Chapter 2Structure and
Propertiesof Organic Molecules
Organic Chemistry, 8th EditionL. G. Wade, Jr.
Around 1923, Louis de Broglie
suggested that the properties of
electrons in atoms are better explained
by treating the electrons as waves
rather than particles.
Chapter 2
Wave Properties of Electrons
• Standing wave vibrates in fixed location.• Wave function, , mathematical
description of size, shape, orientation.• Amplitude may be positive or negative.• Node: amplitude is zero.
Wave Interactions• Linear Combination of Atomic Orbitals
on different atoms produce molecular orbitals (MOs)
on the same atom give hybrid orbitals.• Conservation of orbitals.• Waves that are in phase add together.
Amplitude increases.• Waves that are out of phase cancel out.
Bonding• When 2 atoms bond, their electrons shield the + charged nuclei from each other.• The interatomic distance where the forces of
attraction and repulsion are balanced is called the bond length.
• What forces keep the bond at the optimal length?
Bonding (another view)
Sigma () Bonding
bonds are the most common in o-chem.
bonds are cylindrically symmetrical.
Sigma () Bonding
• Electron density lies between the nuclei.
• A bond may be formed by s-s, p-p, s-p, or hybridized AO overlap.
• In o-chem, all single bonds are bonds!
Sigma Bonding MOs in the H2 Molecule
Sigma Anti-Bonding MOs in H2
Chapter 2
H2 MO Diagram
sigma bond →
Bond From p-p Overlap in Cl2
Constructive overlap along the sameaxis forms a sigma bond.
Chapter 2
Bond From s-p Overlap in HCl
Question: What is the predicted shape for the bonding MO and the antibonding MO of the HCl molecule?
Pi Bonds – from overlap of two p orbitals that are
perpendicular to the line connecting the 2 nuclei.
Pi () Bonds
• e- density centered above and below the line connecting the 2 nuclei.
• A double bond = 1 + 1 bond• A triple bond = 1 + 2 bonds• There must always be a bond before
there can ever be a bond.
Multiple Bonds• A double bond (2 pairs of shared
electrons) consists of a sigma bond and a pi bond.
Two artist conceptions of the pi bond in ethylene.
Multiple Bonds• A triple bond (3 pairs of shared electrons)
consists of a sigma bond and two pi bonds.
Two artist conceptions of the pi bond in acetylene.
Molecular Geometry
• Bond angles in organic molecules cannot be explained with simple s and p orbitals.
• VSEPR theory explains bond angles: electrons repel each other so they get as far away from each other as possible.
• Hybridized atomic orbitals are lower in energy because electron pairs are farther apart.
• Hybridization is LCAO within one atom, just prior to bonding.
Molecular Geometry• VSEPR theory says electron pairs, both
bonding pairs and lone pairs, repel each other.
• To determine molecular geometry…Determine the Steric number
Molecular Geometry• From the Steric number… • Predict the hybridization of the central atom
• If the Steric number is 4, then it is sp3
• If the Steric number is 3, then it is sp2
• If the Steric number is 2, then it is sp
sp3 Hybrid Orbitals• Found in C-H, C-O, C-N, O-H and N-H bonds in O-
chem• 4 electron pairs, steric number = 4• Tetrahedral electron pair geometry• 109.5° bond angle
sp3 Example: Methane
sp3 Geometry• The molecular geometry can be different
from the electron group geometry!
sp2 Hybrid Orbitals• Found in C=C, C=O and C=N double bonds in o-chem• Steric number = 3• Trigonal planar geometry about sp2 atoms• 120° bond angle
sp2 Example: ethylene
H2C=CH2
Problem: Determine the steric number, hybridization, electron group geometry and molecular geometry for this imine.
Remember, Steric number = # of sigma bonds + # lone pairs.
sp Hybrid Orbitals• Found in CC and CN
triple bonds in O-chem• Steric number = 2• Linear electron pair
geometry• 180° bond angle
sp Example: Acetylene
HCΞCH
Problem: Determine the Steric number, the hybridization, the electron group geometry, and the molecular geometry for the following molecules.
• BeH2
• CO2
Predict the Steric number, hybridization, electron group geometry, and bond angle for
each atom in the following molecules:
Caution! You must start with a good Lewis structure!
• NH2NH2
• CH3-CC-CHO
CH3 C
OCH2
_
Prob 2-3. a) 2-butene, CH3CH=CHCH3
b) CH3CH=NH
Prob. 2-4. CH3-CN
• Single bonds rotate freely.
• Double bonds cannot rotate unless the bond is broken.
Rotation Around C-C Bonds
Isomers – same molecular formula, but different arrangement
of atoms
• Constitutional (or structural) isomers differ in their bonding sequence.
• Stereoisomers differ only in the arrangement of the atoms in space.
Constitutional or Structural Isomers
CH3 O CH3 and CH3 CH2 OH
CH3
CH3
and
Stereoisomers
C CBr
CH3
Br
H3CC C
CH3
Br
Br
H3Cand
cis - same side trans - across
Cis-trans isomers are also called geometric isomers.There must be two different groups on the sp2 carbon.
C CH3C
H H
HNo cis-trans isomers possible
Prob 2-10. Which of the following show
cis-trans isomerism?
CH3
H CH3
CH3CH3
CH3
H
a. CHF=CHF b. F2C=CH2
f.d. e.
Prob 2-11. Determine which are a) the same compound, b) structural isomers, c) cis-trans isomers, or
d) not the same compound.
H
HBr
Br Br H
H Br
Cl
HHCl
H
H ClCl
H
HH
H
H
HBr
Br Br H
Br H
CH3
CH3
CH3
CH3
d. c.
e. j.
Good test question!
Bond Dipole Moments• are due to differences in
electronegativity.• depend on the amount of charge and
distance of separation.• In debyes, x (electron charge) x d(angstroms)
Molecular Dipole Moments• Depend on bond polarity and bond
angles. • Vector sum of the bond dipole moments.
Chapter 2
Effect of Lone PairsLone pairs of electrons contribute to the dipole moment.
Intermolecular Forces• Strength of attractions between molecules
influence physical properties: m.p., b.p., and solubility, especially for solids and liquids.
• Classification depends on structure.Dipole-dipole interactions – polar moleculesLondon dispersion forces – nonpolar moleculesHydrogen bonding – molecules containing O-H or N-H bonds
Dipole-Dipole Forces• Between polar molecules.• Positive end of one molecule aligns with
negative end of another molecule.• Lower energy than repulsions, so net
force is attractive.• Larger dipoles cause higher boiling
points and higher heats of vaporization.
Chapter 2
Dipole-Dipole Forces
Why do isobutylene and acetone have such different
MP and BPs?
London Dispersion Force• Between nonpolar molecules.• Temporary dipole-dipole interactions.• Larger atoms are more polarizable.• Branching lowers b.p. because of
decreased surface contact between molecules.
Chapter 2
London Dispersion Forces
Hydrogen Bonding• Strong dipole-dipole attraction.• Organic molecule must have N-H or O-H.• The hydrogen from one molecule is
strongly attracted to a lone pair of electrons on the other molecule.
• O-H more polar than N-H, so stronger hydrogen bonding.
H-Bonding Examples
Boiling Points and Intermolecular Forces
CH3 CH2 OHethanol, b.p. = 78°C
CH3 O CH3
dimethyl ether, b.p. = -25°C
trimethylamine, b.p. 3.5°C
N CH3H3C
CH3
propylamine, b.p. 49°C
CH3CH2CH2 N
H
H
ethylmethylamine, b.p. 37°C
N CH3CH3CH2
H
CH3 CH2 OH CH3 CH2 NH2
ethanol, b.p. = 78° C ethyl amine, b.p. = 17 ° C
Prob 2-17. For each pair of compounds, say which you expect to have the highest
bp and why.
b) CH3(CH2)6CH3 or CH3(CH2)5CH2OH
d) HOCH2(CH2)4CH2OH or (CH3)3CCH(OH)CH3
e) (CH3CH2CH2)2NH or (CH3CH2)3N
Polarity Effects on Solubility
• “Like dissolves like.”Polar solutes dissolve in polar solvents.Nonpolar solutes dissolve in nonpolar solvents.
• Ionic solutes dissolve in _?_ solvents.• Molecules with similar intermolecular
forces will mix freely.
Ionic Solutes Dissolve in
Polar Solvents
Hydration releases energy.Entropy increases.
Ionic Solutes Do Not Dissolve in
Nonpolar Solvents
Nonpolar Solutes Dissolve in
Nonpolar Solvents
Nonpolar Solutes Do Not Dissolve in Polar
Solvents
Classes of Organic Compounds
• Classification based on functional group.• Three broad classes
I. Hydrocarbons – just carbon & hydrogenII. Compounds containing oxygenIII. Compounds containing nitrogen
Hydrocarbons
• Alkane: single bonds, sp3 carbonsCycloalkane: carbons form a ring
• Alkene: double bond, sp2 carbonsCycloalkene: double bond in ring
• Alkyne: triple bond, sp carbons• Aromatic: contains a benzene ring
Organic Compounds Containing Oxygen
• Alcohol: R-OH• Ether: R-O-R'• Aldehyde: RCHO
• Ketone: RCOR’• Carboxylic Acids & Derivatives
CH3CH2 C
O
H
CH3 C
O
CH3
Carboxylic Acids and Their Derivatives
• Carboxylic Acid: RCOOH• Acid Chloride: RCOCl• Ester: RCOOR'• Amide: RCONH2
C
O
OH
C
O
Cl
C
O
OCH3C
O
NH2
Organic Compounds Containing Nitrogen
• Amines: RNH2, RNHR', or R3N• Amides: RCONH2, RCONHR, RCONR2
• Nitriles: RCN
N
O
CH3CH3 C N
Identify the Functional Groups
Vitamin E
cocaine
hydrocortisone
penicillin
Good test question!
OLD EXAM QUESTION
O
NH2
HO
C
HO
O
HO
C CH3
O
C
O
O CH3
Consider the molecule above. Identify the indicated functional groups.How many of the carbons in the molecule are sp3 hybridized? How many of the carbons in the molecule are sp2 hybridized? How many primary (1) carbon atoms in the molecule? How many of the oxygens are sp2 hybridized? How many non-bonded (lone) pairs of electrons in the entire molecule?
End of Chapter 2