Logic Puzzle #11: Pairwise Correlation Confidence … Solution

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

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

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

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

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

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.