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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 55b

Tributyltin hydride (Bu3SnH) is used synthetically to reduce alkyl halides, replacing a halogen atom with hydrogen. ­Free-radical initiators promote this reaction, and free-radical inhibitors are known to slow or stop it. Your job is to develop a mechanism, using the following reaction as the example.

The following bond-dissociation enthalpies may be helpful: 

b. Calculate values of ΔH for your proposed steps to show that they are energetically feasible. (Hint: A trace of Br2 and light suggests it’s there only as an initiator, to create Br• radicals. Then decide which atom can be abstracted most favorably from the starting materials by the Br• radical. That should complete the initiation. Finally, decide what energetically favored propagation steps will accomplish the reaction.)

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Step 1: Begin by analyzing the initiation step. The trace amount of Br2 and light (hv) suggests that Br2 undergoes homolytic cleavage to form two bromine radicals (Br⋅). This is the initiation step, and the bond-dissociation enthalpy for Br—Br is 190 kJ/mol. Write the equation for this step: Br2 → 2 Br⋅.
Step 2: Determine the most favorable atom abstraction by the Br⋅ radical. The bromine radical can abstract a hydrogen atom from Bu3SnH, forming Bu3Sn⋅ and HBr. The bond-dissociation enthalpy for Bu3Sn—H is 310 kJ/mol, and for H—Br is 366 kJ/mol. Write the equation for this step: Br⋅ + Bu3SnH → Bu3Sn⋅ + HBr.
Step 3: Analyze the propagation steps. The Bu3Sn⋅ radical can abstract a halogen atom (X⋅) from the alkyl halide (R—X), forming Bu3SnX and an alkyl radical (R⋅). The bond-dissociation enthalpy for Bu3Sn—Br is 552 kJ/mol, and for R—X (alkyl halide bond) is given as 309 kJ/mol. Write the equation for this step: Bu3Sn⋅ + R—X → Bu3SnX + R⋅.
Step 4: The alkyl radical (R⋅) can then abstract a hydrogen atom from Bu3SnH, regenerating Bu3Sn⋅ and forming the alkane (R—H). This completes the propagation cycle. Write the equation for this step: R⋅ + Bu3SnH → R—H + Bu3Sn⋅.
Step 5: Calculate the ΔH values for each step to confirm energetic feasibility. Use the bond-dissociation enthalpies provided to calculate ΔH for each reaction step. For example, ΔH = bonds broken - bonds formed. Ensure that the overall ΔH for the propagation steps is negative, indicating that the reaction is energetically favorable.

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Key Concepts

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

Free Radical Mechanism

A free radical mechanism involves the formation and reaction of free radicals, which are highly reactive species with unpaired electrons. In the context of the reaction with tributyltin hydride, the initiation step generates bromine radicals (Br·) from Br2 under light exposure. These radicals can then abstract hydrogen from the alkyl halide, leading to the formation of a new radical and the corresponding alkyl bromide.
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Bond Dissociation Enthalpy (BDE)

Bond dissociation enthalpy is the energy required to break a specific bond in a molecule, resulting in the formation of free radicals. Understanding BDE is crucial for predicting the feasibility of the reaction steps. For instance, comparing the BDEs of the bonds involved in the reaction helps determine which bonds are more easily broken, guiding the selection of the most favorable hydrogen abstraction during the propagation steps.
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Energetic Feasibility and ΔH Calculation

Energetic feasibility refers to whether a reaction can occur based on the energy changes associated with bond breaking and forming. The change in enthalpy (ΔH) for each step can be calculated using the bond dissociation enthalpies provided. A negative ΔH indicates that the reaction is exothermic and energetically favorable, which is essential for validating the proposed mechanism and ensuring that the reaction can proceed under the given conditions.
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Related Practice
Textbook Question

When ethene is treated in a calorimeter with H2 and a Pt catalyst, the heat of reaction is found to be –137 kJ/mol (–32.7 kcal/mol), and the reaction goes to completion. When the reaction takes place at 1400 K, the equilibrium is found to be evenly balanced, with Keq =1.  Compute the value of ΔS for this reaction.

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Textbook Question

When healthy, Earth’s stratosphere contains a low concentration of ozone (O3) that absorbs potentially harmful ­ultraviolet (UV) radiation by the cycle shown at right.

Chlorofluorocarbon refrigerants, such as Freon 12 (CF2Cl2), are stable in the lower atmosphere, but in the stratosphere they absorb high-energy UV radiation to generate chlorine radicals.

The presence of a small number of chlorine radicals appears to lower ozone concentrations dramatically. The following reactions are all known to be exothermic (except the one requiring light) and to have high rate constants. Propose two mechanisms to explain how a small number of chlorine ­radicals can destroy large numbers of ozone molecules. Which of the two mechanisms is more likely when the concentration of chlorine atoms is very small?

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Textbook Question

When dichloromethane is treated with strong NaOH, an intermediate is generated that reacts like a carbene. Draw the structure of this reactive intermediate, and propose a mechanism for its formation.

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Textbook Question

When a small amount of iodine is added to a mixture of chlorine and methane, it prevents chlorination from occurring. Therefore, iodine is a free-radical inhibitor for this reaction. Calculate ΔH° values for the possible reactions of iodine with species present in the chlorination of methane, and use these values to explain why iodine inhibits the reaction. (The I―Cl bond-dissociation enthalpy is 211 kJ/mol or 50 kcal/mol.)

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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).

a. Draw the transition state for the rate-limiting step of each of these reactions, showing how a bond to hydrogen or ­deuterium is being broken in this step.

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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).

b. Monochlorination of deuterioethane (C2H5D) leads to a mixture containing 93% C2H4DCl and 7% C2H5Cl. Calculate the relative rates of abstraction per hydrogen and deuterium in the chlorination of deuterioethane.

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