Spectroscopy (MS. IR 1H NMR) workthrough #2. 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 recognizable (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 1o amime should also be
apparent both occuring between 3300–3500 and a nice NH2 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 2o amine is a bit more tricky to spot, occurring at about 3420 cm-1. So if I think I've spotted a 1o
or 2o 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 3o 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 3o 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)
Right... lets get cracking. TABULATE YOUR data from the table.
Peak
Location (cm-1)
Absorption
Assignment / notes
A
about 3440
Medium strong.
This is in the same place as an alcohol but
doesn't have the characteristic broadness. But
we are told the sample is in CCl4 so to assign it
an alcohol is fine because (having a low
concentration of material) dissolved in CCl4
solvent less hydrogen bonding occurs which
sharpens the peak. The important point was the
location of the peak fits.
B
>3000
Weak
v. weak peaks. looks like alkene, there are strong
peaks at 1500 and 1600 also, so it's possible
there's an alkene. A quick look at the NMR
shows a complex multiplet between 7 and 8 so
these peaks previously thought to be alkene are
probably C=C-H stretches in an aromatic ring
C
<3000
v.weak
Really lousy alkane groups (f any, but there does
look like there is some CH alkane peaks).
D
2800 & 2700
weak
Look like CHO aldehydes peaks (although not on
the CIE correlation table)
E
1700
strong
C=O confirming aldehyde. A quick look at the
NMR gives a peak between 9 and 10 ppm, so
definately an aldehyde.
F
Approx 1600
Strong
C=C aromatic?
G
Approx 1500
Strong
C=C aromatic?
H
Approx 1350
Really
strong
The strongest peak in the spectrum. Not
immediately obvious. It doesn't fit in with any of
the peaks on the CIE correlation table but the
closest one is C-O. The C=C is much further
away.
I
Approx 1140
Strong
C-O possibly.
At this stage I'm beginning to run flat of ideas and I don't see anything instantly helpful. The NMR
conforms an aromatic and aldehyde group.
The aromatic ring as a relative number of 3 in it. In the aldehyde we see raltive H count of 1
compared to the aromatics. It's doubtful there are two aldehydes which in turn would require 6
aromatic H's and that would require at least two aromatic rings or a larger single aromatic ring, so
It's worth a shot to go with: One aldehyde group and 3 aromatic protons, so we are dealing with a tri
substituted ring. This would give 77-2 on the mass spec. Worryingly we don't see anything significat
at 75 :( but if the ring was tri substituted, then it's not beyond the realm of imagineion to suggest
those three groups are making the ring fragment. One group is probably an electron withdrawing
aldehyde group (assuming the aldehyde isn't on the end of a chain that's attached to the ring!)
The 1H NMR tells us that a peak at about 6ppm gets smaller with addition of D2O. I'll assume this is
the slightly unusual shape alcohol we assigned at about 3440cm-1 on the IR.
So we think have C6H3, CHO and OH thus far. Lets tie them together in the simplest way possibile...
?
O
H
OH
C6H3, CHO and OH gives us a mass of 12x7 + 16x2 + 2 +3 = 121
Now, mass spec:
I've already got 8 C's from my proposed molecule already hence I'll have an (M+1)+ peak from 13C
isotopes. I think that (M+1)+ peak is at 153. I'll take 152 as the molecular mass then. There is a
slightly smaller peak at 151 to the left, so that's going to be the molecule minus a H atom. Maybe
from the aldehyde or the alcohol group (note: aromatic alcohols have their molecular ion surviving
more than aliphatic ones). I can see lots of peak clusters, so this is variations of various fragments.
Clustering is quite common with fragmented aromatic rings.
As molecular ion = 152, I am missing a mass of 152 - 121 from what I think is a third substituent on
the ring. i.e. I'm 'missing' a mass of 31 units. That'll be 2 carbons maximum, but then I'll have 11 H's
left over and that's too much. Instead of having two C's in that mass 31, then maybe I have an
oxygen. A quick look at the IR does allow me to think there is a C-O. C-O will take up 28 mass units
leaving me with 3 units of mass, which can easily be stuck on the atoms. I might have CH2-OH
coming from the ring or CH3-O (a 'methoxy' group) and I have no info of TWO peaks diminishing on
the 1H NMR, so I'll go with the CH3-O coming from the ring for now.
This gives me a working molecule to test...
H3C
O
O
H
OH
Peak
Location (cm-1)
Assignment / notes
p
153
(M+1)+
Q
152
M+
R
151
(M-H)+
R
151
(M-H)+
137
Loss of 15 mass units from M+. This could be the molecule
minus a CH3 group.
123
Loss of a mass of 30 from M+. This could be the CHO
breaking off along with loss of an additional H broken off
from the OH
114
Loss of 38 from the molecule, or loss of some lesser mass
from other fragments. Not easy to assign this one in terms of
the 'side' groups coming off the ring, so it's could be a
fragment containing a part of the fragmented ring with one
or two of the other groups on it. Say M+ - CH=COH = 152 24+16+2 = 152 - 42. So close. Maybe the fragment is C=C-O =
24+16 = 40. No so maybe another fragment... This one isn't
easy to assign so I'll skip over it for now...
81
This is M+ - 71 mass units. It's fair to say the molecules has
split almost exactly in half. to get mass 81 with remainder
71. 81=36+12+16+3+17=84 = too heavy...
24+1+32+12+2 = 71
O-C-C-C-CHO = 1+16+12+12+!2+12+1+16 = 71 so maybe 81
is the remaining atoms, but it's not easy to come up with a
fragment so easily.
65
51
39
27
15
v. small but suggests CH3
Assigning plausible M.S. fragments from the proposed structure is proving difficult, so I'll jump back
to the NMR.
Peak
Location
(ppm)
Relative
Integration
height
Coupling /
splitting
pattern
Assignment
Ha
about 9.8
1
singlet
An aldehyde.
Hb
about
7.42
1
Cannot
easily
determine
splitting
Two aromatic H's are stated to be in here. They
must have the same environment as they have the
same  value.
Hc
about
7.42
1
Cannot
easily
determine
splitting
Same as above.
Hd
about
7.05
1
singlet
An single an aromatic H in a different environment
from the other two (see above). Seems to be split
into 2 therefore next to one other H - one of the H's
above.
He
about
6.39
1
singlet
Alcohol (peak diminishes on D2O and little or no
evidence of COOH or amine on other spectra)
Hf
about 4
1
singlet
CH3 on a O -O-CH3 (aromatic methoxy)
Perhaps then we could have variations on the ring pattern. e.g.
H3C
O
O
H
OH
H3C
H3C
O
O
O
H3C
O
H
H
O
O
HO
H
HO
OH
O
O
H3C O
O
H
H
OH
H3C O
OH
H
OH
H3C
O
H3C O
O
H
OH
O
H
H3C O
O
H3C O
H
HO
OH
Which of these structures is closest to the aromatic H's pattern? Hint... some of the information in
the complex aromatic region may be overlapping...
Maybe you can't deduce the exact ring substitution, but in an A-level exam you would probably get
max marks IF you proposed any one of them because this exercise was very hard with four
functional groups. Admittedly, rings are pretty tricky to get correct. In a CIE exam you'll probably be
given a much more friendly molecule - probably aliphatic, but if you did get a ring it will probably be
more simple than the above. It will probably me monosubstituted, 1,4-disubstituted (hence
symmetrical!) or some other combination of symmetry 1,3,5-trisubstituted.
The actual molecule is the very last one drawn. We couldn't get good fragments because we didn't
have the right substitution on the ring. Go back now and try and assign fragments.