Finding the HOMO and LUMO• Given the order of the energies of molecular

orbitals of organic compounds, we can make some simple generalizations that will help us locate quickly the HOMO and LUMO for a given molecule. What are those generalizations? What should we look for?

• Using those guidelines, find the HOMO and LUMO for the following species:

— σ* C–H — σ* C–C — σ* C–X

— π* C=C— π* C=X

— C nonbonding orbital (lone pair OR carbocation)

— X nonbonding orbital (lone pair)

— π C=C— π C=X

— σ C–H — σ C–C — σ C–X

Molecules with Several Possible Frontier Orbitals• Most organic molecules have several possible HOMO’s and several

possible LUMO’s. Consider the following molecules. Why can’t we simply specify a single HOMO or a single LUMO for these species? Identify all the possible HOMO’s and LUMO’s in these species.

Arrows: Non-bonding HOMO and Non-Bonding LUMO

• For the following species: 1) Identify the possible HOMO’s and LUMO’s2) Select the likely nucleophile and electrophile from these two species 3) Show how these species will react using curved arrows4) Predict the immediate product of that reaction

Arrows: Non-bonding HOMO and Antibonding LUMO

• For the following species: 1) Identify the possible HOMO’s and LUMO’s2) Select the likely nucleophile and electrophile from these two species 3) Show how these species will react using curved arrows4) Predict the immediate product of that reaction

Predicting Reactions Using Frontier Orbitals

• Here are two species you have never seen before. Using Frontier Molecular Orbital Theory (FMO Theory), predict the first step of the reaction between these two species. Draw the curved-arrow mechanism that shows how they react, and predict the immediate product of that reaction.

Functional Groups

Functional Groups

• Atoms or groups of atoms within a molecule that show a characteristic set of physical and chemical properties.

• Why are functional groups important?• Allow the division of organic compounds into classes• Exhibit characteristic chemical reactions• Basis for naming organic compounds

• A major topic for organic chemistry is to study the structure and properties of functional groups.

• A list of major functional groups is on the inside front cover.

Alcohols

• Contain an —OH (hydroxyl) group bonded to a carbon atom with four single bonds

• Classified as primary (1°), secondary (2°), or tertiary (3°) depending on the number of carbon atoms bonded to the carbon bearing the —OH group

Amines

• Contain an amino group, a nitrogen atom bonded to one, two, or three carbon atom(s) by single bonds

• Classified into primary (1°), secondary (2°), and tertiary (3°) amines

Aldehydes and Ketones

• Contain a carbonyl (C═O) group

Carboxylic Acids

• Contain a carboxyl (—COOH) group

Carboxylic Esters

• Derivatives of carboxylic acids in which the carboxyl hydrogen is replaced by a carbon-containing group

• Commonly known as esters

Carboxylic Amides

• Derivatives of carboxylic acids in which the —OH of the carboxyl group is replaced by an amine

• Commonly known as amides

Summary of the Introduction of Functional Groups

Name Structure Example IUPAC Name

Alcohol

Amine

Aldehyde

Ketone

Carboxylic Acid

Carboxylic Ester

Carboxylic Amide

Acids and Bases

Arrhenius Acids and Bases• In 1884, Svante Arrhenius proposed the following definitions for acids and

bases• Acid - Dissolves in water to produce H+ ions• Base - Dissolves in water to produce OH− ions• Original definition of an acid is slightly modified to take into account the

interactions of H+ with water molecules• Today, it is known that an H+ ion reacts immediately with a water

molecule to give a hydronium ion and hydration of the hydronium ion gives H5O2

+

Brønsted-Lowry Acids and Bases • One of the most useful theories of acidity and basicity is the Brønsted-Lowry

Theory. In this theory, acids and bases are defined as follows:

A Brønsted Acid is a species that can donate a proton (H+). A Brønsted Base is a species that can accept a proton.

• Every Brønsted acid, therefore, must have a conjugate base. What are some examples of Brønsted acids? For each one, identify its conjugate base.

Brønsted Acid Conjugate Base

• When a Brønsted acid reacts with a Brønsted base, a proton is transferred from the acid to the base. These reactions are called proton-transfer reactions. Give some examples of proton-transfer reactions using the above list of acids and bases.

Brønsted-Lowry Bases with Two or More Receptor Sites

Effect of Resonance

Pi Electrons as Brønsted-Lowry Bases

Strengths of Acids and Bases• Strength of an acid is expressed by an equilibrium constant

• Acid dissociation (ionization) of acetic acid is given by the following equation:

• Equilibrium expression for the dissociation of any uncharged acid can be written in a general form

3eq

–

2

+2 3HA

[H O ][A ][HA][H O]

+ H O A + H O

K+ -

=

Ka and pKa

• In this reaction, water is a solvent and the concentration of water can be treated as a constant equal to 1000 g/L, or approximately 55.6 mol/L

• The constants can be combined to define a new constant called an acid dissociation constant, or Ka

• Acid strengths are expressed as pKa (−log10Ka) since dissociation constants for most acids are numbers with negative exponents

• Larger the value of pKa, weaker the acid and vice versa• Weaker the acid, the stronger its conjugate base and vice versa

3a eq 2

[H O ][A ][H O][HA]

K K+ -

= =

Strengths of Acids and Bases: Ka and pKa• All proton-transfer reactions are reversible (at least in principle). In general,

though, one direction of the reaction will predominate over the other (that is, the proton transfer will tend to be unequal). For instance, consider the reaction between the ammonium ion and the hydroxide ion. Do you have a sense of whether this reaction proceeds mainly to the right or mainly to the left?

• Because so much chemistry takes place in water, we typically use water as a “standard reference” for the strengths of acids and bases. For any Brønsted acid HA, the reaction with water will reach some equilibrium:

The equilibrium constant of this reaction is Ka = * Just as “pH” means “the negative log of the H+ concentration,” we use the term “pKa” to refer to “the negative log of the Ka” How can we express that fact mathematically?

• Fill in the blanks: “A strong acid will have a ______________________ pKa” “A strong base will have a conjugate acid with a ____________________ pKa”

pKa Values for Common Acids

• Here are some important pKa values. You should memorize these values, at least to the nearest 5 pKa units:

• For each of the acids, fill in the conjugate base.

• We typically define a strong acid as an acid that is at least as strong as H3O+, and a strong base as a base that is at least as strong as OH–. Identify the strong acids and strong bases in the above list.

Table 4.1 - pKaValues for Some Organic and Inorganic Acids

Calculating Equilibrium Constants for Acid-Base Reactions

• Consider the following acid-base reaction:

• Equilibrium constant for this reaction

• Upon taking the negative logarithm of each side of this equation, the expression is

+

Acid Base Conjugate base Conjugate acid

of HA of B

HA + B A + BH

-

+

eq[A ][BH ][HA][B]

K-

=

+ ++ +3 3 HA

eq + +3 3 BH

[H O ] [A ][H O ][A ][BH ] [BH ][HA][B] [H O ] [HA] [B][H O ]

KKK

--

+= ´ = ´ =

+eq HA BHp = p – pK K K

Calculating Equilibrium Constants for Acid-Base Reactions

• Equilibrium constant for an acid-base reaction can be calculated using these steps if one knows the pKa values of each acid in the equilibrium:

• Look up the pKa values of the acid and conjugate acid involved in the reactants and the products, respectively, of an acid-base reaction

• Subtract the pKa of the conjugate acid (product) from the pKa of the acid (reactant)

• Because pKeq = −log10Keq, take the antilog of −pKeq to arrive at Keq

Using pKa to Predict Acid-Base Equilibria

• Consider, again, the reaction between the ammonium ion and the hydroxide ion. What will be the predominant species present in this system at equilibrium?

• Can you calculate the equilibrium constant for this reaction? How?

• What does the value of the equilibrium constant tell you about the concentrations of reactants and products at equilibrium?

Calculating Equilibrium Constants for Acid-Base Reactions

• Consider the following acid-base reaction:

• Equilibrium constant for this reaction

• Upon taking the negative logarithm of each side of this equation, the expression is

+

Acid Base Conjugate base Conjugate acid

of HA of B

HA + B A + BH

-

+

eq[A ][BH ][HA][B]

K-

=

+ ++ +3 3 HA

eq + +3 3 BH

[H O ] [A ][H O ][A ][BH ] [BH ][HA][B] [H O ] [HA] [B][H O ]

KKK

--

+= ´ = ´ =

+eq HA BHp = p – pK K K

Calculating Equilibrium Constants for Acid-Base Reactions

• Equilibrium constant for an acid-base reaction can be calculated using these steps if one knows the pKa values of each acid in the equilibrium:

• Look up the pKa values of the acid and conjugate acid involved in the reactants and the products, respectively, of an acid-base reaction

• Subtract the pKa of the conjugate acid (product) from the pKa of the acid (reactant)

• Because pKeq = −log10Keq, take the antilog of −pKeq to arrive at Keq

Position of Equilibrium in Acid-Base Reactions

• In an acid-base reaction, the position of equilibrium favors reaction of the stronger acid and stronger base to form the weaker acid and weaker base

• pKeq = 4.76 − 9.24 = −4.48• Keq = 3.0 ×104

• One can conclude that the equilibrium for the reaction between acetic acid and ammonia lies to the right