 I'm sure you are having great time in understanding interpretive spectroscopy. I once again welcome you all to MSP Lecture Series on interpretive spectroscopy. In my last lecture, I started discussion on topics such as mass spectrometry. When I concluded my last lecture, I was talking about resolution mass spectrometry. So let me continue from where I had stopped. So resolution of a mass spectrometer. So let us consider molecular weight of M equals 2000 and R equals 2000, then delta M equals 1, the ratio. So M by Z peaks at 1999 or 2001 can be resolved. This is called unit resolution. On the other hand, if you have M equals 200 and say R equals 2000, then delta M will be 0.1, the ratio. So now spectrometry will distinguish masses such as 200.1 and 199.9. So using HRMS, high resolution can be achieved. That means we call high resolution mass spectra is abbreviated as HRMS. For example, if we consider two iso-electronic gases such as CO and N2, we know the mass of 12 carbon is 12.000 and 14 carbon is 14.0031 and 16 O is 15.9949. So the exact mass, if we consider for N2, it would be 28.0062 and exact mass of carbon monoxide is 27.9949. The difference delta M is 0.0113. So in HRMS, here we have 1234. In HRMS, up to three decimal places is common and can distinguish CO and N2 and resolution can be given up to six decimal places. So that is the strength of HRMS. No matter how much the minute difference is there in the mass between two species that can be understood and analyzed in HRMS. So now let's look into unit mass molecular ion and isotope peaks. In order to elucidate the structure, unit resolution for the molecular ion, together IR and NMR data is sufficient to know the molecular formula. So sometime, although we get vital information about molecular weight and the fragments through ionization, it's always advisable to take some more information from spectroscopic methods such as NMR as well as IR and if needed from UVC by spectroscopy as well. So now let us consider an example of C7H7NO. So here we have 12 carbon atoms are there and 7 hydrogen atoms are there and 14 1 nitrogen and 1 oxygen is there. The unit mass is 121 here. So unit mass can be higher if molecules contain heavier isotopes such as 13 carbon. The abundance is 1.11% and also 2H very low abundance 0.016 that's deuterium and then 15N is about 0.38% and 17O is even less 0.04%. The unit mass will be greater if you consider some of these isotopes as well. If you consider let us say 13C along with 12 carbon and 1H7 14N 16O we will get M plus 1 isotope because we are adding 113C in place of 112. So unit mass will become 122 instead of 121. So other isotopes also contribute to the intensity of M plus 1 peak. For example, you can also have a combination like this 12 carbon 7, 2H 1H6 14N 16O or you can have 15N 12C7 1H7 or you can also have 17O. So that means all these will contribute towards the intensity of M plus 1 peak where the mass unit mass will be 122 rather than 121. So peak height ratio and reason for observing M plus 1, M plus 2 peaks we should understand now the presence of 13C isotope as I showed you in my previous slide in a molecular species cause an additional peak 1 unit to the rate of the M plus peak and that is called M plus 1 peak. So instead of 121 we saw 122 further due to the addition of 113 carbon in place of 112 carbon. So the presence of chlorine or bromine autumn in a compound causes two peaks in the molecular ion region that is M plus and M plus 2 peaks depending on that particular ion contains 35 chlorine or 37 chlorine same is true in case of bromine as well. So here so that means when we have halides we can anticipate M plus and M plus 2 peaks depending upon whether we have 35 chlorine or 37 chlorine or bromine also with the difference of two units. So now in a molecule if only CHN, O, F, P and IR present approximate percentage of M plus 1 intensity can be calculated as follows for the same molecule we are considering with unit mass of 121 in absence of any of these low abundant isotopes. Now for example if you consider percentage M plus 1 1.11 into number of carbon atoms plus 0.38 into number of nitrogen atoms should be considered. So that means if you take here the values are given here 1.11 into 7 and then 0.38 into N. So it gives around 8.15 this comes very close to this value here. So this is actually I simulated for this one and I obtained this data and clearly it shows about M plus 2 peak here. So you can get the value calculated here and we are getting 8.08 here. So now the relative abundance of common elements I have listed here. Carbon of course 99% I would say is 12 carbon only 1.11% is 13 carbon and the relative abundance is 100 and this is 1.11 and then 1H is 100 and here it is 2H deuterium is 0.016 and in case of nitrogen 14N is 100 relative abundance whereas for 15N is 0.38 and similarly for oxygen 16O it is 100% 17O it is 0.004 and also apart from 17 more we also have 18 more that is 0.20 relative abundance. In case of fluorine we have only one isotope 19 of that is 100% abundant in case of silicon again we have 3 isotopes 28 silicon relative abundance is 100% 29 silicon 5.10 and then 30 silicon 3.35 in case of phosphorus the only isotope is 31P which is 100% and in case of sulfur again we have 3 isotopes 32 sulfur 100% and 33 sulfur 0.78 and then 34 years 4.40 and in case of chlorine we have 35 chlorine 100% and 37 chlorine 32.5% and in case of bromine we have 79 bromine 100% and 81 bromine we have 98% and then in case of 127 iodine we have only one which is 100% so the contribution from this one is M and contribution from this one is M plus 1 and contribution from this one would be M plus 2 so this clearly shows why we see apart from M we will see M plus 1 peaks and in some cases we also see M plus 2 peaks because of the different type of isotopes in different abundance present in the sample so the isotope peaks are useful in determining the molecular formula so an intense M plus 2 peak indicate the presence of elements such as sulfur, silicon, chlorine and bromine so that means in our sample when we are subjecting to mass analysis if we have silicon, sulfur, chlorine and bromine we will see an intense peak due to M plus 2 unit so while analyzing and interpreting mass spectra one should always check for M plus 2 and M plus 4 peaks and other higher isotope peaks as well and one should look into relative intensities to know whether that is present or not so now the relative intensity of the M plus 2 peak in the mass spectrum of the compound shown here so indicate the presence of two sulfur atoms the moment you see this molecular formula isotopic abundance is 148.0380 moment we see this one it indicates the two sulfur atoms are there and then mass to charge ratio again this I have taken simulated spectrum from that one this data was obtained and you can see here M plus 2 peak having the fraction intensity of about 8.27 so that is seen here 9.7, this is 8.27 and then isotope abundance is M by Z 148 for M is 100% M by Z 150 is M plus 2 this is about 9.7% here so now mass spectrum of pentane is shown here and in this one one should try to identify different peaks this 72 is for this one if you calculate the molecular weight here 60 plus 12 higher than atoms this is 72 and then we have one at 57 we should try to identify what it is due to so of course from here you remove one CH3 here so then you will get 57 that means one CH3 is missing here and here one CH3 and one CH2 is missing so like that this fragments would give you how the given sample is fragmenting out and at the end we get a 29 this is CH3, CH2 so this 29 is for ethyl group so this is how we can analyze and we can simplify mass spectra this is a simplified mass spectrum of pentane here so now let us try to understand little bit more about these isotopes and their influence on unit peak and shifting that one to M plus 1 or M plus 2 so M plus 1 peak is caused by the presence of the 13C isotope in the molecule 13C is a stable isotope of carbon makes up 1.11% of all carbon atoms that means basically if we have 100 molecules are there out of 100 molecules one molecule will be having 13C whereas other will be 12C or if we have 100 atoms are there out of 100 atoms we have 99 atoms 12 carbon and one atom is 13C if you take methane CH4 one in every 100 will have 13C rather than 12C so one in every 100 molecules will have a mass of 17 rather than 16 so therefore mass spectrum will be consisting of two lines due to molecule ions 13CH4 plus as well as 12CH4 plus the line at M by Z value of 17 will be much smaller in height than the line at M by Z value of 16 because the 13C isotope is only 1.11% so there will be one heavier ion 17 for every 99 lighter 116 as a result the M plus 1 peak is much smaller than the M plus peak so the moment we see that one why M plus 1 peak is much smaller than M plus peak M plus peak represent the isotope which is abundant maximum and whereas the other one for the smaller isotope present in the molecule so now look into one problem here a gas was known to contain only elements among 1H, 12C, 14N and 16V the gas showed a molecular ion peak at mass to charge ratio of 28.0312 in high resolution mass spectrum identify the gas so here the information what we needed is the atomic weight of these four elements so H is 1.0078 12 carbon is 12.00 14N is 14.0031 and in case of 16V it is 15.9949 then if we look into all possible gases possibly N2 we can think of CO and C2H4 N2 will be 28.0062 and CO will be 27.9949 and C2H4 will be 28.0312 so workout will give a mass of 28.0312 so that means here 0.0312, 28.0312 this is for C2H4 so immediately we can say that this HRMS this molecular ion peak is due to ethylene gas so in the mass spectrum of dodecahedrine C20H20 approximate ratio of the peaks at M by Z is 260 and 261 so identify the peak so here only carbon atoms are there and C20H21 and if we consider only one carbon 1.1 into 20 will be 22 so other will be 100 so if you take the ratio of these two it will be 100 by 22 is approximately 5 so the ratio should be 5 is to 1 is the answer so that means the approximate ratio of the peaks at is question mark this should be 5 is to 1 so this how some of the simple problems can be understood and solved without any problem so now another example here in the mass spectrum of 1 to dichloroethane the ratio of peaks at mass to charge values are 9800 and 102 if the ratio of this one is 9 is to 6 is to 1 explain how so when we look into chlorine the two isotopes are there 35 chlorine and 37 chlorine they are in the ratio of 3 is to 1 that is 75 percent and 25 percent ratio so then we write all possible combination of these two isotopes in dichloroethane so one is 35Cl and 35Cl both here is 98, 75 both so that is the reason it is 3 into 39 and whereas here we have a two combination one is 35Cl and 37Cl other one is 37Cl and 35Cl so now we have 3 into 1 and plus 1 into 3 that is equal to 6 and then we have both of them are 37 here it is 102 and this is only 1 so that means this gives the ratio of 9 into 6 is to 1 so we can tell 98 and 102 will be in 9 is to 6 is to 1 ratio because of the possible molecules having this type of distribution of isotopes in these three are four different type of molecules so now another example is there so among C6H7Ns C6H7NO2 and C7H8FN and as well as C8H15N so molecular rate of all of them as 125 which one will show electron impact mass spectral rate of 125 plus 125 for M plus 55 percent and 126 M plus 1 and then 3.65 127 M plus 2 2.35 percent so now if you consider this is 125.19 and this is 125.13 and this one is 125.14 and this one is 125.21 among all the closest 1 to 125 is 125.13 so it corresponds to this one you can just check molecular rate of 125 if you consider the unit mass in all these things this comes to closer to this one and hence the given peak is due to C6H7NO2 so let me stop here and come up with more information about mass spectrometry and also more interesting problems in my next lecture until then have an excellent time thank you