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

H C R

H

H

H C R

H

R

R C R

H

R

R C R

R

R

primary secondary tertiary quaternary

Classification of atoms• H atoms are classified based on the type of carbon to

which they are attached

H C R

H

H

H C R

H

R

R C R

H

R

R C R

R

R

primary secondary tertiary There are no quaternary H's

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:

H C R

OH

H

H C R

OH

R

R C R

OH

R

primary secondary tertiary

CH3 CH CH2

OH

CH3CH3 CH2 CH2 CH2 OH

CH3HO

Classification of amines and amides• Amines and amides are classified based on the number of

C atoms bonded to the N

R N H

H

R N H

R

R N R

R

primaryamine

secondaryamine

tertiaryamine

C N H

H

C N R

H

C N R

R

primaryamide

secondaryamide

tertiaryamide

O

R

O

R

O

R

Classification of amines and amides• Classify these functional groups:

N

H NH2

C

O

N CH3

CH3

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 CH2 CH2 CH2 CH3

n-pentane, b.p. = 36°C

CH3 CH

CH3

CH2 CH3

isopentane, b.p. = 28°C

C

CH3

CH3

CH3

H3C

neopentane, b.p. = 10°C

Boiling Points and Intermolecular Forces

CH3 O CH3CH3 CH2 OH

dimethyl ether, b.p. = -25°C ethanol, b.p. = 78°C

N CH3H3C

CH3

CH3CH2CH2 N

H

HN CH3CH3CH2

H

ethanol, b.p. = 78°C ethyl amine, b.p. 17°C

trimethylamine, b.p. 3.5°C propylamine, b.p. 49°C ethylmethylamine, b.p. 37°C

CH3 CH2 OH CH3 CH2 NH2

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

• 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

Let’s consider carbon…

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-

• 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.

Formation of methane

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

+ 1 unhybridized p orbital• s + px + py + pz → 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 bond

• s bond and bond result in sharing four electrons (C=C)• Shorter and stronger than single bond in ethane

• 4 atomic orbitals → 2 equivalent hybrid orbitals

+ 2 unhybridized p orbitals• s + px + py + pz → 2 sp + 2 p

• Shape = linear (bond angle = 180º)• Remaining p orbitals are perpendicular on y-axis and z-axis

sp hybridization

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 bonds• s bond and two 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 3 , 1s p 2 , 2s p

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?

N

N

O

O O

O

H

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)

C N

H

HH

H

C N

H

HH

H

or

ResonanceHybrid• 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

C N

H

HH

H

C N

H

HH

H

or

C N

H

HH

H+ +

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

C N

H

HH

H

C N

H

HH

H

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 more1. 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?

C N

H

HH

H

C N

H

HH

H

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

or

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:

C

O

CH3CH3

C

O

CH3CH3

C

O

CH3CH3

A B C

C

O

CH3CH3

-

+

Patterns in Resonance StructuresO

C

C C

C

C C

C

Examples

O

CH2 CH CH CH3

CH2 CH CH CH3

ExamplesO

H

NH2

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.