 It's a pleasure to welcome you all to MSP lecture series on interpretive spectroscopy. So I'm discussing now mass spectrometry since last few lectures. So today let me continue from where I had stopped. Let us begin with a simple problem such as this one. An organic compound exhibits M plus, M plus 2 cation and the M plus 4 cation peaks in the intensity ratio of 1 is to 2 is to 1 in the mass spectrum. And also shows a singlet at 7.49 ppm in the 1HNMR spectrum recorded in CDC M3. So identify the compound. You have here four options are there, 1, 2 dichlorobenzene, 1, 4 dibromobenzene, 1, 2 dichlorobenzene and also 1, 2 dibromobenzene. So now you have four options are there and obviously when halogens are there such as chlorine and bromine, M plus 2 and also M plus 4 peaks are expected. So at this moment just by looking into it without looking into the NMR data it may be difficult to identify in absence of NMR you can anticipate this compound to be one of these all four. Of course if you have mass we should be able to identify in absence of mass just if the information given about M plus 2 peak and M plus 4 peak intensity that is 1 is to 2 is to 1. You may think that at least these two will come very close to it but 1HNMR shows only one signal. So let us look into it. Let me draw the structure first all structures first let us analyze 1HNMR data here. So just if you look into these two chloro compound would be very similar 1, 4 and 1, 2. So let us analyze Bromo first when we look into it we can see clearly here you can do rotation like this or you can do rotation like this. You can see that all four hydrogen atoms are identical as a result it would show a single resonance 7.49 and the other hand here these two are different and these two are different you can say this one A, A, X, X something A to X to spin system. So here it is simple so that means since the value given here is 7.49 so without any hesitation we can tell the compound is 1, 4 Di Bromobenzene it also shows M plus 4 cationic peak here. So this is very simple. Now let us look into another little complicated not very simple problem mass fragments of IRCL cation in mass spectrometry shows 3 mass peaks at M by Z mass to charge ratio 226, 228, 230 and also natural abundance of iridium chlorine are given here 191 iridium is 37% whereas 193 iridium is 67% and similarly 35 and 37 chlorine abundance are 76 and 24% respectively. The intensity of the mass peaks are in the order 1, 2, 3, 4 are given we have to identify the right intensity of these peaks the right ratio of these peaks we should identify which are appearing at 226, 228 and 230. So now actually I had simulated the spectrum for this one. So now we see here 3 peaks here you can see here 3 peaks are there at around 231, 230 and then around 228 and also around 226 if you just look into this from here if you compare the abundance with respect to the mass so this one this one this one are in the ratio of this. So this is the information one can get directly from the mass spectrum of IRCL but on the other hand how to calculate this one so for this is little bit it appears like tedious but it is not really so I have highlighted here let us see whether this ratio we can arrive using the calculations now we know that we have iridium 2 isotopes and chlorine 2 isotopes are there 191 iridium 37, 193 iridium 63% and then chlorine 35, 76 and let us designate them as x, y, a and b. Now if you consider this one possibilities 3.7 divided by 10 to make it simplified 3.7 x plus 6.3 y plus 7.6 a and 2.4 b and now if I expand this one we get the values like this one can be simplified 28.12 x a by m and then part 7.82 due to m plus 2 and this also m plus 2 and this is m plus 4. Now if you go for again making it simplified 28.2 for m and 56.7 c is m plus 2 and this one for m plus 4 you just simplify and then make it less complicated multiply all this by 1.76 we get 49.4926.6 so now the ratio we are getting and just if you look into the ratio here it exactly comes 49.5 is to 100 is to 26.6 so the correct option is a so this is how you should be able to do it just look into it once again and try to make yourself comfortable in solving such problems of course if we have one isotope it's very easy but if we have two atoms with the different isotopes with the different abundance one has to go for these two there may be one more complicated ones where we have three having three different type of isotopes just if I find more examples I would come back with such examples in my lectures at the end of this lecture series so now let's look into the information we can get from mass spectra for example if we have a simple hydrocarbon like CNH2N plus 2 what information we can get by recording mass spectrum for this molecule so the different type of atom present in the molecule that comes directly by looking to the molecular formula we get and also the number of each atom present in the molecule that information also one can get it if it is an organic molecule it also gives information about saturation unsaturation presence of cyclic groups and also other functional groups for example if you consider CNH2N plus 2 it has n plus 1 pairs of hydrogen atoms for example if I take C4 H10 is there this is n n plus 1 would be equals 5 5 pairs of hydrogens are there so that means if the molecular formula is CNH2N plus 2 there are n plus 1 pairs of hydrogen atoms are there presence of a ring a double bond reduces the number of hydrogen atoms pairs by one if there is a ring or there is a double bond it reduces the number of hydrogen pairs by one for example if look into cyclohexane if you look into cyclohexane the H12 is there and instead of it is normal hexane it should be 6 14 so now if you see here n plus 1 will be here 14 should be there n plus 1 will be 7 by 2 equals 7 should be there but 6 are there so it reduces by 1 similarly when double bond is there it takes away 2 hydrogen atoms on both the carbon atoms so again it will be reduced by 1 pair so in case of organic molecules as I mentioned molecular formula gives number of atoms and their types the saturation unsaturation cyclic acyclic whether the cyclic is aromatic or non-aromatic this information we should look into then if it is not readily possible we have to use empirical formula that is to find out hydrogen deficiency that is called index of hydrogen deficiency which is given by this simple formula c plus 1 minus half h minus half x plus half n where x is other heteroatom like oxygen or halogen index is the sum of the number of rings double bonds and twice the number of multiple bonds so now if you consider saturated hydrocarbon such as c and h plus 2 n plus 2 we can apply this index of hydrogen deficiency and we shall see so this formula says n plus 1 and half half into 2 n plus 2 this is equals 0 so that means here hydrogen deficiency is not there that is 0 that means you can say if 0 value is there we can say that the organic molecule is saturated so now let us examine you for different compounds here let us look into chloro propane we have 3 carbons are there so c plus 1 will be 4 and then 7 hydrogens are there 7 by 2 and then 1 chlorine is there here so half so then this is 0 again this is a saturated compound so then we will look into this one 4 3 carbons are there 4 and 6 hydrogens are there so 6 by 2 3 of course oxygen we are not considering anything then it says that there is one double bond in this one of course this is an aldehyde now look into this one here 3 chlorine atoms are there and if you just look into index 5 carbon means it is c plus 1 is 6 and then we have 9 hydrogen atoms are there 9 by 2 and then 3 chlorines are there 3 by 2 it is 0 again this is a saturated organic compound when we look into this one here cyclohexane 6 carbon atoms are there so c plus 1 would be 7 and then 12 hydrogen atoms are there 12 by 2 is 6 so 1 ring is there here so this one and similarly when we look into c 6 6 7 will be c plus 1 and then 3 pairs are there 4 that means here 1 ring and 3 double bonds are there 1 ring is there and 1 2 3 double bonds are there so here the hydrogen deficiency index is 4 here so this is how we can calculate the hydrogen deficiency index and then we can start applying that molecular formula to arrive at the right structure of the molecule once the data is obtained from different fragments and ion peak is parent ion peak is obtained from mass spectrum so now let us look into the factors which governs the hydrogen deficiency index so it is very easy to find out the hydrogen deficiency so involves very simple considerations in a acyclic hydrocarbons in acyclic hydrocarbons n carbon atoms have n plus 1 pairs of hydrogen if you look into c n h 2 n plus 2 so n carbon atoms would have n carbon atoms will have n plus 1 hydrogen atoms 2 carbon have 3 pairs index is 0 no rings are double bonds so it's very easy to identify saturated hydrocarbons any added divalent autumn shall not alter this number if it does not form a ring or double bond this is very important any added divalent autumn shall not alter this number if it does not form a ring or double bond for example O is there OH is there so only one it doesn't alter for ethanol and dimethyl ether have 3 pairs of hydrogen index is 0 so that means as long as it doesn't form a ring or double bond this divalent atoms do not alter this hydrogen deficiency index that's the reason in the previous case we didn't consider anything for oxygen if x replaces h formula should have both h and x since every nitrogen adds an extra h to the molecule half n must be added that's the reason we're adding half n and then for example methylamine and dimethylamine trimethylamine etc so this is applicable only if n and p make three bonds as long as they're trivalent whatever I said holds good and s and o should have two bonds and h and x should make only one bond now let us look into the general principles involved in the fragmentation of ions in electron impact mass spectrometry so molecular collisions at very low pressure is very rare in electron impact mass spectrometry the mass spectral fragments are unimolecular decompositions the extensive fragmentation in electron impact is due to the instability of radical cations and excess of energy associated with electron impact methodology and most of the fragments are even electron cations formed due to bond cleavage for example if m plus is there and when it undergoes fragmentation it can give p plus and p product ions and ms fragmentation is governed by the stability of the product ions that in turn is based upon certain chemical principles so what are those chemical principles let's look into it one by one stability of carbocation follows the order tertiary carbocation is more stable than a secondary carbocation which in turn more stable than a primary carbocation and fragmentation is less likely for stride chain compounds when we are considering stride chain compounds fragmentation is less likely and this fragmentation increases with increase in branching so linear molecules are likely to fragment more and it increases with increase in the degree of branching so that means you can see less fragmentation and very simple spectrum in case of linear stride chain compounds like linear stride chain hydrocarbons like pentane, decane etc and linear molecules are likely to fragment more and it increases with increase in the degree of branching for example if you consider like this and it can are radical and it can give something like this carbocation how good saturated rings saturated rings lose their side chains if the saturated rings have some side chains they readily lose them and then a radical would come out and we get some cation something like this due to resonance allylic carbocation formation is favored here the resonance what happens it favors the formation of allylic carbocation for example if you consider here the fragmentation happens between alpha beta and then radical comes out and then we get this allylic carbocation here and of course you can see here the resonance is there and beta cleavage is very probable in alkyl substituted aromatic compounds just look into this compound with side chain here this is the possible side for cleavage and our radical comes out and we get a cation like this and we end up getting some alkyl substituted aromatic compounds will cleave in this way and C-C bond next to heterotoms is prone to cleavage C-C bond next to heterotoms is prone to cleavage provided we have some heterotoms in the side chain for example we consider here we have a heterotom and then the side chain C-C bond next to this one is this one so here the cleavage is anticipated again a radical would come out and we get something like this because here this is going to stabilize this one the carbocation is stabilized by pi donation from y if pi is let us say n o or something like that so they can readily stabilize them through pi donation and the elimination of neutral and stable molecules such as carbon monoxide water and ammonia h c and h 2 s is another possibility during the cleavage for example if you consider something like this here c o can come out or if you consider something like this here h 2 c o can come out so these are all possible possibilities which leads to neutral species along with cations or cations radicals. Prearrangements often resist or compete with fragmentation that means they resist bond cleavage if the activation energies are very low so electronization produces ions with low internal energy so only simple rearrangements should be considered in case of electron ionization whereas in case of electron impact this is other way around cleavage is often associated with elimination of small stable neutral molecules such as c o h 2 o n h 3 and h s let us look into now the rearrangements that happens when a molecule is subjected to mass spectrometer through electron impact or chemical ionization during electron impact rearrangements occur instead of bond cleavages if they require low activation energy since electron impact produces ions of low internal energy only simple rearrangements are anticipated such rearrangements are often accompanied by migration or elimination of species such as h 2 o n 2 c o c o 2 or sometimes olefins or even alcohols. So now let us look into the rearrangements for example if you take this species here the cleavage happened either here and also here in that case what happens you can see the elimination of ethylene and here you get a cationic radical of this type and again this cationic radical of this type can also give a ether radical and then c o 2 so c o 2 elimination from a carbocation also can happen during rearrangement process. Now let us look into representative electron impact mass spectra of organic compounds considering starting from simple saturated hydrocarbons to aromatic hydrocarbons and related compounds with functionalities. So now to begin with let us consider saturated hydrocarbons mass spectrum of linear hydrocarbons display molecular ion peak which provides the molecular weight so that means it's very straightforward saturated hydrocarbons provide molecular ion peak and which can directly give us information about the molecular weight and m ch3 peak is always weak and due to the detachment of ch2 fragment so there will be the loss of 15 mass units for every fragmentation so that means if the fragmentation happens you can see every fragment will be losing 14 mass units due to the elimination of ch2 ch2 has 12 plus to 14. So you can see here hexadecane molecular weight is 226 here you can see out of 16 you can see very nicely all fragments are shown here up to c2 fragment here and it's a very interesting one but it's very simple to analyze. Let us look into more such examples in my next lecture now it's saturated hydrocarbon we look into it now we should go for unsaturated hydrocarbon and then aromatic groups like that. So let's continue discussion on mass spectra of different type of organic molecules in my next lecture until then have an excellent time thank you.