AULA 4

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Chapter 4 
©2010, Prentice Hall 
Organic Chemistry, 7th Edition 
L. G. Wade, Jr. 
The Study of 
Chemical Reactions 
Chapter 4 2 
Introduction 
•  Overall reaction: reactants → products 
•  Mechanism: Step-by-step pathway. 
•  To learn more about a reaction: 
  Thermodynamics 
  Kinetics 
Chapter 4 3 
Chlorination of Methane 
•  Requires heat or light for initiation. 
•  The most effective wavelength is blue, which is 
absorbed by chlorine gas. 
•  Many molecules of product are formed from 
absorption of only one photon of light (chain 
reaction). 
Chapter 4 4 
The Free-Radical Chain 
Reaction 
•  Initiation generates a radical intermediate. 
•  Propagation: The intermediate reacts with 
a stable molecule to produce another 
reactive intermediate (and a product 
molecule). 
•  Termination: Side reactions that destroy 
the reactive intermediate. 
Chapter 4 5 
Initiation Step: Formation of 
Chlorine Atom 
 A chlorine molecule splits homolytically into 
chlorine atoms (free radicals). 
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). 
Chapter 4 7 
Propagation Step: Product 
Formation 
The methyl free radical collides with another 
chlorine molecule, producing the organic 
product (methyl chloride) and regenerating the 
chlorine radical. 
Chapter 4 8 
Overall Reaction 
Chapter 4 9 
Termination Steps 
•  A reaction is classified as a termination step when 
any two free radicals join together producing a 
nonradical compound. 
•  Combination of free radical with contaminant or 
collision with wall are also termination steps. 
Chapter 4 10 
More Termination Steps 
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 as 
atoms not radicals. 
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 to 
completion.” 
Chapter 4 13 
Free Energy Change 
  ΔG = (energy of products) - (energy of reactants) 
  ΔG is the amount of energy available to do work. 
  Negative values indicate spontaneity. 
 ΔGo = -RT(lnKeq) = -2.303 RT(log10Keq) 
 where R = 8.314 J/K-mol and T = temperature in 
kelvins. 
Chapter 4 14 
Factors Determining ΔG° 
Free energy change depends on: 
  Enthalpy 
 ΔH°= (enthalpy of products) - (enthalpy of reactants) 
  Entropy 
 ΔS° = (entropy of products) - (entropy of reactants) 
 ΔG° = ΔH° - TΔS° 
Chapter 4 15 
Enthalpy 
•  ΔHo = heat released or absorbed during 
a chemical reaction at standard 
conditions. 
•  Exothermic (-ΔH) heat is released. 
•  Endothermic (+ΔH) heat is absorbed. 
•  Reactions favor products with lowest 
enthalpy (strongest bonds). 
Chapter 4 16 
Entropy 
•  ΔSo = change in randomness, disorder, 
or freedom of movement. 
•  Increasing heat, volume, or number of 
particles increases entropy. 
•  Spontaneous reactions maximize 
disorder and minimize enthalpy. 
•  In the equation ΔGo = ΔHo - TΔSo the 
entropy value is often small. 
Chapter 4 17 
Calculate the value of Δ G° for the chlorination of methane.
Δ G° = –2.303RT(log Keq)
Keq for the chlorination is 1.1 x 1019, and log Keq = 19.04
At 25 °C (about 298 ° Kelvin), the value of RT is
 RT = (8.314 J/kelvin-mol)(298 kelvins) = 2478 J/mol, or 2.48 kJ/mol
Substituting, we have
 Δ G° = (–2.303)(2.478 kJ/mol)(19.04) = –108.7 kJ/mol (–25.9 kcal>mol)
This is a large negative value for Δ G°, showing that this chlorination has a large driving force that 
pushes it toward completion.
Solved Problem 1 
Solution 
Chapter 4 18 
Bond-Dissociation Enthalpies 
(BDE) 
•  Bond-dissociation requires energy (+BDE). 
•  Bond formation releases energy (-BDE). 
•  BDE can be used to estimate ΔH for a reaction. 
•  BDE for homolytic cleavage of bonds in a 
gaseous molecule. 
  Homolytic cleavage: When the bond breaks, each atom gets 
one electron. 
  Heterolytic cleavage: When the bond breaks, the most 
electronegative atom gets both electrons. 
Chapter 4 19 
Homolytic and Heterolytic 
Cleavages 
Chapter 4 20 
Enthalpy Changes in Chlorination 
CH3-H + Cl-Cl  CH3-Cl + H-Cl 
Bonds Broken ΔH° (per Mole) Bonds Formed ΔH° (per Mole) 
Cl-Cl +242 kJ H-Cl -431 kJ 
CH3-H +435 kJ CH3-Cl -351 kJ 
TOTALS +677 kJ TOTAL -782 kJ 
ΔH° = +677 kJ + (-782 kJ) = -105 kJ/mol 
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 while 
the concentration of the products decrease. 
•  A rate equation is also called the rate law and 
it gives the relationship between the 
concentration of the reactants and the 
reaction rate observed. 
•  Rate law is experimentally determined. 
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. 
Chapter 4 23 
Activation Energy 
•  The value of k depends on temperature as 
given by Arrhenius: 
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. 
Chapter 4 24 
Activation Energy (Continued) 
•  At higher temperatures, more molecules have 
the required energy to react. 
Chapter 4 25 
Energy Diagram of an Exothermic 
Reaction 
•  The vertical axis in this graph represents the potential 
energy. 
•  The transition state is the highest point on the graph, 
and the activation energy is the energy difference 
between the reactants and the transition state. 
Chapter 4 26 
Rate-Limiting Step 
•  Reaction intermediates are stable as long 
as they don’t 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 the 
slowest, therefore rate-determining for the 
entire reaction. 
Chapter 4 27 
Energy Diagram for the 
Chlorination of Methane 
Chapter 4 28 
Rate, Ea, and Temperature 
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 
Chapter 4 29 
Consider the following reaction:
This reaction has an activation energy (Ea) of +17 kJ/mol (+4 kcal/mol) and a Δ H° 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 barrier 
is made to be 17 kJ higher in energy than the reactants.
Solved Problem 2 
Solution 
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).Chapter 4 31 
Primary, Secondary, and Tertiary 
Hydrogens 
Chapter 4 32 
Chlorination Mechanism 
Chapter 4 33 
Bond Dissociation Energies for 
the Formation of Free Radicals 
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 
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 
Chapter 4 36 
Stability of Free Radicals 
•  Free radicals are more stable if they are 
highly substituted. 
Chapter 4 37 
Chlorination Energy Diagram 
•  Lower Ea, faster rate, so more stable 
intermediate is formed faster. 
Chapter 4 38 
Rate of Substitution in the 
Bromination of Propane 
Chapter 4 39 
Energy Diagram for the 
Bromination of Propane 
Chapter 4 40 
Hammond Postulate 
•  Related species that are similar in 
energy are also similar in structure. 
•  The structure of the transition state 
resembles the structure of the closest 
stable species. 
•  Endothermic reaction: Transition state 
is product-like. 
•  Exothermic reaction: Transition state is 
reactant-like. 
Chapter 4 41 
Energy Diagrams: Chlorination 
Versus Bromination 
Chapter 4 42 
Endothermic and 
Exothermic Diagrams 
Chapter 4 43 
Radical Inhibitors 
•  Often added to food to retard spoilage 
by radical chain reactions. 
•  Without an inhibitor, each initiation step 
will cause a chain reaction so that many 
molecules 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. 
Chapter 4 44 
Radical Inhibitors (Continued) 
•  A radical chain reaction is fast and has many 
exothermic steps that create more reactive radicals. 
•  When an inhibitor reacts with the radical, it creates a 
stable intermediate, and any further reactions will be 
endothermic and slow. 
Chapter 4 45 
Carbon Reactive Intermediates 
Chapter 4 46 
Carbocation Structure 
•  Carbon has 6 electrons, positively charged. 
•  Carbon is sp2 hybridized with vacant p orbital. 
Chapter 4 47 
Carbocation Stability 
Chapter 4 48 
Carbocation Stability 
(Continued) 
•  Stabilized by alkyl 
substituents in two ways: 
 1. Inductive effect: Donation 
of electron density along the 
sigma bonds. 
 2. Hyperconjugation: 
Overlap of sigma bonding 
orbitals with empty p orbital. 
Chapter 4 49 
Free Radicals 
•  Also electron-deficient. 
•  Stabilized by alkyl substituents. 
•  Order of stability: 
3° > 2° > 1° > methyl 
Chapter 4 50 
Stability of Carbon Radicals 
Chapter 4 51 
Carbanions 
•  Eight electrons on 
carbon: 6 bonding 
plus one lone pair. 
•  Carbon has a 
negative charge. 
•  Destabilized by alkyl 
substituents. 
•  Methyl >1° > 2 ° > 3 °�
Chapter 4 52 
Carbenes 
•  Carbon is neutral. 
•  Vacant p orbital, so can be electrophilic. 
•  Lone pair of electrons, so can be nucleophilic. 
Chapter 4 53 
Basicity of Carbanions 
•  A carbanion has a negative charge on its 
carbon atom, making it a more powerful base 
and a stronger nucleophile than an amine. 
•  A carbanion is sufficiently basic to remove a 
proton from ammonia. 
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. The 
carbon atom is sp2 hybridized with trigonal geometry. 
•  A carbene has both a lone pair of electrons and an empty p 
orbital, so it can react as a nucleophile or as an electrophile.