How to Interpret Spectra
or
What to do when confronted with a bunch of spectral data
This is not necessarily a stand alone guide. More detail on some or all of these points can
be found in the outlines for chapter 13 (mass spec and IR) and chapter 14 (NMR)
1. First, do I have a formula? If so, I can use the rules for determining the Index of
Hydrogen Deficiency (sometimes called degrees of unsaturation). What this will
tell me is how many double bonds or rings I have. If i have no double bonds or
rings, I can eliminate a lot of possibilities:
No carbonyl (C=O), so no ketones, aldehydes, esters, amides, or carboxylic
acids.
No alkenes, no alkynes, no rings, no nitro, no nitrile
Index of Hydrogen Deficiency (from Chapter 13 notes):
A. Rules
1.
2.
3.
4.
5.
for every “n” carbons, 2n+2 hydrogens should be present
for every Nitrogen, add 1 additional hydrogen
for every group VI element (O, S, etc.), no change in H required
for every halogen, reduce the number of H’s required by 1
the difference between the number of H’s required using these
rules and the actual # H’s present in formula and divide that
number by two
6. this is the number of pi bonds and/or rings present
7. if four or more, possibility of an aromatic
For example; the formula C10H13NO3
1. 10 C X 2 + 2 = 22 H’s should be present if saturated
2. 1 N brings total to 23 H’s needed
3. 3 O’s, but still need 23 H’s
4. No halogens, so still need 23 H’s
5. you have 13 H’s, so (23-13)/2 =5
6. there are five pi bonds and/or rings present
7. 5 is at least 4 or more, so aromatic ring may be present – look for
protons in the H-NMR in the region of 6.5-8 for conformation. If no
H-NMR, you might find conformation in the fingerprint region of
NMR, but this is sometimes difficult to do (and not required of you at
this point).
2. Second, look at the IR. Do we see any obvious peaks (absorptions) in the
functional group region (4000-1400 cm-1)?
a. Is there a broad –OH of an alcohol or broader –OH of a carboxylic
acid?
b. There’s almost certainly sp3 C-H’s slightly below 3000 cm-1, but
are there any sp2 C-H’s or sp C-H’s slightly above 3000 cm-1?
any aldehyde sp2 C-H slightly below the sp3 C-H?
c. Is there a single peak of an N-H or a double for an NH2 in the 33003500 region?
d. Is there a strong C=O absorption of a carbonyl in the 1600-1800
range?
e. Is there a CC or CN triple bond in the area 2100 to 2300?
f. Although it’s a little low in the functional group region, we might be
able to see a C=C peak of an alkene around 1600 or maybe the dual
peaks of an aromatic C=C at 1600 and 1400-1500.
Do not try to do too much with the IR. Give a quick look for these functional groups
(above) and then move on to your next piece of information to solve the puzzle.
Unless the molecule is extremely simple, no one solves it from the IR alone. Just
give a quick look now, you can always come back and mine more details from the IR
later, if needed.
example: (continuing our C10H13NO3 problem)
In this IR, I see the broad –O-H centered around 3400 cm-1, but not so broad that
it extends less than 3000 wavenumbers, so it is an alcohol and not a carboxylic acid. I
see plenty of sp3 C-H stretch just below 3000 cm-1, but I can also see some sp2 C-H
stretch poking out of the shoulder of the O-H at 3100. A more experienced eye might
catch the dual aromatic absorptions at 1600 and 1510 wavenumbers but don’t worry too
much if you missed them on your first few times trying this. However, you definitely
should have caught the O-H and the sp2 C-H.
3. NMR is usually next on my list. There are four pieces of information from the
NMR (actually, there are five, but coupling is only intelligible on the larger grant
size machines, so we are going to stick with the four universal ones).
a. The number of signals will tell you how many different kinds of H’s
that you have. Keep in mind that sometimes, signals can overlap,
especially in the region where the aromatic H’s live (6.5-8 ppm).
b. then I usually look at the peak splitting using the n+1 rule. Anything
above a quartet may be difficult to discern and may only be readable
as a multiplet. The splitting pattern can be determined by the number
of equivalent H’s. Use this table.
# adjacent equivalent H’s
0
1
2
3
4
#multiple peaks in signal
(n+1)
singlet (s)
doublet (d)
triplet (t)
quartet (q)
quintet
relative peak intensities
1
1:1
1:2:1
1:3:3:1
1:4:6:4:1
WARNING: Depending on the resolution of the machine, anything above a quartet may
only be visible as a multiplet. Also, the above pattern in the table is for the number of
equivalent H’s. If a signal has two or more sets of non-equivalent H’s the splitting
pattern gets more complex. For example, in
O
c
H2
C
C
H3C
a
C
b H2
d
CH3
The 3 “a” Hydrogens have no adjacent H’s, so this signal would only have one
peak (it would be a singlet).
The 2 “b” Hydrogens have two equivalent H’s (the “c” H’s), therefore 3 peaks in
this signal, or a triplet
The 3 “d” Hydrogens have two equivalent H’s (the “c” H’s again), therefore 3
peaks in this signal, or a triplet
The 2 “c” Hydrogens are split by the “b” Hydrogens and by the “d” Hydrogens.
“b” and “d” are not equivalent, so it is not a simple sextet ((3+2 = n)+1 = 6). NO! That
would be too simple. Instead, the “c” Hydrogens are split into a triplet by the “b”
Hydrogens and then that triplet is further split into a quartet by the “d” Hydrogens, giving
rise to a triplet x quartet = twelveplet, which we just call a multiplet because the
resolution won’t be good enough to see twelve peaks in that signal.
c. Integration: the splitting (described above) tells you how many
adjacent Hydrogens but integration tells you how many Hydrogens in
the signal. This is a relative ratio. All H’s should show up in the
NMR.
d. Finally, there is chemical shift. From integration, I can determine if it
is a CH3, a CH2, or a CH. From splitting, I can tell whether these H’s
are next to a CH3, CH2, CH or perhaps more than one of those things.
Chemical shift will tell me what is in the vicinity of this CHx that is
not another CHx (That information in given in the splitting). It can tell
me whether this CHx is on a benzene ring (chem shift 6.5-8) or O
(alcohol, ester, or ether) or halogen or carbonyl or what.
example: (continuing our C10H13NO3 problem)
a
b
c
e
f
d
This particular NMR does not have the integration line on it. However, you can follow
along in your book. This is problem #71c and the NMR is on p. 635. What I would do
next is prepare a table telling me number of signals, splitting, integration, chemical shift,
what it is likely to be, and what it is next to (from chemical shift and splitting).
peak
chem shift
split
integration†
what is it?
next to?
a
b
8.1
7.3
d
d
~7 mm
~7 mm
2H
2H
2H on
2H on
benzene
benzene
para
para
substituted substituted
c
3.7
t
~7 mm
2H
CH2
d
2.8
t**
~7 mm
2H
CH2
e
2.5
s
~4 mm
1H
OH
f
1.6*
m
~14 mm
4H
2 CH2’s
next to
one CH2
next to
one CH2
next to no
H’s
next to
many H’s
* the resolution in your text is actually much better than this and you can see two separate
multiplets here. This NMR spectra is not so clear, so we will go with 1 signal here.
** again, this is more clear in your text NMR
† this is like, so totally from your text, as the NMR above has NO integration line
How do I tell how many H’s corresponds to what height on the integration line? There
are several ways to tell here. First the easiest is to remember from the IR that we have an
alcohol. Alcohols are always singlets on the NMR and we have a singlet here which is
half the height of most of the other peaks. The singlet is an OH, then rest must be CH2’s
or two equivalent benzene CH’s. A second way to tell how many H’s in each peak is to
look at the benzene ring splitting here. That symmetrical doublet of doublets tells you
that you have a disubstituted benzene ring and each of those signals belongs to 2 H’s. If
7 mm is equivalent to two H’s, then 4 mm is one H and 14 mm must be 4 H’s. Even if
you did not get any of those clues, it is still fairly easy to determine what the integration
is by using a conversion factor. It is known that we have 13 H’s (from the formula). The
total height of the NMR integration is 46 mm (7+7+7+7+4+14). Therefore 13 H’s = 46
mm. Divide 46 mm/13 H’s and each H is ~3.5 mm tall.
From signal ‘c’, I know that this is a CH2 next to another CH2 and next to no other H’s.
From signal ‘d’, I know that this is a CH2 next to another CH2 and next to no other H’s.
At first, I might be tempted to put ‘c’ and ‘d’ together, but that would mean that the two
CH2’s in signal ‘f’, which are next to many H’s don’t have enough H’s to be next to
them. Instead, I make a butyl with these four CH2’s, putting ‘c’ and ‘d’ on the ends. I
now have these pieces:
H
H2
C
OH ,
C
H2
H
H2
C
C
H2
,
H
H
From the formula, I can see that I have one N and 2 O’s unaccounted. There are no spare
H’s to make an amine (and it’s not in the IR), but the real key is that any molecule needs
at least two end pieces (unless it is a cyclic), and right now there is only one end piece
(the –OH). The other end piece must be a nitro (-NO2) by process of elimination. The
nitro would also take care of my last degree of unsaturation from my formula. Now I
have two end pieces and two middle pieces. the two middle pieces have to connect
together, but how to tell whether the –OH goes on the benzene ring or the butyl piece?
According to our table of chemical shifts, an aromatic –OH falls about 4-7 ppm
downfield from TMS while an alkyl –OH is around 2-5 ppm. The chemical shift of this
OH is only 2.5 ppm, so the –OH is most likely on the butyl. Here is the structure of our
molecule:
H a
H
b
d
H2
C
C
H2
O2N
Ha
H
c
f
f
H2
C
C
H2
OH
e
b
Again, the two –CH2-‘s listed here as “f” are not exactly equivalent. They should
be two separate signals, and they are in the problem on p635. But with this NMR I got
off the web, they appear as one signal. This shows you difference that a stronger magnet
(i.e. a $100K in grant money) can make in taking NMR spectra. I will give you stuff
similar to what is in your text to analyze rather than this.
4. Mass spectrometry: From this data, I can tell a few things quickly. I can
also get more detailed information with more digging.
a. The molecular ion (M) peak gives the molecular wt of the compound. This
can be useful if you have an empirical formula which needs to be made
into a molecular formula.
b. An M+2 peak of equal intensity to the M peak tells you Br, and you may
be able to track the Br into the smaller fragments, though Br usually
breaks off first.
c. An M+2 peak a third as intense as the M peak indicates Cl. Chlorides
usually fragment such that the Cl breaks off or a break between the α C
and a β C.
d. Alcohols and ethers also do a lot of the breaking apart between the α C
and any β C because this results in stable cation radical fragments. In
addition, alcohols with γ CH’s may lose H2O as a fragment – an obvious
loss of 18 from the molecular ion.
O
CH3
O
CH2
+
CH3
alpha cleavage
e. Ketones with at least one γ CH may undergo McLafferty rearrangement.
When this happens, an alkene breaks off the molecule. A good indication
of this is a fragment breaking off the molecular ion with a molecular
weight that is a multiple of fourteen (general formula for an alkene is
CnH2n – keep in mind that n has to be at least two- can’t have a one carbon
alkene).
H
O
CH2
O
C
CH2
C
H
+
H3C
H3C
C
H2
CH2
McLafferty Rearrangement
example: (continuing our C10H13NO3 problem)
H2C
CH2