Spectroscopy (MS. IR 1H NMR) workthrough #1. Mike Allan, KYUEM, 7th Nov 2015.
MS:
IR:

1H NMR:
First I like to look at IR. To get a feel of the molecule. OH from alcohols, OH from carboxylic acids should be instantly recogniseable (I can just 'see' the shape - not needing to look for their wavenumber). Then I look for C=O at about 1700cm -1 . It's my estimation that this will be the MOST
LIKELY type of compound you'll get for analysis in an exam. At this stage, bear that: just for this 1700
+/- peak alone, it could be a ketone, aldehyde, carboxylic acid, acid chloride, acyl chloride or ester. ketones, esters and anhydrides are a bit tricky to spot.
The Infrared peaks of aldehydes are usually distinct and easy to spot quite quickly. two weak to medium peaks at about 2800+/- 20, and 2700. These are not on your correlation tables. If I think I have an aldehyde, or I couldn't remember the 2800 and 2700 peaks, I'll take a quick look at the 1H
NMR to see if I have a peak at 9-10 ppm which will confirm of negate the presence of an aldehyde.
Then I'd look for C=C (alkenes or aromatics) and if I'm not sure if I have an aromatic then I instantly look at the 1H NMR to see if I have complex multiplets at about 7.5ppb (the 1H NMR 'magic number' for aromatics') and then I will know rapidly if it's an alkene or aromatic -> I won't have to struggle to identify C=C-H 's from the IR.
The weak to medium absorption of the ' double fangs' belonging to a 1 o amime should also be apparent both occuring between 3300–3500 and a nice NH
2
STRONG bending absorption at 1550

to 1650 cm -1 (although this latter peak isn't mentioned on your IR correlation tables). The single fang of a 2 o amine is a bit more tricky to spot, occurring at about 3420 cm -1 . So if I think I've spotted a 1 o or 2 o amine, I'll take a quick look at the MS to see if the molecular ion is an odd number. If I do see an odd number I'll be pleased and assume only 1 N containing amine group is been given (and not a higher odd number of N's e.g. 3,5,7,9... which would also give an odd number on the MS). If I see the molecular ion is a positvie number then I'll be unhappy because it may mean the milecule contains two N groups and that means the milecule will take on an extra dimension of difficulty in interpreting it, but I will try to be 'clever' and imagine that the potential 'diammine' molecule will be symmetrical - because such molecules will be much easier to interpret. If CIE gave a non-symmetrical diammine, then that would be incredibly 0mean of them, and there would be little point in them being that mean. C-N absorptions from 3 o amines are found 1200 to 1350 cm -1 which is likely to be a
'busy' area of the spectrum but these peaks are probably going to be be quite strong, helping to assign one peak as a 3 o amine IF NEED BE! but again these peaks are NOT on your correlation tables so you will NOT be expected to have to identify them. The extra info I'm providing here is to try and help you make progress if you are struggling by employing the simpler approach.
Sometimes people just these tips and can rapidly apply them hence can come up with a structure a lot more quickly. If the extra info/tips are not useful to you, then simply ignore them.
(Scroll to next page)

E
F
C
D
Right... lets get cracking. TABULATE YOUR data from the table.
Peak Location (cm -1 ) Absorption Assignment / notes about 3500 weak Will only analyse this if I can't make progress elsewhere.
? Not so clear what this is. An overtone of the strong 1700 frequency?
A <3000 Weak
B 1700 Strong
Looks like alkane, cuy unusually weak. Not a large alkane group???
A C=O without doubt.
Not an COOH, not an aldehyde (not peaks are
2700 and 2800)
So ketone, ester or anhydride
1434
1421
1363
1223
Med
Med
Strong
Strong
??
??
C-O ? Ester? Anhydride?
C-O ? Ester? Anhydride?
Data books correlation tables gives the C-O stretch of alcohols, ethers, esters to be at 1040–1300 and describe it as a strong peak. If we see a peak that corresponds to that, then really we should assign a suitable peak in that region to C-O EVEN THOUGH IT MAY BE WRONG. We can ignore this peak later and say it was probably a low energy bend or stretch from some other group. So my advice is assign this peak if necessary, but only use it if you believe you have to. There bay be evidence of esters somewhere in the supplemental information of the question. If there is no indication in the question that it's an ester them it's not too bad if you want to ignore using this peak after it's been assigned.
Perhaps you should only tablulate peaks that you can identify. I've put some peaks in there which I may pick up ata a later stage of the analysis if I need to (i.e. if I get stuck later).
I'm now going to apply OCCAMS RAZOR" - It will be best to assume the simplest case scenario first
(and if that doesn't work out, I'll choose the next most simple case and so on.) The simplest suggestion at this stage is that the molecules is a ketone. I'll keep in mind it may be an ester or anhydride.

Now I think it's a god time to have a very quick scan of the 1H NMR. I only see one peak (there is a slight shoulder to the left of that large peak at about 2.2ppm , but for now I'll ignore it, as I want to keep things as simple as I can, in order to make some initial structure, and once I have a candidate structure, it will be fairly easy and quick to modify it if necessary so that it fits ALL the data). So it only contains 1 type of H environment. Actually, I can tell from the integration height that indeed I only have one 1H NMR peak (otherwise I'd be given two integrations heights for two peaks!!!)
Now the mass spec...
I already know I have a C=O group. That has a mass of 12+16 = 38!!!
Peak Location (cm -1 ) Assignment / notes p
Q
R
S
59
58
43
15
A v. small peak. Could this be the molecular ion peak? Well, the m/z is odd and there is not IR evidence of an amine, so
I'll discard it as being a molecular ion for an amine. Is it an alcohol? alcohol molecular ions are often small or nonexistent -> NO it can't be a molecular ion peak for an alcohol as the peak is an odd number . This leaves the last possibility.... it is an (M+1) + peak from a 13 C isotope?? - we don't have much mass in this compound (59-38)/12 = 2 carbons, plus the one in C=O bond, so that's a max of 3C, so the M+1 peak should be only 3.3% the height of the peak immediately to the left, i.e. 58. Hummm... the peak at 59 seems slightly large for just 3.3%
This peak being the (M+1) + is the strongest candidate but doesn't inspire any confidence with me so I'll skip it an only return to it later if need be.
The leading candidate for the molecular ion. It's a big peak.
It's a reasonable assumption to imagine at A-level this peak is a molecular ion.
A 'magic number' for MS. Very likely to be a (CH
3
C=O) + group, i.e. an acyl group. Acyl groups are reasonable stable fragments. This peak is the BASE PEAK, and acyl groups are often responsible for base peaks. It fits well with us having a
C=O group too. molecules commonly fragments next to
C=o's
A 'magic number' for MS. Very likely to be a (CH
3
) + group.
The difference in mass between my assumed molecular ion peak, 58 and 15 is 43. This suggests we had an acyl group with a methyl attached.
CH
3
C=O and CH
3
seems to be the only fragments we have and add up to the molecular ion, which suggests the molecules is propanone (or acetone as it's sometimes called)

Looking at the 1H NMR, we only get one H environment. So propanone fits the bill. There is NO
SPLITTING of that peak so the C-H group present in the molecule is NOT adjacent to any other H. This again fits the bill. The integration height is 1 -> given there's only one type of H. The 6 H in the two isolated CH
3
groups will naturally give a simplest whole number of ratio of 1.
We need to check other things, such as the v. important location of the H containing group on the
1H NMR. The peak is at about 2.2ppm. Usually methyl groups bonded to a hydrocarbon group appear at about 1ppm. Our proposed CH
3
groups appear downfield at 2.2. If this is a CH
3
group then it's been shifted downfield by about 1.2 units. This makes sense to have the peak at 2.2 if it's bonded to a C=O which is highly electron withdrawing and de-shields the e- from the CH bonds in the methyl causing a shift downfield. Indeed on the correlation table itself we see " CH3–C=O" being listed at
2.2 to 3 ppm so that's nice. So perhaps one should have approx 2.2 as a magic number for acyl groups given that acyls are quite common!
So I'm strongly confident now that the molecule is propanone.
The peak at m/z=59 was most likely an (M+1) + peak from a 13 C even though it looked worryingly large. The IR peak at 1363 cm -1 is still tricky to assign. I think we can escape with ignoring it. It's probably a C-C bend or something like that.
Note we haven't used 13 C spectroscopy, yet we already are confident on what the molecule is.
The 13 C spectrum will show two peaks. One for the identical 13 C peaks (which will be overlapping) and one for the C=O peak which will be downfield on the 13 C spectrum.
I would normally tabulate the 1H NMR peaks, but as there was only one, I couldn't be bothered. For two or more peaks, then do tabulate your data.
End of walkthrough 1.