Philosophy to Chemistry to Elucidation

Logic Puzzle #2: How to link 3 Fragments

The goal of this puzzle is to logically combine a set of fragments using valence and NMR information.

In this puzzle, three fragments are correlated through 2-3J coupling responses (represented by a green arrow) that were extracted from a 1H-13C HMBC data (spectrum not shown). The carbon atoms with the 13C chemical shifts displayed in blue indicate the presence of an adjacent heteroatom. Based on these criteria, what 'complete' fragment(s) supports the data and is there anything missing?

In order to accommodate these restrictions, a logical fit is to consider a trivalent atom, e.g. nitrogen.

Logic Puzzle #1: The Missing Link

A great skill to master is the capability to conceptualize a fragment or structure directly off a spectrum without resorting to paper-and-pen work. This skill is learnt through lots of practice. Whenever partial information is available, an elucidator can conjure up a mental image of possibilities and should it be required instinctively hunt for any missing data.

In the following example, a set of fragments including 13C and 1H chemical shifts and long-range coupling information were extracted from an HMBC experiment (not shown). The green arrows represent the 2-3J coupling responses between the 3 equivalent methyl groups and the carbonyl’s quaternary carbon. Based on these restrictions, what fragment(s) support the data and is there anything missing?

To accommodate these restrictions, three potential fragments, assigned A, B and C, are shown below. Fragment A can be disregarded on the basis of the carbon valence. Fragment B is not a good candidate because the CH3 chemical shifts do not support the presence of an adjacent heteroatom. Fragment C seems to be the most logical choice. However, there is a missing quaternary carbon. The next step is to re-evaluate the NMR data in search of a weak 13C signal at ~40 ppm.

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.

Qualitative Elucidation via TLC Staining

TLC (Thin Layer Chromatography) offers a simple approach to identifying a solvent(s) for separating out a mixture, monitoring a reaction to completion, etc. Furthermore, the TLC plates can be stained to identify the presence or absence of various functional groups such as amines, ketones, etc. This qualitative experiment does offer various drawbacks.

The process of staining typically requires the items shown in the illustration below.

 The following websites offer more insight into the process of staining.

http://webpub.allegheny.edu/employee/s/smurphre/research/visualization.pdf

Org Prep Daily: http://orgprepdaily.wordpress.com/2006/09/27/tlc-staining-solutions/

Curly Arrow Blog: http://curlyarrow.blogspot.com/2006/11/lets-talk-about-tlcs-part-1-vanillin.html

Re-evaluating the data from MS and NMR … Part 4

With any type of data, there is an inherent risk of misinterpretation. My advice to elucidators is to consider multiple solutions and examine each one thoroughly. In the end, the answer to any problem set lies in tying together the bits of information in hopes of understanding the bigger picture.

Recap of the problem: The ESI+ MS shows a single [M+H]+ at m/z 102 allowing a maximum carbon count of 8. The 13C NMR shows there to be 12 carbons. How can the data from the MS and NMR present such different results for the same unknown?

The data from both the 13C NMR and DEPT-135 spectra are consistent with a mixture of two similar compounds at approximately a 1:1 ratio.

A mixture with an ESI+ MS exhibiting a single molecular ion indicates that the compounds in the mixture differ by a proton. For example, the mixture comprises of one compound with an R-NH2 group and the other with an R-NH3+ group. Some example amine/aminium mixtures are shown below.

Re-evaluating the data from MS and NMR … Part 2

Whenever data appear to contradict each other, an instinctive reaction to this problem is to collect more data. Collecting more data can help to understand the problem and/or complicate the matter. Remember the model for Elucidation Evolution? Maximize data extraction (MDE) while minimizing data collection (MDC).

Recap of the problem: The ESI+ MS shows a single [M+H]+ at m/z 102 allowing a maximum carbon count of 8. The 13C NMR shows there to be 12 carbons. How can the data from the MS and NMR present such different results for the same unknown?

The 1H NMR spectrum below is complicated due to significant peak overlap. As such, it does not offer any further insight into the problem.

Is there any data (or a different interpretation) that can assist in deciphering what the unknown is and thus explain why the MS and NMR data appear to contradict each other?

Re-evaluating the data from MS and NMR … Part 1

Structure elucidators will routinely use data from multiple techniques such as MS and NMR to build a proposed structure(s). When dealing with data from multiple techniques, the issue may arise that the data seem to contradict each other. In these cases, it is best to step back and re-evaluate the data from a different angle.

The ESI+ MS data below shows a prominent [M+H]+ ion at m/z 102 and its sodiated adduct. The maximum number of carbons possible for the ion is 8 (= 102 / 12). The 13C NMR spectrum below shows 12 carbons signals, all aliphatic and no quaternary carbons.

 

 

Assuming no issues with the instruments, how the data was acquired or how the sample was prepared, the lingering issue is how can the data from the MS and NMR present such different results for the unknown?

Determining the Site of Modification … Part 2

Peak matching involves the process of comparing spectral data from a parent or starting material to an unknown compound. (The unknown compound can be referred more specifically as the product, impurity, degradant, metabolite, etc.). The similarities between the data indicate regions that have not changed while the differences indicate regions of change.

The full scan MS data in Part 1 does not offer enough spectral information to eliminate one of the candidate structures. The next step is to examine MS/MS data.

The ESI+ product ion spectra (MS/MS) for both the Parent (m/z 226) and Metabolite (m/z 228) are shown below. The spectra share a common fragment at m/z 121 and differ in the fragments at m/z 134/136 and 149/151.

Based on the fragments shown below, the suspected site of hydrogenation is on the carbonyl for metabolite A.

 

Determining the Site of Modification … Part 1

In past blogs, I have ascribed the process of peak matching as an integral part of structure elucidation. In this series, peak matching is demonstrated with the use of MS data.

The scheme below shows a parent compound (2-oxo-N,2-diphenylacetamide) with two possible sites for hydrogenation. The potential modified compounds are 2-hydroxy-N,2-diphenylacetamide and 2-hydroxy-1-phenyl-2-(phenylamino)ethanone labeled A and B, respectively.

The ESI+ full scan MS below belong to the parent compound (top) and the modified compound (bottom). With the given data, can the suspected site of modification be determined?

 

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.