Let's talk about analyzing and diagramming a mass spectrum for more details on labeling
the molecular ion please see the other video I've posted on mass spec that talks about the nitrogen
rule and how to detect chlorine and bromine and and the loss of water from an alcohol and isotope
effects and all that good stuff for picking the molecular ion it's the highest non-isotopic mass
in your spectrum what we're going to do though is we're going to focus on the major high mass after
we've labeled the molecular ion we're going to focus on the major high mass peaks and we're going
to label those with their mass the largest in a cluster we're going to ignore low mass
fragments that are below the lower of one of the two the lower of the half mass of the molecular
ion or lower than the base peak whichever one's the lowest we're going to ignore the fragments
below that because they're likely fragments of fragments and then we're going to label those
fragments with how they differ from the molecular ion so let's do an example right quick so here's a
spectrum and we're going to label the molecular ion is the highest non isotopic peak so this I
don't see any evidence of chlorine I don't see any evidence of bromine for the M and M + 2
I don't- and let's look and see what this is this is 86 okay very small carbon 13 isotope peak
here that we can ignore and then we've got major fragments are they odd or even well 71, 60 our 59, 58
okay so there's an even and then we have a 1-2-3 43 so mostly odd and one even here that we have
to take into account okay so we've got odd here and an even molecular ion this would mean no nitrogen
but here is 86 so 43 would be halfway and there's the base peak so we're going to ignore everything
underneath the base peak and consider only the upper ones so here we have M minus 15 that's
reasonable here we've got M minus 18 so if we added 18 there excuse me 28, 28 there so 86
M minus 28 this should tell me McClafferty right here that's some an evidence of a McClafferty
rearrangement and that's something you can look at in the text and see McClafferty
rearrangements but then we have M minus 43 here as well okay so 43 less than the molecular ion and
so when we diagram it we diagram not only what mass it is but we diagram what we lose from that
mass and then we can venture a guess that this is loss of methyl that this is loss of ethylene here
28 and for 43 we can run a series of options here but 43, 15 + 14 is 29 plus 14 more would be
minus a propyl group so 28 and 5 would give us 43 so that could be loss of a propyl group
right there so we can venture a guess as to what we're missing but that's how we diagram find the
molecular ion label the fragment masses see these little ones here they're not the highest in the
group we only care about the highest mass in the group in the high range of the spectrum so we're
analyzed and we're ready to go now what do we do well we look for evidence of hetero atoms we can
use the IR we can look for M minus 18 which is loss of water indicating an OH or an oxygen in the
compound methoxy would be M minus so oxygen weighs 16, methyl weighs 15, so 31 might indicate loss
of a methoxy or an ethoxy here with 44 so we can look there easiest place to go for oxygen though is
the IR look for a CO double bond and OH and and something like that okay chlorine, bromine
isotope effects we've talked about previously none present here nitrogen we can look at the IR for
amines we can look at the IR for a nitrile but mainly we would look at the odd molecular
ion for nitrogen okay and we can then find the molecular formula by subtracting anything we
found previously and divide the left over by 13 so here I don't see any you know tremendous evidence
of oxygen other than this tends to happen when a McClafferty tends to happen when there's an ester
or a ketone so we might have an oxygen in there let's just start with what if it's all carbon
and hydrogen so what we're going to do is we're going to say okay our molecular weight molecular
ion's 86 let's divide that by 13 and see how many Cs and Hs we can get so we do that 86 divided by
13 and we get 6 with a remainder so that means C6H6 and let's see what the remainder is 13 times
6 is 78, 86 minus 78 gives us eight leftover so plus 8 there so our all carbon and hydrogen
one possibility for the molecular formula C6H- 8 plus 6 is 14 okay so are we reasonable and we have to decide
that by calculating the degrees of unsaturation so the formula for that as we've seen before in the
isomers video is 1 plus the number of carbons plus the number of nitrogens over 2 minus the number
of things that bond once like hydrogens and halogens the sum of those divided by 2 so here
we'd have the degrees of unsaturation is equal to 1 plus 6 minus 7 is 0 degrees of unsaturation so
that's reasonable if we were negative degrees of unsaturation that would be unreasonable if we had
more unsaturation than we had carbons that would be very unreasonable to have for a compound because
if we remember degrees of unsaturation are pi bonds or rings so very difficult to have more
pi bonds and we have carbons in a molecule over also or rings for that matter as well so that we
have no degrees of unsaturation according to the calculation if it is a hydrocarbon
but I'm going to tell you that there's evidence of McClafferty and evidence of a CO double bond
in the IR and so we need to account for one oxygen and the way we're going to do that
is we're going to say well let's exchange a CH4 which weighs 16 which is also the same as an
oxygen let's exchange a CH4 for an oxygen so I'm going to go C5 take away a carbon take away 4
hydrogens H10 and then give us an oxygen that's going to weigh the same thing it's going to weigh
86 and then we can see if this is reasonable degrees of unsaturation is equal to 1 plus 5 minus
10 over 2 is 5 that's one degree oxygen doesn't count and if we saw a CO in the spectrum
that's very reasonable okay something we could see all right so both of those would be reasonable but
because we saw evidence in the IR we have C5H10O we could say well 5 carbons should mean this is
5.5% of the height of the molecular ion and that looks pretty reasonable hard to tell
we need exact values to check that but that looks pretty reasonable we could also explain
the different peaks here and I know what the compound is and it turns out it's 1-2-3-4-5 it's
2-propanone which would allow for a McClafferty rearrangement from this orientation to eliminate
an ethylene M minus 28 so that is very reasonable for this mass spectrum here okay and so we'll
talk more about rearrangements and cleavage mechanisms in a different video but that's
how we generally analyze a spectrum we go through our steps we eventually
get we label everything with molecular ion the high mass peaks, venture a guess at the
fragments and then we look at the evidence for hetero atoms in the IR and the fragment
patterns and the isotope patterns then we find the molecular formula the all carbon
hydrogen by dividing the molecular weight by 13 taking the remainder and then we venture a
guess at other hetero atoms by looking at the IR and our other evidences and seeing if they
agree with each other and when they do if they are reasonable we can look for more evidence
in our spectrum and we'll keep working examples to get better and better at it.