The Study of Chemical Reactions Organic Chemistry, 7th Edition L. G. Wade, Jr. Chapter 4 The Study of Chemical Reactions Copyright © 2010 Pearson Education, Inc.

Introduction Overall reaction: reactants  products Mechanism: Step-by-step pathway. To learn more about a reaction: Thermodynamics Kinetics Chapter 4

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

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

Initiation Step: Formation of Chlorine Atom A chlorine molecule splits homolytically into chlorine atoms (free radicals). Chapter 4

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

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

Overall Reaction Chapter 4

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

More Termination Steps Chapter 4

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

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

Free Energy Change DG = (energy of products) - (energy of reactants) DG is the amount 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 in kelvins. Chapter 4

Factors Determining G Free energy change depends on: Enthalpy H= (enthalpy of products) - (enthalpy of reactants) Entropy S = (entropy of products) - (entropy of reactants) G = H - TS Chapter 4

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 lowest enthalpy (strongest bonds). Chapter 4

Entropy DSo = 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 DGo = DHo - TDSo the entropy value is often small. Chapter 4

Solved Problem 1 Solution Calculate the value of DG° for the chlorination of methane. Solution D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 Copyright © 2006 Pearson Prentice Hall, Inc. Substituting, we have DG° = (–2.303)(2.478 kJ/mol)(19.04) = –108.7 kJ/mol (–25.9 kcal>mol) This is a large negative value for DG°, showing that this chlorination has a large driving force that pushes it toward completion. Chapter 4

Bond-Dissociation Enthalpies (BDE) Bond-dissociation requires energy (+BDE). Bond formation releases energy (-BDE). BDE can be used to estimate H 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

Homolytic and Heterolytic Cleavages Chapter 4

Enthalpy Changes in Chlorination CH3-H + Cl-Cl  CH3-Cl + H-Cl Bonds Broken DH° (per Mole) Bonds Formed 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 Chapter 4

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

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

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

Activation Energy (Continued) At higher temperatures, more molecules have the required energy to react. Chapter 4

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

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

Energy Diagram for the Chlorination of Methane Chapter 4

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

Solved Problem 2 Solution Consider the following reaction: This reaction has an activation energy (Ea) of +17 kJ/mol (+4 kcal/mol) and a DH° of +4 kJ/mol (+1 kcal/mol). Draw a reaction-energy diagram for this reaction. Solution 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. Copyright © 2006 Pearson Prentice Hall, Inc. Chapter 4

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

Primary, Secondary, and Tertiary Hydrogens Chapter 4

Chlorination Mechanism Chapter 4

Bond Dissociation Energies for the Formation of Free Radicals Chapter 4

Solved Problem 3 Solution Tertiary hydrogen atoms react with Cl• about 5.5 times as fast as primary ones. Predict the product ratios for chlorination of isobutane. Solution There are nine primary hydrogens and one tertiary hydrogen in isobutane. Copyright © 2006 Pearson Prentice Hall, Inc. (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 Chapter 4

Solved Problem 3 (Continued) Solution 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. Copyright © 2006 Pearson Prentice Hall, Inc. Chapter 4

Stability of Free Radicals Free radicals are more stable if they are highly substituted. Chapter 4

Chlorination Energy Diagram Lower Ea, faster rate, so more stable intermediate is formed faster. Chapter 4

Rate of Substitution in the Bromination of Propane Chapter 4

Energy Diagram for the Bromination of Propane Figure: 04_10.jpg Title: Energy Diagram for the Bromination of Propane Caption: Reaction-energy diagram for the first propagation step in the bromination of propane. The energy difference in the transition states is nearly as large as the energy difference in the products. Notes: Chapter 4

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

Energy Diagrams: Chlorination Versus Bromination Chapter 4

Endothermic and Exothermic Diagrams Chapter 4

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

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

Carbon Reactive Intermediates Chapter 4

Carbocation Structure Carbon has 6 electrons, positively charged. Carbon is sp2 hybridized with vacant p orbital. Chapter 4

Carbocation Stability Chapter 4

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

Free Radicals Also electron-deficient. Stabilized by alkyl substituents. Order of stability: 3 > 2 > 1 > methyl Chapter 4

Stability of Carbon Radicals Chapter 4

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

Carbenes Carbon is neutral. Vacant p orbital, so can be electrophilic. Lone pair of electrons, so can be nucleophilic. Chapter 4

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. Figure: 04_16-02UN.jpg Title: Reactivity of a Carbanion Caption: Like amines, carbanions are nucleophilic and basic. A carbanion has a negative charge on its carbon atom, however, making it a more powerful base and a stronger nucleophile than an amine. For example, a carbanion is sufficiently basic to remove a proton from ammonia. Notes: Carbanions are a stronger base than amines, so they can deprotonate amines easily. Chapter 4

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. Figure: 04_16-06UN.jpg Title: Carbenes as Reaction Intermediates Caption: Carbenes are uncharged reactive intermediates containing a divalent carbon atom. The simplest carbene has the formula CH2 and is called methylene, just as a –CH2- group in a molecule is called a methylene group. One way of generating carbenes is to form a carbanion that can expel a halide ion. For example, 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. An unshared pair of electrons occupies one of the sp2 hybrid orbitals, and there is an empty p orbital extending above and below the plane of the atoms. 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. Notes: The reactivity of carbenes lies in its orbitals. The carbon of carbenes have sp2 hybrid orbitals. Two of the hybrid orbitals are bonded to other atoms, one of the sp2 orbitals contains the lone pair of electrons so they can act as a nucleophile. Perpendicular to the plane of the sp2 orbitals is an unoccupied p orbital which allows them to act as an electrophile. Chapter 4