4.3 UNSATURATION NUMBER 139
4.3 UNSATURATION NUMBER
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, C6H12, with hexane, C6H14.) 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 Ccarbon atoms is 2C+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:
U=
!
2C+
2
2-H
!
=number of rings +multiple bonds (4.5)
For example, cyclohexene, C6H10, 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—CH2CH2CH2CH3, has the same formula as
cyclohexene: C6H10.
How does the presence of other elements affect the unsaturation number calculation? You
can readily convince yourself from common examples (for instance, ethanol, C2H5OH) 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 Hrepresent 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, H3CLNH2, has 2C+3 hydrogens.) Therefore,
if Nis the number of nitrogens, the formula for the unsaturation number becomes
U=(4.7)
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) C3H4Cl4(b) C5H8N2
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 C20H34O2. Certain structural evidence suggests that
the compound contains two methyl groups and no carbon–carbon double bonds. Give one
2C+2C+2 +N-H
!!!
2
2C+2 -H
!
2
!
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