Organic Chemistry, 7e (Wade) Chapter 4 The Study of Chemical Reactions 1) __________ is the study of reaction rates. Answer: Kinetics Diff: 1 Section: 4.2 2) What is meant by the mechanism of a chemical reaction? Answer: The mechanism of a reaction is the complete, step-by-step description of exactly which bonds break, which bonds form, and the order in which these events occur in the transformation of reactants into products. Diff: 2 Section: 4.2 3) The chlorination of methane is characterized by a high quantum yield. Explain what this means. Answer: Many molecules of product are formed for every photon of light which is absorbed by the reaction mixture. Diff: 3 Section: 4.2 4) Within the visible spectrum, it has been experimentally determined that blue light is the most effective in initiating the chlorination of methane. What is the mechanistic significance of this observation? Answer: Blue light is effectively absorbed by chlorine but not by methane. Thus the initiating step is most likely the production of two chlorine atoms from molecular chlorine. Diff: 3 Section: 4.2 5) Which of the following is not a possible termination step in the free radical chlorination of methane? A) ∙CH3 + Cl2 → CH3Cl + Cl∙ B) ∙CH3 + Cl∙ → CH3Cl C) ∙CH3 + ∙CH3 → CH3CH3 D) ∙CH3 + wall → CH3-wall E) Cl∙ + wall → Cl-wall Answer: A Diff: 2 Section: 4.3 1 6) In the first propagation step of the free radical chlorination of methane, which of the following occurs? A) Cl2 dissociates B) A chlorine radical abstracts a hydrogen C) A carbon radical reacts with Cl2 D) A carbon radical reacts with a chlorine radical E) Two chlorine radicals combine Answer: B Diff: 2 Section: 4.3 7) In the reaction of Cl2 with ethane and UV light, which of the following reactions would be a chain termination event(s)? I) Cl∙ + CH3-CH3 → CH3-CH2-Cl + H∙ II) Cl∙ + CH3-CH3 → CH3-H2C∙ + HCl III) Cl∙ + CH3-H2C∙ → CH3-CH2-Cl IV) Cl2 + CH3-H2C∙ → CH3-CH2-Cl + Cl∙ V) Cl2 + UV light → C l∙ + Cl∙ A) reaction V B) reactions I and IV C) reactions III and IV D) reactions I and II E) reaction III Answer: E Diff: 2 Section: 4.3 8) In the reaction of Cl2 with ethane and UV light, which of the following reactions would be a propagation event(s)? I) Cl∙ + CH3-CH3 → CH3-CH2-Cl + H∙ II) Cl∙ + CH3-CH3 → CH3-H2C∙ + HCl III) Cl∙ + CH3-H2C∙ → CH3-CH2-Cl IV) Cl2 + CH3-H2C∙ → CH3-CH2-Cl + Cl∙ V) Cl2 + UV light → C l∙ + Cl∙ A) reactions I and V B) reactions II, III and IV C) reactions I and IV D) reactions II and IV E) reactions I, II and IV Answer: D Diff: 2 Section: 4.3 2 3 11) Species with unpaired electrons are called __________. A mechanistic step that does not yield a free-radical stops the chain.3 13) Provide the two propagation steps in the free-radical chlorination of ethane. Answer: CH3CH3 + Cl∙ → CH3CH2∙ + HCl CH3CH2∙ + ClCl → CH3CH2Cl + Cl∙ Diff: 2 Section: 4. Answer: In order for a free-radical chain to propagate. Such a step is known as a termination step.3 12) Explain how the termination step in a free-radical chain reaction stops the chain.9) Which of the following is a propagation step in the free radical chlorination of dichloromethane? A) ∙ CHCl2 + Cl2 → CHCl3 + Cl∙ B) ∙ CHCl2 + Cl∙ → CHCl3 C) CH2Cl2 + Cl∙ → CHCl3 + H∙ D) Cl2 + UV light → 2 Cl∙ E) ∙ CHCl2 + ∙ CHCl2 → CHCl2CHCl2 Answer: A Diff: 2 Section: 4.3 10) When the reaction between methane and chlorine is photochemically initiated. Answer: Cl-Cl + photon (hν) → 2 Cl∙ Diff: 2 Section: 4. Diff: 1 Section: 4. each mechanistic step must yield a free-radical species as one of its products. which of the following compounds cannot be formed through a termination reaction? A) CH3Cl B) HCl C) CH3CH3 D) Cl2 Answer: B Diff: 3 Section: 4.3 14) Write an equation to describe the initiation step in the chlorination of methane.3 3 . Answer: radicals or free radicals Diff: 1 Section: 4. Diff: 2 Section: 4. stepwise mechanism for the following reaction. then: A) Keq < 0 B) Keq = 0 C) 0 < Keq < 1 D) Keq = 1 E) Keq > 1 Answer: C Diff: 1 Section: 4. Answer: CH3Cl + Cl∙ → ∙ CH2Cl + HCl ∙ CH2Cl + Cl-Cl → CH2Cl2 + Cl∙ CH2Cl2 + Cl∙ → ∙ CHCl2 + HCl ∙ CHCl2 + Cl-Cl → CHCl3 + Cl∙ CHCl3 + Cl∙ → ∙ CCl3 + HCl ∙ CCl3 + Cl-Cl → CCl4 + Cl∙ Diff: 3 Section: 4. carbon tetrachloride is one of the components of the final reaction mixture. if ΔG° is greater than zero. Answer: Diff: 3 Section: 4.4 4 .3 18) For a given reaction.3 16) When light is shined on a mixture of chlorine and chloromethane. What experimental conditions should be used to favor the production of chloromethane over the other chlorinated products? Answer: Make sure the molar ratio of methane to chlorine is relatively large.3 17) Write a detailed. Propose a series of mechanistic steps which explain this observation.15) Chlorination of methane can result in a mixture of chlorinated products. 15% B) 0.43. If the ΔG° of this reaction is -0.0. the Keq is __________ and the % conversion is __________. the Keq is __________ and the % conversion is __________. A) 0. If the ΔG° of this reaction is +1. 84% Answer: A Diff: 2 Section: 4. 50% D) 2.18.43. B) The reaction must be endothermic. 50% D) 2.4 5 . 30% C) 1.4.19) If ΔG° for a given reaction at 25°C is less than zero.5 kcal/mol. 50% D) 2.18.3.4 21) Consider the reaction of A being converted into B at 25°C. C) Keq is greater than zero. 84% Answer: D Diff: 2 Section: 4.0.4.5 kcal/mol.4 22) Consider the reaction of A being converted into B at 25°C. which of the following statements also correctly describes this reaction at this temperature? A) The reaction must be exothermic. 70% E) 5. E) Both B and C are true. 30% C) 1. 15% B) 0.3.18. 84% Answer: B Diff: 2 Section: 4.4. Answer: C Diff: 2 Section: 4. 70% E) 5.3. 70% E) 5. 30% C) 1. If the ΔG° of this reaction is +0. the Keq is __________ and the % conversion is __________. 15% B) 0. D) Both A and C are true.43.4 20) Consider the reaction of A being converted into B at 25°C.0 kcal/mol. A) 0. A) 0.0. Calculate the equilibrium concentration of D at 25°C.+ CH3Cl ↔ CH3OH + ClGiven this information which of the following statements must be true? (R = 8. Answer: A Diff: 3 Section: 4. in other words this reaction proceeds to near completion as written.40M E) 1. R = 1.4 6 .2 at 25°C.0 kcal/mol. given that the starting concentrations of A and B are 2M and that ΔG° for the reaction is 1. HO.40M B) 0.0 kJ/mol Diff: 2 Section: 4. what Keq results if at equilibrium 80% of A has become B? Answer: 4 Diff: 2 Section: 4.4 24) Consider the following substitution reaction with a ΔG° value of -91.00M D) 1.314 J/K∙ mol] Answer: -2.1 kJ/mole.4 26) Given a K of 2. [R = 8.4 25) Given a reaction in which reactant A is converted only to product B at 25°C.314 J/K∙ mol] Answer: 2.60M Answer: B Diff: 3 Section: 4.987 cal/mol∙K. calculate the corresponding ΔG° in kJ/mol.45 at 25°C. left to right under standard conditions B) The Keq at 25°C for this reaction is very small (<1). calculate the corresponding ΔG° in kJ/mol.315 J/mole K) A) The Keq at 25°C for this reaction is very large.4 27) Given a K of 0.23) Assume the reaction A + B → C + D proceeds to equilibrium. A) 0. In other words there is more product than at 25°C D) At 250°C the equilibrium concentration of products and reactants is nearly the same E) Both A and C are correct.0 kJ/mol Diff: 2 Section: 4. [R = 8. in other words this reaction does not proceed from left to right but rather is favored from right to left under standard conditions C) At 250°C the equilibrium concentration is shifted right in favor of the products (CH3OH and Cl-).60M C) 1. 4 31) For the reaction A + B → C + D.314 J/K∙ mol] Answer: 25 Diff: 2 Section: 4.e. If at equilibrium at 25°C the concentration of A is 20% of the initial concentration of A.4 29) Given a ΔG° of 0. What is the corresponding equilibrium constant at 25°C? R = 1.82 kcal/mol Diff: 2 Section: 4. D) The entropy term makes a greater contribution to ΔG° in endothermic reactions.5 7 .28) Given a ΔG° of -8. Answer: 4648 Diff: 3 Section: 4.0 kJ/mol at 25°C. [R = 8.4 30) Consider the transformation of A to B (i. [R = 8. determine the value of ΔG° (in kcal/mol) for this reaction.4 32) Which is a measure of the randomness of a system? A) entropy B) enthalpy C) free energy D) halogenation E) stoichiometry Answer: A Diff: 1 Section: 4.987 cal/mol∙K.987 cal/mol∙K. R = 1. E) The entropy term always makes a more significant contribution to ΔG° than does the enthalpy term. calculate the percentage of the axial conformer at 25°C .. B) The entropy term makes a greater contribution to ΔG° at high temperatures. A → B). calculate the corresponding K . ΔG° = -5.314 J/K∙ mol] Answer: 42% Diff: 2 Section: 4. Answer: -0.8 kJ/mol at 25°C for the equilibrium shown below. Answer: B Diff: 2 Section: 4.5 33) Which of the following statements correctly describes the contribution of ΔS° to ΔG°? A) The entropy term makes a greater contribution to ΔG° at low temperatures.00 kcal/mol. C) The entropy term makes a greater contribution to ΔG° in exothermic reactions. weaker bonds are broken and stronger bonds are formed.5 35) Which of the following correctly expresses the standard Gibbs free energy change of a reaction in terms of the changes in enthalpy and entropy? A) ΔG° = ΔH° .5 38) In an exothermic reaction. then: A) ΔS° < 0 B) ΔS° > 0 C) ΔH° < 0 D) ΔH° > 0 E) both B and D Answer: C Diff: 2 Section: 4.TΔS° B) ΔG° = ΔH° + TΔS° C) ΔG° = ΔS° .5 8 .5 36) Which of the following is true for the initiation step of a free radical chlorination reaction? A) ΔH° > 0 and ΔS° > 0 B) ΔH° > 0 and ΔS° < 0 C) ΔH° < 0 and ΔS° > 0 D) ΔH° < 0 and ΔS° < 0 E) ΔH° = 0 and ΔS° = 0 Answer: A Diff: 2 Section: 4. Diff: 1 Section: 4.5 37) Which of the following is true for the termination step of a free radical chlorination reaction? A) ΔH° > 0 and ΔS° > 0 B) ΔH° > 0 and ΔS° < 0 C) ΔH° < 0 and ΔS° > 0 D) ΔH° < 0 and ΔS° < 0 E) ΔH° = 0 and ΔS° = 0 Answer: D Diff: 2 Section: 4.34) If a reaction is exothermic.TΔH° D) ΔG° = ΔS° + TΔH° E) none of the above Answer: A Diff: 2 Section: 4. are stronger bonds broken and weaker bonds formed or are weaker bonds broken and stronger bonds formed? Answer: In an exothermic reaction. 5 41) Predict the signs of ΔG° and ΔS° in the reaction of cyclohexene with H2 to form cyclohexane. or impossible to predict? Explain your reasoning.5 43) The hydrogenation of acetylene to produce ethane is shown below.39) If stronger bonds are formed and weaker bonds are broken. Is ΔS° for this reaction positive. less than zero.5 40) Does one expect ΔS° in a propagation step of the free-radical chlorination of methane to be greater than zero. Answer: ΔH° < 0 and ΔS° < 0 Diff: 2 Section: 4. negative.5 42) Predict the sign of ΔS° in the combustion of propane. two reactant molecules generate two product molecules. then the reaction is __________. Answer: ΔS°>0 Diff: 2 Section: 4. C2H2 (g) + 2H2 (g) → C2H6 (g) Answer: ΔS° < 0. or approximately equal to zero? Briefly explain your choice. Answer: Exothermic Diff: 2 Section: 4. Therefore. This similarity in the number of molecular species means that the disorder in the reaction is neither greatly increased nor diminished. Diff: 2 Section: 4.5 9 . and randomness. Answer: The propagation steps of the free-radical chlorination of methane are shown below. one expects ΔS° in either propagation step to be approximately equal to zero. Diff: 2 Section: 4. The three reactant molecules have significantly more freedom of motion. than does the single molecule of product into which they are converted. CH4 + Cl∙ → CH3∙ + H–Cl CH3∙ + ClCl → CH3Cl + Cl∙ In each of the steps. A) homolytically B) heterolytically C) so as to produce the more stable pair of ions D) via hydrogenation E) none of the above Answer: A Diff: 1 Section: 4.6 46) Energy is __________ when bonds are formed and is __________ when bonds are broken. therefore.44) The bond dissociation energy is the amount of energy required to break a bond __________. exothermic Answer: A Diff: 2 Section: 4.6 45) Predict the enthalpy (ΔH) value for the theoretical reaction below. A-B 63 + C-D 88 → A-C 47 + B-D 96 A) +8 Kcal/mol. endothermic D) +8 Kcal/mol. The bond dissociation energy for each bond in Kcal/mol is shown below each reactant and product.6 10 . A) released / consumed / exothermic B) released / consumed / endothermic C) consumed / released / exothermic D) consumed / released / endothermic E) consumed / released / isothermic Answer: B Diff: 2 Section: 4. exothermic C) +16 Kcal/mol. bond dissociation energies are always __________. and indicate whether it is endothermic or exothermic. endothermic B) -8 Kcal/mol. 6 48) Which compound has the smaller bond dissociation energy for its carbon-chlorine bond. CH3Cl or (CH3)3CCl? Explain your reasoning. Cl–Cl = 58 kcal/mol. Tertiary alkyl radicals are more stable than primary due to hyerconjugative or inductive stabilization.6 11 . Diff: 2 Section: 4. H–Cl = 103 kcal/mol) hv light H2 + Cl2 hv light A) H2 2 HCl H∙ + H∙ H∙ + Cl2 → Cl∙ + HCl H∙ + Cl∙ → HCl hv light B) Cl2 Cl∙ + Cl∙ Cl∙ + H2 → H∙ + HCl H∙ + Cl2 → HCl + Cl∙ hv light C) H2 H∙ + H∙ H∙ + Cl2 → Cl∙ + HCl Cl∙ + H2 → HCl + H∙ hv light D) Cl2 Cl∙ + Cl∙ Cl∙ + H2 → H∙ + HCl H∙ + Cl∙ → HCl Answer: B Diff: 3 Section: 4. Answer: The carbon-chlorine bond of (CH3)3CCl has the smaller bond dissociation energy. The carbon radical that results upon bond cleavage is a tertiary radical.47) Which of the presented mechanisms would be the most energetically favorable and thus the most likely mechanism to actually occur for the following free radical chain reaction? (bond dissociation energies -.H-H = 104 kcal/mol. CH3-Br CH3CH2-Br (CH3)2CH-Br (CH3)3C-Br 70 68 68 65 These data show that the carbon-bromine bond is weakest when bromine is bound to a __________.7 12 .49) Consider the bond dissociation energies listed below in kcal/mol. A) methyl carbon B) primary carbon C) secondary carbon D) tertiary carbon E) quaternary carbon Answer: D Diff: 1 Section: 4. estimate the ΔH° for the propagation step (CH3)2CH∙ + Cl2 (CH3)2CHCl + Cl∙. CH3CH2CH2–H (CH3)2CH–H Cl–Cl H–Cl CH3CH2CH2–Cl (CH3)2CH–Cl 98 95 58 103 81 80 A) -22 kcal/mol B) +22 kcal/mol C) -40 kcal/mol D) +45 kcal/mol Answer: A Diff: 2 Section: 4.7 50) Given the bond dissociation energies below (in kcal/mol). must be determined experimentally Answer: E Diff: 2 Section: 4.7 13 . CH3COCH3 + Cl2 → CH3COCH2Cl + HCl A) rate = [CH3COCH3] B) rate = [Cl2] C) rate = [CH3COCH3][Cl2] D) rate = [CH3COCH3][Cl2]1/2 E) cannot be determined from stoichiometry. This homolytic bond cleavage yields a more stable secondary radical. choose the correct rate law.51) Given the chlorination of acetone shown below. calculate the overall ΔH° for the following reaction: (CH3)3CH + Br2 → (CH3)3CBr + HBr (CH3)3C–H 91 (CH3)3C–Br 65 Br–Br 46 H–Br 88 CH3–Br 70 Answer: -16 kcal/mol Diff: 2 Section: 4. (CH3)2CH–H vs. CH3CH2–H Answer: (CH3)2CH–H has the smaller bond dissociation energy.7 53) Given the bond dissociation energies below (in kcal/mol).7 52) Of the two C–H bonds shown. which has the smaller bond dissociation energy? Explain your choice. Diff: 2 Section: 4. B) The reaction is first order in CH3CH2OH.54) Consider the reaction (CH3)3CBr + CH3CH2OH → (CH3)3COCH2CH3 + HBr. A-B + C-D → A-C + B-D Answer: Rate = kr[AB][CD] Diff: 2 Section: 4.8 14 . Which of the following correctly describes the kinetics of this reaction? A) The reaction is third order in (CH3)3CBr. (CH3)2CHCl → (CH3)2CH+ Answer: rate = k[(CH3)2CHCl] Diff: 2 Section: 4. Experimentally one finds that if the concentration of (CH3)3CBr is tripled. the rate doubles. One also finds that if the concentration of CH3CH2OH is doubled.8 56) Consider the elementary step in the solvolysis of isopropyl chloride shown below and write the rate equation for this step. the rate of the reaction triples. C) The reaction is second order overall. the rate of the reaction is unchanged.8 + Cl- 57) The following reaction occurs readily: CH3Br + I– CH3I + Br–. D) The reaction is first order overall. What is the rate equation for this reaction? Answer: rate = k[CH3Br][I-] Diff: 2 Section: 4.8 55) Given that the theoretical reaction below was found to be second order and bimolecular. Experimentally one finds that if the concentration of I. Also if the concentration of CH3Br is halved. provide a rate equation for the reaction. E) none of the above Answer: D Diff: 2 Section: 4.is doubled. the rate is halved. 9 59) __________ is the minimum kinetic energy reacting molecules must possess to overcome the repulsions between their electron clouds when they collide. B) The rate of a chemical reaction is directly related to the Ea and that increasing the temperature will alter the Ea for that reaction.10 61) What term describes the highest-energy structure in a molecular collision which leads to reaction? Answer: transition state Diff: 1 Section: 4. C) Increasing the temperature of a chemical reaction increases the number of particles in the reaction that have the minimum energy required to meet the Ea. E) both C and D Answer: E Diff: 2 Section: 4. Answer: Diff: 3 Section: 4.58) The Arrhenius equation mathematically models which of the following statements? A) The rate of a chemical reaction increases exponentially with increasing concentration of reactants.10 15 . Estimate ΔH° for the reaction C → D. D) The rate of a chemical reaction is exponentially related to the Ea and relatively small differences in the Ea can dramatically affect the reaction rates of similar reactions at the same temperature.9 60) Provide the structure of the transition state in the first propagation step of the free radical chlorination of ethane. The activation energy of this conversion is 3 kcal/mol. The energy difference between D and the transition state of the reaction is 7 kcal/mol. Answer: -4 kcal/mol Diff: 2 Section: 4.10 62) Consider the conversion of C to D via a one-step mechanism. Answer: The activation energy or Ea Diff: 1 Section: 4. Make sure to show how the given energy differences are consistent with your sketch. sketch a reaction-energy diagram for this reaction.10 16 .63) Consider the reaction: CH3CH2∙ + Br2 → CH3CH2Br + Br∙ . Answer: Diff: 2 Section: 4. Given that the reaction is endothermic by 5 kcal/mol and that the energy difference between G and the transition state for the process is 15 kcal/mol. sketch a reaction-energy diagram for this reaction. Answer: Diff: 2 Section: 4. Label the axes and show Ea and ΔH° on your drawing.10 64) Consider the one-step conversion of F to G. Given that this reaction has an activation energy of +6 kcal/mol and a ΔH° of -22 kcal/mol. which step is the rate-limiting step? Explain your reasoning. The conversion of Y to Z is therefore rate-limiting. there is no rate-limiting step. ΔH° = 13 kcal/mol Ea = 10 kcal/mol. Answer: The conversion of Y to Z has a higher Ea than does the conversion of X to Y. Answer: A Diff: 2 Section: 4. B) The reaction of B to C. E) The most exothermic step is rate-limiting. ΔH° = -20 kcal/mol Which of the three steps is rate-limiting? A) The reaction of A to B. A→B B→C C→D Ea = 15 kcal/mol. D) All three steps occur at the same rate.11 66) Consider the conversion of X to Z through the sole intermediate Y. C) The reaction of C to D.65) Consider the three-step mechanism for the reaction of A through intermediates B and C to produce D shown below. Diff: 2 Section: 4.11 17 . ΔH° = -6 kcal/mol Ea = 2 kcal/mol. Given the reaction-energy diagram shown below. There is a linear relationship between the number of these collisions and the rate constant of the reaction. Answer: B Diff: 2 Section: 4. ΔH° = -20 kcal/mol What's the enthalpy difference between reactant A and intermediate C? Answer: +7 kcal/mol Diff: 2 Section: 4.12 18 .67) Consider the three-step mechanism for the reaction of A through intermediates B and C to produce D shown below.12 69) Which of the following correctly expresses the standard Gibbs free energy change of a reaction in terms of the reaction temperature (T) and equilibrium constant (K)? A) ΔG° = e-K/RT B) ΔG° = eK/RT C) ΔG° = RTlnK D) ΔG° = -RTlnK E) none of the above Answer: D Diff: 2 Section: 4.12 70) The difference in energy between reactants and the transition state is known as __________. ΔH° = -6 kcal/mol Ea = 2 kcal/mol. Diff: 2 Section: 4. D) the activation energy increases.11 68) The rate of a reaction typically increases as the temperature increases because: A) the A term in the Arrhenius equation increases.12 71) Explain the significance of the frequency factor A in the Arrhenius equation. ΔH° = 13 kcal/mol Ea = 10 kcal/mol. C) the activation energy decreases. E) the molecules make more collisions with the wall of the reaction vessel. For a given reaction A is a constant. Answer: the activation energy or Ea Diff: 1 Section: 4. B) the fraction of molecules with kinetic energy greater than Ea increases. A→B B→C C→D Ea = 15 kcal/mol. Answer: The frequency factor accounts for the number of collisions between reacting molecules and the fraction of these collisions with the proper orientation for the reaction to happen. 8 times faster. how many tertiary hydrogens are present? A) 0 B) 1 C) 2 D) 3 E) 4 Answer: D Diff: 1 Section: 4.13 75) The major monobrominated product which results when ethylcyclohexane is subjected to free radical bromination is: A) a primary bromide B) a secondary bromide C) a tertiary bromide D) a quaternary bromide E) bromomethane Answer: C Diff: 1 Section: 4.72) Explain the significance of the exponential factor e-Ea/RT in the Arrhenius equation. Diff: 3 Section: 4.987 cal/mol∙K. Answer: The reaction will occur about 2. Diff: 2 Section: 4.13 19 . Answer: The exponential factor e-Ea/RT in the Arrhenius equation corresponds to the fraction of collisions in which the reacting molecules have the appropriate activation energy to react at a given temperature.12 74) In the hydrocarbon shown below.12 73) Given an activation energy of 15 kcal/mol. R = 1. use the Arrhenius equation to estimate how much faster the reaction will occur if the temperature is increased from 100°C to 120°C. 13 77) How many distinct dichlorination products can result when isobutane is subjected to free radical chlorination? A) 1 B) 2 C) 3 D) 4 E) 6 Answer: C Diff: 1 Section: 4.13 79) How many distinct monochlorinated products can result when cyclopentane is subjected to free radical chlorination? A) 1 B) 2 C) 3 D) 4 E) 5 Answer: A Diff: 1 Section: 4.76) Which of the halogens below undergoes free radical halogenation with ethane most rapidly? A) fluorine B) chlorine C) iodine D) bromine E) pyridine Answer: A Diff: 1 Section: 4.13 20 .13 78) How many distinct monochlorinated products can result when isobutane is subjected to free radical chlorination? A) 1 B) 2 C) 3 D) 4 E) 5 Answer: B Diff: 1 Section: 4. . from most stable to least stable).80) Rank the free radicals (I-III) shown below in order of decreasing stability (i. A) I > III > II B) II > III > I C) I > II > III D) II > I > III E) III > II > I Answer: A Diff: 2 Section: 4.13 21 .13 81) Which H atom in the molecule shown will be most readily abstracted by a bromine radical? A) Ha B) Hb C) Hc D) Hd E) He Answer: C Diff: 2 Section: 4.e. 0 times more reactive Answer: D Diff: 3 Section: 4. If we assign the primary H's a reactivity of 1.6 times more reactive D) 5.1 times more reactive C) 4.6 times more reactive E) 3.3-dimethylbutane and 2-chloro-2.13 83) A certain free radical chlorination reaction of 2. Answer: Diff: 2 Section: 4.3-dimethylbutane leads to the production of two major monochlorinated products 1-chloro-2.3-dimethylbutane.4-trimethylpentane is subjected to free radical bromination. what is the relative reactivity of the tertiary H's.0 times more reactive B) 5.13 22 . it was determined that the product distribution was 51. A) 5.0.3% 2-chloro-2.7% 1-chloro-2.2.13 84) Provide the major organic product that results when 2.3dimethylbutane.3-dimethylbutane and 48. When run through a Gas chromatograph.82) How many secondary hydrogens are present in the hydrocarbon below? A) 2 B) 6 C) 7 D) 8 E) 16 Answer: B Diff: 2 Section: 4. Answer: 1-chloro-2. Answer: Diff: 2 Section: 4.13 86) Given that tertiary H atoms react with a chlorine radical about 5.3. Answer: Diff: 3 Section: 4. estimate the ratio of the two monochlorinated products that result when 2. hv (CH3)3CCH2CH3 + Br2 Answer: (CH3)3CCHBrCH3 Diff: 2 Section: 4.2-dimethylpropane Diff: 1 Section: 4. C7H16.13 88) Write the structures of all of the monobromination products of 1. 48% Diff: 3 Section: 4.3-dimethylbutane.3-dimethylbutane undergoes free radical chlorination.13 89) Predict the major monobromination product in the following reaction.5 times faster than primary ones.1.85) When compound I.3-dimethylbutane. Two of the products were primary alkyl halides and the other was a secondary alkyl halide.13 87) What C5H12 isomer will give only a single monochlorination product? Answer: (CH3)4C or neopentane or 2. was treated with chlorine and light it yielded 3 monochlorination products that could be separated by chromatography. 2-chloro-2. Provide a possible structure for compound I.13 23 . 52%.3tetramethylcyclobutane. 13 93) What is the name of the major monobrominated product which results when 3-methylpentane is subjected to Br2/hν conditions? Answer: 3-bromo-3-methylpentane Diff: 2 Section: 4.13 94) What is the name of the major monobrominated product which results when methylcyclohexane is subjected to Br2/hν conditions? Answer: 1-bromo-1-methylcyclohexane Diff: 2 Section: 4. Diff: 2 Section: 4. CH2=CHCH2∙. the lowest energy reaction pathways go through secondary free radical intermediates to produce secondary alkyl bromides (2-bromopentane and 3-bromopentane). from least stable to most stable).e. Answer: Diff: 2 Section: 4. CH3∙.13 91) List the following radicals in order of increasing stability (i.13 24 .13 92) Free radical bromination of pentane results in poor yields of 1-bromopentane. while cyclopentane can be readily brominated under similar conditions to yield bromocyclopentane. Offer an explanation. CH3CH2∙.90) Predict the major monobromination product in the following reaction. (CH3)2CH∙ Answer: CH3∙ < CH3CH2∙ < (CH3)2CH∙ < (CH3)3C∙ < CH2=CHCH2∙ Diff: 2 Section: 4. All of the hydrogen atom abstractions of cyclopentane lead to the same secondary radical which eventually leads to bromocyclopentane. Answer: In the bromination of pentane. (CH3)3C∙.. Thus 1-bromopentane is a very minor product. Answer: ClCH2C(CH3)2CH2CH3. 15. the transition state is closer to the reactants in structure. and estimate the relative amount of each in the mixture of monochlorinated products. Answer: C Diff: 2 Section: 4.8%. 45. E) Transition states are molecular species of finite lifetime whose properties can be probed through free radical reactions. the relative reactivity of the 1°: 2°: 3° hydrogens is approximately 1:82:1600. D) The structure of the transition state in an organic reaction is always modeled on the structure of the reactants leading to that transition state. 96.95) When 1. Provide the structure of each monochlorination product. (CH3)3CCH2CH2Cl. the transition state is closer in energy to the products. Estimate the amount of each monobromination product.1.3-tetramethylcyclobutane is brominated at 125°C.0.3.14 25 .13 98) Which of the following statements is the best statement of the Hammond Postulate? A) In an endothermic reaction.3% Diff: 3 Section: 4. secondary bromide.8: 5. C) Related species that are similar in energy are also similar in structure.13 96) The relative reactivity of the 1°: 2°: 3° hydrogens of (CH3)3CCH2CH3 in free radical chlorination is 1: 3.5% Diff: 3 Section: 4. Diff: 3 Section: 4. 38. 3. Answer: primary bromide.9%. (CH3)3CCHClCH3.5%.13 97) What is the relative reactivity of 2° vs 1° hydrogens in the free radical bromination of nbutane if the ratio of 1-bromobutane to 2-bromobutane formed is 7:93? Answer: The 2° hydrogens are 20 times more reactive than the 1° ones. B) In an exothermic reaction. Using the Hammond Postulate this means the transition state more closely resembles a carbon radical than a bromine radical. The carbon-hydrogen bond is almost completely broken and the hydrogen-bromine bond completely formed in the transition state.14 100) The hydrogen atom abstraction step in the free radical bromination of methane is endothermic. Use the Hammond Postulate to speculate on the extent of bond formation and bond cleavage in the transition state. Use the Hammond Postulate to speculate on the extent of bond formation and bond cleavage in the transition state. Answer: Diff: 3 Section: 4.14 101) The hydrogen atom abstraction step in the free radical chlorine of methane is exothermic. The carbonhydrogen bond has broken very little and the hydrogen-chlorine bond has formed very little in the transition state. Diff: 3 Section: 4. Using the Hammond Postulate this means the transition state more closely resembles a bromine radical than a carbon radical. Answer: The endothermic nature of this step means that the energy of the transition state is closer to that of the step's products than to its reactants. Answer: The exothermic nature of this step means that the energy of the transition state is closer to that of the step's reactants than to its products.99) Draw an energy diagram for a two step reaction where the structure of the transition state of the rate determining step most closely resembles the starting material and the overall reaction is exothermic.14 26 . Diff: 3 Section: 4. this means that the transition state for the bromination is product-like (ie.16 105) How do alkyl substituents stabilize a carbocationic center to which they are attached? A) Through an inductive donation of electron density to the cationic center.16 104) Which of the following reactive intermediates can best be described as both nucleophilic and strongly basic? A) carbanions B) carbocations C) carbenes D) free radicals E) alkanes Answer: A Diff: 1 Section: 4. radical-like) while the transition state for the chlorination is reactant-like. This is not true in the chlorination case where the transition state possesses little radical character. Answer: The first propagation step in free radical bromination is endothermic while the analogous step in free radical chlorination is exothermic. From the Hammond Postulate. C) Through hyperconjugation.102) Use the Hammond Postulate to explain why free radical brominations are more selective than free radical chlorinations. D) both A and C E) both B and C Answer: D Diff: 2 Section: 4. B) Through an inductive removal of electron density from the cationic center. The product-like transition state for bromination has the C–H bond nearly broken and a great deal of radical character on the carbon atom.14 103) Which of the following is a carbene? A) CH2=CHOB) CH3CH2∙ C) :CCl2 D) CH3CH2+ E) NCOAnswer: C Diff: 1 Section: 4. The energy of this transition state reflects most of the energy difference of the radical products.16 27 . Diff: 2 Section: 4. Diff: 2 Section: 4.106) Which of the following reactive intermediate species maintains sp3 hybridization? A) methyl carbanion B) dibromocarbene C) tertiary carbocation D) secondary alkyl radical E) both B and C Answer: A Diff: 2 Section: 4. A) an alkane B) an alkene C) an alkyne D) an aromatic ring E) a carbanion Answer: B Diff: 2 Section: 4.16 28 . Answer: Diff: 2 Section: 4.16 107) When two carbenes collide.16 108) Provide the structure of the carbene that results when diazomethane (shown below) decomposes.16 109) What accounts for the relatively high stability of the benzyl radical? Answer: The unpaired electron is delocalized through a conjugated pi system. they immediately dimerize to give __________. 16 113) Describe the hybridization of the cationic center and predict the CCC bond angle in (CH3)3C+.16 29 .16 112) When acetaldehyde (CH3CHO) is deprotonated.16 111) When Br radical reacts with 1-butene (CH3CH2CH=CH2). Answer: CH3 HCH=CH2 ↔ CH3CH=CH H2 C C Diff: 2 Section: 4. Answer: Diff: 3 Section: 4. Diff: 3 Section: 4.110) Add a lone pair of electrons to the following structure to create the most reactive (least stable) carbanion intermediate. Draw the major resonance contributing forms of this anion. the resulting anion is stabilized by resonance. the hydrogen atom which is preferentially abstracted is the one which produces a resonance stabilized radical. Draw the major resonance contributing forms of this radical. Answer: Diff: 2 Section: 4. therefore the CCC bond angle is 120°. Answer: Cationic center is sp2 hybridized.