WELCOME TO MODERN ORGANIC CHEMISTRY

Chapter 4 The Study of Chemical Reactions Organic Chemistry, 5 th Edition L. G. Wade, Jr.

A DESCRIPTION OF STRUCTURES AN ENERGIES OF STARTING MATERIALS AND PRODUCTS OF A REACTION AS WELL AS ANY REACTION INTERMEDIATES IN ADDITION, ALL OF THE TRANSITION STATES (ENERGY MAXIMA) SEPARATING THE REACTANTS FROM THE PRODUCTS (ENERGY MINIMA) MUST BE DETERMINED. WHAT IS A REACTION MECHANISM

? 1. ISOLATE THEM 2. TAKE THEIR SPECTRA a. IR - FUNCTIONAL GROUPS b. NMR - environment of hydrogen and carbon atoms c. MS - actual MW d. X-RAY PRODUCT DETERMINATION

ORTEP

STRUCTURE OF TRANSITION STATE? CAN'T MEASUREMENT DIRECTLY MEASURE THE RATE LAW TELLS NUMBER AND KIND OF MOLECULES INVOLVED IN TS CERTAIN ENZYMES RECOGNIZE TS - THAT IS THEY BOND WITH IT!!

HOW FAR DOES A REACTION GO?   G o =  H o  S o = -2.3xRxTxLogKeq HOW FAST DOES A REACTION GO? KINETICS RATE IS INVERSELY PROPORTIONAL TO ACTIVATION ENERGY (E A )

Chlorination of Methane Requires heat or light for initiation. The most effective wavelength is blue, which is absorbed by chlorine gas. High quantum yield Lots of product formed from absorption of only one photon of light (chain reaction). Some ethane is formed THESE FACTORS ARE CHARACHERISTICS OF FREE- RADICAL REACTIONS

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. =>

Initiation Step A chlorine molecule splits homolytically into chlorine atoms (free radicals) => THE CHLORINE ATOM IS ELECTROPHILIC - LEWIS ACID SEEKS AN ELECTRON TO REGAIN OCTET OF ELECTRONS USUALLY ABSTRACTS A HYDROGEN ATOM

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). => Methyl radicals is also electrophilic -seeks e - Methyl radical is a member of the highly reactive intermediates gang

Propagation Step (2) The methyl free radical collides with another chlorine molecule, producing the other product (methyl chloride) and regenerating the chlorine radical. => chlorine atom then attacks another methyl step 1 and 2 repeat over and over again UNTIL

Overall Reaction =>

Termination Steps Collision of any two free radicals Combination of free radical with contaminant or collision with wall. Can you suggest others? =>

Equilibrium constant K eq = [products] [reactants] For chlorination K eq = 1.1 x Large value indicates reaction “goes to completion.” =>

Free Energy Change  G = free energy of (products - reactants), amount of energy available to do work. Negative values indicate spontaneity.  G o = -RT(lnK eq ) where R = cal/K- mol and T = temperature in kelvins Since chlorination has a large K eq, the free energy change is large and negative.

Factors Determining  G  Free energy change depends on –enthalpy –entropy  H  = (enthalpy of products) - (enthalpy of reactants)  S  = (entropy of products) - (entropy of reactants)  G  =  H  - T  S 

Enthalpy  H o = heat released or absorbed during a chemical reaction at standard conditions. Exothermic, (-  H), heat is released. Endothermic, (+  H), heat is absorbed. => Reactions favor products with lowest enthalpy (strongest bonds). =>

Entropy  S o = change in randomness, disorder, freedom of movement. Increasing heat, volume, or number of particles increases entropy. Spontaneous reactions maximize disorder and minimize enthalpy. In the equation  G o =  H o - T  S o the entropy value is often small. =>

PROBLEMS POSSIBLE ANSWERS A. 0 B. -35c. +35 Which entropy value shown below is correct? A. 0 B. -18 c. +18

Bond Dissociation Energy Bond breaking requires energy (+BDE) We can use BDE to estimate  H for a reaction. => Bond formation releases energy (-BDE) Table 4.2 gives BDE for homolytic cleavage of bonds in a gaseous molecule.

Determination of delta H for overall reaction 104 kcal 58 kcal -84 kcal -103 kcal 162 kcal -187kcal -25 kcal EXOTHERMIC

INDIVIDUAL STEPS 104 kcal-103 kcal 58 kcal -84 kcal  THIS IS THE OFFICIAL MECHANISM RDS

Which is more likely? Estimate  H for each step using BDE => kcal +20 kcal

Kinetics Answers question, “How fast?” Rate is proportional to the concentration of reactants raised to a power. Rate law is experimentally determined. =>

Reaction Order For A + B  C + D, rate = k[A] a [B] b –a is the order with respect to A –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. The value of k depends on temperature as given by Arrhenius: ln k = -E a + lnA RT =>

Activation Energy Minimum energy required to reach the transition state. At higher temperatures, more molecules have the required energy. =>

Reaction-Energy Diagrams For a one-step reaction: reactants  transition state  products A catalyst lowers the energy of the transition state. =>

Energy Diagram for a Two-Step Reaction Reactants  transition state  intermediate Intermediate  transition state  product =>

Rate-Determining Step Reaction intermediates are stable as long as they don’t collide with another molecule or atom, but they are very reactive. Trapping agents are used to determine intermediate Transition states are at energy maxima. The reaction step with highest Ea will be the slowest, therefore rate-determining for the entire reaction.

Rate, E a, and Temperature =>

Conclusions With increasing E a, 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). =>

Chlorination of Propane There are six 1  H’s and two 2  ’s. We expect 3:1 product mix, or 75% 1-chloropropane and 25% 2- chloropropane. Typical product mix: 45% 1-chloropropane and 55% 2- chloropropane. Therefore, not all H’s are equally reactive. => 1  C 2  C

Reactivity of Hydrogens To compare hydrogen reactivity, find amount of product formed per hydrogen: 45% 1-chloropropane from 6 hydrogens and 55% 2-chloropropane from 2 hydrogens. 45%  6 = 7.5% per primary H and 55%  2 = 27.5% per secondary H Secondary H’s are 27.5%  7.5% = 3.7 times more reactive toward chlorination than primary H’s. =>

Chlorination of Isobutane methylpropane 36%64% reactivity of 3 o = 36/1 = 36% per 1 H reactivity of 1 o = 64/9 = Rel reactive of 3 o / 1 o = 36/7.1 = 5.1 RelativeReactivity = 3 o > 2 o > 1 o

Free Radical Stabilities Energy required to break a C-H bond decreases as substitution on the carbon increases. Stability: 3  > 2  > 1  > methyl  H(kcal) 91, 95, 98, 104 =>

Chlorination Energy Diagram Lower E a, faster rate, so more stable intermediate is formed faster. =>

There are six 1  H’s and two 2  ’s. We expect 3:1 product mix, or 75% 1-bromopropane and 25% 2-bromopropane. Typical product mix: 3% 1-bromopropane and 97% 2-bromopropane !!! Bromination is more selective than chlorination. => 1  C 2  C Bromination of Propane

To compare hydrogen reactivity, find amount of product formed per hydrogen: 3% 1- bromopropane from 6 hydrogens and 97% 2- bromopropane from 2 hydrogens. 3%  6 = 0.5% per primary H and 97%  2 = 48.5% per secondary H Secondary H’s are 48.5%  0.5% = 97 times more reactive toward bromination than primary H’s. => Reactivity of Hydrogens

Bromination Energy Diagram Note larger difference in E a Why endothermic? =>

Bromination vs. Chlorination =>

Endothermic and Exothermic Diagrams =>

Hammond Postulate Related species that are similar in energy are also similar in structure. The structure of a transition state resembles the structure of the closest stable species. Transition state structure for endothermic reactions resemble the product. Transition state structure for exothermic reactions resemble the reactants. =>

Radical Inhibitors Often added to food to retard spoilage. 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. =>