wade7 lecture 04
TRANSCRIPT
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Chapter 4
Copyright 2010 Pearson Education, Inc.
Organic Chemistry, 7th Edition
L. G. Wade, Jr.
The Study of
Chemical Reactions
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Chapter 4 2
Introduction
Overall reaction: reactants products
Mechanism: Step-by-step pathway.
To learn more about a reaction:
Thermodynamics
Kinetics
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Chapter 4 3
Chlorination of Methane
Requires heat or light for initiation.
The most effective wavelength is blue, which isabsorbed by chlorine gas.
Many molecules of product are formed fromabsorption of only one photon of light (chainreaction).
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Chapter 4 4
The Free-Radical Chain
Reaction Initiation generates a radical intermediate.
Propagat ion: The intermediate reacts with
a stable molecule to produce anotherreactive intermediate (and a product
molecule).
Terminat ion: Side reactions that destroythe reactive intermediate.
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Chapter 4 5
Initiation Step: Formation of
Chlorine Atom
A chlorine molecule splits homolytically into
chlorine atoms (free radicals).
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Chapter 4 6
Propagation Step: Carbon
Radical
The chlorine atom collides with a methane
molecule and abstracts (removes) an H,forming another free radical and one of the
products (HCl).
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Chapter 4 7
Propagation Step: Product
Formation
The methyl free radical collides with another
chlorine molecule, producing the organicproduct (methyl chloride) and regenerating the
chlorine radical.
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Chapter 4 8
Overall Reaction
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Chapter 4 9
Termination Steps
A reaction is classified as a termination step whenany two free radicals join together producing anonradical compound.
Combination of free radical with contaminant orcollision with wall are also termination steps.
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Chapter 4 10
More Termination Steps
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Chapter 4 11
Lewis Structures of Free
Radicals
Free radicals are unpaired electrons. Halogens have 7 valence electrons so one of them
will be unpaired (radical). We refer to the halides asatoms not radicals.
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Chapter 4 12
Equilibrium Constant
Keq = [products][reactants]
For CH4 + Cl2
CH3Cl + HCl
Keq = [CH3Cl][HCl] = 1.1 x 1019
[CH4][Cl2]
Large value indicates reaction goes tocompletion.
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Chapter 4 13
Free Energy Change
DG = (energy of products) - (energy of reactants)
DG is theamount of energy available to do work.
Negative values indicate spontaneity.
DGo = -RT(lnKeq) = -2.303 RT(log10Keq)
where R= 8.314 J/K-mol and T= temperature inkelvins.
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Chapter 4 14
Factors Determining DG
Free energy change depends on:
Enthalpy
DH= (enthalpy of products) - (enthalpy of reactants)
Entropy
DS = (entropy of products) - (entropy of reactants)
DG = DH - TDS
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Chapter 4 15
Enthalpy
DHo = heat released or absorbed during
a chemical reaction at standard
conditions. Exothermic (-DH) heat is released.
Endothermic (+DH) heat is absorbed.
Reactions favor products with lowestenthalpy (strongest bonds).
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Chapter 4 16
Entropy
DSo = change in randomness, disorder,
or freedom of movement.
Increasing heat, volume, or number ofparticles increases entropy.
Spontaneous reactions maximize
disorder and minimize enthalpy.
In the equation DGo = DHo - TDSo the
entropy value is often small.
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Chapter 4 17
Calculate the value ofDG for the chlorination of methane.
DG =2.303RT(logKeq)
Keq for the chlorination is 1.1 x 1019, and log Keq = 19.04
At 25 C (about 298 Kelvin), the value ofRTis
RT= (8.314 J/kelvin-mol)(298 kelvins) = 2478 J/mol, or 2.48 kJ/mol
Substituting, we have
DG = (2.303)(2.478 kJ/mol)(19.04) =108.7 kJ/mol (25.9 kcal>mol)
This is a large negative value forDG, showing that this chlorination has a large driving force that
pushes it toward completion.
Solved Problem 1
Solution
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Chapter 4 18
Bond-Dissociation Enthalpies
(BDE) Bond-dissociation requires energy (+BDE).
Bond formation releases energy (-BDE).
BDE can be used to estimate DHfor a reaction. BDE for homolytic cleavage of bonds in a
gaseous molecule. Homo lyt ic cleavage: When the bond breaks, each atom gets
one electron. Heteroly t ic cleavage: When the bond breaks, the most
electronegative atom gets both electrons.
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Chapter 4 19
Homolytic and Heterolytic
Cleavages
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Chapter 4 20
Enthalpy Changes in Chlorination
CH3-H + Cl-Cl CH3-Cl + H-Cl
Bonds Broken DH (per Mole) Bonds Formed DH (per Mole)
Cl-Cl +242 kJ H-Cl -431 kJ
CH3-H +435 kJ CH3-Cl -351 kJ
TOTALS +677 kJ TOTAL -782 kJ
DH = +677 kJ + (-782 kJ) = -105 kJ/mol
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Chapter 4 21
Kinetics
Kinetics is the study of reaction rates.
Rate of the reaction is a measure of how the
concentration of the products increase whilethe concentration of the products decrease.
A rate equation is also called the rate law andit gives the relationship between the
concentration of the reactants and thereaction rate observed.
Rate law is experimentally determined.
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Chapter 4 22
Rate Law
For the reaction A + B C + D,
rate = kr[A]a[B]b
a is the order with respect to A
b is the order with respect to B
a + b is the overall order
Order is the number of molecules of that
reactant which is present in the rate-determining step of the mechanism.
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Chapter 4 23
Activation Energy
The value ofkdepends on temperature asgiven by Arrhenius:
RTEa
Aek
/
r
where A = constant (frequency factor)
Ea = activation energy
R= gas constant, 8.314 J/kelvin-mole
T= absolute temperature
Ea is the minimum kinetic energy needed to react.
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Chapter 4 24
Activation Energy (Continued)
At higher temperatures, more molecules havethe required energy to react.
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Chapter 4 25
Energy Diagram of an Exothermic
Reaction
The vertical axis in this graph represents the potentialenergy.
The transition state is the highest point on the graph,and the activation energy is the energy differencebetween the reactants and the transition state.
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Chapter 4 26
Rate-Limiting Step
Reaction intermediates are stable as long
as they dont collide with another molecule
or atom, but they are very reactive. Transition states are at energy maximums.
Intermediates are at energy minimums.
The reaction step with highest Ea will be theslowest, therefore rate-determining for the
entire reaction.
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Chapter 4 27
Energy Diagram for the
Chlorination of Methane
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Chapter 4 28
Rate, Ea,and Temperature
X + CH4 HX + CH3
X Ea(per Mole) Rate at 27 C Rate at 227 C
F 5 140,000 300,000
Cl 17 1300 18,000
Br 75 9 x 10-8 0.015
I 140 2 x 10-19 2 x 10-9
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Chapter 4 29
Consider the following reaction:
This reaction has an activation energy (Ea) of +17 kJ/mol (+4 kcal/mol) and a DH of +4 kJ/mol (+1
kcal/mol). Draw a reaction-energy diagram for this reaction.
We draw a diagram that shows the products to be 4 kJ higher in energy than the reactants. The barrieris made to be 17 kJ higher in energy than the reactants.
Solved Problem 2
Solution
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Chapter 4 30
Conclusions
With increasing Ea, rate decreases.
With increasing temperature, rate
increases. Fluorine reacts explosively.
Chlorine reacts at a moderate rate.
Bromine must be heated to react. Iodine does not react (detectably).
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Chapter 4 31
Primary, Secondary, and Tertiary
Hydrogens
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Chapter 4 32
Chlorination Mechanism
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Chapter 4 33
Bond Dissociation Energies for
the Formation of Free Radicals
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Chapter 4 34
Tertiary hydrogen atoms react with Cl about 5.5 times as fast as primary ones. Predict the product
ratios for chlorination of isobutane.
There are nine primary hydrogens and one tertiary hydrogen in isobutane.
(9 primary hydrogens) x (reactivity 1.0) = 9.0 relative amount of reaction
(1 tertiary hydrogen) x (reactivity 5.5) = 5.5 relative amount of reaction
Solved Problem 3
Solution
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Chapter 4 35
Even though the primary hydrogens are less reactive, there are so many of them that the primary
product is the major product. The product ratio will be 9.0:5.5, or about 1.6:1.
Solved Problem 3 (Continued)
Solution
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Chapter 4 36
Stability of Free Radicals
Free radicals are more stable if they are
highly substituted.
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Chapter 4 37
Chlorination Energy Diagram
LowerEa, faster rate, so more stableintermediate is formed faster.
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Chapter 4 38
Rate of Substitution in the
Bromination of Propane
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Chapter 4 39
Energy Diagram for the
Bromination of Propane
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Chapter 4 40
Hammond Postulate
Related species that are similar inenergy are also similar in structure.
The structure of the transition state
resembles the structure of the closeststable species.
Endotherm ic react ion: Transition state
is product-like. Exotherm ic react ion: Transition state is
reactant-like.
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Chapter 4 41
Energy Diagrams: Chlorination
Versus Bromination
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Chapter 4 42
Endothermic and
Exothermic Diagrams
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Chapter 4 43
Radical Inhibitors
Often added to food to retard spoilageby radical chain reactions.
Without an inhibitor, each initiation stepwill cause a chain reaction so that manymolecules will react.
An inhibitor combines with the free
radical to form a stable molecule. Vitamin E and vitamin C are thought to
protect living cells from free radicals.
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Chapter 4 44
Radical Inhibitors (Continued)
A radical chain reaction is fast and has manyexothermic steps that create more reactive radicals.
When an inhibitor reacts with the radical, it creates astable intermediate, and any further reactions will beendothermic and slow.
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Chapter 4 45
Carbon Reactive Intermediates
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Chapter 4 46
Carbocation Structure
Carbon has 6 electrons, positively charged.
Carbon is sp2 hybridized with vacantp orbital.
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Chapter 4 47
Carbocation Stability
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Chapter 4 48
Carbocation Stability
(Continued)
Stabilized by alkyl
substituents in two ways:
1. Induc t ive effect: Donation
of electron density along the
sigma bonds.
2. Hyperconjugat ion:
Overlap of sigma bonding
orbitals with emptyp orbital.
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Chapter 4 49
Free Radicals
Also electron-deficient. Stabilized by alkyl substituents.
Order of stability:3 > 2 > 1 > methyl
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Chapter 4 50
Stability of Carbon Radicals
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Chapter 4 51
Carbanions
Eight electrons oncarbon: 6 bonding
plus one lone pair. Carbon has anegative charge.
Destabilized by alkylsubstituents.
Methyl >1 > 2 > 3
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Chapter 4 52
Carbenes
Carbon is neutral.
Vacantp orbital, so can be electrophilic. Lone pair of electrons, so can be nucleophilic.
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Chapter 4 53
Basicity of Carbanions
A carbanion has a negative charge on its
carbon atom, making it a more powerful baseand a stronger nucleophile than an amine.
A carbanion is sufficiently basic to remove aproton from ammonia.
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Chapter 4 54
Carbenes as Reaction
Intermediates
A strong base can abstract a proton from tribromomethane(CHBr3) to give an inductively stabilized carbanion.
This carbanion expels bromide ion to give dibromocarbene. Thecarbon atom is sp2 hybridized with trigonal geometry.
A carbene has both a lone pair of electrons and an emptyporbital, so it can react as a nucleophile or as an electrophile.