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Chapter 4 The Study of Chemical Reactions Jo Blackburn Richland College, Dallas, TX Dallas County Community College District  2003,  Prentice Hall Organic.

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Presentation on theme: "Chapter 4 The Study of Chemical Reactions Jo Blackburn Richland College, Dallas, TX Dallas County Community College District  2003,  Prentice Hall Organic."— Presentation transcript:

1 Chapter 4 The Study of Chemical Reactions Jo Blackburn Richland College, Dallas, TX Dallas County Community College District  2003,  Prentice Hall Organic Chemistry, 5 th Edition L. G. Wade, Jr.

2 Chapter 42 Tools for Study To determine a reaction’s mechanism, look at:  Equilibrium constant  Free energy change  Enthalpy  Entropy  Bond dissociation energy  Kinetics  Activation energy =>

3 Chapter 43 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). =>

4 Chapter 44 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. =>

5 Chapter 45 Initiation Step A chlorine molecule splits homolytically into chlorine atoms (free radicals) =>

6 Chapter 46 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). =>

7 Chapter 47 Propagation Step (2) The methyl free radical collides with another chlorine molecule, producing the other product (methyl chloride) and regenerating the chlorine radical. =>

8 Chapter 48 Overall Reaction =>

9 Chapter 49 Termination Steps Collision of any two free radicals Combination of free radical with contaminant or collision with wall. Can you suggest others? =>

10 Chapter 410 Equilibrium constant K eq = [products] [reactants] For chlorination K eq = 1.1 x 10 19 Large value indicates reaction “goes to completion.” =>

11 Chapter 411 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 = 1.987 cal/K-mol and T = temperature in kelvins Since chlorination has a large K eq, the free energy change is large and negative. =>

12 Chapter 412 Problem Given that -X is -OH, the energy difference for the following reaction is -1.0 kcal/mol. What percentage of cyclohexanol molecules will be in the equatorial conformer at equilibrium at 25°C? =>

13 Chapter 413 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  =>

14 Chapter 414 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). =>

15 Chapter 415 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. =>

16 Chapter 416 Bond Dissociation Energy Bond breaking requires energy (+BDE) Bond formation releases energy (-BDE) Table 4.2 gives BDE for homolytic cleavage of bonds in a gaseous molecule. We can use BDE to estimate  H for a reaction. =>

17 Chapter 417 Which is more likely? Estimate  H for each step using BDE. 104 103 58 84 => 104 84 58 103

18 Chapter 418 Kinetics Answers question, “How fast?” Rate is proportional to the concentration of reactants raised to a power. Rate law is experimentally determined. =>

19 Chapter 419 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 =>

20 Chapter 420 Activation Energy Minimum energy required to reach the transition state. At higher temperatures, more molecules have the required energy. =>

21 Chapter 421 Reaction-Energy Diagrams For a one-step reaction: reactants  transition state  products A catalyst lowers the energy of the transition state. =>

22 Chapter 422 Energy Diagram for a Two-Step Reaction Reactants  transition state  intermediate Intermediate  transition state  product =>

23 Chapter 423 Rate-Determining 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 E a will be the slowest, therefore rate-determining for the entire reaction. =>

24 Chapter 424 Rate, E a, and Temperature =>

25 Chapter 425 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). =>

26 Chapter 426 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: 40% 1-chloropropane and 60% 2-chloropropane. Therefore, not all H’s are equally reactive. => 1  C 2  C

27 Chapter 427 Reactivity of Hydrogens To compare hydrogen reactivity, find amount of product formed per hydrogen: 40% 1-chloropropane from 6 hydrogens and 60% 2-chloropropane from 2 hydrogens. 40%  6 = 6.67% per primary H and 60%  2 = 30% per secondary H Secondary H’s are 30%  6.67% = 4.5 times more reactive toward chlorination than primary H’s. =>

28 Chapter 428 Predict the Product Mix Given that secondary H’s are 4.5 times as reactive as primary H’s, predict the percentage of each monochlorinated product of n-butane + chlorine. =>

29 Chapter 429 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 =>

30 Chapter 430 Chlorination Energy Diagram Lower E a, faster rate, so more stable intermediate is formed faster. =>

31 Chapter 431 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

32 Chapter 432 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

33 Chapter 433 Bromination Energy Diagram Note larger difference in E a Why endothermic? =>

34 Chapter 434 Bromination vs. Chlorination =>

35 Chapter 435 Endothermic and Exothermic Diagrams =>

36 Chapter 436 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. =>

37 Chapter 437 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. =>

38 Chapter 438 Reactive Intermediates Carbocations (or carbonium ions) Free radicals Carbanions Carbene =>

39 Chapter 439 Carbocation Structure Carbon has 6 electrons, positive charge. Carbon is sp 2 hybridized with vacant p orbital. =>

40 Chapter 440 Carbocation Stability Stabilized by alkyl substituents 2 ways: (1) Inductive effect: donation of electron density along the sigma bonds. (2) Hyperconjugation: overlap of sigma bonding orbitals with empty p orbital. =>

41 Chapter 441 Free Radicals Also electron- deficient Stabilized by alkyl substituents Order of stability: 3  > 2  > 1  > methyl =>

42 Chapter 442 Carbanions Eight electrons on C: 6 bonding + lone pair Carbon has a negative charge. Destabilized by alkyl substituents. Methyl >1  > 2  > 3   =>

43 Chapter 443 Carbenes Carbon is neutral. Vacant p orbital, so can be electrophilic. Lone pair of electrons, so can be nucleophilic. =>

44 Chapter 444 End of Chapter 4


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