Philosophy to Chemistry to Elucidation

Logic Puzzle #13: Extra Correlations on the HSQC Spectrum … Solution

2011-03-01T06:03:00.000-05:00

The intensity of a correlation in relation to other correlations on a 2D NMR spectrum can be just as important in a structure elucidation as assessing the position and phase of a correlation. The example below presents one such case.

From the 1H-13C HSQC in the previous post, the correlations differ significantly in intensity. The weaker correlations can be attributed to long-range couplings, more specifically, 2J coupling responses. In addition to the 1H integrals, the carbons at 121 and 134 ppm are probably CH’s (4 CH’s in total) and probably not CH2’s. Two potential fragments are shown below and here.

I would like to send out a special thanks to Sarah T. for discussing the idea with me.

Logic Puzzle #13: Extra Correlations on the HSQC Spectrum

2011-02-22T10:01:00.000-05:00

A 1H-13C HSQC is frequently used to assist in assigning a carbon as CH, CH2 or CH3 (additional blog here). The purpose of this puzzle is to examine a non-standard assignment.

The 1H-13C HSQC below shows 2 correlations for each carbon at 121 and 134 ppm. Given also the information on the 1H integrals on the F2 axis, are the carbons at 121 and 134 ppm really CH2’s?

Blog on Hiatus from February 7 to 18, 2011

2011-02-09T11:49:00.000-05:00

It’s been a busy week with work and I'm expecting the same for next week. The blog P2C2E will be on hiatus for 2 weeks from February 7 to 18. Posts will resume the following week.

Stay tuned for more logic puzzles.

It’s Been Three Years … Already

2011-02-01T07:29:00.000-05:00

February 1, 2011 marks the third year anniversary of the weblog P2C2E. This milestone is dedicated to you the loyal readers. Thank you.

I am happy to write that the blog has grown linearly in site visits per year.

Although the number of posts has dropped, due to workload and travel, the number of email subscribers has not. Readership comments have been fairly constant. The top blog in 2010 reached 368 visits.

The statistics are based on results from Google Feedburner and Typepad.

Logic Puzzle #12: Torsion Angle and Coupling … Solution

2011-01-26T09:12:00.000-05:00

Couplings can be affected by the torsion angle. The couplings can be expressed mathematically with a cos θ relationship. At certain values of θ, the couplings are expected to be relatively weaker to non-existent.

Based on the torsion angle, fragment B is expected to exhibit a prominent coupling between the red and gold nuclei. Fragment A, exhibiting a torsion angle of 90º, will generally lead to a weak or non-evident coupling on an NMR spectrum.

At the initial stages of an elucidaton of an unknown structure, the torsion angle(s) is usually not known. As such, an elucidator must be prepared to expect the unexpected coupling(s).

Logic Puzzle #12: Torsion Angle and Coupling

2011-01-19T05:35:00.000-05:00

For a simple case, the torsion angle (or dihedral angle) is described as the angle between 4 contiguous atoms or 3 successive bonds. In NMR, the magnitude of the coupling is directly related to the torsion angle between the vicinal nuclei (see the Karplus equation for more details).

Below are two animations, A and B, for identical fragments differing only in the torsion angle. Assuming the following fragments exhibit a rigid geometry, which torsion angle would generally contribute to a prominent coupling between the red and gold nuclei?

A (torsion angle at 90º) B (torsion angle at 35º)

Logic Puzzle #11: Pairwise Correlation Confidence … Solution

2011-01-11T06:41:00.000-05:00

Long range 2D NMR experiments do not necessarily provide information about all the connectivities. The following structure elucidation problem set is one such example.

Based on the 1H-13C HMBC shown below, there is no evident correlation between the 1H at 5.31 ppm and the 13C at 21.1 ppm. Note the green box describes the region of interest.

The correlation between the methyl 1H at 2.15 ppm and 13C at 84.7 ppm (indicated by the purple arrow and the red bonds below) is a weak 4J coupling (also denoted as a W-coupling or M-coupling). The coupling arises from the individual methyl protons rotating and interacting through a W-relationship (or M-)with the carbon. This coupling is not evident for the methine 1H at 5.31 ppm and the 13C at 21.1 ppm.

Thank you Serge for your comment.

Logic Puzzle #11: Pairwise Correlation Confidence

2011-01-04T09:18:00.000-05:00

Typical for long range 2D NMR experiments, spectral data may exhibit more than one correlation for two coupled nuclei (e.g. A to B and B to A). The pairwise correlations offer an extra degree of confidence in the interpretation.

For the following fragment, an 1H-13C HMBC correlation exists for the 1H 2.15 ppm to 13C 84.7 ppm (represented by the purple arrow in the diagram below).

Is the pairwise correlation for the 1H 5.31 ppm to 13C 21.1 ppm evident?

A special thanks goes to Gene M. for pointing me to the data.

Blog on Hiatus from December 13 to January 3

2010-12-13T07:43:00.000-05:00

As the year comes to an end, I am off to my final business trip of the year in Washington, D.C. As such, I will be taking a little break from blogging from December 13 to January 3. Posts will resume in the New Year.

I would like to wish the loyal readers of P2C2E a Happy Holidays.

Logic Puzzle #10: Deciphering the Fragment Pattern using 2D NMR Data … Solution

2010-12-07T05:46:00.000-05:00

Like any new process, it takes some practice to extract, understand and convert the information presented from a set of experimental NMR datasets into a fragment.

The only fragment that can accommodate the set of restrictions from a 1H-13C HSQC and HMBC is 2,3-dimethylbutane-1,1-diyl. The green arrows illustrate the 2-3JCH coupling responses extracted from an HMBC experiment.

Logic Puzzle #10: Deciphering the Fragment Pattern using 2D NMR Data

2010-11-30T05:35:00.000-05:00

The goal of this puzzle is to conceptualize a fragment(s) from the given information.

In the following example, a set of protonated sp3 carbons were extracted from an HSQC experiment (not shown). The green arrows represent the 2-3JCH coupling responses extracted from an HMBC experiment. Based on these restrictions, what fragment(s) supports the data?

Note there is an open valence off one of the carbon atoms.

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

2010-11-25T05:44:00.000-05:00

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 and³⁰Si 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,¹³C³³Sand 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

2010-11-17T06:46:00.000-05:00

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.

Blog on Hiatus from November 1 to 12

2010-10-29T10:29:00.000-04:00

I will be on the road again for the next 2 weeks with more business travel. This time, I will be touring a variety of Universities and Colleges in Texas and Ohio. Unfortunately for the P2C2E readers, the blog will be on hiatus for the next 2 weeks. Posts will resume the following week.

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

2010-10-26T07:03:00.000-04:00

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.

Blog on Hiatus for the week of October 18 to October 22

2010-10-22T07:34:00.000-04:00

Since I just returned home from a forensic conference (SOFT), the blog P2C2E will be on hiatus for this week from October 18 to October 22. Posts will resume the following week.

Logic Puzzle #8: Strong Coupling or Peak Overlap? … Solution

2010-10-15T11:34:00.000-04:00

There are two approaches to deciding which scenario best fits the NMR data in the puzzle. A structure elucidator can provide evidence that one scenario is more probable than an other and/or eliminate one scenario on the grounds of insufficient/contradictory data to support it.

For the following 1H NMR spectrum, if the integral of the doublet at 7.42 ppm equates to 1 H atom from a single structure, then the 1H signals at 7.58-7.59 ppm integrating to 2.57 (and assuming all signals are accounted for) relate to overlapping signals (probably a doublet overlapping with a singlet) from a mixture of compounds at varying concentrations.

Another approach, as Adolfo has nicely provided, the magnitude of the strong coupling demonstrated by the signals at 7.58-7.59 ppm indicates a nearby partner on the left side circa 7.70 ppm. Since this appears not to be the case, then by process of elimination, the signals at 7.58-7.59 ppm relate to two overlapping multiplets.

Logic Puzzle #8: Strong Coupling or Peak Overlap?

2010-10-04T09:56:00.000-04:00

The goal of this puzzle is to distinguish between strong coupling and peak overlap.

For the following 1H NMR spectrum, are the 1H signals at 7.58-7.59 ppm exhibiting strong coupling or are they two overlapping multiplets or both?

Logic Puzzle #7: Almost Missed It … Solution 2

2010-09-28T08:02:00.000-04:00

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

2010-09-20T11:28:00.000-04:00

‘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

2010-09-13T10:10:00.000-04:00

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?

Will the Real Ring System Please Stand Up?

2010-09-09T09:54:00.000-04:00

Based on experience, fused ring systems on an unknown structure tend to conjure up variations of odd connections. The possibilities can seem endless.

The diagram, shown below, samples a subset of fused ring systems for a given dataset. It is important to note that although certain systems may be more appealing to the eye, it is imperative for an elucidator to evaluate each variation prior to accepting or rejecting the proposed candidate.

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

2010-08-31T07:59:00.000-04:00

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.

In Memory of an Uncle

2010-08-24T05:51:00.000-04:00

My uncle, Aldo Moser, passed away due to complications from a yearlong battle with pancreatic cancer. I would like to dedicate this post to him.

I would like to recount a nice story about working side-by-side with my uncle, who worked as a licensed electrician.

About 6 years ago, I remember being awakened on an early Saturday morning by some hammering. It turned out that uncle Aldo and my dad were in the process of installing new electrical outlets and needed to chisel out a portion of the house foundation to fit the outlet. Curious to learn how to do the work, I had asked if I could be of any assistance. Uncle Aldo handed me the chisel and hammer and took the time to explain what needed to be done. With the patience of a gentle teacher, uncle Aldo offered tips on chiselling out the concrete, setting the outlet in place and hooking up the wires for the outlet.

The picture below is of my uncle in January 2010 enjoying the challenge of putting together a jigsaw puzzle.

“The best and most beautiful things in the world cannot be seen or even touched. They must be felt within the heart.” Helen Keller

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

2010-08-12T06:40:00.000-04:00

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?