1. CHE2201, Chapter 6 Learn, 1 An Overview of Organic Reactions Chapter 6 Suggested Problems -

2. CHE2201, Chapter 6 Learn, 2 6.1 Enthalpy Enthalpy (ΔH or q) is the heat energy exchange between the reaction and its surroundings at constant pressure Breaking a bond requires the system to absorb energy. WHY? Show on Figure 6.1 how much energy is needed

3. CHE2201, Chapter 6 Learn, 3 6.1 Enthalpy Bonds can break homolytically or heterolytically Bond dissociation energy (BDE) or ΔH for bond breaking generally represents the energy associated with HOMOlytic cleavage

4. CHE2201, Chapter 6 Learn, 4 More BDEs in table 6.1 6.1 Bond Dissociation Energies

5. CHE2201, Chapter 6 Learn, 5 6.1 BDEs Explain how heat energy is exchanged between the reaction (system) and the solution (surroundings) for each scenario below 1.H and F free radicals come together to form bonds 2.A C–Br bond is broken 3.A strong bond is broken and a weak bond is formed 4.A weak bond is broken and a strong bond is formed

6. CHE2201, Chapter 6 Learn, 6 6.1 BDEs Most reactions involve multiple bonds breaking and forming. The energy associated with each bond that breaks and forms must be considered If during a chemical reaction the temperature of the reaction solution DECREASES, what can be said about the relative potential energies, stabilities, and bond strengths for the reactants and products If during a chemical reaction, the temperature of the reaction solution INCREASES, what can be said about the relative potential energies, stabilities, and bond strengths for the reactants and products

7. CHE2201, Chapter 6 Learn, 7 6.1 Enthalpy ΔH Match the reaction coordinate diagrams below with the statements below –The reaction causes the surrounding temp. to DECREASE –The reaction causes the surrounding temp. to INCREASE

8. CHE2201, Chapter 6 Learn, 8 6.1 Enthalpy ΔH For a chemical reaction, why is the sign (+/-) for ΔH important? If you were performing a chemical reaction in a lab, what experimental considerations might you make if you knew ΔH = + If you were performing a chemical reaction in a lab, what experimental considerations might you make if you knew ΔH = -

9. CHE2201, Chapter 6 Learn, 9 6.2 Entropy ΔS Although most reactions are EXOthermic, there are many ENDOthermic reactions that occur Enthalphy and entropy must BOTH be considered when predicting whether a reaction will occur ENTROPY (ΔS) can be thought of as molecular disorder, randomness, or freedom Entropy may most accurately be thought of as the number of states that a molecule’s energy can be distributed over

10. CHE2201, Chapter 6 Learn, 10 6.2 Entropy ΔS If the energy of molecules can be distributed in a higher number of vibrational, rotational, and translational states, the sample will have a greater entropy. Molecules exhibit vibrational, rotational, and translational motion. Explain HOW Which of the two molecules below should have greater entropy, and WHY?

11. CHE2201, Chapter 6 Learn, 11 6.2 Entropy ΔS Consider the ENTROPY change for the following process The number of possible translational distributions for the molecules increases with increasing volume. HOW?

12. CHE2201, Chapter 6 Learn, 12 6.2 Entropy ΔS The total entropy change will determine whether a process is spontaneous (favors the forward direction) If ΔS tot is positive, the process is spontaneous. What if ΔS tot is negative? When the volume of a gas expands to fill a container, what should ΔS surr be? For chemical reactions, we must consider the entropy change for both the system (the reaction) and the surroundings (the solvent usually)

13. CHE2201, Chapter 6 Learn, 13 6.2 Entropy ΔS For each of the reactions below, predict the sign for ΔS sys Consider how a change in a molecule’s structure affects the number of possible translational, rotational, and/or vibrational distributions for the molecules? Why does ΔS surr depend on ΔH sys ?

14. CHE2201, Chapter 6 Learn, 14 6.2 Entropy ΔS How would the following conditions affect spontaneity (the degree to which the reaction is product favored)? 1.The reactions are exothermic 2.The reactions are highly endothermic 3.The reactions are slightly endothermic

15. CHE2201, Chapter 6 Learn, 15 6.3 Gibbs Free Energy ΔG We know that the spontaneity of a process depends only on ΔS tot ΔS sys can be measured or estimated ΔS surr depends on ΔH sys Plug in

16. CHE2201, Chapter 6 Learn, 16 6.3 Gibbs Free Energy ΔG Multiply both sides by -T » or How will the sign (+ or -) for ΔG affect spontaneity?

17. CHE2201, Chapter 6 Learn, 17 6.3 Gibbs Free Energy ΔG Consider the example reaction Predict the sign (+ or -) for ΔS sys In the reaction, two pi bonds are converted into two sigma bonds. Predict the sign (+ or -) for ΔH sys Predict the sign (+ or -) for ΔS surr Predict the sign (+ or -) for ΔG How will the spontaneity of the reaction depend on temperature?

18. CHE2201, Chapter 6 Learn, 18 6.3 Gibbs Free Energy ΔG If a process at a given temperature is calculated to have a (-) ΔG, the process is exergonic It will be spontaneous and favor the products Note that G is plotted rather than H Does the value for ΔG tell us about the rate of the reaction?

19. CHE2201, Chapter 6 Learn, 19 6.3 Gibbs Free Energy ΔG If a process at a given temperature is calculated to have a (+) ΔG, the process is endergonic It will be NONspontaneous and favors the reactants What does it mean exactly to favor the reactants?

20. CHE2201, Chapter 6 Learn, 20 6.4 Equilibria Consider an exergonic process with a (-) ΔG. Will every molecule of A and B be converted into products? –No, an equilibrium will eventually be reached –A spontaneous process will simply favor the products meaning there will be more products than reactants –The greater the magnitude of a (-)ΔG, the greater the equilibrium concentration of products

21. CHE2201, Chapter 6 Learn, 21 6.4 Equilibria Why doesn’t an exergonic process react 100% to give products? Why will some reactants remain? –The diagram shows one unit of A react with one unit of B –In reality, moles of reactants are present –How will concentrations of A, B, C, and D change as the reaction progresses?

22. CHE2201, Chapter 6 Learn, 22 6.4 Equilibria In any reaction, collisions are necessary As [A] and [B] decrease collisions between A and B will occur less often As [C] and [D] increase, collisions between C and D will occur more often –The more often C and D collide, the more often collisions will occur with enough free energy for the reverse reaction to take place Recall that equilibrium is dynamic and occurs when the forward and reverse reaction rates are equal

23. CHE2201, Chapter 6 Learn, 23 6.4 Equilibria Equilibrium is also the state with the lowest free energy overall Why does the equilibrium mixture have the least G? How would S tot compare for a mixture of all reactants and products versus only products?

24. CHE2201, Chapter 6 Learn, 24 6.4 Equilibria An equilibrium constant (K eq ) is used to show the degree to which a reaction is product or reactant favored K eq, ΔG, ΔH, and ΔS are thermodynamic terms. They do not describe reaction kinetics. What is the difference?

25. CHE2201, Chapter 6 Learn, 25 6.4 Equilibria What trends do you notice in table 6.2?

26. CHE2201, Chapter 6 Learn, 26 6.5 Kinetics Recall that a (-) sign for ΔG tells us a process is product favored (spontaneous) That does NOT tell us anything about the RATE or kinetics for the process Some spontaneous processes are fast such as explosions. Can you think of other examples? Some spontaneous processes are slow such as C (diamond)  C (graphite). Can you think of other examples?

27. CHE2201, Chapter 6 Learn, 27 6.5 Kinetics The rate of a reaction tells us how many molecules are reacting in a given period of time Give some examples for typical reaction rate units Recall that for reactions to take place, reactants must collide with sufficient force. WHY?

28. CHE2201, Chapter 6 Learn, 28 6.5 Kinetics The reaction rate (the number of collisions that will result in product production in a given period of time) is affected by multiple factors 1.The concentrations of the reactants 2.The Activation Energy 3.The Temperature 4.Geometry and Sterics 5.The presence of a catalyst How will an increase in [reactant] generally affect the reaction rate? WHY?

29. CHE2201, Chapter 6 Learn, 29 6.5 Rate Law Equations To quantify how much the reactant concentration affects the rate of reaction, the Rate Law equation is used The degree to which a change in [reactant] will affect the Rate is known as the order. The order is represented by x and y in the Rate Law equation

30. CHE2201, Chapter 6 Learn, 30 6.5 Rate Law Equations Consider a generic reaction that is known to be first order with respect to A and zero order with respect to B: A + B  C + D –Write the appropriate Rate Law –How should he rate change if [A] were doubled? –How should he rate change if [B] were doubled?

31. CHE2201, Chapter 6 Learn, 31 6.5 Rate Law Equations Consider a generic reaction that is known to be first order with respect to A and first order with respect to B: A + B  C + D –Write the appropriate Rate Law –How should he rate change if [A] were doubled? –How should he rate change if [B] were doubled?

32. CHE2201, Chapter 6 Learn, 32 6.5 Rate Law Equations Consider a generic reaction that is known to be second order with respect to A and first order with respect to B: A + B  C + D –Write the appropriate Rate Law –How should he rate change if [A] were doubled? –How should he rate change if [B] were doubled? How are the orders (exponents) determined for a Rate Law?

33. CHE2201, Chapter 6 Learn, 33 6.5 Factors that Affect Rates Locate the Activation Energy in figure 6.13? Why must the free energy (G) increase before the products can be formed? Why would E act be different for different reactions? Free energy (G)

34. CHE2201, Chapter 6 Learn, 34 6.5 Factors that Affect Rates Temperature is a measure of a system’s average kinetic energy Would you expect there to be a temperature below which the reaction rate is zero and above which the reaction rate instantaneous for all molecules? WHY or WHY NOT? Free energy (G)

35. CHE2201, Chapter 6 Learn, 35 6.5 R Factors that Affect Rates Why does a lower E act result in a greater reaction rate? Free energy (G) Free energy (G)

36. CHE2201, Chapter 6 Learn, 36 6.5 Factors that Affect Rates Why does increasing the temperature increase the rate of reaction?

37. CHE2201, Chapter 6 Learn, 37 6.5 Factors that Affect Rates How might geometry and sterics affect the reaction rate? How might the presence of a catalyst affect the reaction rate? Free energy (G)

38. CHE2201, Chapter 6 Learn, 38 6.6 Energy Diagrams Distinguish between kinetics and thermodynamics Free energy (G) Free energy (G)

39. CHE2201, Chapter 6 Learn, 39 For the energy diagram below, which pathway do you think is favored? WHY? 6.6 Kinetics vs Thermodynamics Will a decrease in temperature affect which pathway is favored? Will an increase in temperature affect which pathway is favored? Free energy (G)

40. CHE2201, Chapter 6 Learn, 40 6.6 Kinetics vs Thermodynamics For the energy diagram below, which pathway is kinetically favored? Which pathway is thermodynamically favored? How can temperature be used to control which set of products predominate? Free energy (G)

41. CHE2201, Chapter 6 Learn, 41 6.6 Transition States vs Intermediates Free energy (G)

42. CHE2201, Chapter 6 Learn, 42 6.6 Transition States A transition state occurs at an energy maxima Transition states exist for a fleeting moment; they cannot be isolated or directly observed Why are transition states so unstable? Free energy (G)

43. CHE2201, Chapter 6 Learn, 43 6.6 Intermediates An intermediate occurs at an energy minima Intermediates often exist long enough to be observed because bonds are NOT in the process of breaking or forming Free energy (G)

44. CHE2201, Chapter 6 Learn, 44 6.6 The Hammond Postulate Two points on an energy diagram that are close in energy should be similar in structure Free energy (G)

45. CHE2201, Chapter 6 Learn, 45 For each of the diagrams below, will the transition state structure look more like the reactants or the products? 6.6 The Hammond Postulate Free energy (G) Free energy (G)

46. CHE2201, Chapter 6 Learn, 46 6.6 The Hammond Postulate Draw a reaction coordinate diagram for the generic exergonic reaction sequence below. Label the axes, reactants, products, intermediates, and transition states A  B + C C + D  E Net reaction: A + D  B + E How would the diagram look different if it were endergonic?

47. CHE2201, Chapter 6 Learn, 47 6.7 Nucleophiles and Electrophiles A major focus in this course is on predicting reaction products for ionic reactions and explaining HOW such reactions work Ionic or polar reactions result from the force of attraction between opposite charges Ionic reactions are also guided by the octet rule Consider how methyl chloride and methyl lithium might react

48. CHE2201, Chapter 6 Learn, 48 6.7 Nucleophiles When an atom carries a formal or partial negative charge and an available pair of electrons, it is considered a nucleophile It will love to attack a nucleus. WHY? Explain how the molecules below are nucleophiles What is the difference between a nucleophile and a Lewis Base? NOTHING!

49. CHE2201, Chapter 6 Learn, 49 6.7 Electrophiles When an atom carries a formal or partial positive charge and can accept a pair of electrons, it is considered a electrophile It will love available electrons. WHY? Explain how the molecules above are electrophiles What is the difference between an electrophile and a Lewis Acid? NOTHING!

50. CHE2201, Chapter 6 Learn, 50 6.7 Electrophiles Label all of the nucleophilic and electrophilic sites on the following molecule

51. CHE2201, Chapter 6 Learn, 51 6.8 Mechanisms and Arrow Pushing We use arrows to show how electrons move when bonds break and form It will be a huge benefit in this course to master the skill of arrow pushing There are four main ways that electrons move in ionic reactions 1.Nucleophilic Attack 2.Loss of a Leaving Group 3.Proton Transfers (Acid/Base) 4.Rearrangements

52. CHE2201, Chapter 6 Learn, 52 6.8 Nucleophilic Attack When you identify a nucleophilic site and an electrophilic site, the arrow shows the nucleophile attacking The tail of the arrow starts on the electrons (- charge) The head of the arrow ends on a nucleus (+ charge) The electrons end up being sharing rather than transferred

53. CHE2201, Chapter 6 Learn, 53 6.8 Nucleophilic Attack Nucleophilic attack may appear to occur in two steps The alcohol is the nucleophile in this example. It attacks a carbon with a δ+ charge The second arrow shows the flow of negative charge. WHY is it necessary? The second arrow could be thought of as a resonance arrow. HOW?

54. CHE2201, Chapter 6 Learn, 54 6.8 Loss of a Leaving Group Loss of a leaving group occurs when a bond breaks and one atom from the bond takes BOTH electrons For the molecule below, draw the structure that will result after the leaving group is gone Which arrow shows the loss of a leaving group? What is the purpose of the other arrows?

55. CHE2201, Chapter 6 Learn, 55 6.8 Proton Transfers Recall from Chapter 3 that a base is protonated when it uses a pair of electrons to take an H + from the acid. The acid retains its electron pair A group can also be deprotonated (sometimes shown by writing –H + over the reaction arrow) or

56. CHE2201, Chapter 6 Learn, 56 6.8 Proton Transfers Multiple arrows may be necessary to show the complete electron flow when a proton is exchanged Such electron flow can also be thought of as a proton transfer combined with resonance

57. CHE2201, Chapter 6 Learn, 57 Carbocations can be stabilized by neighboring groups through slight orbital overlapping called hyperconjugation 6.8 Carbocation Rearrangements

58. CHE2201, Chapter 6 Learn, 58 6.8 Carbocation Rearrangements Hyperconjugation and induction can both be invoked to explain the stability trend below. HOW? If a carbocation can INTRAmolecularly rearrange to become more stable, it will likely do so before reacting with a nucleophile. WHY?

59. CHE2201, Chapter 6 Learn, 59 6.8 Carbocation Rearrangements Two types of carbocation rearrangement are common – Hydride shift – Methyl shift Shifts can only occur from an adjacent carbon. WHY? Do the shifts above make the carbocation more stable?

60. CHE2201, Chapter 6 Learn, 60 6.9 Combining Arrow Pushing Patterns Classify each step in the following mechanism

61. CHE2201, Chapter 6 Learn, 61 6.9 Combining Arrow Pushing Patterns Many times a single step in a mechanism will include more than one arrow pushing pattern Identify the patterns below There are hundreds of mechanisms that involve these key patterns

62. CHE2201, Chapter 6 Learn, 62 6.10 Arrow Pushing Rules To draw reasonable mechanisms, a few key rules should be followed 1.The arrow starts ON A PAIR OF ELECTRONS (a bonded pair or a lone pair) –Don’t make the mistake of starting an arrow on a nucleus! Both arrows below are incorrect. WHY?

63. CHE2201, Chapter 6 Learn, 63 6.10 Arrow Pushing Rules A few key rules should be followed 2.The arrow ends ON A NUCLEUS (electrons become a lone pair) or between two NUCLEI (electrons move into position to become a bond)

64. CHE2201, Chapter 6 Learn, 64 6.10 Arrow Pushing Rules A few key rules should be followed 3.Avoid breaking the octet rule. NEVER give C, N, O, or F more than 8 valence electrons

65. CHE2201, Chapter 6 Learn, 65 6.10 Arrow Pushing Rules A few key rules should be followed 4.Draw arrows that follow the 4 key patterns we outlined –The arrow below is unreasonable. WHY?

66. CHE2201, Chapter 6 Learn, 66 6.10 Arrow Pushing Rules Fill in necessary arrows for the reaction below

67. CHE2201, Chapter 6 Learn, 67 6.10 Arrow Pushing Rules For each mechanism below, identify the unreasonable arrows and redraw the mechanism with correct arrows

68. CHE2201, Chapter 6 Learn, 68 6.11 Carbocation Rearrangements When you encounter a carbocation, you must consider all possible rearrangements (Hydride and methyl shifts) 1.Identify all adjacent carbons 2.Identify all –H and –CH 3 groups on the adjacent carbons that are capable of shifting

69. CHE2201, Chapter 6 Learn, 69 In this case, a hydride shift will result in a more stable tertiary carbocation 6.11 Carbocation Rearrangements When you encounter a carbocation, you must consider all possible rearrangements (Hydride and methyl shifts) 3.Imagine each of the groups shifting to see which yields the most stable resulting carbocation

70. CHE2201, Chapter 6 Learn, 70 6.11 Carbocation Rearrangements Complete the same analysis for the molecule below 1.Identify all adjacent carbons 2.Identify all –H and –CH 3 groups capable of shifting 3.Determine which shift yields the most stable carbocation Recall that allylic carbocations are especially stable

71. CHE2201, Chapter 6 Learn, 71 6.12 Reversible and Irreversible Reaction Arrows Why are some reactions drawn as equilibria and others as essentially irreversible? The question of reversibility is both a kinetic and a thermodynamic question. HOW?

72. CHE2201, Chapter 6 Learn, 72 6.12 Reversible and Irreversible Reaction Arrows Consider nucleophilic attack Draw a mechanism for the reverse reaction How do we judge whether it will be reversible?

73. CHE2201, Chapter 6 Learn, 73 6.12 Reversible and Irreversible Reaction Arrows If the attacking nucleophile is also a good leaving group, it will be a reversible attack – The reverse reaction will have a relatively low transition state energy (kinetically favored) – The reactants and products of the reaction will be similar in energy allowing significant quantities of both to exist at equilibrium (thermodynamic equilibrium)

74. CHE2201, Chapter 6 Learn, 74 6.12 Reversible and Irreversible Reaction Arrows If the attacking nucleophile is a poor leaving group, it will essentially be an irreversible attack – The reverse reaction will have a relatively HIGH transition state energy (kinetically disfavored) – The products will be much lower in energy so an insignificant quantity of reactant will remain at equilibrium

75. CHE2201, Chapter 6 Learn, 75 6.12 Reversible and Irreversible Reaction Arrows Consider loss or a leaving group Draw a mechanism for the reverse reaction How do we judge whether it will be reversible?

76. CHE2201, Chapter 6 Learn, 76 6.12 Reversible and Irreversible Reaction Arrows Consider proton transfer We analyzed this in detail in chapter 3 using pK a values How do we judge reversiblity?

77. CHE2201, Chapter 6 Learn, 77 6.12 Reversible and Irreversible Reaction Arrows Consider proton transfer If the pK a difference is 10 units or more, it is generally considered irreversible

78. CHE2201, Chapter 6 Learn, 78 6.12 Reversible and Irreversible Reaction Arrows Carbocations are generally considered irreversible thermodynamically. WHY?

79. CHE2201, Chapter 6 Learn, 79 6.12 Reversible and Irreversible Reaction Arrows When considering thermodynamic equilibrium, in addition to comparing relative energies, Le Châtelier’s principle must also be considered. WHY is the reaction below irreversible?

80. CHE2201, Chapter 6 Learn, 80 Additional Practice Problems Consider electrocyclic ring opening Predict the sign (+ or -) for ΔS sys In the reaction, 1 pi bond is converted into 1 sigma bond. Predict the sign (+ or -) for ΔH sys Predict the sign (+ or -) for ΔS surr Predict the sign (+ or -) for ΔG How will the spontaneity of the reaction depend on temperature?

81. CHE2201, Chapter 6 Learn, 81 Additional Practice Problems Explain why an equilibrium mixture is always lower in free energy than pure products or pure reactants.

82. CHE2201, Chapter 6 Learn, 82 Additional Practice Problems Reactants A and B can react by two different pathways. Pathway 1 is thermodynamically favored, and pathway 2 is kinetically favored. Draw a reaction coordinate diagraph to illustrate the relative energies and explain which pathway will be favored at low temperatures versus higher temperatures.

83. CHE2201, Chapter 6 Learn, 83 Additional Practice Problems Propose a mechanism for the reaction below that explains its Rate Law. 2A + B  C + DRate=k[A] 2

84. CHE2201, Chapter 6 Learn, 84 Additional Practice Problems Label all of the nucleophilic and electrophilic sites on the following molecules

85. CHE2201, Chapter 6 Learn, 85 Additional Practice Problems Draw a mechanism for a generic nucleophilic attack followed by a proton transfer. Draw a mechanism for a generic loss of leaving group followed by a carbocation rearrangement.

86. CHE2201, Chapter 6 Learn, 86 Additional Practice Problems For each of the mechanistic steps below, identify what is incorrect about it.

87. CHE2201, Chapter 6 Learn, 87 Extra Slides

88. CHE2201, Chapter 6 Learn, 88 For a reaction to proceed as written, the energy of the products must be lower than the energy of the reactants. –ie. Energy must be released. The energy change that occurs during a chemical reaction is called the Gibbs free energy (  G). –(  G = G products – G reactants ) This energy is released on the favored side of an equilibrium reaction –For a favorable reaction,  G has a negative value, meaning energy is lost by the chemical system and released to the surroundings, usually as heat – exergonic. –For an unfavorable reaction,  G has a positive value, meaning that energy is absorbed by the chemical system from the surroundings – endergonic. Free Energy and Equilibrium

89. CHE2201, Chapter 6 Learn, 89 The standard free energy change at 1 atm pressure and 298 K is  Gº The relationship between free energy change and an equilibrium constant is:   Gº = - RT ln K eq where –R = 8.314 J/(K. mol) or 1.987 cal/(K. mol) –T = temperature in Kelvin –ln K eq = natural logarithm of K eq Numeric Relationship of Keq and Free Energy Change

90. CHE2201, Chapter 6 Learn, 90 Relationship between K eq and  G If K eq > 1, energy is released to the surroundings (exergonic reaction) If K eq < 1, energy is absorbed from the surroundings (endergonic reaction)

91. CHE2201, Chapter 6 Learn, 91 The reaction of ethylene (ethene) with HBr has K eq = 7.1 x 10 7. What is  Gº in kJ/mol at 298K?  Gº = - RT ln K eq where –R = 8.314 J/(K. mol) –T = 298 K –ln K eq = natural log of K eq = ln (7.1 x 10 7 ) = 18.08 A Problem  Gº = - RT ln K eq = - [8.314 J/(K. Mol)] (298 K) (18.08) = - 44,800 J/mol = - 44.8 kJ/mol

92. CHE2201, Chapter 6 Learn, 92 Free Energy Change - Enthalpy Component  G made up of an enthalpy term and an entropy term  G o =  H o –T  S o Enthalpy change, (  H), also called heat of reaction -Measures the change in total bonding energy during a reaction.  H = negative -Products have less energy than reactants -Products more stable, have stronger bonds -Heat is released (exothermic)  H = positive -Products have more energy than reactants -Products less stable, have weaker bonds -Heat is absorbed (endothermic)

93. CHE2201, Chapter 6 Learn, 93 Free Energy Change - Entropy Component  G made up of an enthalpy term and an entropy term  G o =  H o –T  S o Entropy change, (  S) -Measures the change in molecular randomness during a reaction.  S = positive -Increase in entropy -More freedom of movement; more randomness -eg. A B + C  S = negative -Decrease in entropy -Freedom of movement restricted; less randomness -eg. A + B C

94. CHE2201, Chapter 6 Learn, 94 Free Energy Change,  G For the reaction of ethene with HBr at RT, the approximate values are: What does this data say about the bond energies? The entropy? Does the reaction proceed as written? Do both of these favor the reaction proceeding as written?