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.
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.
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.
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 13Ca+1X + …)
2. Contribution of Isotope = Number of Atoms * % abundance / (100 - % abundance)
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.
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.
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.
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.
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.
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?
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.
‘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.
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?
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.
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.
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
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?
Due to some business travel, the blog P2C2E will be on hiatus for 2 weeks from July 26 to August 6. Posts will resume the following week.
Stay tuned for more logic puzzles.
The goal of this puzzle is to determine the respective benzene ring systems that would exhibit the following 1H NMR spectrum.
The number of substituted benzene ring systems that would exhibit the following 1H NMR spectrum is 3 (assuming no repeating units). The keys to solving this puzzle lie with the determination of which multiplets are coupled in combination with the integral information.
Starting with the easy set of multiplets, M08 and M07 are coupled to each other. The second order effects (evident by tilting towards each other) and the similar coupling constants (roughly +/- 0.2 Hz) support this claim. Along side the integral ratio of 2:2, this pattern is intrinsic for a para-substituted benzene ring. Next, the multiplets M01, M03 and M05 are coupled. The 1:2:2 integral ratio and identical coupling constants support a mono-substituted benzene ring. Finally, M02, M04 and M06 are coupled. The 1:1:1 integral ratio and meta-coupling pattern (post#2 link) indicate the presence of a 1,2,4-trisubstituted benzene ring system.
The goal of this puzzle is to determine the respective benzene ring systems that would exhibit the following 1H NMR spectrum.
For the following aromatic region of the 1H NMR spectrum, how many benzene ring systems are present?
The goal of this puzzle is to determine the respective benzene ring systems that would exhibit the following 1H NMR spectrum.
For the following aromatic region of the 1H NMR spectrum, how many benzene ring systems are present?
The goal of this puzzle is to resolve the ambiguity exhibited within a 2D NMR spectrum and thus provide the correct signal correlation. Although this exercise may seem a trivial one, it is important to go over the rationale when correlating one signal to another.
For the following 1H-13C HSQC-DEPT NMR spectrum it is important to note that the 2 correlations are phased positively (red) and thus represent either a CH or CH3 group and not a CH2. Next, one must ensure that the carbons at 61.5 and 62.2 ppm are 1 carbon each. Although this detail is not certain, we will assume this to be the case. Finally, it is best to start with the easy part first. The 1H signal at 3.50 ppm is correlated to the 13C signal at 62.2 ppm. By process of elimination, one can conclude that the 1H signal at 2.73 ppm is correlated to the 13C signal at 61.5 ppm (see this post for more details).
Ambiguity in correlating 1D and 2D NMR data can routinely occur. Some extra steps that can help avoid this issue are re-aligning the data and/or re-processing the ‘raw’ data with different parameters.
The goal of this puzzle is to resolve the ambiguity exhibited within a 2D NMR spectrum and thus provide the correct signal correlation. Although this exercise may seem a trivial one, it is important to go over the rationale when correlating one signal to another.
For the following 1H-13C HSQC-DEPT NMR spectrum it is important to note that the 2 correlations are phased positively (red) and thus represent either a CH or CH3 group and not a CH2. Next, one must ensure that the carbons at 61.5 and 62.2 ppm are 1 carbon each. Although this detail is not certain, we will assume this to be the case. Finally, it is best to start with the easy part first. The 1H signal at 3.50 ppm is correlated to the 13C signal at 62.2 ppm. By process of elimination, one can conclude that the 1H signal at 2.73 ppm is correlated to the 13C signal at 61.5 ppm (see this post for more details).
Ambiguity in correlating 1D and 2D NMR data can routinely occur. Some extra steps that can help avoid this issue are re-aligning the data and/or re-processing the ‘raw’ data with different parameters.
The goal of this puzzle is to resolve the ambiguity exhibited within a 2D NMR spectrum and thus provide the correct signal correlation.
The following 1H-13C HSQC-DEPT NMR spectrum shows two one-bond correlations linked to the 1H signals 2.75 and 3.50 ppm and two closely spaced 13C signals at 61.5 and 62.2 ppm. Does the 1H signal at 2.75 ppm correlate to the 13C signal at 61.5 ppm or the one at 62.2 ppm?
Note: the blue line was added to help align the correlation to the F1 domain.
The goal of this puzzle is to resolve the ambiguity exhibited within a 2D NMR spectrum and thus provide the correct signal correlation.
The following 1H-13C HSQC-DEPT NMR spectrum shows two one-bond correlations linked to the 1H signals 2.75 and 3.50 ppm and two closely spaced 13C signals at 61.5 and 62.2 ppm. Does the 1H signal at 2.75 ppm correlate to the 13C signal at 61.5 ppm or the one at 62.2 ppm?
Note: the blue line was added to help align the correlation to the F1 domain.
The goal of this puzzle is to determine how the molecular formulae of the intermediates may assist in reasoning out the final product.
In this puzzle, let’s consider the following one-pot synthetic reaction (solvents and additional reactants are not shown). The molecular formula (MF) and the RDBE information are also presented. The chemical reaction illustrates a C11 compound reacting to form a C9 compound and subsequently a C15 compound. The reaction continues and produces an unknown compound with a MF of C24 H23 N1 O2. Based on the given information, how can the MF of the unknown compound be explained?
The unknown comprises of 24 carbon atoms and so if you add the carbons from the 2nd intermediate with the carbons from the 3rd intermediate you arrive at the 24 carbons (9+15=24). This is also evident for the nitrogen atom count (0+1=1) and the RDBE count (5+9=14). This is not the case for the hydrogen (10+15=25) and oxygen (2+1=3) atoms. However, if the loss of a H2O molecule is considered, the unknown can be a combination of the 2nd and 3rd intermediate.
The goal of this puzzle is to determine how the molecular formulae of the intermediates may assist in reasoning out the final product.
In this puzzle, let’s consider the following one-pot synthetic reaction (solvents and additional reactants are not shown). The molecular formula (MF) and the RDBE information are also presented. The chemical reaction illustrates a C11 compound reacting to form a C9 compound and subsequently a C15 compound. The reaction continues and produces an unknown compound with a MF of C24 H23 N1 O2. Based on the given information, how can the MF of the unknown compound be explained?
The unknown comprises of 24 carbon atoms and so if you add the carbons from the 2nd intermediate with the carbons from the 3rd intermediate you arrive at the 24 carbons (9+15=24). This is also evident for the nitrogen atom count (0+1=1) and the RDBE count (5+9=14). This is not the case for the hydrogen (10+15=25) and oxygen (2+1=3) atoms. However, if the loss of a H2O molecule is considered, the unknown can be a combination of the 2nd and 3rd intermediate.
With the warm weather and some good football/soccer matches upon us, the blog P2C2E will be on hiatus for 2 weeks from June 7 to 18. Posts will resume the following week.
Stay tuned for more logic puzzles.
With the warm weather and some good football/soccer matches upon us, the blog P2C2E will be on hiatus for 2 weeks from June 7 to 18. Posts will resume the following week.
Stay tuned for more logic puzzles.
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.
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.
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.
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.
There are many advantages in working with a 1H-13C HSQC-DEPT spectrum over a 13C DEPT-135 and a 1H-13C HSQC (see Post 1 & Post 2). In most cases, a 1H-13C HSQC-DEPT is more valuable than either one of those experiments.
The aliphatic region of a 1H-13C HSQC-DEPT is spectrum below. Two coincidental carbon signals are overlapping at 38.51 ppm, one pertaining to a CH while the other a CH2 group. The DEPT-135, attached to the F1 domain, exhibits a weak 13C signal that can easily be misconstrued, for example as an impurity, if not for the extra information from the 2D NMR experiment.
There are many advantages in working with a 1H-13C HSQC-DEPT spectrum over a 13C DEPT-135 and a 1H-13C HSQC (see Post 1 & Post 2). In most cases, a 1H-13C HSQC-DEPT is more valuable than either one of those experiments.
The aliphatic region of a 1H-13C HSQC-DEPT is spectrum below. Two coincidental carbon signals are overlapping at 38.51 ppm, one pertaining to a CH while the other a CH2 group. The DEPT-135, attached to the F1 domain, exhibits a weak 13C signal that can easily be misconstrued, for example as an impurity, if not for the extra information from the 2D NMR experiment.
Instruments can fail mechanically and when they do fail, it is important to recognize the signs. For NMR data, any irregularities in the baseline can indicate an instrument issue. The best strategy to minimize instrument failures is to perform regular maintenance and collect data for standards with well documented results prior to any data collection.
The legs used to support the NMR magnet are cushioned by lifts to reduce excessive floor vibrations. If one of the lifts is not performing as it should, then the acquired data will exhibit some vibrational noise. The noise can affect the analysis of the data.
Below are two 1H NMR spectra for the same sample magnified by a factor of ten. The spiky baseline (somewhat symmetrical too) in the top spectrum is a result of a malfunctioning lift on one of the legs. The bottom spectrum does not exhibit these spikes; it was collected from the same NMR instrument with the same sample under identical conditions but the lift was repaired.
For the case where all the legs’ lifts are disabled, Glenn Facey’s blog shows the resulting spectrum.
I would like to give a special thanks to
Instruments can fail mechanically and when they do fail, it is important to recognize the signs. For NMR data, any irregularities in the baseline can indicate an instrument issue. The best strategy to minimize instrument failures is to perform regular maintenance and collect data for standards with well documented results prior to any data collection.
The legs used to support the NMR magnet are cushioned by lifts to reduce excessive floor vibrations. If one of the lifts is not performing as it should, then the acquired data will exhibit some vibrational noise. The noise can affect the analysis of the data.
Below are two 1H NMR spectra for the same sample magnified by a factor of ten. The spiky baseline (somewhat symmetrical too) in the top spectrum is a result of a malfunctioning lift on one of the legs. The bottom spectrum does not exhibit these spikes; it was collected from the same NMR instrument with the same sample under identical conditions but the lift was repaired.
For the case where all the legs’ lifts are disabled, Glenn Facey’s blog shows the resulting spectrum.
I would like to give a special thanks to
Part 1 presented a challenge to determine an experiment to distinguish two very similar products from each other, namely 3-methyl-5-(pyridin-2-yloxy)pyridine and 5'-methyl-2H-1,3'-bipyridin-2-one. The products have identical formula weights and the LC/MS and 1H NMR are too similar to draw any conclusion from.
The first step is to determine what is different between the two products and then identify an experiment specifically designed to focus on that difference. The obvious difference between the two products is the position of the oxygen atom—an ester group verse a carbonyl group. An FT-IR experiment, as commented by the reader Felipe A., can be used to sort out the products.
Other experiments can include the use of reducing agents, 15N NMR, 1H -13C HMBC, 1D NOE, 1H-1H TOCSY, MS2, etc. Note free water, acids and sample concentration can inhibit the use of some of these experiments.
A 13C NMR experiment may appear to be another good choice when trying to identify a carbonyl group. However, the carbonyl is part of a conjugated system and so the 13C chemical shift is expected around 160 ppm, which also happens to be expected for the 13C chemical shift of the O-C=N group on the other product.
Part 1 presented a challenge to determine an experiment to distinguish two very similar products from each other, namely 3-methyl-5-(pyridin-2-yloxy)pyridine and 5'-methyl-2H-1,3'-bipyridin-2-one. The products have identical formula weights and the LC/MS and 1H NMR are too similar to draw any conclusion from.
The first step is to determine what is different between the two products and then identify an experiment specifically designed to focus on that difference. The obvious difference between the two products is the position of the oxygen atom—an ester group verse a carbonyl group. An FT-IR experiment, as commented by the reader Felipe A., can be used to sort out the products.
Other experiments can include the use of reducing agents, 15N NMR, 1H -13C HMBC, 1D NOE, 1H-1H TOCSY, MS2, etc. Note free water, acids and sample concentration can inhibit the use of some of these experiments.
A 13C NMR experiment may appear to be another good choice when trying to identify a carbonyl group. However, the carbonyl is part of a conjugated system and so the 13C chemical shift is expected around 160 ppm, which also happens to be expected for the 13C chemical shift of the O-C=N group on the other product.
Many organic chemists—if not all—check to see if a synthetic reaction is complete via TLC and LC/MS and/or 1H NMR. At the same time, the chemists are using the analytical data to verify that the final product is what they intended on making. In some cases, LC/MS and 1H NMR do not adequately distinguish one potential product from another. It then becomes a question of identifying a technique(s) that can clearly verify the correct product.
The chemical structures shown below (3-methyl-5-(pyridin-2-yloxy)pyridine and 5'-methyl-2H-1,3'-bipyridin-2-one) are two possible products for a synthetic reaction. They have an identical formula weight (FW) and a nearly identical MS and 1H NMR (not shown). What other experiments can a chemist/spectroscopist propose that will assist in identifying the correct structure and thus distinguish the ester from the carbonyl product?
I would like to give a special thanks to David C. Adams for proposing the idea.
Many organic chemists—if not all—check to see if a synthetic reaction is complete via TLC and LC/MS and/or 1H NMR. At the same time, the chemists are using the analytical data to verify that the final product is what they intended on making. In some cases, LC/MS and 1H NMR do not adequately distinguish one potential product from another. It then becomes a question of identifying a technique(s) that can clearly verify the correct product.
The chemical structures shown below (3-methyl-5-(pyridin-2-yloxy)pyridine and 5'-methyl-2H-1,3'-bipyridin-2-one) are two possible products for a synthetic reaction. They have an identical formula weight (FW) and a nearly identical MS and 1H NMR (not shown). What other experiments can a chemist/spectroscopist propose that will assist in identifying the correct structure and thus distinguish the ester from the carbonyl product?
I would like to give a special thanks to David C. Adams for proposing the idea.
In this final installment of the series on Drug Development, we examine stage 4 and the effort involved in Drug Manufacturing and Process. The 4th stage begins with a large scale production of the new drug, followed by formulation studies and then ending with regulatory reviews of the entire process before the drug can be marketed and sold.
The new drug is batched produced through a scale-up synthesis of the active drug component. At this point, new impurities may surface and thus warrant further investigations into its toxicity effects. Each impurity must be identified, elucidated, re-synthesized and re-tested to ensure all safety precautions were taken.
On the formulation side, further studies are done to ensure that the active drug ingredient and its impurities are combined with an excipient that is compatible and that the dosage upon intake is consistent. In addition, stress tests are performed on the mixture to check for any harmful degradation products that may occur during storage. These studies tend to overlap with the Drug Trials set in stage 3.
For good measures, all the analyses and tests are scrutinized by a team of internal and external experts. They verify that all the correct procedures were applied and the data is consistent with what is intended to be sold.
In this final installment of the series on Drug Development, we examine stage 4 and the effort involved in Drug Manufacturing and Process. The 4th stage begins with a large scale production of the new drug, followed by formulation studies and then ending with regulatory reviews of the entire process before the drug can be marketed and sold.
The new drug is batched produced through a scale-up synthesis of the active drug component. At this point, new impurities may surface and thus warrant further investigations into its toxicity effects. Each impurity must be identified, elucidated, re-synthesized and re-tested to ensure all safety precautions were taken.
On the formulation side, further studies are done to ensure that the active drug ingredient and its impurities are combined with an excipient that is compatible and that the dosage upon intake is consistent. In addition, stress tests are performed on the mixture to check for any harmful degradation products that may occur during storage. These studies tend to overlap with the Drug Trials set in stage 3.
For good measures, all the analyses and tests are scrutinized by a team of internal and external experts. They verify that all the correct procedures were applied and the data is consistent with what is intended to be sold.
Drug Trials commence with in vitro and in vivo experiments on nonhuman subjects and observing the effects of the candidate drug(s). These are commonly referred to as the pre-clinical trials. Once safe and effective thresholds have been set, the candidate drug(s) can be administered to human subjects and thus begin the clinical trials. The clinical trials are divided into various Phases I, II, III, IV.
In a concerted effort, preformulation studies look at the candidate’s solubility, stability, crystal properties, etc. to optimize the method of delivery. As testing nears completion (or soon after the Drug Design stage is complete), an application for a new drug is submitted to an internal regulatory board and a patent is filed.
Drug Trials commence with in vitro and in vivo experiments on nonhuman subjects and observing the effects of the candidate drug(s). These are commonly referred to as the pre-clinical trials. Once safe and effective thresholds have been set, the candidate drug(s) can be administered to human subjects and thus begin the clinical trials. The clinical trials are divided into various Phases I, II, III, IV.
In a concerted effort, preformulation studies look at the candidate’s solubility, stability, crystal properties, etc. to optimize the method of delivery. As testing nears completion (or soon after the Drug Design stage is complete), an application for a new drug is submitted to an internal regulatory board and a patent is filed.
The main focus surrounding the Drug Design stage (see diagram in Part 1) is to optimize the hit compound(s) and produce analogues that will increase the activity at the target site. Lead optimization is accomplished through minor modifications of the hit compound.
In Drug Design, medicinal chemists aim to build a diverse library of compounds using the hit compound(s) as a blue print for synthesis. On the flip side, an increase in the compound’s toxicity can have it rejected from the library. For this purpose, many pharmaceutical companies apply at this stage an early ADME/Tox (Absorption, Distribution, Metabolism, Elimination/Excretion and Toxicity) screening.
The main focus surrounding the Drug Design stage (see diagram in Part 1) is to optimize the hit compound(s) and produce analogues that will increase the activity at the target site. Lead optimization is accomplished through minor modifications of the hit compound.
In Drug Design, medicinal chemists aim to build a diverse library of compounds using the hit compound(s) as a blue print for synthesis. On the flip side, an increase in the compound’s toxicity can have it rejected from the library. For this purpose, many pharmaceutical companies apply at this stage an early ADME/Tox (Absorption, Distribution, Metabolism, Elimination/Excretion and Toxicity) screening.
