Transcript Sigma 180

STRUCTURE
Dr. Clower
CHEM 2411
Spring 2014
McMurry (8th ed.) sections 1.6-1.11, 2.10-2.11, 20.2, 2.4-2.6,
3.5-3.7, 4.3-4.9, 7.2, 7.6
Topics
• Structure
• Physical Properties
• Hybridization
• Resonance
• Acids and Bases
• Conformations of Alkanes and Cycloalkanes
• Unsaturation
• Alkene Stability
Structure
• Drawing organic structures
• Sigma (s) and pi (p) bonds
• Single bonds = 2e- = one sigma bond
• Double bonds = 4e- = one sigma bond and one pi bond
• Triple bonds = 6e- = one sigma bond and two pi bonds
• Which bond is shortest? Longest? Weakest? Strongest?
• Remember formal charges
Ionic Structures
• Be on the lookout for metals (cations) and ions
• Example: NaOCH3
• This is a Na+ cation and a CH3O- anion
• Example: NH4Cl
• This is a NH4+ cation and a Cl- anion
Classification of atoms
• C atoms can be classified as:
• Primary (1º) = C bonded to 1 other C
• Secondary (2º) = C bonded to 2 other C
• Tertiary (3º) = C bonded to 3 other C
• Quaternary (4º) = C bonded to 4 other C
Classification of atoms
• H atoms are classified based on the type of carbon to
which they are attached
Classification of alcohols
• Alcohols are classified based on the type of carbon to
which the -OH is bonded
• Classify these alcohols as 1º, 2º or 3º:
Classification of amines and amides
• Amines and amides are classified based on the number of
C atoms bonded to the N
Classification of amines and amides
• Classify these functional groups:
Electronegativity and Bond Polarity
• Electronegativity
• Ability of atom to attract shared electrons (in a covalent bond)
• Most electronegative atom = F
• Differences in electronegativity determine bond polarity
• Bond polarity
• How electrons are shared between nuclei
• Equal sharing of electrons = nonpolar; unequal = polar
Bond Polarity
• Example: C─O
• What atom is more electronegative (C or O)?
• More EN atom has partial negative charge (d-)
• Less EN atom has partial positive charge (d+)
• Arrow shows direction of polarity
• Nonpolar bonds
• Any atom with itself
• C─H
Molecular Dipole Moment
• Overall electron distribution within a molecule
• Depends on bond polarity and bond angles
• Vector sum of the bond dipole moments (consider both
magnitude and direction of individual bond dipole moments)
• Lone pairs of electrons contribute to the dipole moment
• Symmetrical molecules with polar bonds = nonpolar
Intermolecular Forces
• Strength of attractions between molecules
• Based on molecular polarity
• Influence physical properties (boiling point, solubility)
1. Dipole-dipole interactions
2. Hydrogen bonding
3. London dispersions (van der Waals)
1. Dipole-Dipole Interactions
• Between polar molecules
• Positive end of one molecule
aligns with negative end of
another molecule
• Lower energy than repulsions
• Larger dipoles cause higher
boiling points
2. Hydrogen Bonding
• Strongest dipole-dipole attraction
• H-bonded molecules have higher boiling points
• 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
3. London Dispersion Forces
• van der Waals forces
• Exist in all molecules
• Important with nonpolar compounds
• Temporary dipole-dipole interactions
• Molecules with more surface area have stronger
dispersion forces and higher boiling points
• Larger molecules
• Unbranched molecules
CH3
CH3
CH2
CH2
CH2
n-pentane, b.p. = 36°C
CH3
CH3 CH CH2 CH3
isopentane, b.p. = 28°C
CH3
H3C
C
CH3
CH3
neopentane, b.p. = 10°C
Boiling Points and Intermolecular Forces
CH3
CH2
OH
ethanol, b.p. = 78°C
CH3CH2
H3C N CH3
CH3
trimethylamine, b.p. 3.5°C
CH3
CH2
CH3
O CH3
dimethyl ether, b.p. = -25°C
CH3CH2CH2
N CH3
H
ethylmethylamine, b.p. 37°C
OH
ethanol, b.p. = 78°C
CH3
N H
H
propylamine, b.p. 49°C
CH2
NH2
ethyl amine, b.p. 17°C
Solubility and Intermolecular Forces
• Like dissolves like
• Polar solutes dissolve in polar solvents
• Nonpolar solutes dissolve in nonpolar solvents
• Molecules with similar intermolecular forces will mix freely
Example
• Which of the following from each pair will have the higher
boiling point?
(a)
CH3CH2CH2CH3
CH3CH2CH2OH
(b)
CH3CH2NHCH3
CH3CH2CH2NH2
(c)
CH3CH2CH2CH2CH3
CH3CH2CH(CH3)2
Example
• Will each of the following molecules be soluble in water?
(a)
CH3CO2H
(b)
CH3CH2CH3
(c)
CH3C(O)CH3
(d)
CH2=CHCH3
Structure of Organic Molecules
• Previously:
• Atomic/electronic structure
• Lewis structures
• Bonding
• Now:
• How do atoms form covalent bonds?
• Which orbitals are involved?
• What are the shapes of organic molecules?
• How do bonding and shape affect properties?
Linear Combination of Atomic Orbitals
• Bonds are formed by the combination of atomic orbitals
containing valence electrons (bonding electrons)
• Two theories:
• Molecular Orbital Theory
• Atomic orbitals of two atoms interact
• Bonding and antibonding MO’s formed
• Skip this stuff
• Valence Bond Theory (Hybridization)
• Atomic orbitals of the same atom interact
• Hybrid orbitals formed
• Bonds formed between hybrid orbitals
Let’s consider carbon…
• How many valence electrons? In which orbitals?
• So, both the 2s and 2p orbitals are used to form bonds
• How many bonds does carbon form?
• All four C-H bonds are the same
• i.e. there are not two types of bonds from the two different orbitals
• How do we explain this?
Hybridization
Hybridization
• The s and p orbitals of the C atom combine with each
other to form hybrid orbitals before they combine with
orbitals of another atom to form a covalent bond
• Three types we will consider:
• sp3
• sp2
• sp
sp3 hybridization
• 4 atomic orbitals → 4 equivalent hybrid orbitals
• s + px + py + pz → 4 sppp → 4 sp3
• Orbitals have two lobes (unsymmetrical)
• Orbitals arrange in space with larger lobes away from one
another (tetrahedral shape)
• Each hybrid orbital holds 2e-
Formation of methane
• The sp3 hybrid orbitals on C overlap with 1s orbitals on 4 H
atoms to form four identical C-H bonds
• Each C–H bond strength = 439 kJ/mol; length = 109 pm
• Each H–C–H bond angle is 109.5°, the tetrahedral angle.
Motivation for hybridization?
• Better orbital overlap with larger lobe of sp3 hybrid orbital
then with unhybridized p orbital
• Stronger bond
• Electron pairs farther apart in hybrid orbitals
• Lower energy
Another example: ethane
• C atoms bond by overlap of an sp3 orbital from each C
• Three sp3 orbitals on each C overlap with H 1s orbitals
• Form six C–H bonds
• All bond angles of ethane are tetrahedral
• Both methane and ethane have only single bonds
• Sigma (s) bonds
• Electron density centered between nuclei
• Most common type of bond
• Pi (p) bonds
• Electron density above and below nuclei
• Associated with multiple bonds
• Overlap between two p orbitals
• C atoms are sp2 or sp hybridized
Bond rotation
• Single (s) bonds freely rotate
• Multiple (p) bonds are rigid
sp2 hybridization
• 4 atomic orbitals → 3 equivalent hybrid orbitals
• s + px + py + pz
+ 1 unhybridized p orbital
→ 3 spp + 1 p = 3 sp2 + 1 p
• Shape = trigonal planar (bond angle = 120º)
• Remaining p orbital is perpendicular to hybrid orbitals
Formation of ethylene (C2H4)
• Two sp2-hybridized orbitals overlap to form a C─C s bond
• Two sp2 orbitals on each C overlap with H 1s orbitals (4 C ─ H)
• p orbitals overlap side-to-side to form a p bond
• s bond and p bond result in sharing four electrons (C=C)
• Shorter and stronger than single bond in ethane
sp hybridization
• 4 atomic orbitals → 2 equivalent hybrid orbitals
• s + px + py + pz
+ 2 unhybridized p orbitals
→ 2 sp + 2 p
• Shape = linear (bond angle = 180º)
• Remaining p orbitals are perpendicular on y-axis and z-axis
Formation of acetylene (C2H2)
• Two sp-hybridized orbitals overlap to form a s bond
• One sp orbital on each C overlap with H 1s orbitals (2 C─H)
• p orbitals overlap side-to-side to form two p bonds
• s bond and two p bonds result in sharing six electrons (C≡C)
• Shorter and stronger than double bond in ethylene
Summary of Hybridization
Hybridization of C
sp3
sp2
sp
Example
Methane, ethane
Ethylene
Acetylene
# Groups bonded to C
4
3
2
Geometry
Tetrahedral
Trigonal planar
Linear
Bond angles
~109.5
~120
~180
Types of bonds to C
4s
3s, 1p
2s, 2p
C-C bond length (pm)
154
134
120
C-C bond strength (kcal/mol)
90
174
231
Hybridization of Heteroatoms
• Same theory
• Look at number of e- groups to determine hybridization
• Each lone pair will occupy a hybrid orbital
• Ammonia:
• N’s orbitals (sppp) hybridize to form four sp3 orbitals
• One sp3 orbital is occupied by the lone pair
• Three sp3 orbitals form bonds to H
• H–N–H bond angle is 107.3°
• Water
• The oxygen atom is sp3-hybridized
• The H–O–H bond angle is 104.5°
Example
• Consider the structure of
thalidomide and answer
the following questions:
a) What is the hybridization of each oxygen atom?
b) What is the hybridization of each nitrogen atom?
c) How many sp-hybridized carbons are in the molecule?
d) How many sp2-hybridized carbons are in the molecule?
e) How many sp3-hybridized carbons are in the molecule?
f)
How many p bonds are in the molecule?
Example
• Consider the structure of 1-butene:
a) Predict each C─C─C bond angle in 1-butene.
b) Which carbon-carbon bond is shortest?
c) Draw an alkene that is a constitutional isomer of 1-butene.
Resonance
• Multiple Lewis structures for one molecule
• Differ only in arrangement of atoms
• Example: CH2NH2+ ion
• These are resonance structures/forms
• Valid Lewis structures (obey Octet Rule, etc.)
• Same number of electrons in each structure
• Atoms do not move
• Differ only in arrangement of electrons (lone pair and p electrons)
Resonance
Hybrid
• These structures imply that the C─N bond length and
formal charges are different
• Actually not true; these structures are imaginary
• Molecule is actually one single structure that combines all
resonance forms
• Resonance hybrid
• Contains characteristics of each resonance form
• More accurate and more stable than any single resonance form
• Lower energy (more stable) because of charge delocalization
Electron Movement
• Electrons move as pairs
• Can move from an atom to an adjacent bond, or from
bonds to adjacent atoms or bonds
• Use curved arrows to show e- motion (electron pushing)
• Start where electrons are, end where electrons are going
• Connect resonance forms with resonance arrow
• This is not an equilibrium arrow
Contribution to Hybrid Structure
• Resonance forms do not necessarily contribute equally to
the resonance hybrid
• They are not necessarily energetically equivalent
• More stable structures contribute more
1. Filled valence shells
2. More covalent bonds
3. Least separation of unlike charges (if applicable)
4. Negative charge on more EN atom (if applicable)
• Which of these is the major contributor to the resonance
hybrid?
Benzene
• Resonance structures:
• Curved arrows?
• Is one structure more stable (contribute more)?
• Resonance hybrid:
• All carbon-carbon bonds are the same length
• Somewhere between C─C and C=C
Acetone
• Resonance structures:
• Curved arrows?
• Which structure is the major contributor?
• Which is the minor contributor?
• Are any structures not likely to form?
• Resonance hybrid:
Patterns in Resonance Structures
Examples
CH2
CH
CH
CH3
CH2
CH
CH
CH3
O
Examples
O
NH2
H
H Br
Acids and Bases
• Two types in organic chemistry
1. Brønsted-Lowry
• Acid = proton (H+) donor
• Base = proton acceptor
• Some molecules can be both (e.g. water) = amphoteric
• Reaction will proceed from stronger acid/base to weaker acid/base
• Acid strength measured by pKa
• Stronger acid = lower pKa
Acids and Bases
• You can predict acid strength without a pKa value
• Strong acids have weak conjugate bases
• Weak conjugate bases are stable structures
• Have negative charge on EN atom (within a period)
• Have negative charge on a larger atom (within a group)
• Negative charge delocalized by resonance
Example
• Which is the stronger acid in each pair?
a) H2O or NH3?
b) HBr or HCl?
c) CH3OH or CH3CO2H?
Acids and Bases
2. Lewis
• Acid = electron pair acceptor, “electrophile”
• Base = electron pair donor, “nucleophile”
• Lewis acid react with Lewis base form a new covalent bond
Lewis Acids
• Incomplete octet (e.g. CR3+, BX3), or
• Polar bond to H (e.g. HCl), or
• Carbon with d+ due to polar bond (e.g. CH3Cl)
Lewis Bases
• Nonbonded electron pair (anything with O, N, anions)
Lewis Bases
• If there is more than one possible reaction site (more than
one atom with a lone pair), reaction occurs so that the
more stable product is formed.