Download Developing Lewis structures for organic molecules 1) Draw the full

Developing Lewis structures for organic molecules
1) Draw the full structure of the molecules with the connectivity suggested by
the formula such that all valences are met.
2) Count the number of valance electrons in the entire molecule(equal to the
group number in the Periodic Table).
3) From the number in 2, subtract the bonding electrons (two times the no. of
lines in your structure). Any electrons that are unplaced are added to atoms
that lack a full octet (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.
*H needs only two atoms, B and Al need only six.
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.
CH3NH2
CH3OH
CH2O
H3CCN
H3CNO2
H2COH+
HCONH2
(CH3O)2CH+
H2CNN
H3CSOCH3
H2NCONH2
[CH2CHCH2]+
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
H2C
H
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
H
C
O
H
bond to form a lone pair:
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) When comparing resonance structures with the same number of bonds (as in part
i above) 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 cases where there are resonance structures with the same number of formal
charges, the favoured structure will be that in which the negative charge is on the more
electronegative atom or positive charge is 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 Properties
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
Cl
C
C
N
MgBr
C
H
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
Increasing e
nergy
We have a picture of where electrons are found in atoms: in atomic orbitals of various
types.
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:
H
H
protons
In orbital terms...
two overlapping 1s H
atomic orbitals
overlap constructively to
form a new orbital with
enhanced e density
between the two nuclei.
We can imagine the HF molecule arising from the following orbital overlap...
F
and H2O?
F
H
y
O
x
z
CH4
y
x
y
z
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 orbitals to form two bonds to the H atoms. Consider the
superimposition of the 2s and 2px orbitals on the Be atom...
Two possible combinations...
These are called
"hybrid" orbitals.
In this case they
are sp hybrids.
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
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.
2
3
4
Geometry
Hybridization
Unhybridized
Ethane
H
H
H
C
H
C
C
H
H
H
H
C
H
H
H
H
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
Resulting in two π bonds, mutually perpendicular.
C
C
C
H
Reactive Carbon-centered Intermediates
H3C
H3C
H3C
H2C
Steric
Number
Hybzn:
H
H
C
C
H
H
Carbocation
H
H
C
H
H
H
H
H
Carbanion
Radical
Carbene
Heteroatoms...
H
O
O
H
N
R
H
H
Halogens...
X
C
C
N
C
R
C
N
CH3OH
C
O
C
O
H
H2C=O
H
H
H
H
C
O
C
N
H
C
N
Hybridization and Resonance Structures
Frequently, different resonance structures predict different geometries about the same
atom...
O
O
C
H
N
CH3
C
H
N
CH3
CH3
N is sp3
C
N
C
N
CH3
N is sp2
O
C
N
H
Remember: resonance structures show how electrons are delocalized, electrons are
delocalized via π-bonds, π-bonds are formed by the overlap of atomic p orbitals.
Pyrrole
H
H
H
H
H
N
N
N
N
N
N
H
N
H
Molecular Orbital (MO) Theory - the bonding model for adults
We have looked at bonds as being the result of overlapping atomic orbitals
and/or hybrid atomic orbitals. MO theory is similar. We take all of the valence
atomic orbitals of the atoms in a given molecule and use them to derive a set of
new orbitals for a given molecule.
The principles:
• the number of molecular orbitals in a molecule must equal the number of valence
atomic orbitals of the constituent atoms
• like all orbitals, MOs can hold two electrons at most.
• MOs can be delocalized - where atomic orbitals are found around an atom,
molecular orbitals can be spread over the entire molecule
• MOs can be σ or π
• molecular orbitals can be bonding, non-bonding or anti-bonding
• a chart showing the relative energies of the MOs to each other and to the AOs
from which they are derived is known as an MO diagram.
Consider the H2 molecule: we have ...
In phase:
=
Out of phase:
=
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 for diatomics, 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.
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
=
Why are these bonding and antibonding orbitals so important, especially if the antibonding
orbitals are empty?
H2C
CH2
Energy
π∗2p
LUMO
2p
π2p
HOMO
The HOMO is the ...
The LUMO is the ...
For any molecule, the HOMO and LUMO are the "Frontier Molecular Orbitals".
When molecules react, their bonds are rearranged. In the MO model, that means that
MOs must change and the MOs that are involved are the FMOs because:
MO theory is very useful for examining how electron delocalization occurs in molecules.
ELECTRON DELOCALIZATION OCCURS THROUGH PI BONDS so 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...
π∗2p
π∗2p
πΝΒΜΟ2p
πΝΒΜΟ2p
2p
2p
π2p
π2p
Benzene
The benzylic system.
C
H
H
C
H
H
Intermolecular Forces
D, donor must be O or N.
A, acceptor must be O or N.
D
Hydrogen bond
A
H
O
Ion-ion
C
Na
CH3
O
H
Ion-dipole
O
Na
H
Dipole-dipole
O
H
+
C
H
Dispersion or London Forces
H3C
e
e
CH3
e
H3C
H3C
e
e
e
CH3
CH3
H3C
H3C
e
e
CH3
CH3
C
Formula
1
2
3
4
5
6
7
8
9
10
16
17
18
19
Name
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
B.P.
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane
Octane
Nonane
Decane
Hexadecane
Heptadecane
Octadecane
Nonadecane
M.P.
-162
-88.5
-42
0
36
69
98
126
151
174
287
302
317
330
18
22
29
32
CH3
CH
CH3
CH3
CH
CH3
CH3
1/r6
CH3
CH
CH
CH3
CH3
MW (g/mol)
compound
MP/BP
58
BP 0°C
60
BP 7°C
64.5
BP 12°C
60
BP 97°C
60
BP 116°C
60
MP 58°C
O
LiO2CCH3
The compounds:
H2
C
C
H2
H3C
H2
C
H3C
O
CH3
C
H
O
H2
C
CH3
H3C
CH3
H2
C
C
H2
OH
H3CCH2
Cl
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
H
O
Intermolecular
H
0.270
Intramolecular
0.263
O
O
O
H
H
O
O
H
O
H
N
N
0.288
0.304
O
H
H
N
0.310
N
H
N
O
N
N
NH2
H
N
N
H
O
N
NH
O
O
H
N
HN
N
HN
N
N
N
H
H
Adenine
Thymine
Tender Loving Chromatography
Silica gel: the solid phase adsorbent.
A dilute solution of a compound or mixture is spotted
onto a TLC plate with a capillary.
The plate is then placed upright in a tank containing a
small amound of some solvent or solvent mixture
(known as the eluant) such that only the very bottom
of the plate is in the solvent and that the spots are not
covered. Capillary action of the adsorbent draws the
eluant upward. When the eluant hits the spot, the
compounds are carried with it depending on the
relative strength of their interaction with the eluant or
the adsorbent -- the partitioning.
elution
S
OH
S
M
S
S
S
S
O
M
S
S
OH
OH
OH
OH
S
S
The compounds:
1
C
H2
H3C
H2
C
4
H3C
O
CH3
C
CH3
5
CH3
O
H
Hexane
(non-polar)
Effect of
Eluant
Polarity
2
LiO2CCH3
3
O
H2
C
H2
C
H3C
H2
C
C
H2
6
H3CCH2
OH
Toluene
(aromatic
non-polar)
Chloroform
(somewhat
polar)
Ethyl
Acetate
(polar)
1 2
1 2
1 2
Ethanol
(very polar)
1
CH2OH
Br
2
1 2
Cl
1 2
SM
Prod
RM
SM
Prod
RM
SM
Prod
AS or
DP
AS
DP
AS or
DP
AS
DP
Applications of TLC
RM