Organic Chemistry, 5th Edition H C——H L G. Wade Jr H Chapter 4 The study of Chemical Reactions Jo blackburn Richland College, Dallas, TX Dallas County Community College District C 2003. Prentice hall
Chapter 4 The Study of Chemical Reactions Jo Blackburn Richland College, Dallas, TX Dallas County Community College District ã 2003, Prentice Hall Organic Chemistry, 5th Edition L. G. Wade, Jr
H Tools for Study + C—H H To determine a reactions mechanism look at >Equilibrium constant >Free energy change >Enthalpy >Entropy >Bond dissociation energy >Kinetics >Activation energy Chapter 4
Chapter 4 2 Tools for Study • To determine a reaction’s mechanism, look at: ØEquilibrium constant ØFree energy change ØEnthalpy ØEntropy ØBond dissociation energy ØKinetics ØActivation energy =>
H Chlorination of Methane HA heat or light H-C-H CI H--C-CI HCl Requires heat or light for initiation The most effective wavelength is blue, which is absorbed by chlorine gas Lots of product formed from absorption of only one photon of light (chain reaction Chapter 4
Chapter 4 3 Chlorination of Methane • Requires heat or light for initiation. • The most effective wavelength is blue, which is absorbed by chlorine gas. • Lots of product formed from absorption of only one photon of light (chain reaction). => C H H H H + Cl2 heat or light C H H H Cl + HCl
H Free-Radical chain reaction C——H Initiation generates a reactive 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
Chapter 4 4 Free-Radical Chain Reaction • Initiation generates a reactive 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. =>
H Initiation Step C——H H A chlorine molecule splits homolytically into chlorine atoms(free radicals) :C: C:+ photon(hv) Cl t Cl. Chapter 4
Chapter 4 5 Initiation Step A chlorine molecule splits homolytically into chlorine atoms (free radicals) => Cl Cl + photon (h) Cl + Cl
H Propagation Step (1)HN The chlorine atom collides with a methane molecule and abstracts (removes)a H, forming another free radical and one of the products(Hcl) H +:C HC·+H-Cl Chapter 4
Chapter 4 6 Propagation Step (1) The chlorine atom collides with a methane molecule and abstracts (removes) a H, forming another free radical and one of the products (HCl). C H H H H + Cl C H H H + H Cl =>
H Propagation Step(2) C——H The methyl free radical collides with another chlorine molecule, producing the other product (methyl chloride) and regenerating the chlorine radical H H H- Cl-Cl H Chapter 4
Chapter 4 7 Propagation Step (2) The methyl free radical collides with another chlorine molecule, producing the other product (methyl chloride) and regenerating the chlorine radical. C H H H + Cl Cl C H H H Cl + Cl =>
Overall reaction H + C—H H iC: Cl+ photon(hv) Cl +: Cl +:Cl HC·+H-Cl H ClCl HC-Cl+Cl· H H—C-H+C-Cl—>H-C-C1+H-Cl= H Chapter 4
Chapter 4 8 Overall Reaction C H H H H + Cl C H H H + H Cl C H H H + Cl Cl C H H H Cl + Cl C H H H H + Cl Cl C H H H Cl + H Cl => Cl Cl + photon (h) Cl + Cl
Termination Steps H C-h Collision of any two free radicals Combination of free radical with contaminant or collision with wall H HC°+Cl H-C—Cl H Can you suggest others? Chapter 4
Chapter 4 9 Termination Steps • Collision of any two free radicals • Combination of free radical with contaminant or collision with wall. C H H H + Cl C H H H Cl Can you suggest others? =>
H Equilibrium constant H Keg= products reactants For chlorination Keg =1.1 x 1019 Large value indicates reaction "goes to completion.” Chapter 4
Chapter 4 10 Equilibrium constant • Keq = [products] [reactants] • For chlorination Keq = 1.1 x 1019 • Large value indicates reaction “goes to completion.” =>