 Once again, welcome you all to MSB lecture series on interpretive spectroscopy. In my last couple of lectures, I have been discussing on mass spectrometry and discussing about different type of organic molecules having different functionalities, how they fragment, how the molecular ion look like and what are the incremental degradation of molecules and all those things. Let's continue from where I had stopped. Let's look into mass spectra of amides. So in case of amides, we do get intense molecular ion peaks. Primary amides show a strong peak at MZ44 due to H2NCO cation and this is how the fragmentation happens initially. This bond cleaves here and we get a species of mass M by Z equals 44 and also beta cleavage is quite possible and it dominates when we are considering secondary amides having more than three carbon-carbon bonds in this R group and then that leads to eventually the carbon monoxide cleavage to form something like this, this species, this is the base peak it becomes and M by Z equals 30 here. You can see here, bond breaks here, this is called beta cleavage and then this radical comes out and after that one in the subsequent step in from this radical cation, this bond breaks, Cn bond breaks and then this species come out and this is formed here. So this is a typical fragmentation of amides and let's look into mass spectra of aromatic amides. In case of aromatic amides, loss of NH2 radical results in resonance stabilized benzoil cation that also eventually cleaves to form phenyl cation. For example, you can see this is the base peak for benzamide and then one NH2 comes out and then eventually CO also comes out and we get 77, this is for phenyl cation. So this is how steps involved in the degradation or decomposition or fragmentation of benzamide here and base peak can be clearly seen here. Now let's look into mass spectra of nitro compounds. In nitro compounds initially elimination of NO2 is possible or NO is possible. In those cases, what we get is characteristic peak with M by Z value of 92 or 108 and if NO2 is lost, we get 92 and in case NO is lost, we get 108 which eventually gives a phenyl cation similar to what we saw in case of amides after CO elimination. For example, initially it can break here or it can break here. When it breaks here, we get NO, NO2 will come out if it breaks here and then we get anilinium cation and from that one HCN comes out and we get C5H5 and then if NO is eliminated and then we get this radical cation and this one eliminates one CO to form this one here phenyl cation. So this is also very similar to what we saw in case of aromatic amides. Now let's look into mass spectra of sulfur compounds. So sulfur containing compounds can be readily identified by characteristic M plus 2 peaks due to the presence of 4.4 percent of 34 S isotope here. The size of M plus 2 peaks provides information about the number of sulfur atoms. So this is one advantage with sulfur, invariably we see M plus 2 characteristic peaks. Thiols such as oralkyl SH form fragments similar to corresponding alcohols. The fragmentation is very similar to considering alkyl alcohols and beta cleavage results in the elimination of this cation with M by Z value of 47 and again primary thiols eliminate H2S to give a strong M minus 34 peak which in turn eliminates ethylene and primary and secondary thiols lose readily branches if there are any. And cleavage of sulfates are very similar to beta cleavage we come across in case of ethers. Let's look into one example here for example let's look into this compound here. Firstly cleavage happens next to beta and then we get the elimination of C4H9 radical that results in the formation of this mass fragment with M by Z 117 and then it further decomposition happens this are the possible cleavage pass either it can H migration can happen or CS bond can break and we can get a species here and also eventually what we are getting is this species cation with M by Z value. 47 you can see here this is the base peak here from this one what we lose is this 47 goes off and then we get 117 where we have this portion this breaks to give this one but initially this one would break another CH2M to give 103 peak and then from 103 peak this goes off to give the 71 that means basically one goes here and then this migration takes place and then this portion will be there C5H10 that's what it shows here and also you can see the relative intensities of peaks here M is 174100 percent M plus 1 is 11.33 percent and M plus 2 is 4.6 so you can see here. Let's look into mass spectra of halogen compounds characteristic isotopic patterns can be readily recognized in case of halogenated compounds similar to what we saw in case of sulfur compounds compounds containing one chlorine atom will have M plus 2 peak with one-third of the intensity of M for example M 37 Cl by M 35 Cl is 32.5 by 100 roughly it is one-third of the intensity. So M plus 2 peak in a compound with one bromine atom will have the same intensity as that of M and compounds containing two halogen atoms so then it can be Cl2, ClBr or Br2 display a distinct M plus 4 peak from which we should be able to distinguish and we can immediately tell or we can get information about the presence of halogens in the molecules. If we have instead of chlorine or bromine if we have fluorine or iodine they have only one isotopes this is 19 of 100 percent abundant it's 127 100 percent abundant. This chart what I have given I would tell you about molecular and peaks for bromide and chloride compounds the application of isotope contribution is limited by weak molecular ion peaks but the fragments containing halides can be readily recognized. So you can see here in case of chlorine and Cl2 when you have Cl3 and also when you have Br is there and BrCl is there BrCl2, BrCl3, Br2, Br2Cl, Br2Cl2, Br2Cl3, Br3Cl. So all these combination is shown here typical ion peaks containing bromide and chlorides. So this is a very useful chart that I have shown here a guide to interpreting mass spectra for example if a methyl cation is coming out you know what the m by z value is 15 or ethyl is coming 29 so or if you are getting a nitrogen carbon fragment this is coming so all this information is there and also some vital information is also given at the left hand side of the chart here and also the sample how mass is detected how mass spectrometry typical mass spectrometry instrument works also shown here this is a useful chart got it from this site here so please go through it this is very useful site for as a reference. So let us look into some more spectra here this is the mass spectrum of ferrocein and of course this one I simulated and collected the data here and 100 percent abundant one can be seen here 269 and 270 and this is for diacetyl ferrocein here the base peak should be around 170 271 and then it fragments out I have another one to show here and the different masses I have shown here just look into it what fragment is coming out whether first acetyl group is coming out and then eventually cp group comes out and it is getting stabilized with eventually cpfe plus it will become that finally we get it but here we should go further to see that one that is the typical peak we observed the end in case of ferrocein if the number of atoms with more than one isotope increases in a molecule the distribution becomes more complex for example if you have too many atoms having many isotopes in the system then the distribution becomes more complex and also understanding becomes more difficult so I have shown here one such ccl4 here of course you should remember when you are considering ccl4 we are considering 35cl, 37cl and also 12c and 13c and ccl3 it shows the pattern something like this and if you have ccl2 it shows and ccl it shows 2 and cl it shows 2 I think I have more examples so let us look into fulerine c60 despite fulerine not having halogens or sulphurs shows 4 ions with significant abundances so 1 2 3 4 here and then if you look into the parent peak because it is 60 and molecular weight will be 720 and this is how it is split with m peak 720 m plus 1 m plus 2 and m plus 3 of course the combination also I have shown here this is 100 percent and this you can see combination of 13c and 12 and here we have 13c2 and then 12 and here we have 13c3 here and base peak comes here plus m plus 1 is shown here so what would happen if the number of carbon atoms in the molecule increases then what kind of molecular peaks we get to see is shown here for example if molecular m is 100 and if let us say we have 10 carbon atoms then we will see m plus 1 and 11 in case of and then 27, 55 and 100 and if you have m plus 2 in case of this one c25 we will see 3 and 15 and 49 and similarly m plus 3 we will see one in case of c50 and above and c90 it is 60 and m plus 4 one in case of c90 and that means when you have 90 carbon atoms we will see here 3 so this shows give some information about higher analogs of further in molecules so this shows typical isotopic pattern for compounds containing different elements the first one is about main group elements of course when hydrogen is there it is very simple carbon is there we will see 2 this because of 13c and 12c and nitrogen we will see because of 14n and 15n and oxygen we will see 16 and 17 and then fluorine only one because 19f is 100% abundant the same thing is to in case of phosphorus 31 per 100% abundant sulfur we have 32 34 and 33 so we see 3 here and in case of cl again we have 35 and 37 so this is the typical isotopic pattern one should remember whenever we look into the mass spectra having these elements in the molecule so now similar chart I have given here for transfer metals for example ruthenium this is the typical pattern and rhodium again 103 we have 100% we don't see anything else and palladium we have this and silver we have and then in case of cadmium it's a little bit more complex and indium we have 2 and tin again it's a bit more complex and in case of anti-money we will see 2 isotopic patterns 2 peaks here so isotope patterns of polyatomic ions are calculated using binomial expansion as I mentioned if you have more atoms having more number of isotopes then how to calculate the isotopic patterns we see in the molecular n peak that can be simply understood by using the binomial expansion this is a binomial expansion we use here and what is a a b a or c a etc are the fractional abundances of the isotopes of x a y a and z a of element a and the n number of atoms of a present in the ion n number of i present in the ion so then represent similarly for m atoms of b it continues like this if you have more than 2 or 3 different atoms having different isotopes simply we can calculate using this formula so now let us look into one example here this is di carbonyl dichloro iridium anion iridium has 2 isotopes 191 that's about 37% and 193 that is 63% and similarly we have 2 isotopes of chlorine 35 cl is 76% or 77% and 37 cl is 24% the contributions of minor isotopes of carbon is 13c 1% and oxygen 17 very minute 0.00038% and also 18 o is about 0.008% or one can neglect this oxygen isotopes such as o 17 and o 18 following table what I have shown is shows how to calculate the relative abundances of each of the peaks in the isotope pattern with mass spectrum of this one mass spectrum also I will be showing you in the next slide for example one can start writing ion composition one should not do any mistake first you can consider 191 iridium and 35 cl2 both are 35 and co2 is there and this is 317 and fractional abundance we have to take 0.37 for this one and 0.76 for 35 chlorine and here the total value will be 0.214 and this will constitute 43% and then if you consider 193 iridium and instead of 191 now we are considering second case 193 iridium and same 35 chlorine we are considering both of them and here this is 319 because it increases and then by 2 and now 0.63 is considered for 193 iridium and then 0.76 continues so when you sum up we will get 0.364 this constitutes 100% and then we can take the combination now take 191 iridium and take the combination of both 35 chlorine and 37 chlorine so here the fractional abundance can be calculated like this so into 2 we have to use because the ion intensity is made up of contribution from 35 and 37 cl so here it is 0.135 it is a small quantity and then of course in the same combination we can also have 193 35 and 37 so this also comes here similarly we have taken the fractions here and some of these two will be coming around 50% and then another one is now change the iridium to 193 to 191 and consider both cl and that is 323 and then we get this calculation done and it shows 0.021 and now we can consider 193 iridium and both 37 so that amounts to this value and under the end we get the value of 7% so that means basically we can see three peaks with almost 1 is to 2 is to 1 ratio in the spectrum of this one iridium dichlorodicarbonyl anion so you can see here this shows three so one is coming around about 43% one is coming around and one is coming around 50% close to and then other is 100% you can see the peaks here so whatever the calculation we have made here is reflected in the spectrum we have observed here so this how one can calculate and get to know how to interpret the molecular ion peak or fragment peaks in this fashion if you have more isotopes and this is a very beautiful mass spectrum of hexa phenyl tin compound this is a cationic compound and then here I have identified here the major peaks here and of course this comes something like this very beautiful one this is again simulated using simple software I simulated this one and also collected all the data for all the peaks observed here and this is a very useful periodic table this gives about the entire period table and then all possible isotopes and their abundance everything is listed in this period table so whenever you get time you just go through it for example cadmium is there and almost all for reference cadmium is taken here and now let us look into one example here inorganic complex ruthenium simin pta cl2 pta is phosphatriazide adamantane is a monophosphine here this is bound to ruthenium and this is the simine we call this is methyl 4 isopropyl benzene that is called the simine this is taken in methanol in the presence of lithium salt li plus and the left one what I have shown here is a normal scan and resolution is 1000 you see the broader peaks for the weakly bound solvent like here you can see the broad one for weakly bound solvent and the right one is expanded one the resolution of m plus li peak has improved to 3000 and the weakly bound solvent products have disappeared to be replaced by undrizzled peak here so this is again a very interesting mass spectrum of ruthenium two compound ruthenium is in plus two state so now before I conclude this lecture I will show you one more this is a crown ether and I am sure you know how to name the crown ether you start numbering like this one so this one the number of atoms in the ring are 18 crown and six oxygen atoms are there so this is called 18 crown six ether and this can conveniently encapsulate a positive ion such as alkali metals or alkane earth metal ion to impose a octahedral geometry so this one will go up and this will come out and they will be remained in the plane this is the typical mass spectrum of 18 crown six ether the malcher ion peak is 264 from 264 it becomes 221 that means basically it is losing one C2H3O cation and you get 221 from onwards what happens then incremental decrease in the loss of C2H4O C2H4O moiety is coming here and then you can see here incrementally it is losing and at the end eventually we get 45 so if you see here this one we are losing and from here we are losing another one and one more we are losing and then one more we are losing and this becomes the base peak for C2H4OH plus so this is again a typical for crown ether and in case of crown ethers the fragmentation happens with regular decrease in the molecular ion peak with a mass of 44 that corresponds to CH2 OH so let me continue in my next lecture with more examples before I include mass spectrometry and of course I have a lot of problems on mass spectrometry as well so in the end I would try to combine some of these issues related with mass spectrometry with NMR or IR to solve the problems until then have an excellent time thank you.