Lewis Structures (II)
§7.1 Lewis structures for saturated compounds
(i) Write the molecular skeleton:
Chloromethane
Formic acid
H
H
C
O
Cl
H
C
O
H
H
(ii) Assume that all bonds are covalent.
* This assumption is not always accurate, but it works most of the time, especially for
organic compounds.
(iii) Count the available balance
electrons:
(iv) Make single bonds between
two bonded atoms and then
provide each atom with a
complement of eight
electrons:
Chloromethane
3H 1×3= 3
Methanol
4H 1×4= 4
1C 4×1= 4
1 Cl 7 × 1 = 7
14
1C 4×1= 4
1O 6×1= 6
14
Chloromethane
Methanol
H
H
C
H
H
Cl
H
C
O
H
H
(v) Count the electrons in the Lewis structure and compare the result with the number
derived from (iii). If the two numbers are the same, the Lewis structure is correct.
例1.
dimethyl ether
H
H
C
H
例2.
H
O
C
H
methylamine
H
H
H
C
H
N
H
H
例3.
methanethiol
H
H
C
S
H
H
例4.
methylal (or dimethoxymethane)
H
H
C
H
H
O
C
H
H
O
C
H
H
§7.2 Unsaturated groups
When the compound in question is unsaturated, step(v) shows that the number of
electrons in the trial structure is larger than the number of available valence electrons.
For example, ethylene
H
H
H
C
C
H
C
C
H
H
H
H
In such case the trial structure cannot be correct, and it must be modified by
introducing one or more multiple bonds.
(vi) Remove an unshared pair from each of the two adjacent atoms and add a second
bond between them. (Each such operation reduces the number of electrons in the trial
structure by two.)
H
H
H
H
C
C
C
H
H
H
例1.
H
H
C
C
C
H
H
H
formaldehyde
H
C
O
H
例2.
acetonitrile (We have to do the operation in step(vi) twice to make the number
of electrons in the trial structure equal to the number of available valence
electrons.)
H
H
C
C
N
H
例3. formic acid
The double bond could be placed between the carbon atom and either one of the two
oxygen atoms.
O
H
O
C
H
O
H
O
or
C
O
H
H
C
O
H
The second structure is less acceptable since the oxygen atoms have the unfamiliar
valences of one and three respectively.
例4.
acetyl chloride
H
H
C
O
C
Cl
H
Exercises
ex1. propyne
H
H
C
C
C
H
O
H
C
C
C
H
H
ex2. acetone
H
H
H
H
ex3. formamide
O
H
C
N
H
H
ex4. urea
O
H
N
H
C
N
H
H
Exceptions to the octet rule
It is not always possible for all molecules to arrange the electrons in pairs. This is
most obvious for molecules with an odd number of electrons. For example, ClO2 and
NO2 :
ClO2
NO2
Cl
O
N
O
O
bent, partial double bond,
and an angle of 117.5o.
O
bent, partial double bond,
and an angle of 134.25o.
Molecules which contain such unpaired electron are called radicals, usually
categorized reactive species. Following the octet rule, the radical center needs to
complete its valence shell by the acquisition of one more electron, for chemical
stability.
Lewis acids are species containing an atom which is short of a pair of electrons to
make an octet.
Aluminum trichloride
Boron trifluoride
Cl
Cl
F
F
Al
B
Cl
F
Although stable under the right conditions, these Lewis acids are vigorously reactive
toward compounds that have electron pairs to give. Foe example,
BF3
+
N(CH3)3
F
F
CH 3
B
N
F
CH 3
CH 3
Molecules with such lone pairs are called Lewis bases. The reactions in which Lewis
acids react to complete their valence shell are used in organic syntheses to yield
unstable intermediates such as the cations CH3+ and Cl+ which are not easily obtained
by other routes.
AlCl3 + Cl－CH3  AlCl4- + CH3+
AlCl3 + Cl－Cl  AlCl4- + Cl+
A particularly interesting example of the Lewis acid/base classification is the family
of carbenes CX2. These species are Lewis bases since they have a lone pair on carbon,
and also Lewis acids since the carbon atom only has six valence electrons.
Cl
C
Cl
Cl
C
Cl
Cl 2C
CCl 2
§7.3 Large molecules
Organic chemistry is largely the chemistry of functional groups. Often in writing
reaction mechanisms it is only necessary to write the Lewis structure of the functional
groups; for example,
CH 3
H3C
C
O
O
O
H
CH 3
C
,
H
N
CH 3
,
Most organic textbooks do not include the unshared pairs in functional groups; this is
really a unfortunate practice because much of the chemistry of a functional group is
determined by the presence (or absence) of unshared electron pairs.
In order to write Lewis structures of functional groups attached to any alkyl (or
aliphatic, R-) or aryl (or aromatic, Ar-) group, one allows each R- or Ar- attached to
the functional group to bring one electron into the union.
例1.
phenyl methyl ketone (acetophenone)
O
C
例2.
CH 3
benzyldimethylamine
CH 3
H2
C
N
CH 3
例3.
benzaldoxime
H
C
N
O
Exercises
H
ex1. furan
H
H
C
H
C
C
C
H
O
ex2. azobenzene
N
N
ex3. methyl benzimidate
H
N
C
O
CH 3
ex4. benzophenone phenylhydrazone
H
C
N
N
ex5. ethyl crotonate
H 3C
H
H
C
C
O
C
O
H2
C
CH 3
ex6. DMSO, dimethylsulfoxide, (CH3)2SO
§7.4
例1.
Formal charge
nitrobenzene
The nitrogen atom has a formal charge
of +1; the oxygen atom that is bonded to
N by a single bond has a formal charge
of -1.
O
N
O
例2.
pyridine N-oxide
N
例3.
The nitrogen atom has a formal charge
of +1; the oxygen atom has a formal
charge of -1.
O
benzenesulfonic acid
O
S
O
O
H
The sulfur atom has a formal charge of
+2; each oxygen atom that is bonded to
S by a single bond and has three
unshared electron pairs has a formal
charge of -1.
§7.5 Ions
A very number of organic reactions involve ions as intermediates. With very few
exceptions the ions encountered in organic chemistry will have a total charge of +1 or
-1. The way how ions arise in a reaction involves some sort of bond-breaking or
bond-making process.
Cations
例1. Consider the molecule of chloromethane is broken at the C-Cl bond so that the
two electrons constituting the C-Cl bond depart with Cl:
H3C
CH3+
Cl
+
Cl-
Organic chemists use curved arrows to indicate the movement of electrons, and this
process is called “pushing electrons”. Note that the base of the arrow begins at the
original location of the bonding pair of electrons, and that the head of the arrow points
to the destination of the electrons－the chlorine atom. When the head of the arrow has
two barbs, it denotes the movement of a pair of electrons.
If a neutral molecule is cleaved, an excess of electrons (negative charge) will result at
the head of the arrow, and a deficient of electrons (positive charge) will result at the
tail of the arrow. Two ions, one electron-poor and the other electron-rich, result from
the heterolytic (unsymmetrical) bond cleavage.
例2.
Consider the result when a proton, H+, becomes bonded to methanol by way of
one of the unshared electron pair on O:
H+
H3C
O
H
H
H3C
O
H
This time the curved arrow is used to signify bond making. A pair of unshared
electrons on O is pushed toward the region between the oxygen atom and the
hydrogen ion; it becomes an O-H covalent bond. The formal charge distribution on
the resulting structure is predictable from the arrow: electrons are pushed away from
O, leaving it with a positive charge; electrons are pushed toward the hydrogen ion,
neutralizing its erstwhile positive charge.
Anions
例1. Consider the result if the C-Li bind of methyl lithium were broken so that the
two electrons constituting the bond both remain with the carbon:
H
H3C
H
Li
C
+
Li+
H
Compare this example with those in (1)Cations; all are heterolytic cleavages. In this
example the electrons move toward C, generating a carbanion; conversely, in those
examples of (1) the electrons move away from C and produce a carbocation. The
original polarity of the bond influences the direction of heterolysis: the more
electronegative atom takes the electrons.
Free radicals
Covalent bonds can suffer hemolytic (symmetrical) bond breaking, yielding radicals
with a formal charge of zero.
例1.
When di-tert-butyl peroxide breaks homolytically,
CH 3
H3C
C
CH 3
CH 3
O
O
C
CH 3
CH 3
CH 3
2 H3C
C
O
CH 3
Note that the head of the arrow has only one barb signifying the movement of a single
electron, and two such arrows are required to denote hemolytic bond breaking.
Homolytic bond breaking usually requires quite large energies; many free radical
reactions are promoted photochemically. Homolytic cleavage reactions important to
organic chemistry often occur when the covalent bond in question is weak;
oxygen-oxygen (peroxide) bonds are a good example.
例2.
Carbon radicals are usually produced indirectly. An exception is the hemolytic
cleavage of azobis(isobutyronitrile):
CN
H3C
C
CH3
CN
N
N
C
CH3
CN
CH3
2 H3C
C
CH3
+
N
N