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The concept of unsaturation number, a valuable source of structural information about organic compounds. It provides the formula to calculate the unsaturation number based on the molecular formula, including the presence of rings, double bonds, halogens, and nitrogen. Several examples are given to illustrate the calculation.
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An alkene with one double bond has two fewer hydrogens than the alkane with the same car-
bon skeleton. Likewise, a compound containing a ring also has two fewer hydrogens in its
molecular formula than the corresponding noncyclic compound. (Compare cyclohexane or
1-hexene, C
6
12
, with hexane, C
6
14
.) As both examples illustrate, the molecular formula of
an organic compound contains “built-in” information about the number of rings and double
(or triple) bonds.
The presence of rings or double bonds within a molecule is indicated by a quantity called
the unsaturation number, or degree of unsaturation, U. The unsaturation number of a mol-
ecule is equal to the total number of its rings and multiple bonds. The unsaturation number of
a hydrocarbon is readily calculated from its molecular formula as follows. The maximum
number of hydrogens possible in a hydrocarbon with C carbon atoms is 2 C + 2. Because each
ring or double bond reduces the number of hydrogens from this maximum by 2 , the unsatura-
tion number is equal to half the difference between the maximum number of hydrogens and
the actual number H :
!= number of rings + multiple bonds (4.5)
For example, cyclohexene, C
6
10
, has U = [2(6) + 2 - 10]! 2 = 2. Cyclohexene has two de-
grees of unsaturation: one ring and one double bond. A triple bond contributes two degrees of
unsaturation. For example, 1-hexyne, HC'C
2
2
2
3
, has the same formula as
cyclohexene: C
6
10
How does the presence of other elements affect the unsaturation number calculation? You
can readily convince yourself from common examples (for instance, ethanol, C
2
5
OH) that
Eq. 4.5 remains valid when oxygen is present in an organic compound. Halogens are counted
as if they were hydrogens, because halogens are monovalent, and each halogen reduces the
number of hydrogens by 1. Equation 4.5 remains unchanged as long as we let H represent the
total number of hydrogens and halogens:
U = ( H = hydrogens plus halogens) (4.6)
Another common element found in organic compounds is nitrogen. When nitrogen is present,
the number of hydrogens in a saturated compound increases by one for each nitrogen. (For ex-
ample, the saturated compound methylamine, H
3
2
, has 2 C + 3 hydrogens.) Therefore,
if N is the number of nitrogens, the formula for the unsaturation number becomes
Remember that the unsaturation number is a valuable source of structural information
about an unknown compound. This idea is illustrated in Problems 4.9 and 4.10.
PROBLEMS
4.7 Calculate the unsaturation number for each of the following compounds:
(a) C
3
H
4
Cl
4
(b) C
5
H
8
N
2
4.8 Without writing a formula or structure, give the unsaturation number of each of the following
compounds:
(a) 2,4,6-octatriene (b) methylcyclohexane
4.9 A compound has the molecular formula C
20
H
34
O
2
. Certain structural evidence suggests that
the compound contains two methyl groups and no carbon–carbon double bonds. Give one
structure consistent with these findings in which all rings are six-membered. (Many struc-
tures are possible.)
4.10 Which of the following cannot be correct formula(s) for an organic compound? Explain.
(a) C
10
H
20
N
3
(b) C
10
H
20
N
2
O
2
(c) C
10
H
27
N
3
O
2
(d) C
10
H
16
O
2
Except for their melting points and dipole moments, many alkenes differ little in their physi-
cal properties from the corresponding alkanes.
Like alkanes, alkenes are flammable, nonpolar compounds that are less dense than, and insol-
uble in, water. The alkenes of lower molecular weight alkenes are gases at room temperature.
The dipole moments of some alkenes, though small, are greater than those of the corre-
sponding alkanes.
How can we account for the dipole moments of alkenes? Remember that the electron density
in sp
2
orbitals lies closer to the nucleus than it does in sp
3
orbitals (Sec. 4.1A). In other words, an
sp
2
carbon has a greater attraction for electrons. This, in turn, means that an sp
2
carbon is some-
what more electronegative than an sp
3
carbon. As a result, any sp
2
3
carbon–carbon bond has
a small bond dipole (Sec. 1.2D) in which the sp
3
carbon is the positive end and the sp
2
carbon is
the negative end of the dipole.
The dipole moment of cis -2-butene is the vector sum of the H
3
CLC and HLC bond dipoles.
Although both types of bond dipole are probably oriented toward the alkene carbon, there is
good evidence (Problem 4.62, p. 176) that the polarization of the H
3
CLC bond is greater.
This is why cis -2-butene has a net dipole moment.
polarization of electron density
toward trigonal carbon H
3
sp
3
sp
2
3
3
m = 0.25 D
3
2
2
3
m = 0 D
3
2
4
3
boiling point
melting point
density
water solubility
dipole moment
63.4 °C
0.673 g mL
negligible
0.46 D
68.7 °C
0.660 g mL
negligible
0.085 D
2
2
3
3
1-hexene hexane