Chem e2a lecture 2-2011

Chem E-2a Lecture 2Chapter 3

Introduction to Alkenes

• Alkenes are unsaturated hydrocarbons that contain at least one carbon-carbon double bond.

General formula: CnH2n where n = 2, 3, 4…

β-Pinene

Lycopene

β-Carotene

CaryophylleneCampheneThujene

Terpene Natural Products

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Bonding in Alkenes

Reading: Section 3.2

• Construct an MO diagram for the C=C double bond in ethene, H2C=CH2.

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Energy

σC−C

σ∗C−C

C sp2 C sp2

πC=C

π∗C=C

C p C p

C CH

H

H

H

Energies of Molecular Orbitals• Many organic molecules contain carbon, hydrogen, and several more electronegative elements such as O, N. Cl, Br, etc. In general, the energies of the molecular orbitals in such a molecule will have the following pattern: (X represents an electronegative atom)

π C=C

π∗ C=C

σ∗ C–H

σ∗ C–C

σ∗ C–X

π∗ C=X

C nonbonding orbital(lone pair OR carbocation)

X nonbonding orbital(lone pair)

π C=X

σ C–H

σ C–C

σ C–X

Reading: Supplemental Handout, Section 2.33

Energy-Level Diagrams for Simple Molecules

Step 1 Draw a complete Lewis structure for the molecule,including all lone pairs

Step 2 Make a list of all the molecular orbitals in the molecule. Count the orbitals to make sure that you haven’t forgotten any! (How many molecular orbitals must there be?)

Step 3 Arrange the molecular orbitals in orderusing the general order on the previous page.

Step 4 Fill the molecular orbitals with the correct number of electrons.Step 5 Check your energy-level diagram:

- Does it have the correct number of orbitals?- Does it have the correct number of electrons?- Are all the bonding orbitals filled with electrons? (They should be!)- Are all the antibonding orbitals vacant? (They should be!)- Count the total number of filled bonding orbitals: is that equal to the number of bonds in the Lewis structure?

• Construct an approximate energy-level diagram for acetonitrile, CH3CN, using the following steps:


Introduction to Reactions of Alkenes: An Overview

• Addition of H–X to alkenes

• Hydration of alkenes

H X

HH

H

H

X

+

H OH

HH

H

H

OH

+H+ cat.

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What Happens When Two Molecules React?• Consider a general reaction between molecules A and B to yield products C and D:

A + B C + D


• Each of the reacting species (A and B) has many molecular orbitals, filled, and unfilled. What happens when these two molecules interact (when they come close together)?

• There are three basic types of interactions between the orbitals of A and the orbitals of B, depending on whether the orbitals are filled or unfilled. What can we say about these three types of interactions?

Molecule A Molecule B

filled−filled unfilled−unfilled filled−unfilled

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Frontier Orbitals: The Importance of the HOMO and LUMO

• It turns out that the interaction between orbitals that are close in energy is more important than the interaction between orbitals that are far apart in energy. Why is this the case?

• Because of the above observation, we can understand most of the reactivity of organic molecules by examining only a small number of orbitals, known as the frontier orbitals. These are:

HOMO (Highest Occupied Molecular Orbital)LUMO (Lowest Unoccupied Molecular Orbital)

Molecule A Molecule B


Identifying 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?

π C=C

π∗ C=C

σ∗ C–H

σ∗ C–C

σ∗ C–X

π∗ C=X

C nonbonding orbital(lone pair OR carbocation)

X nonbonding orbital(lone pair)

π C=X

σ C–H

σ C–C

σ C–X

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

Br

O


The Shapes of Frontier Orbitals• Draw “cartoon orbitals” to represent the shapes of the specified orbitals in the following molecules:

The HOMO of CH3OH

The HOMO of ethylene (H2C=CH2)

The LUMO of tert-butyl carbocation ((CH3)3C+)

The LUMO of methyl bromide (CH3Br)

The LUMO of formaldehyde (H2C=O)


Arrows: Non-Bonding HOMO + Non-Bonding LUMO

Reading: Sections 1.4, 3.17 Supplemental Handout, Section 2.6

• For the following reaction:1) Identify the possible HOMO’s and LUMO’s2) Select the likely nucleophile and electrophile from these two species3) Show how these species will react using curved arrows4) Predict the immediate product of that reaction

Cl +

CH3

H3C CH3

• Curved arrows:Must start on a pair of electrons (filled orbital -- HOMO).Must point at a place where electrons can go (vacant orbital -- LUMO).

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HO + C Br

H

HH


Arrows: Non-Bonding HOMO + Antibonding LUMO

• For the following reaction:1) Identify the possible HOMO’s and LUMO’s2) Select the likely nucleophile and electrophile from these two species3) Show how these species will react using curved arrows4) Predict the immediate product of that reaction

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Predicting Reactions Using Frontier Orbitals

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• Using Frontier Molecular Orbital Theory (FMO Theory), predict the reaction between these two species. Draw the curved-arrow mechanism that shows how they react and show the product that would result.

C CH +O

Brønsted-Lowry Acids and Bases

• 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


• 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.

• 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.

Strengths of Acids and Bases: Ka and pKa

• 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

• 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?

A strong acid will have a small (negative) pKa.

A strong base will have a conjugate acid with a large (positive) pKa.

pKa Values for Common Acids

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

Conjugate Acid pKa Conjugate Base

CH4

NH3

HC CH

H3C OH

H2O

NH4+

HCN

H3O+

HCl

H3COH

O

48

35

24

16

16

9

9

5

−2

−7


Using pKa Values to Predict Acid-Base Equilibria

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

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

• All proton–transfer reactions are reversible (at least in principle). In general, though, one direction of the reaction will prodominate over the other (that is, the proton transfer will tend to be unequal). Consider the reaction between the ammonium ion and the hydroxide ion. What will be the predominant species present in this system at equilibrium? Do you have a sense of whether this reaction proceeds mainly to the right or mainly to the left?

NH4+ + OH NH3 + H2O

Frontier Orbitals of Proton–Transfer Reactions

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• We can, of course, look at proton–transfer reactions in terms of donor and acceptor involved in the reaction. Can you find the donor and acceptor of the following proton–transfer reactions?

• What generalization can we make about the acceptor in any proton–transfer reaction?

H3C CH3

OH

+ H ClH3C CH3

OH2+ Cl

H3C CH3

OH

+H3C CH3

O+C CH3C C CHH3C


What’s Your Role: Acid or Electrophile?

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• In one of the following reactions, acetone plays the role of an acid; in the other it plays the role of an electrophile. Which is which, and why? Can you draw the curved arrows and identify the donor and acceptor of each reaction?

H3C CH3

O+ HO

H3C CH3

OHO

H3C CH3

O+ HO

H3C CH2

O+ H2O


What’s Your Role: Base or Nucleophile?

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• In one of the following reactions, acetone plays the role of a base; in the other it plays the role of a nucleophile. Which is which, and why? Can you draw the curved arrows and identify the donor and acceptor of each reaction?

H3C CH3

O+ H3O

H3C CH3

OH+ H2O

H3C CH3

O+

H3C CH3

O+

CH3

H3CO

CH3H3C

OCH3

CH3


Factors That Influence Acidity: The Main Atom

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• Examine the pKa’s for each of the following pairs of acids and explain why one acid is stronger than the other.

• Next examine the four acids from the upper-right corner of the periodic table. Do the strengths of these acids follow the trend you would expect? Why or why not?

H OH H F

+ 15.7 + 3.2

H OH H OH2

+ 15.7 − 1.7

H2C N

H

HH

H3C

+ 10

C N HH3C

− 10

H OH H F

+ 15.7 + 3.2

H SH H Cl

+ 7.0 − 7

Factors That Influence Acidity: The Adjacent Groups

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• The acidity of a particular proton can be influenced by adjacent or nearby groups in the molecule. Can you explain the difference in the following pKa’s?

• Whenever a molecule exhibits resonance, and the resonance allows charge to be delocalized, then the charged structure will be more stable than a comparable structure that does not have the delocalized charge. Thus, resonance can stabilize either the conjugate acid or the conjugate base, whichever is charged. Let’s look at some examples:

H3CO H

O

FH2CO H

O

+ 4.76

+ 2.66

H3CO H

O

+ 4.76

H3CO H

CH3

+ 16.5

N H

N H

+ 5.25

+ 9.2

NH3C

H3C

Chem e2a lecture 2-2011

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