CLAYDEN. Organic Chemistry. 2ª edição. Oxford. 2012.
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CLAYDEN. Organic Chemistry. 2ª edição. Oxford. 2012.


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is covered in 
Chapter 7; for the moment, just 
accept that both NO bonds are 
the same.
CHAPTER 3   DETERMINING ORGANIC STRUCTURES70
2069_Book.indb 70 12/12/2011 8:23:37 PM
Why the positions of the peaks vary, and what we can make of this information, will be 
discussed in Chapter 18.
 \u25cf Important absorptions in the double bond region
2000 1500 cm\u201311600170018001900
arenes
1600\u20131500
carbonyl
C O
between
1750 and
1650
(depending
on type)
alkene
C C
1640
(weak)
The strength of an IR absorption depends on dipole moment
If you look back at the X\u2013H regions (3000\u20134000 cm\u22121) of the four spectra on pp. 66\u201367. 
you\u2019ll notice something that at \ufb01 rst sight seems odd. The N\u2013H and O\u2013H absorptions are 
stronger than the C\u2013H absorptions at 3000 cm\u22121, despite there being more C\u2013H bonds in 
these molecules than O\u2013H or N\u2013H bonds. The reason for this is that the strength of an IR 
absorption varies with the change of dipole moment (see the box below for a de\ufb01 nition) 
when the bond is stretched. If the bond is perfectly symmetrical, there is no change in 
dipole moment and there is no IR absorption. Obviously, the C\ufffdC bond is less polar than 
either C\ufffdO or N\ufffdO and its absorption is less intense in the IR. Indeed it may be absent 
altogether in a symmetrical alkene. By contrast the carbonyl group is very polarized, with 
oxygen attracting the electrons away from carbon, and stretching it causes a large change 
in dipole moment. C\ufffdO stretches are usually the strongest peaks in the IR spectrum. O\u2013H 
and N\u2013H stretches are stronger than C\u2013H stretches because C\u2013H bonds are only weakly 
polarized.
Dipole moments
Dipole moment depends on the variation in distribution of electrons along the bond and also its length, which is why 
stretching a bond can change its dipole moment. For bonds between unlike atoms, the larger the difference in electro-
negativity, the greater the dipole moment and the more it changes when stretched. For identical atoms (C\ufffdC, for 
example) the dipole moment, and its capacity to change with stretching, is much smaller. Stretching frequencies for 
symmetrical molecules can be measured using an alternative method known as Raman spectroscopy. This is an 
IR-based technique using scattered light that relies on the polarizability of bonds. Raman spectra are outside the scope 
of this book.
This is a good point to remind you of the various deductions we have made so far about IR 
spectra.
 \u25cf Absorptions in IR spectra
Position of band 
depends on:
reduced mass of atoms 
bond strength
light atoms give high frequency 
strong bonds give high frequency
Strength (intensity) of band 
depends on:
change in dipole 
moment
large dipole moment gives strong 
absorption
Width of band depends on: hydrogen bonding strong H bond gives broad peak
 \u25a0 Contrast the term \u2018strength\u2019 
applied to absorption and to 
bonds. A stronger absorption is 
a more intense absorption. A 
strong bond on the other hand 
has a higher frequency of 
absorption (other things 
being equal).
INFRARED SPECTRA 71
2069_Book.indb 71 12/12/2011 8:23:38 PM
The single bond region is used as a molecular \ufb01 ngerprint
The region below 1500 cm\u22121 is where the single bond vibrations occur. Here our hope that 
individual bonds may vibrate independently of the rest of the molecule is usually doomed to 
disappointment. The atoms C, N, and O all have about the same atomic weight and C\u2013C, 
C\u2013N, and C\u2013O single bonds all have about the same strength.
Single bonds
Pair of atoms Reduced mass Bond strength
C\u2013C 6.0 350 kJ mol\u22121
C\u2013N 6.5 305 kJ mol\u22121
C\u2013O 6.9 360 kJ mol\u22121
In addition, C\u2013C bonds are often joined to other C\u2013C bonds with virtually identical 
strength and reduced mass, and they have essentially no dipole moments. The only one of 
these single bonds of any value is C\u2013O, which is polar enough to show up as a strong 
absorption at about 1100 cm\u22121. Some other single bonds, such as C\u2013Cl (weak and with a 
large reduced mass, so appearing at low frequency), are quite useful at about 700 cm\u22121. 
Otherwise the single bond region is usually crowded with hundreds of absorptions from 
vibrations of all kinds used as a \u2018\ufb01 ngerprint\u2019 characteristic of the molecule but not really 
open to interpretation.
Among those hundreds of peaks in the \ufb01 ngerprint region there are some of a quite differ-
ent kind. Stretching is not the only bond movement that leads to IR absorption. Bending of 
bonds, particularly C\u2013H and N\u2013H bonds, also leads to quite strong peaks. These are called 
deformations. Bending a bond is easier than stretching it (which is easier, stretching or bend-
ing an iron bar?). Consequently, bending absorptions need less energy and come at lower 
frequencies than stretching absorptions for the same bonds. These bands may not often be 
useful in identifying molecules, but you will notice them as they are often strong (they are 
usually stronger than C\ufffdC stretches, for example) and may wonder what they are.
Deformation frequencies
Group Frequency, cm\u20131
CH2 1440\u20131470
CH3 ~1380
NH2 1550\u20131650
Mass spectra, NMR, and IR combined make quick 
identi\ufb01 cation possible
If these methods are each as powerful as we have seen on their own, how much more effective 
they must be together! We shall \ufb01 nish this chapter with the identi\ufb01 cation of some simple 
unknown compounds using all three methods. The \ufb01 rst is an industrial emulsi\ufb01 er used to 
blend solids and liquids into smooth pastes. Its elecrospray mass spectrum shows it has M + H 
with a mass of 90, so an odd molecular weight (89) suggests one nitrogen atom. High-resolution 
mass spectrometry reveals that the formula is C4H11NO.
 \u25a0 A matching \ufb01 ngerprint is 
used to link a suspect to a 
crime, but you can\u2019t interpret a 
\ufb01 ngerprint to deduce the height, 
weight, or eye-colour of a 
criminal. Likewise with the 
\ufb01 ngerprint region: a matching 
\ufb01 ngerprint con\ufb01 rms that two 
compounds are identical, but 
without a \u2018suspect\u2019 you have to 
rely on the rest of the spectrum, 
above 1500 cm\u22121, for analysis.
CHAPTER 3   DETERMINING ORGANIC STRUCTURES72
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50
51 57 63 65 69 73 74 82
91
95 97 1020
20
40
%
60
80
60 70 80 90 100
100 90
m/z
R
el
at
iv
e 
ab
un
da
nc
e
The 13C NMR spectrum has only three peaks so two of the carbon atoms must be the same. 
There is one signal for saturated carbon next to oxygen, and two for other saturated carbons, 
one more down\ufb01 eld than the other.
13C NMR spectrum
160 140 120 100 80
ppm
60 40 20 0
The IR spectrum reveals a broad peak for an OH group with two sharp NH2 peaks just 
 protruding. If we put this together, we know we have C\u2013OH and C\u2013NH2. Neither of these 
carbons can be duplicated (as there is only one O and only one N) so it must be the other two 
C atoms that are the same.
5001000150020002500300035004000
100%
40%
60%
80%
20%
0%
Tr
an
sm
is
si
on
Frequency/cm\u20131
IR spectrum
MASS SPECTRA , NMR, AND IR COMBINED MAKE QUICK IDENT IF ICAT ION POSS IBLE 73
2069_Book.indb 73 12/12/2011 8:23:40 PM
The next stage is one often overlooked. We don\u2019t seem to have much information, but try 
and put the two fragments together, knowing the molecular formula, and there\u2019s very little 
choice. The carbon chain (shown in red) could either be linear or branched and that\u2019s it!
HO
NH2
HO NH2
NH2 OH
OH
NH2
HO NH2
HO
NH2
H2N
OH
branched carbon chainlinear carbon chain
A
B
There is no room for double bonds or rings because we need to \ufb01 t in the 11 hydrogen 
atoms. We cannot put N or O in the chain because we know from the IR that we have the 
groups OH and NH2, which can each be joined only to one other group. Of the seven possi-
bilities only the last two, A and B, are possible since they