In the following reactions, identify the bonds formed and the bonds broken.
(a)

Calculate ∆H° for the following reactions.

(a)

Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.

(a)

Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.

(c)

Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.

(d) I• vs •OH

In the following reactions, identify the bonds formed and the bonds broken.

(c)

Teflon-coated frying pans routinely endure temperatures that would cause polyethylene or polypropylene to oxidize and decompose. Decomposition of polyethylene is initiated by free-radical abstraction of a hydrogen atom by O₂. Bond-dissociation energies of C—H bonds are about 400 kJ/mol, and C—F bonds are about 460 kJ/mol. The BDE of the H—OO bond is about 192 kJ/mol, and the F—OO bond is about 63 kJ/mol. Show why Teflon (Figure 7-5) is much more resistant to oxidation than polyethylene is.

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.

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.)