Well, it's not necessary to induce the entire structure from the mass spectrum, nor do we have the time to do so. You complement the mass spectra with the IR and NMR spectra; this way, it's easier to induce the functional groups present. The mass spectrum is usually only used to determine (1) the molecular mass and (2) the molecular formula (and further confirm the evidence from the other spectra). Furthermore, there is software that helps us narrow down the list of possible structures quickly.

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Ehehe, there are massive databases of fragmentation spectra that you use for matching your experimental mass spectra to spectra of compounds that have been uploadead to libraries. Mass spec is used for studying extremely complex biological systems with thousands upon thousands of compounds present in a single sample, of course you do not identify them by hand like this in such studies. For holistic chemical analysis of complex biological systems, mass spec is by far the most widely used / most powerful tool.

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Thanks, MrKippie. That's a good idea. I will add the link.

Fragmentation processes are covered in a different video--kzfaq.info/get/bejne/kMqhfMtes73FmJ8.html

Those are percentages for the relative abundances of carbon-12 (100%, molar mass of 12.011) and carbon-13 (1.1%, molar mass of 13.011)

Hi Naveen, Sorry about the delay in responding to your question. The video moves rather fast, so it is not obvious that the vertical axis shows the intensity of the peaks as a percentage relative to the tallest peak in the spectrum. The tallest peak also happens to be associated with the molecular ion in this spectrum. If you had the time and a precise ruler you would find that the peak at M+1 is 10.9% as long as the tallest peak. I am sorry that the video does not make that clear (it goes too fast and the scale is not easy to read).

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Hi Shama, The key here is the pattern of strong peaks separated by 2 amu. In the high mass cluster around 236 we find a strong peak at 234, then 236 and then at 238. This must be a result of two atoms that have heavy isotopes that are 2 amu heavier than the most abundant (lighter) isotope. Look at the table of isotopes at 10:00 minutes. Chlorine and bromine look like good candidates. A molecule with a single bromine atom would give two peaks of almost the exact same intensity for the molecular ion, M, and the M+2 peak (a "doublet"). We see three intense peaks in a ratio of 1:2:1. That is what you might expect to see for a molecule with 2 bromine atoms. Let's label the bromine atoms A and B. The probability that any individual molecule with heavy bromine for A is ~50%; the probability that B is a heavy isotope is also ~50%. The probability that we find a molecule with heavy isotopes for both A and B = (50%)•(50%) = 25%. The same is true for the probability that both atoms are light isotopes, 25%. For heavy A and light B, the probability is also 25%. For light A and heavy B, the probability is also 25%. But a molecule that has a heavy version of A and light B has the same mass as a molecule with a light version of A and a heavy B. So, that mass (236 in the example) occurs 50% of the time. Now looking at the range of peaks near the center of the spectrum we see two peaks of almost the same height at 155 and 157, (that doublet--separated by 2 amu again!) that looks like a fragment with one bromine. The table at 10 minutes is very helpful.

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Rosbin Ravanales it’s a bit of a late response but the generally the last peak in the MS will give the Molecular mass

Hi Habiba, If you look at the table of isotope masses for bromine, you find that the relative abundance for the isotope at mass 81 is almost the same as the isotope at mass 79. So, any fragment containing a single bromine will exhibit a pattern of two equally tall peaks separated by 2 amu. We were already suspicious that there were two bromine atoms in the full molecule because of the 1:2:1 pattern of intensities for the high mass cluster where the peaks were separated by 2 amu. Breaking off a fragment and losing one bromine should give a doublet for the two isotopes of the remaining bromine.

Hi Muhammed, Since all of the calculations one does with the peak heights involves taking a ratio of intensities, the peaks can be measured in any units that are convenient as long as they are the same unit. The ratio is dimensionless and the same value whatever the units that the peak heights are in. You can use a ruler and measure the peak heights in millimeters from a paper copy of the graph, if you like.

@@garymabbott4064 thank you sir.

Hi Jessica, I am sorry that I did not see your comment earlier. The vertical axis of the spectrum is sometimes marked relative abundance, but it is also proportional to the ion intensity at a given mass.

Hi Harshit, That is a very good question. Perhaps, I should have addressed that in the video. Let's try it now. Going back to the table of isotopes (in the video) we see that Si would also contribute to the M+2 peak (3.4 % compared to 4.4% for S, but close enough to consider it. We want to be flexible when the intensities are weak, since the relative uncertainty in real data is often bigger with weak signals.) Now, notice that the Si also contributes 5.1 % to the M+1 peak. We need to correct the intensity of the M+1 peak for the contribution of 1 Si atom before calculating the number of carbon atoms, but we see it is bigger than the signal at M+1. (5.1% > 4.2%) If the M+1 peak is the result of a silicon atom, then there is no room for carbon. So, a silicon atom does not seem likely. Make sense?

Thanks for the explaination. So, sir in the nutshell, do we have to consider only those elements whose M+1 intensity percentage in the table is less than the M+1 intensity given in question? Because if (5.1% of 75.2 =3.8) < 4.1 is consider here, we might not be able to put Carbon in the formula.

@@harshitshukla7208 Yes. I think you have it. I want to apologize for sloppiness in my earlier answer. I should have multiplied 5.1% (the contribution to the M+1 peak that a silicon atom would provide) by 75.2 (the intensity of the M peak) to get 3.8 as you have done. That is the Si contribution to M+! in intensity units. That is the number to compare with the signal at M+1. 3.8 is still smaller than 4.1, but it seems to leave little room for carbon. If we go back to correct our calculation for the number of carbon atoms, we must subtract the Si contribution out first. --> [(4.1-3.8)/75.2]x100/1.1 = 0.398 carbon atoms. Much less than 1.0 does not look good. If the data were really noisy we might round up and consider it, one carbon and one silicon, but we now have to go back and account for 75 amu in the molecular ion again. That would also lead to more inconsistencies. Silicon seems unlikely.

Thanks sir.

Think of a molecule with the formula that you have given containing atoms of only the most abundant isotope of each type. (If you don't have a table of isotopes for elements common to organic molecules, you can find one on-line.) For most non-metallic elements, the most abundant isotope tends to be the lightest. Do this for carbon, hydrogen and chlorine separately. Multiply those isotope masses by the number of atoms of that element that appears in each molecule and sum these numbers for all the elements in the molecule.

This method depends on being able to measure the isotope peak intensities. Once in a while I have seen a mass spectrum with only a single peak for the molecular ion. In those cases it was a result of a weak signal. Usually, recording the spectrum again with more sample produces a strong enough isotope peak to use.

@@garymabbott4064,Actually sir, most of the drug spectrums (pubchem) only have one characteristic peak often. As per your answer sir, so technically you are impliciting that we always have small peaks based on relevent isotopic abundance?

@@muhammadzeeshanqasim4017 I agree that some (maybe many) spectra are posted without showing any isotope peaks associated with the molecular ion. There must be some artificial reason for that (such as a plotting or editing decision to display only peaks above an arbitrary minimum intensity). That is unfortunate. Perhaps, the person preparing the plot does not think about the value in providing the isotope information. Certainly, the isotopes are present in the real sample. It would be extraordinary if they were not. There is one other way of finessing the calculation in those cases, although it makes the puzzle much more challenging. The heaviest fragment peak often is the result of the loss of a small piece or molecule. That peak should also exhibit heavy isotope peaks. One can use the same reasoning to derive the chemical composition of this fragment and use the information about what was lost in its formation to arrive at the molecular formula.

@@garymabbott4064 Much appreciated sir for your so kind and comprehensive information. Your this video is quite helpful to me. 1. By the way sir, I wanted to know if we have a certain drug and we need to find its metabolites. We will get the m/z ratio of respective metabolites. How do we get further information about those daughter compounds if they haven't any information in literature. 2. Secondly, if we interpret their chemical formulas by the very method you stated above ( M+1/M)* 100%/1.1% by they dont have any CAS number or no reports in literature, especially related to the metabolites of certain pharmaceuticals. How, we solve such puzzles that these are the exact metabolites. Desperately waiting for your valuable school of thoughts. Thank you sir.

@@muhammadzeeshanqasim4017 Wow. Your questions go well beyond the space that I have for answering (and well beyond my expertise). Let me offer a couple of quick comments and then recommend some resources. Looking for drug metabolites is an ambitious research problem. Some people do preliminary studies (often with electrochemistry) to see how the drug will oxidize in the lab, since many metabolic pathways in the liver start with one or more oxidation steps. (The work of Anna Brajter-Toth's research group is a good example. You might search for their work at her website at the department of chemistry at the University of Florida.) Metabolites are often polar compounds, so this presents two more obstacles. First of all, in order to isolate them (usually from an aqueous medium) one generally uses liquid chromatography (HPLC). (Some good discussion of theory, general guidelines , as well as practical recommendations, can be found at the chromacademy website: www.chromacademy.com .) In order to introduce molecules separated by HPLC into a mass spectrometer, one generally uses electro-spray ionization. This produces molecular species that are charged because of the attachment of a hydrogen ion from the aqueous solution (M+H)+ . The good news is that this is fairly gentle and usually produces lots of "molecular ion". Consequently, one needs to take that into account in determining the molecular formula. Finally, whenever one has a good idea about the identity of compound, the best evidence to support your identification is to obtain an authentic sample (if you are lucky enough to be able to order it from a chemical supplier) and run the authentic compound through your instrument under the same conditions as the drug metabolite solution.

Hi Abdullah, Assuming that the molecule contains carbon, there will be isotope ions containing one carbon 13 that are heavier than the nominal molecular ion. They should appear at M+1. However, whenever the intensity of the molecular ion is very weak, then then the M+1 peak may be too weak to find. That makes the problem tougher. In those cases, we look at a group of ions that is almost as heavy as the molecular ion. A good choice is a strong peak that represents a logical loss of a small fragment. For example, the loss of a methyl group will lead to a peak at 15 amu less than the molecular ion. If this fragment ion and its heavy isotope cousins are intense enough, one can calculate the number of carbon atoms for this fragment. The complete molecule has, of course, one more carbon and 3 more hydrogen atoms. Remember that interpreting a mass spectrum is a process of solving a puzzle. You have to be flexible and look for clues that lead to a self-consistent structure.

Hi Shama, I am not sure where to pick the story up, but assuming you have found a molecular formula, the next thing to do is to think of possible structures that you can draw with that formula. Not every combination of atoms will make a reasonable organic molecule. The double bond equivalents (DBE) rule can help. This is also called the number of degrees of unsaturation or the "rings plus double bonds" rule. It allows you to predict the number of double bonds or rings that a molecule with a given formula must have. If the calculation gives a non-integral number, then you know the formula is no good. So, here is the rule. We take a general formula for a molecule CxHyNzOn (I don't know how to write subscripts on this message system. Sorry. x, y, z, and n are integer subscripts.) This actually this should be extended to other elements. x is the sum of all atoms that are tetravalent (can form 4 single bonds like carbon). So, if we hand a molecule with 3 carbons and 1 silicon atom, x = 3 + 1. Likewise, y is the sum of all atoms that are monovalent like hydrogen. That includes halogens. So, for a molecule with 5 hydrogens 1 chlorine and 1 bromine y = 5 + 1 +1 = 7. Continuing the argument, N is trivalent, so z stands for the sum of all atoms in a formula that are trivalent. Boron would be another (somewhat common) example. Oxygen is divalent. Another divalent atom is sulfur. (I can't think of any other common divalent elements at the moment, unless you are working with organometallic molecules.) So, n is the sum of oxygen and sulfur atoms. (Things get a bit more complicated if nitro- and sulfur functional groups are possible. I suggest looking at organic chemistry texts for those cases.) The calculation is this: DBE = x - y/2 + z/2 + 1. For example, consider the formula C6H7N DBE = 6 - 7/2 + 1/2 + 1 = 4. A value of 4 could be 4 double bonds, but a very common way of getting 4 is an aromatic ring. If you draw a benzene ring as one ring with 3 alternating double bonds, the count for rings plus double bonds is 4. I hope this helps.

No. More appropriate for (3rd year) college level.

There are these situations when the molecular ion is not the tallest peak in the high mass cluster: When the sample is contaminated and more than one compound is being ionized at the same time. We can usually rule this out, if our separation and/or clean up steps are rigorous. However, when the signal is weak background peaks (trace level contaminants and electronic noise) can complicate interpretation. When the compound of interest breaks apart so easily that no molecular ion survives to give a signal. This happens with long alkyl chains, for example. That is a tough puzzle, but one can get a lot of information by applying the same strategy to the cluster of ions at the highest mass with the caveat that this cluster is due to a fragment of the molecular ion is quite possibly an even mass ion. (See Fred McLafferty, Interpretation of Mass Spectra for an in-depth discussion.) When the molecule has multiple chlorine and/or bromine atoms. In those cases the probability of having a molecule with at least one heavy halogen isotope is greater than a molecule with all light isotopes. But these cases are easy to spot, since the tall peaks are spaced by two mass units from each other. Remember that everything heavier than the molecular ion with all light isotopes must be the result of ions with a combination heavy isotopes (and the same chemical structure, namely, the molecule minus one electron). Other ionization methods that we were not considering here also give a different signal for the molecular ion. Electrospray ionization (ESI) is very important method in LC/MS and in desorption ESI (or DESI). These methods lead to much less fragmentation, but form “adducts” (combinations of the molecule plus a hydrogen ion or sometimes a sodium ion). Here the signals in the high mass cluster must be explainable by combinations of isotopes for the structure of the whole molecule plus a hydrogen, that is, (M+H).

2:15 Interpreting Graph

www.sisweb.com/mstools/isotope.htm

Hi Shama, Looking at the high mass cluster of peaks, we see the very tall peak at 128 amu. We suspect that is the mass of the molecular ion. That would mean that the peak at 129 amu must be due to heavy isotopes of the A+1 type. Most commonly that is due to a single carbon-13 in the molecule. The peak at 128 amu also is the tallest peak in the spectrum and, therefore, has a relative intensity of 100%. The table gives us the relative peak intensities. A real instrument may give the intensities in terms of other units, such as counts per second. The peak at 129 amu has a relative intensity of 10.9 %. We use these two values to calculate an estimate of the number of carbon atoms in the molecule from the equation give about 30 seconds earlier in the video, namely, # of carbon atoms = ([M+1]/[M])•(100/1.1) where the numbers in the brackets are the intensities or relative intensities (either works if they are both in the same units). So, # of Carbons = (10.9/100)•(100/1.1) =9.9 or 10 to the nearest whole number.

Hi Kumodgurav, I am going to make some assumptions about the information that you gave me. You are reporting a retention time that suggests that your data is from an LC/MS run. Is it possible that you used an electrospray ionization (ESI) detector in the negative ion mode? ESI frequently produces adduct ions that are a combination of the analyte molecule plus some an ion present in the buffer. The mass difference of 45 amu fits nicely with the idea that the observed parent ion is the neutral drug molecule plus one formate ion [M + HCO2]- See: www.acdlabs.com/blog/common-adduct-and-fragment-ions-in-mass-spectrometry/