Organic Chemistry Lecture Notes PDF

Organic Chemistry, 9th Edition L. G. Wade, Jr. Chapter 4 Lecture The Study of...

Organic Chemistry, 9th Edition L. G. Wade, Jr. Chapter 4 Lecture The Study of Chemical Reactions Chad Snyder, PhD Grace College © 2017 Pearson Education, Inc. © 2014 Pearson Education, Inc.  Introduction Overall reaction: reactants products To learn more about a reaction: – Thermodynamics is the study of the energy changes that accompany chemical and physical transformations. – Kinetics is the study of reaction rates. Mechanism: step-by-step description of how the reaction happens © 2017 Pearson Education, Inc.  Chlorination of Methane It 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). © 2017 Pearson Education, Inc.  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). Terminations are side reactions that destroy the reactive intermediate. © 2017 Pearson Education, Inc.  Initiation Step: Formation of Chlorine Atom A chlorine molecule splits homolytically into chlorine atoms (free radicals). © 2017 Pearson Education, Inc.  Lewis Structures of Free Radicals Free radicals are reactive species with odd numbers of electrons. Halogens have seven valence electrons, so one of them will be unpaired (radical). We refer to the halides as atoms, not radicals. © 2017 Pearson Education, Inc.  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). © 2017 Pearson Education, Inc.  Propagation Step: Product Formation The methyl free radical collides with another chlorine molecule, producing the organic product (methyl chloride) and regenerating the chlorine radical. © 2017 Pearson Education, Inc.  Overall Reaction © 2017 Pearson Education, Inc.  Termination Steps A reaction is classified as a termination step when any two free radicals join together, producing a nonradical compound. Combination of a free radical with a contaminant or collision with a wall are also termination steps. © 2017 Pearson Education, Inc.  More Termination Steps © 2017 Pearson Education, Inc.  Equilibrium Constant © 2017 Pearson Education, Inc.  Free Energy Change DG = (energy of products) – (energy of reactants) DG is the amount of energy available to do work. A reaction with a negative DG is favorable and spontaneous. where R = 8.314 J/K-mol and T = temperature in kelvins. © 2017 Pearson Education, Inc.  Factors Determining G° Free energy change depends on the following: - H° = (enthalpy of products) – (enthalpy of reactants) © 2017 Pearson Education, Inc. © 2017 Pearson Education, Inc.  Enthalpy DH° = 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 the lowest enthalpy (strongest bonds). © 2017 Pearson Education, Inc.  Entropy DS° = 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 DG° = DH° – TDS°, the entropy value is often small. © 2017 Pearson Education, Inc.  Solved Problem 1 Calculate the value of DG° for the chlorination of methane. Solution DG° = –2.303RT(log Keq) Keq for the chlorination is 1.1 x 1019, and log Keq = 19.04 At 25 °C (about 298 K), the value of RT is 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 for DG°, showing that this chlorination has a large driving force that pushes it toward completion. © 2017 Pearson Education, Inc.  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. © 2017 Pearson Education, Inc.  Homolytic and Heterolytic Cleavages © 2017 Pearson Education, Inc.  Enthalpy Changes in Chlorination © 2017 Pearson Education, Inc. © 2017 Pearson Education, Inc.  Kinetics Kinetics is the study of reaction rates. Rate of the reaction is a measure of how the concentration of the products increases while the concentration of the starting materials decreases. A rate equation (also called the rate law) is the relationship between the concentrations of the reactants and the observed reaction rate. Rate law is determined experimentally. © 2017 Pearson Education, Inc.  Rate Law For the reaction A + B C + D, rate = kr[A]a[B]b - where kr is the rate constant - 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. © 2017 Pearson Education, Inc.  Activation Energy The rate constant, kr, depends on the conditions of the reaction, especially the temperature: - 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. © 2017 Pearson Education, Inc.  Temperature Dependence of Ea At higher temperatures, more molecules have the required energy to react. © 2017 Pearson Education, Inc.  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 (Ea) is the energy difference between the reactants and the transition state. © 2017 Pearson Education, Inc.  Rates of Multistep Reactions The highest points in an energy diagram are transition states. The lowest points in an energy diagram are intermediates. The reaction step with the highest Ea will be the slowest step and will determine the rate at which the reaction proceeds (rate-limiting step). © 2017 Pearson Education, Inc.  Energy Diagram for the Chlorination of Methane © 2017 Pearson Education, Inc.  Rate, Ea, and Temperature X + CH4 HX + CH3 © 2017 Pearson Education, Inc.  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). © 2017 Pearson Education, Inc.  Solved Problem 2 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. 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. © 2017 Pearson Education, Inc.  Primary, Secondary, and Tertiary Hydrogens © 2017 Pearson Education, Inc.  Chlorination Mechanism © 2017 Pearson Education, Inc.  Bond Dissociation Energies for the Formation of Free Radicals © 2017 Pearson Education, Inc.  Stability of Free Radicals Highly substituted free radicals are more stable. © 2017 Pearson Education, Inc.  Chlorination Energy Diagram The lower the Ea, the faster the rate, so a more stable intermediate is formed faster. © 2017 Pearson Education, Inc.  Solved Problem 3 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. (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 © 2017 Pearson Education, Inc.  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. © 2017 Pearson Education, Inc.  Rate of Substitution in the Bromination of Propane © 2017 Pearson Education, Inc.  Energy Diagram for the Bromination of Propane © 2017 Pearson Education, Inc.  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 resembles the product. Exothermic reaction: Transition state resembles the reactant. © 2017 Pearson Education, Inc.  Energy Diagrams: Chlorination Versus Bromination © 2017 Pearson Education, Inc.  Endothermic and Exothermic Diagrams © 2017 Pearson Education, Inc.  HINT Free-radical bromination is highly selective, chlorination is moderately selective, and fluorination is nearly nonselective. © 2017 Pearson Education, Inc. Chapter 4 45  Radical Inhibitors These are 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. © 2017 Pearson Education, Inc.  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. © 2017 Pearson Education, Inc.  Reactive Intermediates Reactive intermediates are short-lived species. They are never present in high concentrations because they react as quickly as they are formed. © 2017 Pearson Education, Inc.  Carbocation Structure A carbocation (also called a carbonium ion or a carbenium ion) is a positively charged carbon. Carbon is sp2 hybridized with a vacant p orbital. © 2017 Pearson Education, Inc.  Carbocation Stability More highly substituted carbocations are more stable. © 2017 Pearson Education, Inc.  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 © 2017 Pearson Education, Inc.  Unsaturated Carbocations Unsaturated carbocations are also stabilized by resonance stabilization. If a pi bond is adjacent to a carbocation, the filled p orbitals of the bond will overlap with the empty p orbital of the carbocation. © 2017 Pearson Education, Inc.  Free Radicals Carbon is sp2 hybridized with one electron in the p orbital. Stabilized by alkyl substituents Order of stability: 3° > 2° > 1° > methyl © 2017 Pearson Education, Inc.  Stability of Carbon Radicals More highly substituted radicals are more stable. © 2017 Pearson Education, Inc.  Unsaturated Radicals Like carbocations, radicals can be stabilized by resonance. Overlap with the p orbitals of a p bond allows the odd electron to be delocalized over two carbon atoms. Resonance delocalization is particularly effective in stabilizing a radical. © 2017 Pearson Education, Inc.  Carbanions Eight electrons on carbon: six bonding plus one lone pair Carbon has a negative charge. Carbanions are nucleophilic and basic. © 2017 Pearson Education, Inc.  Stability of Carbanions Alkyl groups and other electron-donating groups slightly destabilize a carbanion. The order of stability is usually the opposite of that for carbocations and free radicals. © 2017 Pearson Education, Inc.  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. © 2017 Pearson Education, Inc.  Carbenes Carbon in carbenes is neutral. It has a vacant p orbital, so it can react as an electrophile. It has a lone pair of electrons in the sp2 orbital, so it can react as a nucleophile. © 2017 Pearson Education, Inc.  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. © 2017 Pearson Education, Inc.  Summary of Reactive Species KNOW © 2017 Pearson Education, Inc.