Download Developing Lewis structures 1) Add up all the valence electrons for

Developing Lewis structures
1) Add up all the valence electrons for all atoms. For ions, add one electron
for each negative charge; subtract one electron for each positive charge.
2) Connect the atoms using lines to represent bonding pairs of electrons. The
central atom is usually that wich is least electronegative. (In most organic
molecules, the carbons atoms make up the frame of the molecule.)
3) Any electrons that are unplaced are added to the outer atoms (usually
heteroatoms in organic molecules) as lone pairs.
4) Atoms lacking octets are completed by using lone pairs from adjacent atoms
to form multiple bonds. Where more than one such pair is available, then all
possibilities must be drawn and evaluated: these are resonance structures.
5) Formal charges on specific atoms are assigned using the formula: FC =
(no. valence electrons - no. bonds to that atom)-(no. lone pair electrons)
6) Atoms below the second row of the periodic table may have more than eight
electrons.
For most (neutral) organic compounds, Lewis structures can be determined simply by
taking the connectivity implied, remembering that there are 4 bonds to carbon, 3 to
nitrogen, 2 to oxygen and 1 to hydrogen and by filling octets.
2-aminobutane
CH2O
2-bromo-3-pentanone
H2CCH2
HCCH
HCONH2
H3CNNN
H2COH+
H3CCN
benzene
H3CNO2
H3CSOCH3
Rules for Drawing and Interpreting Resonance Structures
Resonance structures show how electrons are delocalized within a species. When
drawing and evaluating resonance structures, it is important to keep track of lone
pairs.
1) Only lone pair electrons and multiple-bond electrons (pi-electrons) move from
one resonance structure to another. Atoms remain in exactly the same place.
O
O
C
C
H2C
H
H2C
O
OH
C
C
H
H2C
H
H3C
H
2) From structure to structure, electron pairs move in the following ways:
lone pair to from an adjacent bond:
H
H
C
H
C
O
H
bond to form a lone pair:
H
H
O
O
S
H3C
O
S
CH3
H3C
CH3
bond to form a new bond:
H2C
H
H
C
C
CH2
H2C
CH2
3) Electrons move only to adjacent positions, but more than one pair of electrons can move
from one structure to another.
N
N
O
N
N
O
4) Always check to see what happens to the formal charges on specific atoms.
The net charge on each structure must be the same.
5) Structures that are identical in form are said to be degenerate. Such structures
contribute the same to the overall structure. Resonce "hybrids" are often drawn of
such structures.
H2C
H
H
C
C
CH2
H2C
H
=
C
H2C
CH2
CH2
6) When resonance structures are not degenerate, use the following rules to judge
which structures contribute more to the character of the species:
i) Resonance structures in which second period atoms (C-F) all have octets are
favoured over those with electon deficient atoms--resonance structures that have
more than 8 electrons on such atoms are not valid.
H
H
C
H
H
C
O
O
H
H
ii) Resonance structures with fewer formal charges contribute more to the character
of a species.
H
O
C
H
H
O
N
C
H
H
N
H
iii) In some cases there are resonance structures with the same number of formal
charges. In these cases, the favoured structure will be that in which the negative charge
is on the more electronegative/positive charge on the more electropositive element:
N
N
O
N
N
O
Invalid Resonance Structures
We do not show resonance structures that:
i) have more than eight electrons on a second row element
N
N
O
N
N
O
X
N
N
O
ii) have less than eight electrons on O or N (or similarly electronegative element)
O
C
CH2
O
C
CH2
H
H
iii) have more than two atoms bearing formal charges (excepting polynitros)
H
H
C
C
O
O
O
H
X
O
C
O
O
iv) have any atom with a formal charge greater than +/-1.
O
C
N
O
C
N
X
2O
C
N
HCO2
CH3NO2
H2CN2
HCONH2
Benzene
Bond Length and Bond Strength
Bond
Length
(nm)
Strength
(kJ/mol)
Bond
Length
(nm)
Strength
(kJ/mol)
C—C
0.153
347
C—N
0.147
305
C=C
0.131
611
C=N
0.128
615
C
0.118
837
C
0.114
891
C—H
0.109
414
C—F
0.140
485
C—O
0.143
360
C—Cl
0.179
339
C=O
0.121
740
C—Br
0.197
285
C—S
0.182
C—I
0.216
218
C
N
Polar Covalent Bonds
The periodic trend in Pauling's electronegativity...
O
O
C
C
O
C
O
H
C
H
O
C
N
H
C
O
C
C
N
H
Cl
C
MgBr
C
Li
VSEPR for organic chemists
...The steric number and counting to four.
BeH2
BF3
CH4
H2C=CH2
HCCH
H
CH3CH3
CH3OH
H2CO
H
C
H2C=CHCH3
C
PhNO2
C
H
NH
CH3CN
Bonding
We have a picture of where electrons are found in atoms: in atomic orbitals of various
types.
Increasing energy
The size and energy of orbitals depend on n
and on the atomic number, Z.
2s
2s
2p
This is a single 2p
orbital. It has two lobes
of different phase and a
node at the nucleus.
A reminder about wave addition...
+
2s
1s
Orbitals are wavefunctions and have wave properties:
These two 2s
orbitals are the
same in all
respects except
they have
different phase.
2p
=
-
... and about spatial components.
Any vector in 2D (or 3D) can be expressed as
the vector sum of two component vectors.
y
y
x
x
The vector lies in the xy plane because it has some "xness" and some "yness".
Chemical bonds...
... result from the sharing of electrons between atoms.
electrons
represented
in a Lewis
structure as:
protons
In orbital terms...
two overlapping 1s H
atomic orbitals
We can imagine the HF molecule arising from the following orbital overlap...
F
and H2O?
F
H
H
y
O
x
z
CH4
y
H
H
x
H
y
z
C
H
C
H
z
H
x
H
H
Orbital hybridization
BeH2
This is a linear molecule. Be has a 2s and three 2p orbitals. It needs to use
two of these to form bonds to the H atoms. Consider the superimposition of
the 2s and 2px orbitals on the Be atom...
Two possible combinations...
Note that the two hybrid orbitals share the spatial (along the x axis) and nodal
properties of the atomic orbitals from which they are hybridized.
H
Be
H
The remaining atomic 2p orbitals are still there but vacant.
We can consider each Be—H bond as resulting from the in phase overlap of a
sp hybrid orbital on Be with the 1s atomic orbital on H.
Be
H
BH3 - trigonal planar
We need three hybrid orbitals to form the B—F bonds.
Three sp2 hybrid orbitals can be formed.
They have spatial components in one
plane - that of the atomic p orbitals from
which they are hybridized.
B
Where is the unhybridized orbital?
H
H
B
H
CH4 - tetrahedral
C
C
4 sp3 hybrid orbitals.
The hybridization of any atom can be determined from its geometry:
Steric No.
Geometry
Hybridization
Unhybridized
2
Linear
sp
2 x 2p
3
Trigonal
Planar
sp2
2p
4
Tetrahedral
sp3
Ethane
H
H
H
C
H
C
C
H
H
H
H
C
H
H
H
H
The C—C and C—H bonds of ethane result from the overlap of atomic orbitals
that point directly to each other. The resulting bonding orbital has radial
symmetry. We call this type of bond a σ bond.
One important aspect of this bond's symmetry is that it is possible to rotate one end of
the molecule with respect to the other leading to different molecular conformations.
H
H
C
H
H
C
C
H
H
H
H
H
H
H
C
H
Bonding in Ethylene
H
H
C
12 valence electrons
C
H
H
H
C
C
C
C
H
H
C—C σ bond.
H
H
C
The σ bond framework of ethylene.
C
H
H
H
H
H
C
C
H
H
H
H
C
C
H
H
For this bond to exist, the atomic p orbitals must be parallel. One critically
important result of this is that, unlike σ bonds, free rotation about the bond is no
longer possible without breaking the π bond. This takes substantial energy.
H3C
C
H
C
C
H
H
H
H3C
CH3
H
H
H
C
C
CH3
H
CH3
H3C
C
C
H
H
H
H3C
CH3
H
Bonding in Acetylene
H
C C
H
10 valence electrons.
Each C atom is sp hybridized...
C
H
C
C
H
C
H
H
The σ bond framework.
Each C atom has TWO unhybridized atomic p orbitals...
H
C
C
H
H
C
C
C
H
Allene
H2C
C
CH2
Heteroatoms...
CH3OH
C
O
C
O
H
H2C=O
H
H
H
H
C
O
C
N
H
C
N
Molecular Orbital (MO) Theory
We have looked at bonds as being the result of overlapping atomic orbitals
and/or hybrid atomic orbitals. MO theory is similar. Valence atomic orbitals
are mixed to give a new set of molecular orbitals. For simple diatomic
molecules, MO Theory gives a simple picture. For molecules of more than
two atoms, things get more complicated. In organic chemistry, MO theory is
usually used to help us understand pi bonding and delocalized electron systems.
Let us consider the hydrogen molecule again...
In phase:
=
Problem: we started with two atomic orbitals (capacity four electrons).
We must end up with two molecular orbitals. How?
Out of phase:
=
We can combine other orbital types as well to give MOs of various shapes...
2p and hybrid orbitals:
Out of phase
In phase
The π and π* orbitals
Out of phase
=
In phase
=
Some combinations don't make sense
We construct a picture of the electronic structure of molecules using
correlation diagrams, which indicate the energy and type of molecular
orbitals that are occupied and vacant.
σ∗1s
E
1s
1s
σ1s
The AOs appear on the left (and right) side of the diagram at their corresponding
energy level and the MOs that result from the AOs are in the middle at their
corresponding energy level.
An appropriate no. of valence electrons are added to the diagram following Hund's Rule
and the Pauli exclusion principle to determine which orbitals are occupied.
The bond order between the two atoms is the sum of the no. of bonding electrons the sum of the antibonding electrons divided by two.
The second row diatomics (Li2 to Ne2) all involve combinations of the 2s and 2p atomic
orbitals giving a set of eight MOs.
σ∗2pz
π∗2px π∗2py
2p
2p
σ2pz
E
π2px
π2py
σ∗2s
2s
2s
σ2s
Even for molecules as simple as CH4, MO theory becomes rather complex. It is very
useful, however, for examining how electron delocalization occurs in molecules with
extensive π-bonding. In these cases we look exclusively at the π-MO diagrams.
H2C
CH2
π∗2p
=
2p
π2p
=
H
H2C
C
C
CH2
H
π∗2p
π∗2p
2p
π2p
π2p
The Allyl system
H
C
H2C
CH2
π∗2p
πΝΒΜΟ2p
2p
π2p
Isolobal systems...
Benzene
Reactive Intermediates
H
C
H
C
H
H
H
H
C
H
H
H
The benzylic system.
C
H
H
Intermolecular Forces
In order of decreasing strength...
Hydrogen bonds
Dipole-dipole
London dispersion or Van der Wäals interactions.
C
H
H
Hydrogen Bonding
Hydrogen bonds are relatively strong non-covalent interactions. There are two
components to a hydrogen bond: an O—H or N—H group (donor) and a lone
pair on an electronegative atom (acceptor).
Length (nm)
H
O
O
Intermolecular
H
0.270
O
O
H
O
Intramolecular
0.263
O
H
H
O
O
H
O
H
N
N
0.288
0.304
O
H
H
N
0.310
N
H
N
O
O
H
N
N
N
N
O
N
NH
N
NH2
O
H
N
N
H
HN
HN
N
N
H
H
Adenine
Thymine
Dipole-dipole attractions
O
H3C
+
C
CH3
London Forces
H3C
e
e
CH3
e
H3C
H3C
e
e
e
CH3
CH3
H3C
H3C ee
C
1
2
3
4
5
6
7
8
9
10
16
17
18
19
Formula
CH4
CH3CH3
CH3CH2CH3
CH3(CH2)2CH3
CH3(CH2)3CH3
CH3(CH2)4CH3
CH3(CH2)5CH3
CH3(CH2)6CH3
CH3(CH2)7CH3
CH3(CH2)8CH3
CH3(CH2)14CH3
CH3(CH2)15CH3
CH3(CH2)16CH3
CH3(CH2)17CH3
Name
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane
Octane
Nonane
Decane
Hexadecane
Heptadecane
Octadecane
Nonadecane
B.P.
M.P.
-162
-88.5
-42
0
36
69
98
126
151
174
287
302
317
330
18
22
29
32
e
e
CH3
CH3
CH3
CH
CH3
CH
CH3
CH3
CH3
CH3
CH
CH
CH3
CH3
Pentane
B.P. 36°
Isopentane
B.P. 30°
Neopentane
B.P.10°