule out a CH assignment.

The HSQC, shown below, correlates H-C groups separated by a single bond. Since a carbon correlation to the doublet at 7.78 ppm is absent from the spectrum, one can say with a high degree of certainity that the doublet at 7.78 ppm does not belong to a CH and therefore by the process of elimination the doublet belongs to the NH group.

Tip: There may be additional value to use a database or library to compare 1H NMR spectra of similar compounds.

Tip: Start with the easy stuff to simplify data analysis.

Examining the Molecular Ion on a Mass spectrum

Without any structural information, we can estimate an upper limit on the number of carbons using the m/z for the molecular ion.

For the molecular ion at m/z 386.4 Da (shown in the mass spectrum below), the upper limit on the number of carbons is calculated at 386.4 / 12 = 32.2. Rounding down, we arrive at a maximum of 32 carbons. Therefore, one would expect the carbon count to be within a range of 0 to 32.

To estimate the upper range the # of Carbons ≈

(m/z for the Molecular Ion) / 12

Examining the Molecular Ion on a Mass spectrum

Without any structural information, we can estimate an upper limit on the number of carbons using the m/z for the molecular ion.

For the molecular ion at m/z 386.4 Da (shown in the mass spectrum below), the upper limit on the number of carbons is calculated at 386.4 / 12 = 32.2. Rounding down, we arrive at a maximum of 32 carbons. Therefore, one would expect the carbon count to be within a range of 0 to 32.

To estimate the upper range the # of Carbons ≈

(m/z for the Molecular Ion) / 12

Using 1H NMR couplings to infer structural information

1H-1H couplings arise from influences on a proton’s magnetic field induced by the magnetic field of a neighbouring proton(s). As a proton experiences coupling from a neighbouring proton, so too is the coupling reciprocated to the neighbouring proton(s). By measuring the J-coupling for each splitting pattern, the proton connectivity information can be deduced.

By comparing the J-coupling for each splitting pattern in the 1H NMR below, roughly +/- 0.2 Hz, one can determine which multiplets pair up. Based on the pairing information, we can deduce that the proton is within proximity of its pair.

Coupled protons are:

Multiplets A and C = green line

Multiplets B and C = orange line

Multiplets D and C = purple line.

Multiplets E and C = red line.

Multiplets D and E = black line.

Using 1H NMR couplings to infer structural information

1H-1H couplings arise from influences on a proton’s magnetic field induced by the magnetic field of a neighbouring proton(s). As a proton experiences coupling from a neighbouring proton, so too is the coupling reciprocated to the neighbouring proton(s). By measuring the J-coupling for each splitting pattern, the proton connectivity information can be deduced.

By comparing the J-coupling for each splitting pattern in the 1H NMR below, roughly +/- 0.2 Hz, one can determine which multiplets pair up. Based on the pairing information, we can deduce that the proton is within proximity of its pair.

Coupled protons are:

Multiplets A and C = green line

Multiplets B and C = orange line

Multiplets D and C = purple line.

Multiplets E and C = red line.

Multiplets D and E = black line.

What is the difference between Profile and Centroid MS data?

MS data collected off an instrument is presented as either profile or centroid mode. Shown below are two mass spectra illustrating an ion cluster for profile data and a centroid mass spectrum created from the profile data.

In profile mode, a peak is represented by a collection of signals over several scans. The advantage of profile data is it is easier to classify a signal as a true peak from noise off the instrument.

In centroid mode, the signals are displayed as discrete m/z with zero line widths. The advantage of centroid data is the file size is significantly smaller as there is less information describing a signal.

What is the difference between Profile and Centroid MS data?

MS data collected off an instrument is presented as either profile or centroid mode. Shown below are two mass spectra illustrating an ion cluster for profile data and a centroid mass spectrum created from the profile data.

In profile mode, a peak is represented by a collection of signals over several scans. The advantage of profile data is it is easier to classify a signal as a true peak from noise off the instrument.

In centroid mode, the signals are displayed as discrete m/z with zero line widths. The advantage of centroid data is the file size is significantly smaller as there is less information describing a signal.

How to recognize a doubly charged ion on a Mass spectrum?

On an MS dataset, the x-axis is a measure of the mass to charge (m/z) ratio. Although it is easy to classify a signal as belonging to a singly charged ion (+1 or -1), one needs to be capable of recognizing a multiply charged (+2, -2, +3, etc.) ion as not to misinterpret the MS data. Multiply charged ions are commonly seen in MS data on peptides, proteins, diamines and diacids.

Ions of a multiply charged isotope cluster will appear at a fractional m/z, i.e. much less than one mass unit apart. The A+1 peak of a doubly-charged ion will appear at an m/z 0.5 above the monoisotopic peak.

Illustrated below is a case of an ion cluster from a centroid ES+ spectrum for a compound with a nominal mass of 400 Da. A doubly charged ion, [M+2H]+2, with a nominal mass at 402 Da, will have a signal appear at mass/charge = 402/2 = 201. If the monoisotopic peak, for the compound with all 12C, is present at 201.1, then the signal from the 12C+1 form is seen at 201.6.

How to recognize a doubly charged ion on a Mass spectrum?

On an MS dataset, the x-axis is a measure of the mass to charge (m/z) ratio. Although it is easy to classify a signal as belonging to a singly charged ion (+1 or -1), one needs to be capable of recognizing a multiply charged (+2, -2, +3, etc.) ion as not to misinterpret the MS data. Multiply charged ions are commonly seen in MS data on peptides, proteins, diamines and diacids.

Ions of a multiply charged isotope cluster will appear at a fractional m/z, i.e. much less than one mass unit apart. The A+1 peak of a doubly-charged ion will appear at an m/z 0.5 above the monoisotopic peak.

Illustrated below is a case of an ion cluster from a centroid ES+ spectrum for a compound with a nominal mass of 400 Da. A doubly charged ion, [M+2H]+2, with a nominal mass at 402 Da, will have a signal appear at mass/charge = 402/2 = 201. If the monoisotopic peak, for the compound with all 12C, is present at 201.1, then the signal from the 12C+1 form is seen at 201.6.