Chapter 5

Introduction

• In general, organic chemical reactions are studied by looking at : • what occurs (what kinds of reactions occur ) • how it happens (how reactions occur )

I. Kinds of Organic Reactions

A.

Addition reactions

B.

Elimination reactions

C.

Substitution reactions

D.

Rearrangement reactions

A.

Addition reactions

•

Addition reactions

occur when two reactants combine to form a single new product with no atoms left over.

Two molecules combine

B .

Elimination reactions

•

Elimination reactions

occur when a single reactant splits into two products.

One molecule splits into two

C .

Substitution reactions

•

Substitution reactions

occur when two reactants exchange parts to give two new products.

Parts from two molecules exchange

D .

Rearrangement reactions

•

Rearrangement reactions

occur when a single reactant undergoes a reorganization of bonds and atoms to yield an isomeric product.

A molecule undergoes changes in the way its atoms are connected

Practice Problem

: Classify each of the following reactions as an addition, elimination, substitution, or rearrangement: a. CH 3 Br

+

KOH  CH 3 OH

+

KBr b. CH 3 CH 2 OH  H 2 C=CH 2

+

H 2 O c. H 2 C=CH 2

+

H 2  CH 3 CH 3

II. How Organic Reactions Occur: Mechanisms

A.

Reaction Mechanisms

B.

Radical Reactions

C.

Polar Reactions

A.

Reaction Mechanisms

•

A reaction mechanism

how a reaction occurs.

is a full description of • It describes what takes place at each stage of a chemical transformation • Reactions occur in defined steps that lead from reactant to product • A reaction mechanism describes these steps – the

types

of bonds that are broken and formed, – the

order

and

relative rates

formed, … in which they are broken and

Steps in Mechanisms

• The types of steps in a sequence are classified  A step involves either the breaking or formation covalent bond of a  Steps can occur individually other steps or in combination with • When several steps occur at the same time they are said to be

concerted

Types of Steps in Reaction Mechanisms

• The breaking of a covalent bond may be: • Homolytic • Heterolytic • The formation of a covalent bond may be: • Homogenic • Heterogenic • A functional group may undergo: • Oxidation • Reduction

Homolytic Breaking of a Covalent Bond

• It is a symmetrical  cleavage Each product gets one electron from the bond • Not common in organic chemistry

Heterolytic Breaking of a Covalent Bond

• It is an unsymmetrical  cleavage Both electrons from the bond that is broken become associated with one resulting fragment • A common pattern in reaction mechanisms

Homogenic Formation of a Bond

• It is a symmetrical formation • One electron comes from each fragment • No electronic charges are involved • Not common in organic chemistry

Heterogenic Formation of a Bond

• It is an unsymmetrical formation • One fragment supplies two electrons • One fragment supplies no electrons • Combination can involve electronic charges • Common in organic chemistry

Indicating Steps in Mechanisms

• Curved arrows indicate breaking and forming of bonds  Arrowheads with a ‘radical processes’ ) “half” head (“fish-hook”) indicate homolytic and homogenic steps (called  Arrowheads with a complete head indicate heterolytic and heterogenic steps (called ‘polar processes’ )

Radicals

• A radical (also known as “free radical”): – is a

neutral

chemical species – contains an

odd

number of electrons – has a

single, unpaired

electron in one of its orbitals – is abbreviated R – can be written as

R .

Radicals

• Alkyl groups are abbreviated “R” for radical Example:

Methyl

iodide = CH 3 I

Ethyl

iodide = CH 3 CH 2 I

Alkyl

iodides (in general) = R I • A “free radical” is an “R” group on its own: – CH 3 is a “free radical” or simply “radical” – It has a single unpaired electron, shown as:

CH 3 .

– Its valence shell is one electron short of being complete

B.

Radical Reactions

•

A radical reaction

involves symmetrical bond breaking and bond making • It is

not

as common as polar reactions • It involves species that have an

odd

electrons (i.e.

radicals

) number of • Radicals are

highly reactive

; they react to complete electron octet of valence shell

Radicals can complete a valence-shell octet via: • Radical Substitution Reaction and leaving a new radical : A radical can break a bond in another molecule and abstract an atom with an electron, giving substitution in the original molecule

Radicals can complete a valence-shell octet via: • Radical Addition Reaction : A radical can

add

to an alkene, taking one electron from the double bond and yielding a new radical

Steps in Radical Substitution: Three types of steps 1. Initiation – homolytic formation of two reactive species with unpaired electrons 2. Propagation – reaction with molecule to generate radical 3. Termination – combination of two radicals to form a stable product

An Example of a Radical Substitution Chlorination of methane is • an example of a radical substitution.

• a multistep process.

1. Initiation – homolytic formation of two reactive species with unpaired electrons •

Example

– formation of two reactive Cl

.

from Cl 2 and U.V light radicals The weak Cl-Cl bond is homolytically broken by irradiation with U.V. light

2. Propagation – reaction with molecule to generate radical (chain reaction) •

Example

– reaction of Cl

.

radical with methane to give HCl and CH 3

.

– reaction of CH 3

.

with Cl 2 and Cl

.

radical to give CH 3 Cl

3. Termination – combination of two radicals to form a stable product. •

Example

– CH 3

.

+ CH 3

.

 CH 3 CH 3

Practice Problem

: Alkane chlorination is not a generally useful reaction because most alkanes have hydrogens in many different positions, causing mixtures of chlorinated products to result. Draw and name all monochloro substitution products you might obtain by reaction of 2-methylpentane with Cl 2

Practice Problem

: Radical chlorination of pentane is a poor way to prepare 1-chloropentane, CH 3 CH 2 CH 2 CH 2 CH 2 Cl, but radical chlorination of neopentane, (CH 3 ) 4 C, is a good way to prepare neopentyl chloride, (CH 3 ) 3 CCH 2 Cl. Explain.

C.

Polar Reactions

•

A polar reaction

involves unsymmetrical bond breaking and bond making • It is

more

common than radical reactions • It involves species that have an

even

number of electrons (have

only electron pairs

in their orbitals) • It occurs because of the

attraction

between positive and negative charges on different functional groups

How Polar Reactions occur • Molecules can contain local unsymmetrical electron distributions (polar bonds) – This is due to differences in electronegativities the bonded atoms of – This causes a partial negative charge on an atom and a compensating partial positive charge on an adjacent atom – The more electronegative atom has the greater electron density

Electronegativity of Some Common Elements • The relative electronegativity is indicated • Higher numbers indicate greater electronegativity • Carbon bonded to a more electronegative element has a partial positive charge (  +)

Electrostatic potential maps • Carbon bonded to a more electronegative element has a partial positive charge (  +) • Carbon bonded to a less electronegative element has a partial negative charge (  -) D EN = 3.0-2.5 = 0.5

D EN = 2.5-1.0 = 1.5

Polar bonds can also result from the interaction of functional groups with solvents and with Lewis acids or bases

Example

– The electron-poor character of the carbon atom in methanol is greatly enhanced by protonation of the oxygen atom with an acid

Polarizability • Polarization is a change in electron distribution as a response to change in electronic nature of the surroundings • Polarizability is the tendency to undergo polarization • Polar reactions occur between regions of

high

electron density and regions of

low

electron density

• Larger atoms with more loosely held electrons are more polarizable than smaller atoms with tightly held electrons •

Example

– I is much more polarizable than F

Generalized Polar Reactions • An electrophile , an electron-poor species, combines with a nucleophile , an electron-rich species • The combination is indicated with a curved arrow from nucleophile to electrophile

An electrophile – – – – – is “electron-loving” is an electron-poor species can form a bond by accepting a pair of electrons may be either neutral or positively charged is a Lewis acid

A nucleophile – – – – – is “nucleus-loving” is an electron-rich species can form a bond by donating a pair of electrons may be either neutral or negatively charged is a Lewis base

• Some species can act as an electrophile or a nucleophile depending on the circumstances •

Example

– Water acts as a nucleophile when it donates a pair of electrons, and acts as an electrophile when it donates H +

Polar Reactions vs Radical Reactions

Practice Problem

: Which of the following species is likely to be an electrophile, and which a nucleophile? a. HCl b. CH 3 NH 2 c. CH 3 SH d. CH 3 CHO

Practice Problem

: An electrostatic potential map of boron trifluoride is shown. Is BF 3 likely to be an electrophile or a nucleophile? Draw a Lewis structure for BF 3 , and explain the result.

III. Polar Reactions

A.

An Example of a Polar Reaction

B.

Using Curved Arrows in Polar Reaction Mechanisms

A.

An Example of a Polar Reaction

• The addition reaction of an alkene, such ethylene with HBr, is a typical polar process.

• Alkanes are relatively inert – their valence electrons are tied up in strong s bonds – the bonding electrons are relatively they are sheltered in s inaccessible bonds between nuclei.

because • Alkenes are more reactive – their double bonds have a greater electron single bonds density than – the electrons in the p bond are accessible because they are located above or below to reactants the plane

Reaction between HBr and ethylene is a typical electrophile nucleophile combination of all polar reactions • The p bond is electron-rich, allowing it to function as a nucleophile • H-Br is electron deficient at the H since Br is much more electronegative, making HBr an electrophile • HBr adds to the p part of C-C double bond  H Br

Mechanism of Addition of HBr to Ethylene

• HBr (electrophile) is attacked by p electrons of ethylene (nucleophile) to form a carbocation intermediate and bromide ion • Bromide adds to the positive center of the

carbocation

, which is an electrophile, forming a C-Br s bond • The result is that ethylene and HBr combine to form bromoethane

• All polar reactions occur by combination of an electron-rich site of a nucleophile and an electron-deficient site of an electrophile

Practice Problem

: What product would you expect from reaction of cyclohexene with HBr? With HCl?

Practice Problem

: Reaction of HBr with 2-methylpropene yields 2-bromo-2-methylpropane. What is the structure of the carbocation formed during the reaction? Show the mechanism of the reaction. (CH 3 ) 2 C=CH 2

+

HBr  (CH 3 ) 3 C-Br

B.

Using Curved Arrows in Polar Reaction Mechanisms

• Curved arrows are a way to keep track of changes in bonding in polar reaction • The arrows track “electron movement” – Electrons always move in pairs – Charges change during the reaction • One curved arrow corresponds to one step reaction mechanism in a

Rules for Using Curved Arrows 1.

Electrons move

from

a nucleophilic source ( Nu: )

to

an electrophilic sink ( E ) • The arrow goes from the nucleophilic reaction site to the electrophilic reaction site

Rules for Using Curved Arrows 2.

The nucleophile can be either neutral or negatively charged

Rules for Using Curved Arrows 3.

The electrophile can be either neutral or positively charged

Rules for Using Curved Arrows 4.

The octet rule must be followed

Practice Problem

: Add curved arrows to the following polar reactions to indicate the flow of electrons in each:

Practice Problem

: Add curved arrows to the following polar reactions to indicate the flow of electrons in each: (b) CH 3 ..

..

: H | + H – C – Br  | H CH 3 ..

..

3 : ..

.. -

Practice Problem

: Add curved arrows to the following polar reactions to indicate the flow of electrons in each: (c) H 3 C Cl .. : : | C OCH 3  H 3 C | | C OCH 3 + : ..

.. -

Practice Problem

: Predict the products of the following polar reaction by interpreting the flow of electrons as indicated by the curved arrows:

IV. Describing a Reaction

A.

Equilibria, Rates and Energy Changes

B.

Bond Dissociation Energies

C.

Energy Diagrams and Transition States

D.

Intermediates

A.

Equilibria, Rates and Energy Changes

• Reactions can go either forward or backward to reach equilibrium

a

A +

b

B

c

C +

d

D

K

eq

=

[Products]

_________________

[Reactants] [C] c [D] d =

______________

[A] a [B] b •

K

eq is the equilibrium constant

Magnitudes of Equilibrium Constants

a

A +

b

B

c

C +

d

D • If

K

eq > 1 , then at equilibrium – [C] c [D] d > [A] a [B] b – – most of the material is present as If

K

eq = 10, then [C] c [D] d = 10x[A] a products [B] b • If

K

eq < 1 , then at equilibrium – [A] a [B] b > [C] c [D] d – most of the material is present as reactants – If

K

eq = 0.10, then [A] a [B] b = 10x [C] c [D] d

• If

K

eq is greater than 10 3 , the amount of reactant left over will be barely detectable (less than 0.1%)

Free Energy and Equilibrium • The ratio of products to reactants is controlled by their relative

Gibbs free energy

– This energy is released an equilibrium reaction on the favored side of – The change in Gibbs free energy between products and reactants is written as “ D

G

”

• If

K

eq > 1 , D G o is negative surrounding (

exergonic

; energy is reaction) released to the • If

K

eq < 1 , D G o is positive ; energy is the surrounding (

endergonic

absorbed reaction) from

Numeric Relationship of K eq and Free Energy Change • The relationship between free energy change ( D G º) and an equilibrium constant (

K

eq ) is: D

G º = -

R T

ln

K

eq

where  D

G º

is the standard free energy change at 1 atm pressure, 298 K and 1M concentration   

R T

= 8.315 J /(K x mol) = temperature in Kelvin

K

eq

is the equilibrium constant

Changes in Energy at Equilibrium • Free energy changes ( D G º) can be divided into: – a temperature-dependent entropy term ( T D S º ) – a temperature-independent enthalpy term ( D H º ) • Overall relationship: D

G º

= D

H º

-

T

D

S º

D

G º

= D

H º

-

T

D

S º

where  D

H º

is enthalpy change that is associated with heat given off (exothermic) or absorbed (endothermic)  D

S º

is entropy change that measures the change in the amount of disorder in the system 

T

= temperature in Kelvin

D

G º - Gibb’s free-energy change

D G o < 0 negative exergonic spontaneous D G o > 0 positive endergonic nonspontaneous Favorable

K

eq Unfavorable

K

eq D

H º - Enthalpy change

D H º < 0 negative D H º > 0 positive exothermic endothermic Bonds of products are stronger (more stable) Bonds of products are weaker (less stable) D

S º - Entropy change

D S º < 0 negative Disorder decreases D S º > 0 positive Disorder increases A + B  C A  B + C

Practice Problem

: Which reaction is more favored, one with D G o = -44 kJ/mol or one with D G o = +44 kJ/mol?

Practice Problem

: Which reaction is likely to be more exergonic, one with Keq = 1000 or one with Keq = 0.001?

Practice Problem

: What is the value of D G o at 298 K for reactions where Keq = 1000, Keq = 1, and Keq = 0.001? What is the value of Keq for reactions where D G o = -40 kJ/mol, D G o = 0 kJ/mol, and D G o = +40 kJ/mol?

B .

Bond Dissociation Energies

Bond dissociation energy (

D

): – measures heat change that occurs when a bond is broken by homolysis – is the amount of energy required to break a given bond to produce two radicals in the

gas

phase at 25 o C

• Bond dissociation energy (

D

) is mostly determined by the type of bond, independent of the molecule • The C-H bond in methane requires a net heat input of 105 kcal/mol to be broken at 25 ºC.

• Table 5.3

• Changes in bonds can be used to calculate net changes in heat or enthalpy change ( D H º ) • However, it does have some limitations (it provides no information on D S º, D G º, and rate of the reaction; it refers to molecules in the gas phase and thus it has no direct relevance to solution chemistry).

Calculation of an Energy Change from Bond Dissociation Energies • • When a bond is formed , heat is released ( negative D H When a bond is broken , heat is absorbed ( positive D H ) )

• Solvation is a phenomenon where solvent molecules can surround and interact with dissolved molecules – It weakens bonds and causes large deviations from the gas-phase value of D H o

Practice Problem

: Use the data in Table 5.3 to calculate D H o for the gas-phase radical substitution reaction Br 2 with methane. Is this reaction more exothermic or less exothermic than the corresponding reaction with Cl 2 ?

CH 4 + Br 2  CH 3 Br + HBr

Practice Problem

: Calculate D H o for the following reactions: (a) CH 3 CH 2 OCH 3 + HI  CH 3 CH 2 OH + CH 3 I (b) CH 3 Cl + NH 3  CH 3 NH 2 + HCl

C .

Energy Diagrams and Transition States

A reaction energy diagram – depicts graphically the energy changes that occur during a reaction – follows the reaction course from reactant through transition state to product

The transition state

‡

– is an activated complex occurring at the highest-energy point in a reaction step – is unstable and cannot be isolated

The activation energy ( D G ‡ ) – is the amount of energy needed by reactants to reach the transition state – determines the rate of a reaction at a given temperature: The higher the D G ‡ , the slower the reaction

• The addition reaction of ethylene with HBr involves 1. breaking of the p bond of ethylene and H-Br bond and formation of a new C-H bond 2. formation of a new C-Br bond

Step 1 Step 2

• Reaction energy diagram for Step 1 of of the addition reaction of ethylene with HBr

• The (conceptual) transition-state structure for the first step of the addition reaction of ethylene with HBr : – The p bond between carbons begins to break – The C–H bond begins to form – The H–Br bond begins to break

The Gibb’s free-energy change ( D G º) – is represented in the energy diagram as the difference in level between reactant and product  D G o < 0 ( negative ), reaction is exergonic D G o > 0 ( positive ), reaction is endergonic

The higher the activation energy ( D G ‡ ), the slower the rate of the reaction

A fast exergonic reaction A slow exergonic reaction A fast endergonic reaction A slow endergonic reaction

Practice Problem

: Which reaction is faster, one with D G ‡ kJ/mol or one with D G ‡ = +70 kJ/mol? = +45

D .

Reaction Intermediates

A reaction intermediate or intermediate – is a species formed during a multistep reaction – is neither the reactant nor the final product – lies at an energy minimum between steps on the reaction curve – is more stable than either of the two transition states that neighbor it – is formed briefly during the course of a reaction – cannot be isolated

Formation of a Carbocation Intermediate • HBr, a Lewis acid, adds to the p bond • This produces an intermediate with a positive charge on carbon - a carbocation • This is ready to react with bromide

Carbocation Intermediate Reactions with Anion • Bromide ion adds an electron pair to the carbocation • An alkyl halide is produced • The carbocation is a reactive intermediate

• A multistep reaction – occurs in more than one step – involves reaction intermediates • Each step has – – its own free activation energy ( D G ‡ ) its own Gibb’s free-energy change ( D G º)

• The complete energy diagram for a multistep reaction shows: – – – – – the different steps the transition states the reaction intermediates the free activation energy ( D G ‡ ) of each step the free energy changes ( D G º) associated with the initial reactants, intermediates, and final products

Complete Reaction Energy Diagram for Addition of HBr to Ethylene • Two separate steps, each with its own transition state • Energy minimum between the steps belongs to the carbocation reaction intermediate.

Biological Reactions • Reactions in living organisms follow reaction diagrams too – They take place in very controlled conditions – They are promoted by catalysts that lower the activation barrier • Enzymes are biological catalysts, usually proteins – Enzymes provide an alternative mechanism that is compatible with the conditions of life

A typical reaction energy diagram for a biological reaction • Enzymes lower the activation barrier and provide an alternative mechanism that proceeds through many small steps

Practice Problem

: Sketch a reaction energy diagram for a two step reaction with an endergonic first step and an exergonic second step. Label the parts of the diagram corresponding to reactant, product, and intermediate.

Practice Problem

: Sketch a reaction energy diagram that shows both propagation steps in the radical reaction of chlorine withy methane. Is the overall D G o for this reaction positive or negative? Label the parts of your diagram corresponding to D G o and D G ‡ . CH 4 + Cl 2 Light  CH 3 Cl + HCI

Chapter 5