Transcript Document

Chapter 6
Understanding Organic Reactions
Organic Chemistry, Second Edition
Janice Gorzynski Smith
University of Hawai’i
Prepared by Rabi Ann Musah
State University of New York at Albany
Copyright © The McGraw-Hill Companies, Inc.
Permission required for reproduction or display.
1
Writing Equations for Organic Reactions
• Equations for organic reactions are usually drawn with a
single reaction arrow () between the starting material and
product.
• The reagent, the chemical substance with which an organic
compound reacts, is sometimes drawn on the left side of the
equation with the other reactants. At other times, the reagent
is drawn above the arrow itself.
• Although the solvent is often omitted from the equation, most
organic reactions take place in liquid solvent.
• The solvent and temperature of the reaction may be added
above or below the arrow.
• The symbols “h” and “” are used for reactions that require
2
light and heat respectively.
Writing Equations for Organic Reactions
Figure 6.1 Different ways of writing organic reactions
3
Writing Equations for Organic Reactions
• When two sequential reactions are carried out without
drawing any intermediate compound, the steps are usually
numbered above or below the reaction arrow.
• This convention signifies that the first step occurs before the
second step, and the reagents are added in sequence, not at
the same time.
4
Kinds of Organic Reactions
• A substitution is a reaction in which an atom or a group
of atoms is replaced by another atom or group of atoms.
• In a general substitution, Y replaces Z on a carbon atom.
5
Substitution reactions
• Substitution reactions involve  bonds: one  bond breaks
and another forms at the same carbon atom.
• The most common examples of substitution occur when Z is
a halogen or a heteroatom that is more electronegative than
carbon.
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6
Kinds of Organic Reactions
• Elimination is a reaction in which elements of the
starting material are “lost” and a  bond is formed.
7
Elimination reaction
• In an elimination reaction, two groups X and Y are removed
from a starting material.
• Two  bonds are broken, and a  bond is formed between
adjacent atoms.
• The most common examples of elimination occur when X = H
and Y is a heteroatom more electronegative than carbon.
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8
Kinds of Organic Reactions
• Addition is a reaction in which elements are added to the
starting material.
9
Addition reaction
• In an addition reaction, new groups X and Y are added to
the starting material. A  bond is broken and two 
bonds are formed.
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10
Addition and elimination reactions
• Addition and elimination reactions are exactly opposite.
A  bond is formed in elimination reactions, whereas a 
bond is broken in addition reactions.
11
Bond Making and Bond Breaking
• A reaction mechanism is a detailed description of how bonds
are broken and formed as starting material is converted into
product.
• A reaction can occur either in one step or a series of steps.
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Bond Making and Bond Breaking
• Regardless of how many steps there are in a reaction, there
are only two ways to break (cleave) a bond: the electrons in
the bond can be divided equally or unequally between the two
atoms of the bond.
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Bond Making and Bond Breaking
• Homolysis and heterolysis require energy.
• Homolysis generates uncharged reactive intermediates with
unpaired electrons.
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• Heterolysis generates charged intermediates.
Movement of a single electron and an electron pair
• To illustrate the movement of a single electron, use a halfheaded curved arrow, sometimes called a fishhook.
• A full headed curved arrow shows the movement of an
electron pair.
15
Homolysis and Heterolysis
• Homolysis generates two uncharged species with
unpaired electrons.
• A reactive intermediate with a single unpaired electron is
called a radical.
• Radicals are highly unstable because they contain an
atom that does not have an octet of electrons.
• Heterolysis generates a carbocation or a carbanion.
• Both carbocations and carbanions are unstable
intermediates. A carbocation contains a carbon
surrounded by only six electrons, and a carbanion has a
negative charge on carbon, which is not a very
electronegative atom.
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Reactive intermediates
Figure 6.2 Three reactive intermediates resulting from
homolysis and heterolysis of a C – Z bond
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Radicals, Carbocations and Carbanions
• Radicals and carbocations are electrophiles because they
contain an electron deficient carbon.
• Carbanions are nucleophiles because they contain a carbon
with a lone pair.
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Bond formation
• Bond formation occurs in two different ways.
• Two radicals can each donate one electron to form a twoelectron bond.
• Alternatively, two ions with unlike charges can come together,
with the negatively charged ion donating both electrons to
form the resulting two-electron bond.
• Bond formation always releases energy.
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Types of arrows used in describing
organic reactions
20
Bond Dissociation Energy
• The energy absorbed or released in any reaction, symbolized
by H0, is called the enthalpy change or heat of reaction.
• Bond dissociation energy is the H0 for a specific kind of
reaction—the homolysis of a covalent bond to form two
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radicals.
Bond Dissociation Energy
• Because bond breaking requires energy, bond dissociation
energies are always positive numbers, and homolysis is always
endothermic.
• Conversely, bond formation always releases energy, and thus is
always exothermic. For example, the H—H bond requires +104
kcal/mol to cleave and releases –104 kcal/mol when formed.
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23
Bond Dissociation Energy
• Comparing bond dissociation
comparing bond strength.
energies
is
equivalent
to
• The stronger the bond, the higher its bond dissociation energy.
• Bond dissociation energies decrease down a column of the
periodic table.
• Generally, shorter bonds are stronger bonds.
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Bond Dissociation Energy
• Bond dissociation energies are used to calculate the enthalpy
change (H0) in a reaction in which several bonds are broken
and formed.
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Bond Dissociation Energy - Sample Problem 6.2
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Bond Dissociation Energy
• H° is negative for both oxidations, so both reactions are
exothermic.
• Both isooctane and glucose release energy on oxidation
because the bonds in the products are stronger than the bonds
in the reactants.
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Bond dissociation energies - important limitations
• Bond dissociation energies present overall energy
changes only. They reveal nothing about the reaction
mechanism or how fast a reaction proceeds.
• Bond dissociation energies are determined for reactions
in the gas phase, whereas most organic reactions occur
in a liquid solvent where solvation energy contributes to
the overall enthalpy of a reaction.
• Bond dissociation energies are imperfect indicators of
energy changes in a reaction. However, using bond
dissociation energies to calculate H° gives a useful
approximation of the energy changes that occur when
bonds are broken and formed in a reaction.
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Thermodynamics
• For a reaction to be practical, the equilibrium must favor
products and the reaction rate must be fast enough to form them
in a reasonable time. These two conditions depend on
thermodynamics and kinetics respectively.
• Thermodynamics describes how the energies of reactants and
products compare, and what the relative amounts of reactants
and products are at equilibrium.
• Kinetics describes reaction rates.
• The equilibrium constant, Keq, is a mathematical expression that
relates the amount of starting material and product at
equilibrium.
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Thermodynamics
• The size of Keq expresses whether the starting materials or
products predominate once equilibrium is reached.
• When Keq > 1, equilibrium favors the products (C and D) and the
equilibrium lies to the right as the equation is written.
• When Keq < 1, equilibrium favors the starting materials (A and B)
and the equilibrium lies to the left as the equation is written.
• For a reaction to be useful, the equilibrium must favor the
products, and Keq > 1.
• The position of the equilibrium is determined by the relative
energies of the reactants and products.
• G° is the overall energy difference between reactants and
products.
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Thermodynamics
Figure 6.3 Summary of the relationship between ∆G° and Keq
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Thermodynamics
• G° is related to the equilibrium constant Keq by the following
equation:
• When Keq > 1, log Keq is positive, making G° negative, and
energy is released. Thus, equilibrium favors the products when
the energy of the products is lower than the energy of the
reactants.
• When Keq < 1, log Keq is negative, making G° positive, and
energy is absorbed. Thus, equilibrium favors the reactants when
the energy of the products is higher than the energy of the
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reactants.
Thermodynamics
• Compounds that are lower in energy have increased stability.
• The equilibrium favors the products when they are more stable
(lower in energy) than the starting materials of a reaction.
• Because G° depends on the logarithm of Keq, a small change in
energy corresponds to a large difference in the relative amount
of starting material and product at equilibrium.
(Table 6.3)
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Thermodynamics
34
Thermodynamics
• These equations can be used for any process with two states in
equilibrium. As an example, monosubstituted cyclohexanes exist
as two different chair conformations that rapidly interconvert at
room temperature, with the conformation having the substituent
in the roomier equatorial position favored.
• Knowing the energy difference between two conformations
permits the calculation of the amount of each at equilibrium.
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35
Enthalpy and Entropy
• G° depends on H° and the entropy change, S°.
• Entropy change, S°, is a measure of the change in the
randomness of a system. The more disorder present, the higher
the entropy. Gas molecules move more freely than liquid
molecules and are higher in entropy. Cyclic molecules have
more restricted bond rotation than similar acyclic molecules and
are lower in entropy.
• S° is (+) when the products are more disordered than the
reactants. S° is (-) when the products are less disordered than
the reactants.
• Reactions resulting in increased entropy are favored.
• G° is related to H° and S° by the following equation:
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Enthalpy and Entropy
• This equation indicates that the total energy change is due
to two factors: the change in bonding energy and the
change in disorder.
• The change in bonding energy can be calculated from bond
dissociation energies.
• Entropy changes are important when
The number of molecules of starting material differs
from the number of molecules of product in the
balanced chemical equation.
An acyclic molecule is cyclized to a cyclic one, or a
cyclic molecule is converted to an acyclic one.
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Enthalpy and Entropy
• In most other reactions that are not carried out at high
temperature, the entropy term (TS°) is small compared to the
enthalpy term (H0), and therefore it is usually neglected.
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Energy Diagrams
• An energy diagram is a schematic representation of the energy
changes that take place as reactants are converted to products.
• An energy diagram plots the energy on the y axis versus the
progress of reaction, often labeled as the reaction coordinate, on
the x axis.
• The energy difference between reactants and products is H°. If
the products are lower in energy than the reactants, the reaction
is exothermic and energy is released. If the products are higher
in energy than the reactants, the reaction is endothermic and
energy is consumed.
• The unstable energy maximum as a chemical reaction proceeds
from reactants to products is called the transition state. The
transition state species can never be isolated.
• The energy difference between the transition state and the
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starting material is called the energy of activation, Ea.
Energy Diagrams
• For the general reaction:
• The energy diagram would be shown as:
40
Energy Diagrams
• The energy of activation is the minimum amount of energy
needed to break the bonds in the reactants.
• The larger the Ea, the greater the amount of energy that is
needed to break bonds, and the slower the reaction rate.
• The structure of the transition state is somewhere between the
structures of the starting material and product. Any bond that is
partially formed or broken is drawn with a dashed line. Any atom
that gains or loses a charge contains a partial charge in the
transition state.
• Transition states are drawn in brackets, with a superscript
double dagger (‡).
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Energy Diagrams
Example 1 (Figure 6.4)
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Energy Diagrams
Example 2 (Figure 6.4)
43
Energy Diagrams
Example 3 (Figure 6.4)
44
Energy Diagrams
Example 4 (Figure 6.4)
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Energy Diagrams
Figure 6.5 Comparing ∆H° and Ea in two energy diagrams
46
Energy Diagrams
• Consider the following two step reaction:
• An energy diagram must be drawn for each step.
• The two energy diagrams must then be combined to form an
energy diagram for the overall two-step reaction.
• Each step has its own energy barrier, with a transition state at
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the energy maximum.
Energy diagram for step 1
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Energy diagram for step 2
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Figure 6.6 Complete energy diagram for the two-step conversion
50
Understanding Organic Reactions
Kinetics
• Kinetics is the study of reaction rates.
• Recall that Ea is the energy barrier that must be exceeded
for reactants to be converted to products.
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Kinetics
• The higher the concentration, the faster the rate.
• The higher the temperature, the faster the rate.
• G°, H°, and Keq do not determine the rate of a reaction.
These quantities indicate the direction of the equilibrium and
the relative energy of reactants and products.
• A rate law or rate equation shows the relationship between
the reaction rate and the concentration of the reactants. It is
experimentally determined.
52
Kinetics
• Fast reactions have large rate constants.
• Slow reactions have small rate constants.
• The rate constant k and the energy of activation Ea are inversely
related. A high Ea corresponds to a small k.
• A rate equation contains concentration terms for all reactants in
a one-step mechanism.
• A rate equation contains concentration terms for only the
reactants involved in the rate-determining step in a multi-step
reaction.
• The order of a rate equation equals the sum of the exponents of
the concentration terms in the rate equation.
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Kinetics
• A two-step reaction has a slow rate-determining step, and a fast
step.
• In a multi-step mechanism, the reaction can occur no faster than
its rate-determining step.
• Only the concentration of the reactants in the rate-determining
step appears in the rate equation.
54
Catalysts
• Some reactions do not proceed at a reasonable rate unless a
catalyst is added.
• A catalyst is a substance that speeds up the rate of a reaction. It
is recovered unchanged in a reaction, and it does not appear in
the product.
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Figure 6.7 The effect of a catalyst on a reaction
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Enzymes
• Enzymes are biochemical catalysts composed of amino acids
held together in a very specific three-dimensional shape.
• An enzyme contains a region called its active site which binds an
organic reactant, called a substrate. The resulting unit is called
the enzyme-substrate complex.
• Once bound, the organic substrate undergoes a very specific
reaction at an enhanced rate. The products are then released.
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Applications in
Organic Chemistry
Laboratory
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Heat of Reaction and Hess’s Law
The specific heat (s) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the
substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat
(q) required to raise the temperature of a given quantity (m)
of the substance by one degree Celsius.
C=mxs
Heat (q) absorbed or released:
q = m x s x t
q = C x t
t = tfinal - tinitial
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Heat of Reaction and Hess’s Law
Constant-Pressure Calorimetry
No heat enters or leaves!
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = m x s x t
qcal = Ccal x t
Reaction at Constant P
H = qrxn
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Heat of Reaction and Hess’s Law
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Heat of Reaction and Hess’s Law
The standard enthalpy of reaction (Hrxn0) is the enthalpy of
a reaction carried out at 1 atm.
aA + bB
cC + dD
H0rxn = [ cH0f (C) + dH0f (D) ] - [ aH0f (A) + bH0f (B) ]
H0rxn = S nH0f (products) - S mHf0 (reactants)
Hess’s Law: When reactants are converted to products, the
change in enthalpy is the same whether the reaction takes
place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get
there, only where you start and end.)
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Chemical Kinetics
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a
reactant or a product with time (M/s).
A
B
[A]
rate = t
[A] = change in concentration of A over
time period t
[B]
rate =
t
[B] = change in concentration of B over
time period t
Because [A] decreases with time, [A] is negative.
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Chemical Kinetics
A
B
[A]
rate = t
[B]
rate =
t
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Chemical Kinetics : Breath Analyzer
3CH3CH2OH + 2K2Cr2O7 + 8H2SO4
3CH3COOH + 2Cr2(SO4)3 + 2K2SO4 + 11H2O
[Cr2O72-] a  Absorption
65
Chemical Kinetics : First-Order Reactions
A
k=
product
[A]
rate = t
rate
M/s
=
= 1/s or s-1
M
[A]
[A] = [A]0e−kt
rate = k [A]
[A]
= k [A]
t
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
ln[A] = ln[A]0 - kt
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Chemical Kinetics : Graphical Determination of k
2N2O5
4NO2 (g) + O2 (g)
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Chemical Kinetics : First-Order Reactions
The half-life, t½, is the time required for the concentration of a
reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln
t½ =
[A]0
[A]0/2
k
ln 2
0.693
=
=
k
k
What is the half-life of N2O5 if it decomposes with a rate constant
of 5.7 x 10-4 s-1?
0.693
t½ = ln 2 =
= 1200 s = 20 minutes
-4
-1
k
5.7 x 10 s
How do you know decomposition is first order?
units of k (s-1)
68
Chemical Kinetics : Second-Order Reactions
A
product
[A]
rate = t
rate
M/s
=
= 1/M•s
k=
2
2
M
[A]
1
1
=
+ kt
[A]
[A]0
rate = k [A]2
[A]
= k [A]2
t
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
1
t½ =
k[A]0
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Chemical Kinetics : Zero-Order Reactions
A
product
[A]
rate = t
[A]
=k
t
rate
= M/s
k=
0
[A]
[A] = [A]0 - kt
rate = k [A]0 = k
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
[A]0
t½ =
2k
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Chemical Kinetics : Summary of the Kinetics of
Zero-Order, First-Order and Second-Order Reactions
Order
0
Rate Law
rate = k
1
rate = k [A]
2
[A]2
rate = k
Concentration-Time
Equation
[A] = [A]0 - kt
ln[A] = ln[A]0 - kt
1
1
=
+ kt
[A]
[A]0
Half-Life
t½ =
[A]0
2k
t½ = ln 2
k
1
t½ =
k[A]0
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