Structure and Bonding of Organic Molecules

Orbital TheoryHybridization and Geometry

PolarityFunctional Groups

Orbitals are Probabilities

Bonding in H2

The Sigma ( ) s Bond

Wave Depiction of H2

Molecular OrbitalsMathematical 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.

Chapter 2 8

s and s* of H2

Electron Configurations

Ground State Electron Configurations

px, py, pz

Electron Configuration of Carbon

For CARBON, In the Ground State2 bonding sites, 1 lone pair

ground state

zyx2p

2sH

HH

H

CCH4

sp3 Hybridization 4 Regions of electron Density link

4

tetrahedral geometryexcited state

hybridize

4 identical sp3 orbitalszyx2p

2s4 sigma bonds requires 4 hybrid orbitals

CH4C

H

HH

H 2s

2p x y z

ground state

sp3

Hybridization of 1 s and 3 p Orbitals gives 4 sp3 Orbitals

sp3 is Tetrahedral GeometryMethane

Tetrahderal Geometry

Methane Representations

Ammonia Tetrahedral Geometry

Pyramidal Shape

All Have the Same GeometryAll Have 4 Regions of Electron Density

All are sp3 Hybridized

Orbital Depiction of Ethane, C2H6 , the s bond

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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.

Chapter 2 23

A Model of a Saturated Hydrocarbon

sp2 Hybridization3 Regions of Electron Density

pz

H

HC

H

H

CH2=CH2

C

3 sigma bonds requires 3 hybrid orbitals

2s

2p x y z

ground state

2s

2p x y z3 identical sp2 orbitals

hybridize

excited statetrigonal planar geometry

3sp2 pz

Hybridization of 1 s and 2 p Orbitals – sp2

An sp2 Hybridized Atom

The p bondOverlap of 2 parallel p Orbitals

Ethylene CH2=CH2

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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.

Chapter 2 31

Views of Ethylene, C2H4

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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.

Chapter 2 33

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.

Chapter 2 34

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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.

Chapter 2 35

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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.

Chapter 2 36

Formaldehyde

sp Hybridization2 Regions of Electron Density

py

H C HC

pz

2 sigma bonds requires 2 hybrid orbitals

2s

2p x y z

ground state

2s

2p x y z2 identical sp orbitals

hybridize

excited statelinear geometry

2

The sp Orbital

Acetylene, C2H2, 1 s bond

2 perpendicular p bonds

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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.

Chapter 2 41

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.

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.)

Solved Problem 2

Solution

Chapter 2 42

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

Chapter 2 43

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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.

Chapter 2 44

Lone Pairs and Dipole Moments

Chapter 2 45

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

Chapter 2 46

Dipole–Dipole Interaction

Chapter 2 47

London Dispersions

Chapter 2 48

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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.

Chapter 2 49

Hydrogen Bonds

Chapter 2 50

Rank the following compounds in order of increasing boiling points. Explain the reasons for your chosen order.

Solved Problem 4

Chapter 2 51

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.

Solution

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

The actual boiling points are given here to show that our prediction is correct.

10 °C 58 °C 69 °C 102 °C 138 °C

Chapter 2 52

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.

Chapter 2 53