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. CH3Br + KOH  CH3OH + KBr
b. CH3CH2OH  H2C=CH2 + H2O
c. H2C=CH2 + H2  CH3CH3
II. How Organic Reactions
Occur: Mechanisms
A.
Reaction Mechanisms
B.
Radical Reactions
C.
Polar Reactions
A.
Reaction Mechanisms
• A reaction mechanism is a full description of
how a reaction occurs.
• 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 in which they are broken and
formed, …
Steps in Mechanisms
• The types of steps in a sequence are classified
 A step involves either the breaking or formation of a
covalent bond
 Steps can occur individually or in combination with
other steps
• 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 “half” head
(“fish-hook”) indicate homolytic
and homogenic steps (called
‘radical processes’)
 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 = CH3I
Ethyl iodide = CH3CH2I
Alkyl iodides (in general) = RI
• A “free radical” is an “R” group on its own:
– CH3 is a “free radical” or simply “radical”
– It has a single unpaired electron, shown as: CH3.
– 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 number of
electrons (i.e. radicals)
• Radicals are highly reactive; they react to complete
electron octet of valence shell
Radicals can complete a valence-shell octet via:
• Radical Substitution Reaction: A radical can break a
bond in another molecule and abstract an atom with
an electron, giving substitution in the original molecule
and leaving a new radical
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. radicals
from Cl2 and U.V light
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 CH3.
– reaction of CH3. with Cl2 to give CH3Cl
and Cl. radical
3. Termination – combination of two radicals to form a
stable product.
•
Example – CH3. + CH3.  CH3CH3
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 Cl2
Practice Problem: Radical chlorination of pentane is a poor way to
prepare 1-chloropentane, CH3CH2CH2CH2CH2Cl,
but radical chlorination of neopentane, (CH3)4C,
is a good way to prepare neopentyl chloride,
(CH3)3CCH2Cl. 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 of
the bonded atoms
– 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 (-)
DEN = 3.0-2.5 = 0.5
DEN = 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. CH3NH2
c. CH3SH
d. CH3CHO
Practice Problem: An electrostatic potential map of boron
trifluoride is shown. Is BF3 likely to be an
electrophile or a nucleophile? Draw a Lewis
structure for BF3, 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 inaccessible because
they are sheltered in s bonds between nuclei.
• Alkenes are more reactive
– their double bonds have a greater electron density than
single bonds
– the electrons in the p bond are accessible to reactants
because they are located above or below the plane
Reaction between HBr and ethylene is a typical electrophilenucleophile 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.
(CH3)2C=CH2 + HBr  (CH3)3C-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 in a
reaction mechanism
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) CH3 – O
..
H
|
..
..
+ H – C – Br  CH3 – ..
O – CH3 + :Br:
..
|
H
Practice Problem: Add curved arrows to the following polar
reactions to indicate the flow of electrons in
each:
.. :O:
|
C
(c)
H3C
Cl
:O:
||
C

OCH3
H3C
+
OCH3
..
: ..
Cl -
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
cC + dD
aA + bB
[C]c [D]d
[Products]
Keq =
_________________
[Reactants]
=
______________
[A]a[B]b
• Keq is the equilibrium constant
Magnitudes of Equilibrium Constants
aA + bB
cC + dD
• If Keq > 1, then at equilibrium
– [C]c [D]d > [A]a[B]b
– most of the material is present as products
– If Keq = 10, then [C]c [D]d = 10x[A]a[B]b
• If Keq < 1, then at equilibrium
– [A]a[B]b > [C]c [D]d
– most of the material is present as reactants
– If Keq = 0.10, then [A]a[B]b = 10x [C]c [D]d
• If Keq is greater than 103, 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 on the favored side of
an equilibrium reaction
– The change in Gibbs free energy between
products and reactants is written as “DG”
• If Keq > 1, DGo is negative; energy is released to the
surrounding (exergonic reaction)
• If Keq < 1, DGo is positive; energy is absorbed from
the surrounding (endergonic reaction)
Numeric Relationship of Keq and Free Energy Change
• The relationship between free energy change
(DGº) and an equilibrium constant (Keq) is:
DGº = - RT ln Keq
where
 DGº is the standard free energy change at 1 atm
pressure, 298 K and 1M concentration
 R = 8.315 J /(K x mol)
 T = temperature in Kelvin
 Keq is the equilibrium constant
Changes in Energy at Equilibrium
• Free energy changes (DGº) can be divided
into:
– a temperature-dependent entropy term (TDSº)
– a temperature-independent enthalpy term (DHº)
• Overall relationship:
DGº = DHº - TDSº
DGº = DHº - TDSº
where
 DHº is enthalpy change that is associated with
heat given off (exothermic) or absorbed
(endothermic)
 DSº is entropy change that measures the
change in the amount of disorder in the
system
 T = temperature in Kelvin
DGº - Gibb’s free-energy change
DGo < 0 negative
exergonic
spontaneous
Favorable Keq
DGo > 0 positive
endergonic
nonspontaneous
Unfavorable Keq
DHº - Enthalpy change
DHº < 0 negative
exothermic
Bonds of products are stronger
(more stable)
DHº > 0 positive
endothermic
Bonds of products are weaker (less
stable)
DSº - Entropy change
DSº < 0 negative
Disorder decreases
A+BC
DSº > 0 positive
Disorder increases
AB+C
Practice Problem: Which reaction is more favored, one with
DGo = -44 kJ/mol or one with DGo = +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 DGo at 298 K for
reactions where Keq = 1000, Keq = 1, and
Keq = 0.001? What is the value of Keq for
reactions where DGo = -40 kJ/mol, DGo = 0
kJ/mol, and DGo = +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 25oC
• 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 (DHº)
• However, it does have some limitations (it provides no
information on DSº, DGº, 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 DH )
• When a bond is broken, heat is absorbed (positive DH )
• 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 DHo
Practice Problem: Use the data in Table 5.3 to calculate DHo for
the gas-phase radical substitution reaction Br2
with methane. Is this reaction more
exothermic or less exothermic than the
corresponding reaction with Cl2?
CH4 + Br2 
CH3Br + HBr
Practice Problem: Calculate DHo for the following reactions:
(a) CH3CH2OCH3 + HI  CH3CH2OH + CH3I
(b) CH3Cl + NH3  CH3NH2 + 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 (DG‡)
– 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 DG‡, 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 (DGº)
– is represented in the energy diagram as the difference in
level between reactant and product
 DGo < 0 (negative), reaction is exergonic
DGo > 0 (positive), reaction is endergonic
The higher the activation energy (DG‡), 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 DG‡ = +45
kJ/mol or one with DG‡ = +70 kJ/mol?
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 (DG‡)
– its own Gibb’s free-energy change (DGº)
• The complete energy diagram for a multistep
reaction shows:
–
–
–
–
–
the different steps
the transition states
the reaction intermediates
the free activation energy (DG‡) of each step
the free energy changes (DGº) 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 twostep 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 DGo
for this reaction positive or negative? Label
the parts of your diagram corresponding to
DGo and DG‡.
Light
CH4 + Cl2

CH3Cl + HCI
Chapter 5