As seen in this chapter, hydrocarbons typically do not

Chapter 11, Problem 11.54

(choose chapter or problem)

As seen in this chapter, hydrocarbons typically do not undergo radical iodination in the presence of I2 (compound 2a). Furthermore, radical halogenation (even chlorination) of strained hydrocarbons, like cubane (1), is problematic if the reaction produces a halogen radical (X). An alternative method circumvents both of these issues by replacing the elemental halogen, X2, with a tetrahalomethane, CX4 (J. Am. Chem. Soc. 2001, 123, 18421847). The reaction is initiated by formation of a trihalomethyl radical, CX3. + + ZH 1 I I Z 2 3 4 2a: Z = I 2b: Z = CI3 4a: Z = I 4b: Z = CI3 (a) Propose a propagation cycle for the production of iodocubane (3) from cubane (1) and tetraiodomethane (2b). (b) Provide three possible termination steps. (c) The bond dissociation energy (BDE) for I!CI3 is 192 kJ/mol, while the BDE for H!CI3 is 423 kJ/mol. Using this information, provide a thermodynamic argument that explains why 2b is an effective iodination reagent while 2a is not.