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