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Ch.4 - The Study of Chemical Reactions
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
Chapter 4, Problem 4.20

Use the bond-dissociation enthalpies in Table 4-2 (page 167) to calculate the heats of reaction for the two possible first propagation steps in the chlorination of isobutane. Use this information to draw a reaction-energy diagram like Figure 4-8, comparing the activation energies for formation of the two radicals.

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Step 1: Identify the two possible propagation steps in the chlorination of isobutane. The first propagation step involves the abstraction of a hydrogen atom from either a primary carbon or a tertiary carbon in isobutane by a chlorine radical (Cl•). This results in the formation of either a primary radical or a tertiary radical.
Step 2: Use the bond-dissociation enthalpy values from TABLE 4-2 to calculate the energy required to break the C-H bond in isobutane. For the primary C-H bond, the bond-dissociation enthalpy is 423 kJ/mol. For the tertiary C-H bond, the bond-dissociation enthalpy is 400 kJ/mol.
Step 3: Calculate the energy released when the H-Cl bond is formed in the propagation step. The bond-dissociation enthalpy for the H-Cl bond is 431 kJ/mol. Subtract the bond-dissociation enthalpy of the C-H bond from the energy released by the formation of the H-Cl bond to determine the heat of reaction for each propagation step.
Step 4: Compare the heats of reaction for the two propagation steps. The step involving the tertiary C-H bond will likely have a lower activation energy and be more exothermic due to the stability of the tertiary radical compared to the primary radical.
Step 5: Draw a reaction-energy diagram similar to FIGURE 4-8. Represent the energy difference between the activation energies for the formation of the primary and tertiary radicals, as well as the difference in the overall energy change (ΔH) for the two propagation steps. Highlight that the tertiary radical formation has a lower activation energy and is more thermodynamically favorable.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Bond-Dissociation Enthalpy

Bond-dissociation enthalpy (BDE) is the energy required to break a specific bond in a molecule, resulting in the formation of free radicals. It is a crucial concept in understanding reaction mechanisms, as it helps predict the stability of intermediates and the overall energy changes during a reaction. In the context of chlorination, BDE values for C-H and C-Cl bonds will be essential for calculating the heats of reaction for the propagation steps.
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How to calculate enthalpy using bond dissociation energies.

Propagation Steps in Radical Reactions

Propagation steps are the stages in a radical chain reaction where radicals react with stable molecules to form new radicals, continuing the reaction cycle. In the chlorination of isobutane, the first propagation steps involve the reaction of chlorine radicals with isobutane to form new radicals. Understanding these steps is vital for calculating the heats of reaction and determining the preferred pathway based on activation energies.
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Reaction-Energy Diagram

A reaction-energy diagram visually represents the energy changes during a chemical reaction, illustrating the relative energies of reactants, products, and transition states. It helps in comparing activation energies for different pathways, such as the formation of various radicals in the chlorination of isobutane. By analyzing these diagrams, one can infer the feasibility and rate of reactions based on the energy barriers involved.
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Related Practice
Textbook Question

The following reaction has a value of ΔG° = –2.1 kJ/mol (–0.50 kcal/mol).

CH3Br + H2S ⇌ CH3SH + HBr

b. Starting with a 1 M solution of CH3Br and H2S, calculate the final concentrations of all four species at equilibrium.

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Textbook Question
Deuterium (D) is the hydrogen isotope of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C―D bond is slightly stronger than the C―H bond by 5.0 kJ/mol (1.2 kcal/mol). Reaction rates tend to be slower when a C―D bond (as opposed to a C―H bond) is broken in a rate-limiting step.This effect, called a kinetic isotope effect, is clearly seen in the chlorination of methane. Methane undergoes free-radical chlorination 12 times as fast as tetradeuteriomethane (CD4)Faster: CH4 + Cl⋅ —> CH3Cl + HCl relative rate= 12Slower: CD4 + Cl⋅ —> CD3Cl + DClrelative rate= 1 c. Consider the thermodynamics of the chlorination of methane and the chlorination of ethane, and use the Hammond postulate to explain why one of these reactions has a much larger isotope effect than the other.
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Textbook Question

Free-radical chlorination of hexane gives very poor yields of 1-chlorohexane, while cyclohexane can be converted to chlorocyclohexane in good yield.

a. How do you account for this difference?

b. What ratio of reactants (cyclohexane and chlorine) would you use for the synthesis of chlorocyclohexane?

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Textbook Question

The following reaction has a value of ΔG° = –2.1 kJ/mol (–0.50 kcal/mol).

CH3Br + H2S ⇌ CH3SH + HBr

a. Calculate Keq at room temperature (25 °C) for this reaction as written.

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Textbook Question
The reaction of tert-butyl chloride with methanol(CH3)3C—Cl Tert-butylchloride + CH3—OH methanol —> (CH3)C—OCH3 methyltert-butylether + HCl is found to follow the rate equation rate= Kr[(CH3)3C—Cl] c. What is the kinetic order overall?
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Textbook Question
The bromination of methane proceeds through the following steps:1. Br2 + 2 Br• ΔH° (per mole)/+190 kJ (45 kcal)Ea (per mole)/ 190 kJ (45 kcal)2. CH4 + Br• —> CH3+ HBr +73 kJ (17 kcal) 79 kJ (19 kcal) 3. • CH3 + Br2 —> CH3Br + Br -112 kJ (-27 kcal) 4 kJ (1 kcal) a. Draw a complete reaction-energy diagram for this reaction. b. Label the rate-limiting step.
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