The chemistry of organic
compounds
© E.V. Blackburn, 2011
What is “organic chemistry” and
why is it so important?
• Organic chemistry is the study of the compounds of
carbon.
• Think about how organic compounds affect our daily
life:
• Our clothes – natural and synthetic fibers
• Our medicines
• Our food – carbohydrates, proteins, triglycerides
• Oils, perfumes, paints, plastics, detergents, etc.
© E.V. Blackburn, 2011
What is “organic chemistry”
and why is it so important?
O
OH
OH
O
O
N
H
H
N
alizarin: the first naturally occurring
dye to be synthesized (1868)
indigo: used to dye blue jeans
O
CH 3CH 2OH
ethanol: a fermentation product
© E.V. Blackburn, 2011
“Vital force” theory
Friedrich Wöhler (1828):
NH 4+
NCO
-

H2N-C-NH 2
O
ammonium
cyanate
urea
A slight problem! The ammonium cyanate was
synthesized from bones.......but the Kolbe synthesis of
acetic acid in 1845 put the theory to rest!
© E.V. Blackburn, 2011
Vitamin B12
The synthesis of vitamin B12
was finally completed in
1972 by Woodward and
Eschenmoser after 10 years
of work and the assistance
of roughly one hundred
graduate students.
© E.V. Blackburn, 2011
“A production of amino acids
under possible primitive earth
conditions”
S.L. Miller, Science, 117, 528 (1953)
CH4 + NH 3 + H2O + H2
electric
discharge
amino acids including
glycine and alanine
© E.V. Blackburn, 2011
Proteins
In order to give you an idea of the vastness of organic
chemistry, we will look at proteins, our final topic of study
in CHEM 263!
E. coli contains ~5,000 different chemical compounds
of which 3,000 are proteins.
Man contains ~2,000,000 different proteins.
Biologists believe that there are in excess 10,000,000
of proteins which take part in the process of life!
http://www.wisegeek.com/how-many-proteinsexist.htm
© E.V. Blackburn, 2011
Angiotensin II
Angiotensin II is a blood pressure regulating hormone.
It contains 8 amino acid residues.
It is possible to arrange these in 40,320 different ways
only one of which corresponds to the hormone!
Its structure is actually: Asp-Arg-Val-Tyr-Ile-His-Pro-Phe.
© E.V. Blackburn, 2011
Organic chemistry
A logical subject based on fundamental principles:
Can we predict the reactivity of a group of atoms
based on what we’ve learned in school chemistry
courses?
The answer is YES!
You know a great deal about acetone so let us look
at it.
© E.V. Blackburn, 2011
Acetone
Here is acetone!
But…where are the carbons and hydrogens?
© E.V. Blackburn, 2011
Acetone
Here is acetone!
But…where are the carbons?
© E.V. Blackburn, 2011
Acetone
But…where are the hydrogens?
© E.V. Blackburn, 2011
H3C
C O
H3C
C
C
H H
H C H
C C
H
H
H H
C
C
C
C
H
1o
CH3CH2CH3
H
o
2
H
H H
H
H
H
H
H H
H
H
H
H H
H
3o
H
© E.V. Blackburn, 2011
The chemistry of acetone
© E.V. Blackburn, 2011
Acetone - a base!
O
H
H O+
H
H
O+
+ H2O
Remember the curved-arrow convention. The tail points
to the electron source and the head to the electron
destination.
© E.V. Blackburn, 2011
Acetone - a Lewis acid!
O
d+
NH 3
O
C
NH 3
© E.V. Blackburn, 2011
Models of chemical bonding
These models are based on the premise that atoms
react to produce the electronic configuration of a noble
gas.
© E.V. Blackburn, 2011
Models of chemical bonding:
the ionic bond
Consider the Li - F bond.
Li
F
1s2 2s1
1s2 2s2 2p5
- gives He configuration on loss of e- gives Ne configuration on gain of eLi - e-
Li+
F + e-
F-
“Ions” are the structural units of ionic compounds and
have strong electrostatic forces holding them
together.
© E.V. Blackburn, 2011
Models of chemical bonding:
the covalent bond
When atoms of similar electronegativities are bonded,
complete electron transfer cannot take place. The noble
gas configurations are attained by sharing electrons:
Electronegativity - the ability of an atom to attract electrons.
In the covalent bond, the atoms share electrons.
The structural unit is the “molecule.”
H
H
C
H
H
© E.V. Blackburn, 2011
Lewis structures
1. Find the total number of valence electrons of all of the
atoms.
2. Use pairs of electrons to form bonds between all bonded
pairs of atoms.
3. Distribute the remaining electrons to give each hydrogen
a duet and atoms of the second period an octet.
e.g. CH4
e.g. C2H4
H
H
C
H
H
H
H
C
C
H
H
© E.V. Blackburn, 2011
Lewis structures
(CH3)2CHCH2OH
1. Find the total number of valence electrons of all of
the atoms.
4 x 4 + 10 x 1 + 6 = 32
2. Use pairs of electrons to form bonds between all
bonded pairs of atoms.
28 electrons
© E.V. Blackburn, 2011
Lewis structures
(CH3)2CHCH2OH
3. Distribute the remaining electrons to give each
hydrogen a duet and atoms of the second period an
octet.
© E.V. Blackburn, 2011
Formal charge
It is often necessary to include a formal charge when we
draw a structure. Thus in drawing the hydronium ion we
need to know where the charge is located.
formal = valence - unshared electrons - 0.5 x shared electrons
charge electrons in bonded atom
in bonded atom
in free atom
H3O+ - for oxygen, formal charge = 6 - 2 - 0.5 x 6 = +1
What is the formal charge on nitrogen in the ammonium
ion?
© E.V. Blackburn, 2011
O
O
1
O
2
3
formal = valence - unshared electrons - 0.5 x shared electrons
charge electrons in bonded atom
in bonded atom
in free atom
Formal charge for atom 1 = 6 - 6 (non-bonded e-) - 0.5 x 2 = -1
Formal charge for atom 2 = 6 - 2 - 0.5 x 6 = +1
Formal charge for atom 3 = 6 - 4 - 0.5 x 4 = 0
© E.V. Blackburn, 2011
Resonance
Consider the carbonate ion, CO32-. We can draw three
equivalent structures:
-
O
-
O
-
O
O-
-
O
O
O
O
O-
In reality the ion is perfectly symmetric.
All C-O bond lengths are identical and the negative
charge is delocalized over the three oxygens.
The structure is a hybrid of these three contributing
structures.
© E.V. Blackburn, 2011
The theory of resonance
• Whenever a molecule can be represented by 2 or more
structures which differ only in the arrangement of their
electrons, there may be resonance:
-
O
-
O
-
O
O-
-
O
O
O
O
O-
• The molecule is a hybrid of all the contributing structures
and cannot be adequately represented by any one of
these structures.
© E.V. Blackburn, 2011
The theory of resonance
and
O
H3C
and
O-
O
H3C
O-
OH
and
H3C
H3C
O
O+
OH
???
© E.V. Blackburn, 2011
The theory of resonance
• Resonance is important when these structures are of
about the same stability. For example,
O
H3C
However:
H3C
OO
OH
Oand
H3C
O
OH3C +
OH
• The hybrid is more stable than any of the contributing
structures. This increase in stability is called the
resonance energy.
© E.V. Blackburn, 2011
Benzene
© E.V. Blackburn, 2011
Benzene
© E.V. Blackburn, 2011
Benzene
© E.V. Blackburn, 2011
Benzene
© E.V. Blackburn, 2011
How to draw resonance
structures
CH2=CH-CH 2
CH2=CH-O-CH 3
CH2-CH=CH 2
+
CH2-CH=O-CH 3
© E.V. Blackburn, 2011
How to draw resonance
structures
CH2=CH-CH 2
CH2-CH=CH 2
© E.V. Blackburn, 2011
How to draw resonance
structures - 1,3-dienes
+
-
-
+
H
H
H
H
H
FC = 4 - 0 - 0.5x6 H
= +1
FC = 4 - 2 - 0.5x6
= -1
© E.V. Blackburn, 2011
How to draw resonance
structures - benzene
© E.V. Blackburn, 2011
How to draw resonance
structures
+
CH2=CH-CH 2
+
CH2-CH=CH 2
© E.V. Blackburn, 2011
How to draw resonance
structures
Draw all the resonance structures of:
H
H
-
H
H
+
H
H
H
H
H
H
H
. H
H
© E.V. Blackburn, 2011
Quantum mechanics
In order to understand covalent bonding, we must start
by studying the electronic structure of individual atoms.
Schrödinger calculated mathematical expressions which
describe the motion of electrons.
These “wave equations” give the energy levels available
to electrons as well as the relative probability of finding an
electron associated with a given energy level at any point
in space.
Orbitals are 3-dimensional representations of these
probabilities.
© E.V. Blackburn, 2011
Atomic orbitals
Atomic orbitals are described by three quantum numbers:
The most important is the principal quantum number, n.
It governs the energy of an orbital. It can be equal to any
positive integer except for zero.
The second is the angular momentum quantum number, l,
whose value depends on n: l = 0, 1, 2, …. n-1. It
determines the shape of the orbital.
The third is the magnetic quantum number, ml, which
governs the orientation of the orbital relative to the three
axes. ml = -l, -l + 1 … 0 … l - 1, l.
© E.V. Blackburn, 2011
Atomic orbitals
Orbitals are classified (named) according to their values of
n and l using a number and a letter. The number
represents the value of n and the letter represents the
value of l.
Electrons in an orbital having l = 0 are called s electrons.
Electrons in an orbital having l = 1 are called p electrons.
Electrons in an orbital having l = 2 are called d electrons.
© E.V. Blackburn, 2011
Atomic orbitals
n = 1, l = 0, ml = 0
one 1s orbital
n = 2, l = 0, ml = 0
one 2s orbital
n = 2, l = 1, ml = -1, 0, or +1
three 2p orbitals!
© E.V. Blackburn, 2011
Atomic orbitals
y
z
x
y
z
1s
y
y
x
x
2px
z
2py
z
x
2pz
© E.V. Blackburn, 2011
1s
2s
2p
© E.V. Blackburn, 2011
Phase signs
When the value of a wave equation is calculated for a
particular point in space relative to the nucleus, the
result may be a positive number, a negative number, or
zero.
y
z
node
Y (+)
x
Y (-)
2py
© E.V. Blackburn, 2011
Electron configurations
The aufbau principle: Orbitals of lowest energy are filled
first.
The Pauli exclusion principle: Orbitals can
accommodate a maximum of two electrons but only if they
are of opposite spin.
Hund’s rule: One electron is placed in each degenerate
orbital before adding a second electron to an orbital.
The electronic configuration of carbon is therefore
1s2 2s2 2p1 2p1
© E.V. Blackburn, 2011
Linear combination of atomic
orbitals
+
Representation of the formation of the H-H bond by the
sharing of electrons by the two hydrogens and overlap of
their singly occupied atomic orbitals.
© E.V. Blackburn, 2011
Molecular orbitals
repulsion
no interaction
436 kJ
maximum stability
0.74
The H-H bond
© E.V. Blackburn, 2011
s bond formation
© E.V. Blackburn, 2011
Antibonding molecular orbitals
When two atomic orbitals combine, they form two
molecular orbitals.
We have met the bonding molecular orbital. Here the
atomic orbitals combine by addition. Thus orbitals of the
same phase sign overlap.
+
The antibonding molecular orbital is formed by interaction
of orbitals of opposite phase sign:
+
node
© E.V. Blackburn, 2011
Energy diagram for the hydrogen
molecule
© E.V. Blackburn, 2011
The structure of methane
Ground state electronic configuration of carbon:
1s
2s


2p
 
H
C H
© E.V. Blackburn, 2011
The structure of methane
Promote an electron to give four half filled atomic orbitals:
1s
2s


2p
 

HH
C H
H
© E.V. Blackburn, 2011
Hybrid atomic orbitals
Orbital hybridization is a mathematical approach that
involves the combination of individual orbital wave
functions to obtain wave functions for new orbitals.
These orbitals have, in varying proportions, the properties
of the original orbitals taken separately.
What does this mean for methane?
© E.V. Blackburn, 2011
sp3 hybrid orbitals
1s
2s
2p


  
The four orbitals are mixed (hybridized) to give four new
sp3 hybrid orbitals which are oriented at angles of 109.5o
and are more directional in character:
© E.V. Blackburn, 2011
© E.V. Blackburn, 2011
Ethane - C2H6
H
H C
H
H
C H
H
H
s
H
H
C
C
H
H
s
sp3
H
© E.V. Blackburn, 2011
sp2 hybrid orbitals
The electronic configuration of boron is 1s2 2s2 2p1. What is
the structure of BF3?
1s
2s


F
F
B
2p
 
F
o
120
© E.V. Blackburn, 2011
Structure - ethylene - C2H4
s
H
H
120o
sp2
H
H

120o
H
H
H
H
© E.V. Blackburn, 2011
sp hybrid orbitals
sp hybridization leads to a linear structure. BeH2 is
an example of such a molecule. The 2s orbital and one
of the 2p orbitals are hybridized.
o
180
H Be H
© E.V. Blackburn, 2011
Acetylene - C2H2

o
180
H
C
C
H
s
sp
H
C
C
H
© E.V. Blackburn, 2011
NH3
Ammonia is pyramidal and its bond angles (H-N-H) are
107o.
..
N
H
H
H
© E.V. Blackburn, 2011
H2O
Oxygen has 8 valence electrons in the water molecule:
..
H
O :
H
Its bond angle is 104.5.
© E.V. Blackburn, 2011
Hybrid orbitals and carbon
Using sp, sp2 and sp3 hybrid orbitals of carbon, we can
construct the carbon chains and rings of organic compounds.
The compound below is hystrionicotoxin, a non-protein
based toxin that has been isolated from the skin of the
“Poison Dart Frog”. The Golden Poison Dart frog may have
enough toxin to kill 10 adult humans. Identify the hybrid
orbitals to form the C-C, C-H, C-O and O-H bonds in this
compound. Predict bond angles.
OH
N
H
© E.V. Blackburn, 2011
Functional groups – structural
families
Organic compounds are divided into “families” based
on the groupings of atoms within the molecule.
Lets now look at the “families” that we will encounter
in CHEM 261/263.
© E.V. Blackburn, 2011
Hydrocarbons
These are compounds containing only carbon and
hydrogen atoms.
This “clan” is divided into smaller “family” units:
• alkanes
• alkenes
• alkynes
• aromatic compounds
© E.V. Blackburn, 2011
Alkanes
Alkanes do not have multiple bonds between carbon
atoms.
They are therefore “saturated” compounds.
CH3CH2CH2CH3
butane
© E.V. Blackburn, 2011
Alkenes
Alkenes have a carbon – carbon double bond.
They are therefore said to be “unsaturated” compounds.
-pinene
a component of turpentine
© E.V. Blackburn, 2011
Alkynes
Alkynes have a carbon – carbon triple bond and are
therefore also “unsaturated”.
O
Capillin
an antifungal agent
© E.V. Blackburn, 2011
Benzene – a representative
aromatic hydrocarbon
We will study this group of unstaurated cyclic
hydrocarbons in CHEM 263.
Kekulé structures of benzene
The six  electrons are said to be “delocalized” over the
six carbons of the benzene ring.
© E.V. Blackburn, 2011
Alkyl groups
An alkyl group is the structure obtained when a hydrogen
atom is removed from an alkane.
These groups are named by replacing the -ane suffix of
the corresponding alkane by -yl, hence “alkyl”.
• CH3-
methyl
(Me-)
• CH3CH2-
ethyl
(Et-)
• CH3CH2CH2-
propyl
(Pr-)
“R” is used as a general symbol to represent any alkyl
group.
© E.V. Blackburn, 2011
Phenyl and benzyl groups
CH 2
phenyl
Ar-
benzyl
Ar-CH2-
“Ar” is used as a general symbol to represent any
aromatic ring system.
© E.V. Blackburn, 2011
Functional groups
A group of atoms and their associated bonds that has
about the same chemical reactivity whenever it occurs
in different compounds.
© E.V. Blackburn, 2011
Alkyl halides
R-X alkyl halide
R = alkyl group (substituted or unsubstituted)
H
R
R
X
R
X
H
H
1o
2o
R
R
X
R
3o
© E.V. Blackburn, 2011
carbon classification
• a “primary” carbon is bonded to one other carbon
• a “secondary” carbon is bonded to two carbon atoms
• a “tertiary” carbon is bonded to three carbon atoms
© E.V. Blackburn, 2011
Alcohols and phenols
OH
R OH
alcohols
phenols
© E.V. Blackburn, 2011
O
R O R
ethers
R
H
O
O
R
R
ketones
R
O
R
aldehydes
carboxylic
H acids
O
O
O
R
esters
R
X
acid
halides
R = alkyl or aryl
© E.V. Blackburn, 2011
O
O
R
R
NH 2
amide
O
anhydride
R
O
R C N
nitrile
R NH2
amine
© E.V. Blackburn, 2011
The “hottest” compound
This is probably capsaicin, the spicy constituent of
peppers (capsicum annuum), cayenne pepper, chili
peppers and other capsicum species.
O
N
H
HO
OCH3
Identify the functional groups.
© E.V. Blackburn, 2011
Bond Properties
© E.V. Blackburn, 2011
Bond dissociation energies
Energy is released when bonds are formed.
Energy is absorbed when bonds are broken.
This energy is called the bond dissociation energy, D.
© E.V. Blackburn, 2011
Modes of bond breaking
Homolysis
A
B
A + B
free radicals
Heterolysis
A
B
A + B
ions
© E.V. Blackburn, 2011
Breaking of bonds to carbon
C Z
+ :Z-
+
carbocation
C Z
C:-
+ Z+
carbanion
+Z
C Z
radical
© E.V. Blackburn, 2011
Bond dissociation energies
H for bond homolysis at 25C
Bond broken
H-H
D-D
F-F
Cl - Cl
Br - Br
I-I
H-F
H - Cl
H - Br
H-I
H3C - H
H3C - F
H3C - Cl
H3C - Br
H (kJ/mol)
436
444
159
243
193
151
570
432
366
298
438
452
351
293
Bond broken
(CH3)2CH - H
(CH3)2CH - F
(CH3)2CH - Cl
(CH3)2CH - Br
(CH3)2CH - I
(CH3)2CH - OH
(CH3)2CH - OCH3
(CH3)2CHCH2 - H
(CH3)3C - H
(CH3)3C - Cl
(CH3)3C - Br
(CH3)3C - I
(CH3)3C - OH
(CH3)3C - OCH3
H (kJ/mol)
401
439
339
274
222
385
337
410
390
330
263
209
379
326
© E.V. Blackburn, 2011
Bond dissociation energies
Bond broken
H3C - I
H3C - OH
H3C - OCH3
CH3CH2 - H
CH3CH2 - F
CH3CH2 - Cl
CH3CH2 - Br
CH3CH2 - I
CH3CH2 - OH
CH3CH2 - OCH3
CH3CH2CH2 - H
CH3CH2CH2 - F
CH3CH2CH2 - Cl
CH3CH2CH2 - Br
CH3CH2CH2 - I
CH3CH2CH2 - OH
CH3CH2CH2 - OCH3
H (kJ/mol)
234
380
335
420
444
338
285
222
380
335
410
444
338
285
222
380
335
Bond broken
C6H5CH2 - H
CH2=CHCH2 - H
CH2=CH - H
C6H5 - H
HCC - H
H3C - CH3
CH3CH2 - CH3
CH3CH2CH2 - CH3
CH3CH2 - CH2CH3
(CH3)2CH - CH3
(CH3)3C - CH3
HO - H
HOO - H
HO - OH
CH3CH2O - OCH3
CH3CH2O - H
H (kJ/mol)
368
361
444
464
552
376
355
355
343
351
339
498
377
213
184
436
© E.V. Blackburn, 2011
H - CH3 + Cl - Cl
Cl - CH3 + H - Cl
438 kJ
351 kJ
243 kJ
681 kJ
432 kJ
783 kJ
H = + 438 + 243 - 351 - 432 = -102 kJ
© E.V. Blackburn, 2011
Polar bonds and polar
molecules
A polar bond is a covalent bond between two atoms
of differing electronegativity.
d+ dH F
d+
dO
H H
d+
© E.V. Blackburn, 2011
Dipole moments
A polar molecule is one in which the centres of positive
and of negative charge do not coincide.
This uneven electron distribution is measured by a quantity
called the dipole moment, , measured in units of debye.
The direction of bond and molecule dipoles are often
indicated using an arrow:
positive end
H
F
 = 1.75D
© E.V. Blackburn, 2011
Dipole moments
Cl
Cl
Cl
Cl
Cl
=0
H
H
H
 = 1.86D
© E.V. Blackburn, 2011
Physical properties
Physical properties such as melting point,
boiling point and solubility are dependant on
bond polarity and bond type (ionic or covalent).
© E.V. Blackburn, 2011
Melting point
The temperature at which the thermal energy
of the structural units is great enough to
overcome the intracrystalline forces that hold
them together.
© E.V. Blackburn, 2011
Melting points v structural unit
ionic compounds
High mp. Why?
NaCl has a mp = 801C. Each sodium ion is surrounded
by six chloride ions.....the crystal is very strong and the
structure rigid.
covalent compounds
Lower mp. Why?
The structural unit is the molecule. What are the
intermolecular forces of attraction?
© E.V. Blackburn, 2011
Dipole-dipole attractions
The attraction of the positive ends of polar molecules by the
negative ends of other polar molecules.
Polar compounds therefore have higher mp and bp than
non-polar compounds of comparable molecular mass.
© E.V. Blackburn, 2011
Hydrogen bonding
Hydrogen bonding is an example of a dipole - dipole
interaction in which a hydrogen atom acts as a “bridge”
between two very electronegative atoms. This attractive
force is ~20 kJ/mol.
Ethanol (CH3CH2OH) and dimethyl ether (CH3OCH3) have
the same molecular formula.
However ethanol has a boiling point of 78.3oC whereas that
of the ether is -24.8oC.
© E.V. Blackburn, 2011
van der Waals forces
Methane has no net dipole moment.
However when two molecules come in close contact,
their electron clouds repel one another.
The result is an induced dipole, albeit a small one.
The electron deficient part of one molecule attracts the
electron rich part of another.
The dipoles continually change but the net result is an
attraction between the molecules.
© E.V. Blackburn, 2011
Criteria of purity
All scientists observe systems, make observations
and draw conclusions from their results.
However, we must know for certain exactly what is under
scrutiny. Otherwise any results obtained are simply a
worthless jumble of irreproducible facts.
Even a sample taken from a labeled bottle is no
assurance of purity!
© E.V. Blackburn, 2011
The peroxide effect
HCl
CH3CH=CH 2
CH3CHClCH 3
CH3CH=CH 2
CH3CH=CH 2
CH3CH=CH 2
HI
H2O/H+
HBr
CH3CHICH 3
CH3CH(OH)CH 3
CH3CH2CH2Br
© E.V. Blackburn, 2011
The peroxide effect
HBr
CH3CH=CH 2
CH3CHBrCH 3
absence of
peroxides
HBr
CH3CH=CH 2
CH3CH2CH2Br
peroxides
© E.V. Blackburn, 2011
Melting point
- used to indicate degree of purity and in
identification of a solid
- recrystallize to constant melting point
- a sharp melting point range (<1C) between
appearance of drops of liquid within the sample to the
disappearance of the last trace of solid
- mixed melting points. Salt is often placed on an icy
sidewalk. Why?
© E.V. Blackburn, 2011
Boiling points
Pure liquids which distil without decomposition will
have sharp, constant boiling points and will leave no
residue on distillation to dryness. However, boiling
points are very pressure dependent making
comparisons unreliable.
Some liquid mixtures pose problems – azeotropic
mixtures.
© E.V. Blackburn, 2011
Refractive index
Using an Abbé refractometer, the index can be
measured and compared with that of an authentic
sample or with literature values. An excellent
method of evaluating the purity of a liquid.
© E.V. Blackburn, 2011
Thin layer chromatography
How can chemists prove that a compound is pure
or identical with an authentic sample using
chromatography? We can’t!
If more than one spot or peak is present, we know
that the sample is impure.
The presence of one spot or peak is an indication of
purity (or identity) but it can never constitute a proof
of purity.
© E.V. Blackburn, 2011
Elemental analysis
C
H
O
52.24%
13.05%
34.71%
100g contains 52.24g
13.05g 34.71g
or
52.24mol 13.05mol 34.71mol
12.01
1.008
16.00
4.36mol
12.93mol
2.16mol
C4.36H12.93O2.16
C2H6O
Empirical formula
© E.V. Blackburn, 2011
Mass spectrometry
- used to determine molecular weights
- used to determine molecular formulas
- used in structure elucidation
© E.V. Blackburn, 2011
Infrared spectroscopy
- bonds vibrate with characteristic frequencies.
- absorption of IR energy occurs only when there is a
match between the wavelength of the radiation and the
wavelength of the bond vibration.
- the position of the IR absorption depends on the
strength of the bond, the masses of the bonded atoms
and the type of vibration observed
© E.V. Blackburn, 2011
Bond vibrations
stretching vibrations
A< >B
A > <B
bending vibrations
B
A
B
C
A
C
© E.V. Blackburn, 2011
The 5000 to 1250 cm-1 region
Absorptions between 5000 and 1250cm-1 can usually
be assigned to a specific bond and a specific vibrational
mode.
© E.V. Blackburn, 2011
The finger print region
Absorptions between 1250 and 675cm-1 are associated
with more complex vibrational and rotational modes and
are often characteristic of the molecule as a whole.
This region is therefore called the “finger print region” of
the IR spectrum.
© E.V. Blackburn, 2011
Compound
type
Frequency
(cm-1)
Intensity Vibration
alcohols a) RCH2OH
3600
3400
1050
v
s
s
O-H unassociated
O-H associated
C-O stretching
b) R2CHOH
3600
3400
1150
v
s
s
O-H unassociated
O-H associated
C-O stretching
c) R3COH
3600
3400
1200
v
s
s
O-H unassociated
O-H associated
C-O stretching
1690 - 1750
2720 - 2850
s
m
C=O stretching
C-H stretching
aldehydes
© E.V. Blackburn, 2011
Compound
type
alkanes
Frequency Intensity Vibration
(cm-1)
2850 - 3000
s
C-H stretching
1450 - 1470
s
CH2 and CH3
bending
1370 - 1380
s
CH2 and CH3
bending
720 - 725
m
CH2 and CH3
bending
© E.V. Blackburn, 2011
Compound
type
alkenes a) RCH=CH2
b) R2C=CH2
c) (Z)-RCH=CHR
d) (E)-RCH=CHR
e) R2C=CHR
f) R2C=CR2
Frequency Intensity Vibration
(cm-1)
3080 - 3140
m
=C-H stretching
1800 - 1860
m
overtone
1645
m
C=C stretching
990
s
C-H bending
910
s
C-H bending
3080 - 3140
m
=C-H stretching
1750 - 1800
m
overtone
1650
m
C=C stretching
890
s
C-H bending
3020
w
=C-H stretching
1660
w
C=C stretching
675 - 725
m
C-H bending
3020
w
=C-H stretching
1675
vw C=C stretching
970
s
C-H bending
3020
w
=C-H stretching
1670
w
C=C stretching
790 - 840
s
C-H bending
1670
vw C=C stretching
© E.V. Blackburn, 2011
Frequency
(cm-1)
Compound
type
alkynes
Intensity Vibration
b) RCCR
3300
2100 - 2140
600 - 700
2190 - 2260
s
m
s
vw
C-H stretching
CC stretching
C-H bending
CC stretching
a) fluorides
b) chlorides
c) bromides
d) iodides
1000 - 1350
750 - 850
500 - 680
200 - 500
vs
s
s
s
C-F
C-Cl
C-Br
C-I
a) RCCH
alkyl
halides
© E.V. Blackburn, 2011
Compound
type
Frequency
(cm-1)
Intensity Vibration
amines
3300 - 3500
1180 - 1360
m
N-H stretching
C-N stretching
benzene
ring
3000 - 3100
m
C-H stretching
1600, 1500
690 - 710,
730 - 770
735 - 770
v
C=C stretching
C-H bending,
mono subst.
C-H bending, osubst.
C-H bending, msubst.
C-H bending, psubst
690 - 710,
750-810
810 - 840
© E.V. Blackburn, 2011
Compound
type
carboxylic
acids
Frequency Intensity Vibration
(cm-1)
2500 - 3000 s, b O-H stretching
1700 - 1725
s
C=O stretching,
aliphatic
C=O stretching,
aromatic,
or unstaturated
C-O stretching
O-H bending
1680 - 1700
s
1250
1400, 920
b
esters
1715 - 1740
1050 - 1300
s
s
C=O stretching
C-O stretching, 2
bands
ethers
1070 - 1150
s
C-O stretching
ketones
1690 - 1750
s
C=O stretching
nitriles
2210 - 2260
v
CN stretching
© E.V. Blackburn, 2011
C 7 H8 O
© E.V. Blackburn, 2011
Examining a spectrum
• ~3400 cm-1: O-H and N-H stretch
• ~3100 cm-1: sp and sp2 C-H stretch
• ~2900 cm-1: sp3 C-H stretch
• ~2750 cm-1: aldehyde C-H stretch
• ~2200 cm-1: C≡C and C≡N stretch
• ~1700 cm-1: C=O stretch
• ~1600 cm-1: C=C stretch of alkenes and arenes
• ~1200 cm-1: C-O stretch (limited utility)
© E.V. Blackburn, 2011
Degree of unsaturation
Degree of unsaturation = (2NC - NX + NN – NH + 2)/2
NC = number of carbons
NX = number of halogens
NN = number of nitrogens
NH = number of hydrogens
NH 2
H
Cl
O
© E.V. Blackburn, 2011
C7H6O2
© E.V. Blackburn, 2011
C 7 H5 N
© E.V. Blackburn, 2011
C 7 H9 N
© E.V. Blackburn, 2011