Structure and Properties of Organic Molecules

Report
Structure and Bonding of
Organic Molecules
Orbital Theory
Hybridization and Geometry
Polarity
Functional Groups
Orbitals are Probabilities
Bonding in H2
The Sigma (s) Bond
Wave Depiction of H2
Molecular Orbitals
Mathematical Combination of Atomic Orbitals
Antibonding Molecular Orbital
Destructive Overlap Creates Node
Sigma Bonding
• Electron density lies between the nuclei.
• A bond may be formed by s—s, p—p, s—p, or
hybridized orbital overlaps.
• The bonding molecular orbital (MO) is lower
in energy than the original atomic orbitals.
• The antibonding MO is higher in energy than
the atomic orbitals.
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s and s* of H2
Electron Configurations
Ground State Electron Configurations
px, py, pz
Electron Configuration of Carbon
For CARBON, In the Ground State
2 bonding sites, 1 lone pair
H
CH
4
2p
C
H
H
H
2s
x
y
ground state
z
sp3 Hybridization
4 Regions of electron Density
link
H
CH
4
2p
C
H
2s
H
H
x
y
z
ground state
hybridize
2p
2s
x
y
excited state
z
4 identical sp
3
orbitals
4 sigma bonds requires 4 hybrid orbitals
tetrahedral geometry
4
sp 3
Hybridization of 1 s and 3 p Orbitals
gives 4 sp3 Orbitals
sp3 is Tetrahedral Geometry
Methane
Tetrahderal Geometry
Methane Representations
Ammonia
Tetrahedral Geometry
Pyramidal Shape
All Have the Same Geometry
All Have 4 Regions of Electron Density
All are sp3 Hybridized
Orbital Depiction of Ethane, C2H6 ,
the s bond
Rotation of Single Bonds
• Ethane is composed of two methyl groups bonded by the
overlap of their sp3 hybrid orbitals.
• There is free rotation along single bonds.
© 2013 Pearson Education, Inc.
Chapter 2
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A Model of a Saturated Hydrocarbon
sp2 Hybridization
3 Regions of Electron Density
CH 2 =CH
H
2
2p
H
C
2s
C
H
x
H
y
z
ground state
hybridize
2p
2s
x
y
excited state
z
pz
3 identical sp
2
orbitals
3 sigma bonds requires 3 hybrid orbitals
trigonal planar geometry
3
sp 2 p z
Hybridization of 1 s and 2 p Orbitals
– sp2
An sp2 Hybridized Atom
The p bond
Overlap of 2 parallel p Orbitals
Ethylene CH2=CH2
Multiple Bonds
• A double bond (two pairs
of shared electrons)
consists of a sigma bond
and a pi bond.
• A triple bond (three pairs
of shared electrons)
consists of a sigma bond
and two pi bonds.
© 2013 Pearson Education, Inc.
Chapter 2
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Views of Ethylene, C2H4
Rotation Around Double Bonds?
• Double bonds cannot rotate.
• Compounds that differ in how their substituents are arranged
around the double bond can be isolated and separated.
© 2013 Pearson Education, Inc.
Chapter 2
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Isomerism
• Molecules that have the same molecular
formula but differ in the arrangement of
their atoms are called isomers.
• Constitutional (or structural) isomers differ
in their bonding sequence.
• Stereoisomers differ only in the
arrangement of the atoms in space.
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Constitutional Isomers
• Constitutional isomers have the same chemical formula,
but the atoms are connected in a different order.
• Constitutional isomers have different properties.
• The number of isomers increases rapidly as the number of
carbon atoms increases.
© 2013 Pearson Education, Inc.
Chapter 2
35
Geometric Isomers: Cis and Trans
• Stereoisomers are compounds with the atoms bonded in the same
order, but their atoms have different orientations in space.
• Cis and trans are examples of geometric stereoisomers; they occur
when there is a double bond in the compound.
• Since there is no free rotation along the carbon–carbon double bond,
the groups on these carbons can point to different places in space.
© 2013 Pearson Education, Inc.
Chapter 2
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Formaldehyde
sp Hybridization
2 Regions of Electron Density
H
C
C
2p
H
x
2s
y
z
ground state
hybridize
2p
2s
x
y
excited state
z
py
pz
2 identical sp orbitals
2 sigma bonds requires 2 hybrid orbitals
linear geometry
2
The sp Orbital
Acetylene, C2H2,
1 s bond
2 perpendicular p bonds
Molecular Shapes
• Bond angles cannot be explained with simple s and p
orbitals.
• Valence-shell electron-pair repulsion theory (VSEPR) is
used to explain the molecular shape of molecules.
• Hybridized orbitals are lower in energy because electron
pairs are farther apart.
© 2013 Pearson Education, Inc.
Chapter 2
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Solved Problem 2
Borane (BH3) is not stable under normal conditions, but it has been detected at low
pressure.
(a) Draw the Lewis structure for borane. (b) Draw a diagram of the bonding in this
molecule, and label the hybridization of each orbital. (c) Predict the H–B–H bond
angle.
Solution
There are only six valence electrons in borane. Boron has a single bond to each of the three hydrogen
atoms.
The best bonding orbitals are those that
provide the greatest electron density in the
bonding region while keeping the three
pairs of bonding electrons as far apart as
possible. Hybridization of an s orbital with
two p orbitals gives three sp2 hybrid
orbitals directed 120° apart. Overlap of
these orbitals with the hydrogen 1s orbitals
gives a planar, trigonal molecule. (Note
that the small back lobes of the hybrid
orbitals have been omitted.)
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Summary of Hybridization and
Geometry
Hybrid Orbitals
(# of s bonds)
Hybridization
Geometry
Approximate
Bond Angle
2
s + p = sp
linear
180⁰
3
s + p + p = sp2
trigonal
120⁰
4
s + p + p + p = sp3
tetrahedral
109.5⁰
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Molecular Dipole Moment
• The molecular dipole moment is the vector sum
of the bond dipole moments.
• Depends on bond polarity and bond angles.
• Lone pairs of electrons contribute significantly to
the dipole moment.
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Chapter 2
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Lone Pairs and Dipole Moments
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Intermolecular Forces
• Strength of attractions between molecules
influences the melting point (m. p.), boiling
point (b. p.), and solubility of compounds.
• Classification of attractive forces:
– Dipole–dipole forces
– London dispersions forces
– Hydrogen bonding in molecules with —OH or —
NH groups
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Dipole–Dipole Interaction
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London Dispersions
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Effect of Branching on Boiling Point
• The long-chain isomer (n-pentane) has the greatest
surface area and therefore the highest boiling point.
• As the amount of chain branching increases, the molecule
becomes more spherical and its surface area decreases.
• The most highly branched isomer (neopentane) has the
smallest surface area and the lowest boiling point.
© 2013 Pearson Education, Inc.
Chapter 2
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Hydrogen Bonds
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Solved Problem 4
Rank the following compounds in order of increasing boiling points.
Explain the reasons for your chosen order.
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Solution
To predict relative boiling points, we should look for differences in (1) hydrogen
bonding, (2) molecular weight and surface area, and (3) dipole moments. Except
for neopentane, these compounds have similar molecular weights. Neopentane is
the lightest, and it is a compact spherical structure that minimizes van der Waals
attractions. Neopentane is the lowest-boiling compound.
Neither n-hexane nor 2,3-dimethylbutane is hydrogen bonded, so they
will be next higher in boiling points. Because 2,3-dimethylbutane is more highly
branched (and has a smaller surface area) than n-hexane, 2,3-dimethylbutane will
have a lower boiling point than n-hexane.
The two remaining compounds are both hydrogen bonded, and pentan-1-ol has more area for van der
Waals forces. Therefore, pentan-1-ol should be the highest-boiling compound. We predict the
following order:
neopentane < 2,3-dimethylbutane < n-hexane < 2-methylbutan-2-ol < pentan-1-ol
10 °C
58 °C
69 °C
102 °C
138 °C
The actual boiling points are given here to show that our prediction is correct.
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Polarity Effects on Solubility
• Like dissolves like.
• Polar solutes dissolve in polar solvents.
• Nonpolar solutes dissolve in nonpolar
solvents.
• Molecules with similar intermolecular
forces will mix freely.
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