 Hello everyone, it's a pleasure to welcome you all once again to MSP lecture series on interpretive spectroscopy. In my previous lecture, I started showing you mass spectra of organic molecules. So let us begin looking into more such examples in this lecture. As I mentioned, let me show you some representative and important and also simple one to understand mass spectra of various molecules having different type of functionalities. So let us first consider saturated hydrocarbons, mass spectra of linear hydrocarbons display molecular ion peaks which provides the molecular weight. So MCH3 peak, the first methyl group is always weak and due to the detachment of CH2, fragments will be 14 mass units apart. That means when the fragmentation is there, you can see regular degradation with the loss of 14 mass units due to the detachment of methylene fragments here. So one such example is N hexadecane is there, molecular weight 226, the parent ion peak is here and then we can start here, 16 is not shown. So 14, one means probably a loss to CH2 units and it continues and you can see up to C2 is there. Of course, saturated hydrocarbons and linear saturated hydrocarbons, it is quite very simple to analyze and understand and parent peak can be readily identified. So now let us look into some acyclic saturated hydrocarbons. For example, one such again this is for C6H40, this is for hexade, normal hexade 86 is there, you can see here and then this one is losing becoming C5H11, one CH3 fragment is lost here and then again if you see here 58 is there, this is for C4H9 plus and this is 57 and 58 is also given here, you can just look into it. And then we have 43, 41, 42, 39 and also we have 27. So that means very nicely you can distinguish all these fragments and you can give the right mass for them and then if you make yourself familiar with interpretation in this fashion, it becomes very easy when you are going for unknown samples. So then here you can see all of them I have listed here starting from this is the one after losing one a methyl group 71 and then it is 58 and then 57 is there and then we have 56, 55 and then 43, 42, 41. So that means up to 27 you can see a large number of fragments here but nevertheless it is very easy to interpret here and now let us look into the mass spectrum of branched saturated hydrocarbons. So this cationic radical is always present in branched saturated hydrocarbons and branch fragmentation will also be observed as in 5-methyl pentadecane I have shown here. Here CH3, C4H9 and C10H21, hydrogen loss will give intense CNH2N peaks here. If you look into 5-methyl pentadecane molecular weight is 226, this is the parent peak and then once it loses one methyl group it comes here and when it loses four carbon links and it comes here and then here it is 6, it is 5, 4, 3 that means up to 3 ok. We can see here almost loss of everything except for three carbon we can see the fragments here. Hydrogen loss will give intense CNH2N peaks in case of saturated branched hydrocarbons. Let us look into now saturated cyclic hydrocarbons, the simplest saturated cyclic hydrocarbon become across the cyclohexane here. Here branches are cleaved, if there are any branches are they are cleaved whereas the increase in the relative stability of molecular ion peaks is also observed. More and more molecular species become stabilized here. So ring when cleaved consists of even number of fragments for example C2H4 something like that. So now you can see here molecular ion peak is molecular weight is 84 and then we have 56 here and then 41 here it is for C3H5. So now let us look into cyclic saturated hydrocarbons that is the same thing cyclohexane is there little bit more I have elaborated here giving the peaks the corresponding peaks and the explanation here. So M plus is the molecular ion peak with mass to charge ratio 84 due to C6H12 plus here. The small M plus 1 peak at 85 is due to 113 carbon atom in it that is 13C12C5H12. So now isotope is coming that is the abundance of 13C is 1.1. Then 13C accounts for 1 percent more carbon atoms in the molecule the greater the probability of observing this 13C M plus 1 peak. So that means if more number of carbon atoms are there or more abundant ion peaks of cyclohexane possibilities their observation here intensity of M plus 1 is 6.6. So the most abundant ion peaks of cyclohexane is M by Z is 56 and then this goes to C4H8 plus and also ethylene comes out. The M by Z 56 is base peak the most abundant and stable ion fragment that is the reason you can see 100 percent abundance of this one. It is formed by the elimination of ethylene from the molecular ion the M by Z value of 83, 69, 55 are related to the loss of a mass unit of 14 that is for CH2 from one fragment ion to the next one. You can see here I have listed all of them mass to charge ratio and the fragment which is essentially cation as shown from 69 to all the way that means of this one CH3 here and then you have C4H9 and then we have C4H8, C4H7, C3H7, C3H6, C3H5, C3H3, C2H5 and C2H3. So you can clearly see all these fragments and they are corresponding mass very nicely in this beautiful mass spectrum of cyclohexane. So now let us look into mass spectra of olefins. So allylic cleavage is a possibility in olefins so the moment you look into the mass spectrum of olefin you can see the allylic cleavage and also the fragment having allylic mass. The mass spectrum of m beta might be seen molecular weight is 136 shows clustering of CnH2n plus 1 and CnH2n and also CnH2n minus 1 at intervals of 14 due to the loss of again ethylene fragments. For example, CH3, C4H9, C10, H21 and also hydrogen loss when hydrogen loss is there it will give intense peak of molecular mass CnH2n. So now you can see here one such example is given for beta myrosine molecular weight is 136. So when it loses this radical allylic radical you can see here it comes to 93 here C7H9 and then it becomes C5H9 we have mass of 69 and 67 then we get C4H7 and then C3H5 plus. The fragmentation as I mentioned so here happens that results or it can also happen here giving the allylic chain and other possibilities here when this happens here this is allylic you can get here and in this case you can see mass by charge 67 you can identify here C5H9 here this is the one and then what we have is here 93 we have C7H9 plus. So then here you can see again C5H9 plus so this how it fragments it gives some idea about possibility of thinking the way in which the compound can cleave and form different fragments in a unknown sample. This gives a lot of information and experience to analyze your own spectrum of unknown samples. Now look into aromatic hydrocarbons so aromatic hydrocarbons show little fragmentation ions but lots of molecular ions due to the loss of H dot H radical and also one can also see the loss of H2 as well as CH2 from aromatic compounds and this is the one most common if you just look into the mass spectrum given here of naphthalene here so here molecular weight is 102 here and we can see loss of CH2 and it continues like this. So this is not 108 this is 128 so then loss of we will give you 14 102 we can also see here some fragments and some fragments here. So now let us look into all kind substituted benzene which show interesting features in electron impact mass spectra so methylene group gets into ring and ring expansion takes place to form C7H7 tropolium cation. How that happens for example this is the bond vulnerable for cleavage alpha beta and then you can get a radical like this and you can get something like this and then this one gives a tropolium by this one getting inserted to the ring here so tropolium cation so mass to charge ratio is 91 and then on the other hand through this one hydrogen immigration is there then we get benzylium radical here. You can see the naphthalene spectrum I have shown 128 here 128 is the one and 129 is M plus 1 due to 13 C. Here this shows fragment details of 14 due to the loss of CH2 units and aromatic hydrocarbons show little fragmentation I had mentioned but lots of molecular ions and then this is common so this one is a simulated one and the values I have shown here intensity and also this one M plus 1. Now let us look into alcohols ion peaks tend to be very small in case of alcohols and non existent for tertiary alcohols this ion peaks may not be observed in case of tertiary alcohols beta cleavage to the OH is common resulting in peaks such as CH2 OH plus 31 MECH OH or ME2COH species respectively for primary, secondary and tertiary alcohols CH3 OH, ME2CH OH and ME3COH. So here mass spectrum is given for ethanol here 46 you can see here this is the one and then you can give it is giving CH2 OH and also it is giving C2H5O and also it is giving C2H3 and C2H5 plus C2H2, C2H3O as we have mentioned here so and then eventually it gives the last one CH3 plus at a 15 value. So this is how ethanol shows the mass spectrum and few more spectra I have shown here this for 1 butanol and also this for pentanol. So identify the peaks and maybe calculate and then see the difference between the fragment so that what kind of fragment has come out from here to here and from here to here. You can examine these things leisurely. So now M plus is the molecular ion peak in case of ethanol with mass to charge ratio 46 due to C2H6O plus that is CH3 CH2 OH plus M plus 1 peak at MZ 47 is due to C 13 12 CH6O so that means 130 12 and CH6O so now we see a peak at 1 plus 1 that is 47. So ethanol has two carbon atoms so on average 1 in 50 will have a 13 C carbon atom. The most abundant ion of the molecule under mass spectrometry investigation is usually given an arbitrary abundance value of 100 that is called the base ion peak and all other abundances are measured against or with respect to the base ion peak. So now a little bit more information I have given about this ethanol fragmentation. This explains the principle fragmentation of the mass spectrum of ethanol. This may be helpful in analyzing organic molecules spectra especially taken from either electron impact or chemical ionization. So formation of M by Z equals 43 and 45 ions the reason peak at M by Z 45 is due to the loss of a H actually hydrogen radical from the ionizer ethanol. The parent molecular ion is C2H6O plus which gives C2H5O plus and H comes out and then mass change 46 minus 1 equals 45 so loss of H2 from this ion from the M by Z 43 and that means if you consider C2H6O plus it gives C2H3O plus H2 so then if you consider formation of M by Z equals 31 is formed by session of CC bond in the parent molecular ion for example if you consider again C2H6O plus it gives C H2OH plus and C H3 and mass changes from 46 to 15 is 31 so this is the base peak the most stable fragment and of course you can go back and look into the spectrum and then formation of 28 mass by charge ratio 28 is formed by the elimination of water from the parent molecular ion so for example if you consider C H3C2OH plus and if you remove water you get C2H4 plus this has 28 mass changes 46 minus 18 28 is given. This is the systematic analysis of the spectrum then one should do for every spectra you look into so formation of 29 again 29 is just look into it C H3C2OH that means plus gives C2H5 ethyl radical plus OH OH is 17 so mass will be 29 so that means mass 29 is due to ethyl cation and then formation of 27 IR here if you just take C2H5 plus and then C2H5OH plus so this one so then it gives removes and gets C2H3 plus and then OH comes out so this is for 27 and then formation of M by Z equals 15 ion is formed by cation of C H2OH and C H3 this is for C H3 this one so this how you can completely analyze and give explanation for the every fragment you see in the mass spectrum so now let us look into phenols so molecular ions are more abundant compared to aliphatic alcohols common fragments lost are CO that means minus 28 and C H O minus 29 so mass spectra of substituted phenols differ because of fragmentation of substituents the fragmentation happens in different ways for example if you look into this one here there are two possibilities are there and in one C H3 radical can come out and you can generate a species of molecular weight M by Z 107 or it can get one hydrogen radical coming out to form C H C H3 OH and this species with M by Z value of 121 so these two possibilities are there and a typical example is shown here for this one with molecular weight 122 so this 107 is also shown here and 104 121 is here and then 77 further goes that is due to eventually the phenol radical let us move on to another example mass spectra of aliphatic ethers the simplest one is diethyl ether so here we are considering two ethoxy butane this is the one with molecular weight 102 and possible cleavage sites are here as well as here and also hydrogen migration also possible here for example its molecular weight is 102 when it loses one methyl radical it is 87 when it loses ethyl radical it is 73 when it loses ethoxy radical it is 57 and then we get this 451 so this is the fragmentation pattern here and cleavage can also happen here to ethoxy group and these two are important fragments 87 is shown here and also 73 is shown here okay so you can see the loss of ethyl would give 73 and loss of a methyl radical would give 87 value so now let us look into mass spectra of aromatic ethers consider paradigm ethoxy benzene so here the possible cleavage is OC so methyl radical would come out and you can get a cationic radical and then this can give you cyclopentenium cation here is 65 and then it can also cleave at both sides to release HCHO to give phenyl catalyst H6 plus radical so if the alkyl portion has a longer chain then the cleavage leads to the formation of olefin via hydrogen migration so for example this is the possible side if you have a longer chain in that case what happens ethylene would come out substituted ethylene and we get something like this and eventually this gives and this species here I think this would be enough just go back into whatever the spectra had shown just look into once again analyze yourself and see the fragmentation pattern so that you would be remembering and also it would be very easy to interpret as I mentioned when you look into unknown sample spectra until then so see you in my next lecture and have a good time. Thank you.